http://ballistipedia.com/api.php?action=feedcontributions&user=David&feedformat=atomShotStat - User contributions [en]2020-02-29T07:06:43ZUser contributionsMediaWiki 1.31.6http://ballistipedia.com/index.php?title=Home&diff=1448Home2019-10-07T17:59:37Z<p>David: Made footnotes smaller type</p>
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<div>This site explains and demonstrates statistics for analyzing the precision of projectile weapon systems.<ref name="nuance"><small>Typical examples would be target shooting with a rifle or pistol. Such weapons as shotguns, mortars, and ballistic missiles would have some similar characteristics, but also have factors that are neglected in the discussions and measurements. The wiki will discuss some factors of ballistics, but it is not intended to address all the nuances of internal, external, or terminal ballistics. Rather, the focus is primarily on the analysis of the precision of the whole weapon system which can be observed directly by the relative impact points on a target. Some effort will be made to explore the precision of weapons subsystems.</small></ref><br />
<br />
High level topics, which are good places to start exploring the site, include:<br />
* [[What is Precision?]]: An important explanation of the difference between precision and accuracy as the terms are used in statistics<br />
* [[Describing Precision]]: Units, terms, and relationships<br />
* [[Precision Models]]: Statistical approaches for efficient estimation and inference of precision <br />
* [[Prior Art]]: Reviews of past efforts to address this question<br />
* [[FAQ]]<br />
* [[Ballistic Accuracy Classification]]: A proposed industry standard for determining and describing precision<br />
<br />
= Synopsis =<br />
<br />
When testing a gun, shooter, and/or ammunition the most popular measure is [[Range Statistics#Extreme Spread|Extreme Spread]] or "group size" of a sample of target shots. However Extreme Spread must be used with care since it is frequently and easily abused.<ref><small>Many [[references]] are worth reading for further background on how Extreme Spread is broadly misunderstood. Recommended include [https://www.thetruthaboutguns.com/understanding-rifle-precision/ Understanding Rifle Precision], [https://www.autotrickler.com/blog/thinking-statistically Thinking Statistically], and [http://www.bisonops.com/2019/08/17/rifle-ammunition-load-workup/ Rifle Ammnition Load Workup].</small></ref> As with all measures, the single best measurement is meaningless in isolation. The proper statistical estimator is an "average" (a.k.a., ''expected value'') of the measurement.<br />
<br />
Another consideration is that some measures, such as Extreme Spread, change value when there are more shots in a group. Measures that have such a dependency will be referred to here as ''variant measures''. There are ''[[Describing_Precision#Invariant_Measures|invariant measures]]'', like [[Circular Error Probable]] or Mean Radius, for which the expected values do not change with the number of shots in a group. Instead [[Precision_Models#How_large_a_sample_do_we_need.3F|having more shots only increases the confidence in the measure's value]]. Of course the experimental error of either type of measurement can also be decreased by increasing the sample size (i.e., shooting more groups). <br />
<br />
Furthermore, by first making assumptions about the inherent shot dispersion, then it is possible to use theoretical models to estimate measurements and their precision. The distributions are of two basic types: If the expected values and the expected precision factor for the measurements depend on distributions which have an [[Closed Form Precision|explicit solution]] then the values can be calculated formulaicly. If the values don't have a distribution with a closed form expression then they can be estimated via Monte Carlo approaches.<br />
<br />
[[:Category:Examples|Examples]] of the application of these [[Precision Models|methods]] and [[Measuring Tools|tools]] include:<br />
* Determining how many sighter shots you should take.<br />
* Determining the likelihood of a hit on a particular target by a zeroed shooting system.<br />
* Comparing the inherent precision of different shooting systems.<br />
* Determining which ammunition shoots better in a particular gun.<br />
<br />
----<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Home&diff=1447Home2019-10-07T17:57:57Z<p>David: /* Synopsis */ Added references</p>
<hr />
<div>This site explains and demonstrates statistics for analyzing the precision of projectile weapon systems.<ref name="nuance">Typical examples would be target shooting with a rifle or pistol. Such weapons as shotguns, mortars, and ballistic missiles would have some similar characteristics, but also have factors that are neglected in the discussions and measurements. The wiki will discuss some factors of ballistics, but it is not intended to address all the nuances of internal, external, or terminal ballistics. Rather, the focus is primarily on the analysis of the precision of the whole weapon system which can be observed directly by the relative impact points on a target. Some effort will be made to explore the precision of weapons subsystems.</ref><br />
<br />
High level topics, which are good places to start exploring the site, include:<br />
* [[What is Precision?]]: An important explanation of the difference between precision and accuracy as the terms are used in statistics<br />
* [[Describing Precision]]: Units, terms, and relationships<br />
* [[Precision Models]]: Statistical approaches for efficient estimation and inference of precision <br />
* [[Prior Art]]: Reviews of past efforts to address this question<br />
* [[FAQ]]<br />
* [[Ballistic Accuracy Classification]]: A proposed industry standard for determining and describing precision<br />
<br />
= Synopsis =<br />
<br />
When testing a gun, shooter, and/or ammunition the most popular measure is [[Range Statistics#Extreme Spread|Extreme Spread]] or "group size" of a sample of target shots. However Extreme Spread must be used with care since it is frequently and easily abused.<ref><small>Many [[references]] are worth reading for further background on how Extreme Spread is broadly misunderstood. Recommended include [https://www.thetruthaboutguns.com/understanding-rifle-precision/ Understanding Rifle Precision], [https://www.autotrickler.com/blog/thinking-statistically Thinking Statistically], and [http://www.bisonops.com/2019/08/17/rifle-ammunition-load-workup/ Rifle Ammnition Load Workup].</small></ref> As with all measures, the single best measurement is meaningless in isolation. The proper statistical estimator is an "average" (a.k.a., ''expected value'') of the measurement.<br />
<br />
Another consideration is that some measures, such as Extreme Spread, change value when there are more shots in a group. Measures that have such a dependency will be referred to here as ''variant measures''. There are ''[[Describing_Precision#Invariant_Measures|invariant measures]]'', like [[Circular Error Probable]] or Mean Radius, for which the expected values do not change with the number of shots in a group. Instead [[Precision_Models#How_large_a_sample_do_we_need.3F|having more shots only increases the confidence in the measure's value]]. Of course the experimental error of either type of measurement can also be decreased by increasing the sample size (i.e., shooting more groups). <br />
<br />
Furthermore, by first making assumptions about the inherent shot dispersion, then it is possible to use theoretical models to estimate measurements and their precision. The distributions are of two basic types: If the expected values and the expected precision factor for the measurements depend on distributions which have an [[Closed Form Precision|explicit solution]] then the values can be calculated formulaicly. If the values don't have a distribution with a closed form expression then they can be estimated via Monte Carlo approaches.<br />
<br />
[[:Category:Examples|Examples]] of the application of these [[Precision Models|methods]] and [[Measuring Tools|tools]] include:<br />
* Determining how many sighter shots you should take.<br />
* Determining the likelihood of a hit on a particular target by a zeroed shooting system.<br />
* Comparing the inherent precision of different shooting systems.<br />
* Determining which ammunition shoots better in a particular gun.<br />
<br />
----<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=References&diff=1446References2019-10-07T17:39:34Z<p>David: Added reference to Rifle Ammunition Load Workup</p>
<hr />
<div>'''[[Prior Art]]''' details previous work on the problem of estimating shooting statistics.<br />
<br />
'''[[CEP literature]]''' focuses on the broader body of work related to characterizing Circular Error Probable, which is applicable not only to ballistics but also to fields like navigation and signal processing.<br />
<br />
Following is a complete list of useful References and Prior Art:<br />
<br />
* Bookstaber, David (2014). [http://www.thetruthaboutguns.com/2014/12/daniel-zimmerman/understanding-rifle-precision/ '''Understanding Rifle Precision'''].<br />
<br />
* Danielson, Brent J. (2005). [[Prior_Art#Danielson.2C_2005.2C_Testing_loads|'''Testing Loads''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Grubbs, Frank E. (1964). [[Prior_Art#Grubbs.2C_1964.2C_Statistical_Measures_of_Accuracy_for_Riflemen_and_Missile_Engineers|'''Statistical Measures of Accuracy for Riflemen and Missile Engineers''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Hogema, Jeroen (2005). [[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|'''Shot group statistics''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Hogema, Jeroen (2006). [[Prior_Art#Hogema.2C_2006.2C_Measuring_Precision|'''Measuring Precision''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Kolbe, Geoffrey (2010). [[Prior_Art#Kolbe.2C_2010.2C_Group_Statistics|'''Group Statistics''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Leslia, John E. III (1993). [[Prior_Art#Leslie.2C_1993.2C_Is_.22Group_Size.22_the_Best_Measure_of_Accuracy.3F|'''Is "Group Size" the Best Measure of Accuracy?''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* MacDonald, Adam (2017). [https://www.autotrickler.com/blog/thinking-statistically '''Thinking Statistically'''].<br />
<br />
* Molon (2006). [[Prior_Art#Molon.2C_2006.2C_The_Trouble_With_3-Shot_Groups|'''The Trouble With 3-Shot Groups''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Rifleslinger (2014). [http://artoftherifleblog.com/on-zeroing/2014/02/on-zeroing.html '''On Zeroing'''].<br />
<br />
* Siddiqui, M. M. (1961). [[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|'''Some Problems Connected With Rayleigh Distributions''']]. The Journal of Research of the National Bureau of Standards, Sec. D: Radio Science, Vol. 68D, No. 9.<br />
<br />
* Siddiqui, M. M. (1964). [[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|'''Statistical Inference for Rayleigh Distributions''']]. The Journal of Research of the National Bureau of Standards, Sec. D: Radio Propagation, Vol. 66D, No. 2. (''Summarizes and extends Siddiqui, 1961.'')<br />
<br />
'''''Important Note on Siddiqui''': Siddiqui parameterizes the Rayleigh distribution with <math>\frac{\sigma}{\sqrt{2}}</math>. Therefore, should you endeavor to relate Siddiqui's work to that referenced here and in more modern usage, remember that <math>\sigma_{modern} = \sqrt{2} \sigma_{Siddiqui}</math>.''<br />
<br />
* Taylor, M. S. & Grubbs, Frank E. (1975). [[Prior_Art#Taylor_.26_Grubbs.2C_1975.2C_Approximate_Probability_Distributions_for_the_Extreme_Spread|'''Approximate Probability Distributions for the Extreme Spread''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Triplett, Ben (2019). [http://www.bisonops.com/2019/08/17/rifle-ammunition-load-workup '''Rifle Ammunition Load Workup'''].<br />
<br />
<br />
= Reference Data =<br />
<br />
* [[File:Confidence Interval Convergence.xlsx]]: Shows how precision confidence intervals shrink as sample size increases.<br />
<br />
* [[File:Sigma1RangeStatistics.xls]]: Simulated median, 50%, 80%, and 95% quantiles, plus first four sample moments, for shot groups containing 2 to 100 shots, of: Extreme Spread, Diagonal, Figure of Merit.<br />
<br />
* [[File:SymmetricBivariateSigma1.xls]]: Monte Carlo simulation results validating the [[Closed Form Precision]] math.</div>Davidhttp://ballistipedia.com/index.php?title=References&diff=1445References2019-07-30T17:24:42Z<p>David: </p>
<hr />
<div>'''[[Prior Art]]''' details previous work on the problem of estimating shooting statistics.<br />
<br />
'''[[CEP literature]]''' focuses on the broader body of work related to characterizing Circular Error Probable, which is applicable not only to ballistics but also to fields like navigation and signal processing.<br />
<br />
Following is a complete list of References and Prior Art:<br />
<br />
* Bookstaber, David (2014). [http://www.thetruthaboutguns.com/2014/12/daniel-zimmerman/understanding-rifle-precision/ '''Understanding Rifle Precision'''].<br />
<br />
* Danielson, Brent J. (2005). [[Prior_Art#Danielson.2C_2005.2C_Testing_loads|'''Testing Loads''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Grubbs, Frank E. (1964). [[Prior_Art#Grubbs.2C_1964.2C_Statistical_Measures_of_Accuracy_for_Riflemen_and_Missile_Engineers|'''Statistical Measures of Accuracy for Riflemen and Missile Engineers''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Hogema, Jeroen (2005). [[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|'''Shot group statistics''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Hogema, Jeroen (2006). [[Prior_Art#Hogema.2C_2006.2C_Measuring_Precision|'''Measuring Precision''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Kolbe, Geoffrey (2010). [[Prior_Art#Kolbe.2C_2010.2C_Group_Statistics|'''Group Statistics''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Leslia, John E. III (1993). [[Prior_Art#Leslie.2C_1993.2C_Is_.22Group_Size.22_the_Best_Measure_of_Accuracy.3F|'''Is "Group Size" the Best Measure of Accuracy?''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* MacDonald, Adam (2017). [https://www.autotrickler.com/blog/thinking-statistically '''Thinking Statistically'''].<br />
<br />
* Molon (2006). [[Prior_Art#Molon.2C_2006.2C_The_Trouble_With_3-Shot_Groups|'''The Trouble With 3-Shot Groups''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Rifleslinger (2014). [http://artoftherifleblog.com/on-zeroing/2014/02/on-zeroing.html '''On Zeroing'''].<br />
<br />
* Siddiqui, M. M. (1961). [[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|'''Some Problems Connected With Rayleigh Distributions''']]. The Journal of Research of the National Bureau of Standards, Sec. D: Radio Science, Vol. 68D, No. 9.<br />
<br />
* Siddiqui, M. M. (1964). [[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|'''Statistical Inference for Rayleigh Distributions''']]. The Journal of Research of the National Bureau of Standards, Sec. D: Radio Propagation, Vol. 66D, No. 2. (''Summarizes and extends Siddiqui, 1961.'')<br />
<br />
'''''Important Note on Siddiqui''': Siddiqui parameterizes the Rayleigh distribution with <math>\frac{\sigma}{\sqrt{2}}</math>. Therefore, should you endeavor to relate Siddiqui's work to that referenced here and in more modern usage, remember that <math>\sigma_{modern} = \sqrt{2} \sigma_{Siddiqui}</math>.''<br />
<br />
* Taylor, M. S. & Grubbs, Frank E. (1975). [[Prior_Art#Taylor_.26_Grubbs.2C_1975.2C_Approximate_Probability_Distributions_for_the_Extreme_Spread|'''Approximate Probability Distributions for the Extreme Spread''' &ndash; ''detailed in Prior Art'']].<br />
<br />
= Reference Data =<br />
<br />
* [[File:Confidence Interval Convergence.xlsx]]: Shows how precision confidence intervals shrink as sample size increases.<br />
<br />
* [[File:Sigma1RangeStatistics.xls]]: Simulated median, 50%, 80%, and 95% quantiles, plus first four sample moments, for shot groups containing 2 to 100 shots, of: Extreme Spread, Diagonal, Figure of Merit.<br />
<br />
* [[File:SymmetricBivariateSigma1.xls]]: Monte Carlo simulation results validating the [[Closed Form Precision]] math.</div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1444Closed Form Precision2019-05-24T20:28:03Z<p>David: Undo revision 1443 by David (talk)</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} \equiv c_4 = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula.<br />
<br />
=== Multiple Groups ===<br />
<br />
In general we give up two degrees of freedom for every group center we estimate. So if we are aggregating results from ''g'' sample groups then:<br />
:&nbsp; <math>\hat{\sigma} = c_G(2n+1-2g) \sqrt{\frac{\sum r_i^2}{2n-2g}}</math><br />
<br />
[[Rayleigh_sigma_estimate#Known_true_center|When the true center is known]] then this formula is correct with ''g'' = 0.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - ''g'') degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-g)) \leq \chi_L^2) = \alpha/2, \quad Pr(\chi^2(2(n-g)) \leq \chi_U^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2(n-g))</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2(n-g))</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
Note that for large ''n'' the median of this <math>\chi^2</math> distribution is very close to the MLE: <math>\widehat{\sigma} \approx \sqrt{\frac{\sum r^2}{\chi_{50\%}^2}}</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
(In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the ''expected sample median radius'' of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
I.e., the observed sample median tends to be lower than the true median when ''n'' is small.)<br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1443Closed Form Precision2019-05-24T20:24:27Z<p>David: /* Confidence Intervals */ Had alpha terms in initial formula backwards</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} \equiv c_4 = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula.<br />
<br />
=== Multiple Groups ===<br />
<br />
In general we give up two degrees of freedom for every group center we estimate. So if we are aggregating results from ''g'' sample groups then:<br />
:&nbsp; <math>\hat{\sigma} = c_G(2n+1-2g) \sqrt{\frac{\sum r_i^2}{2n-2g}}</math><br />
<br />
[[Rayleigh_sigma_estimate#Known_true_center|When the true center is known]] then this formula is correct with ''g'' = 0.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - ''g'') degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-g)) \leq \chi_L^2) = 1 - \alpha/2, \quad Pr(\chi^2(2(n-g)) \leq \chi_U^2) = \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2(n-g))</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2(n-g))</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
Note that for large ''n'' the median of this <math>\chi^2</math> distribution is very close to the MLE: <math>\widehat{\sigma} \approx \sqrt{\frac{\sum r^2}{\chi_{50\%}^2}}</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
(In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the ''expected sample median radius'' of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
I.e., the observed sample median tends to be lower than the true median when ''n'' is small.)<br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Prior_Art&diff=1442Prior Art2019-04-26T15:29:24Z<p>David: Updated Jeroen's links</p>
<hr />
<div>= [http://www.public.iastate.edu/~jessie/PPB/Stats/Testing%20loads.htm Danielson, 2005, ''Testing loads''] =<br />
[http://www.public.iastate.edu/~jessie/PPB/Stats/Testing%20loads.htm Brent J. Danielson] suggests that a practical way to assess and compare precision is to shoot many 2-shot groups and measure the extreme spread of each.<br />
* When one simply wants to assess whether one sample is more precise than another that probability is given by the one-tailed T-test for two samples with unequal variance -- i.e., the spreadsheet function <code>=1-TTEST({Sample1},{Sample2},1,3)</code>.<br />
* Using the same data it is possible to determine the precision parameter. It is also noteworthy that the square of the 2-tailed T-Test gives the exact confidence range for which the precision parameters of two samples do not overlap -- i.e., the probability that two samples have different precision parameters is <code>=POWER(1-TTEST({Sample1},{Sample2},2,3), 2)</code>. This is all illustrated using Danielson's own data set in [[Media:DanielsonExample.xlsx]].<br />
<br />
= Grubbs, 1964, ''Statistical Measures of Accuracy for Riflemen and Missile Engineers'' =<br />
[[Media:Statistical Measures for Riflemen and Missile Engineers - Grubbs 1964.pdf|''Statistical Measures of Accuracy for Riflemen and Missile Engineers'', Frank E. Grubbs, 1964]].<br />
<br />
= [http://sport-shooters-ots.com/various/ballist/sgs/sgs.htm Hogema, 2005, ''Shot group statistics''] =<br />
[http://sport-shooters-ots.com/various/ballist/sgs/sgs.htm Jeroen Hogema] provides an accessible proof of the equivalence between the symmetric bivariate normal and Rayleigh distributions. He provides extensive examples, simulations, and applications to scoring and load selection, and begins to address the problem of estimating the Rayleigh parameter.<br />
<br />
= [http://sport-shooters-ots.com/various/ballist/precision/Measuring_precision.htm Hogema, 2006, ''Measuring Precision''] =<br />
[http://sport-shooters-ots.com/various/ballist/precision/Measuring_precision.htm ''Picking the most precise ammo, probably.''] Jeroen Hogema:<br />
* Reproduces Leslie’s 1993 results.<br />
* Confirms that for radius measures it is preferable to incorporate all data at once, not to break them into separate groups.<br />
* For FOM and ES it is best to generate many groups so as to preserve more data points.<br />
* Looks at T-tests for significance and shows very large groups are needed to detect statistically meaningful differences.<br />
<br />
= Leslie, 1993, ''Is "Group Size" the Best Measure of Accuracy?'' =<br />
[[Media:Is_Group_Size_the_Best_Measure_of_Accuracy_by_J.E._Leslie_III.pdf|''Is "Group Size" the Best Measure of Accuracy?'', John "Jack" E. Leslie III, 1993]]. Notes:<br />
* Extreme Spread: Maximum distance between any two shots in group. Note that this effectively only uses two data points.<br />
* Figure of Merit (FoM): Average of the maximum horizontal group spread and the maximum vertical group spread. This uses only 2-4 data points depending on the group. Like Diagonal, FoM becomes more efficient than Extreme Spread for larger group sizes.<br />
* Mean Radius: Average distance to center of group for all shots.<br />
* Radial Standard Deviation: Sqrt (Horizontal Variance + Vertical Variance).<br />
* Found military using RSD and Mean Radius as early a 1918.<br />
<br />
His Monte Carlo analysis shows sample RSD to be most efficient predictor of precision, followed closely by Mean Radius. I.e., they can distinguish between loads of different inherent precision more accurately and using fewer sample shots than the other measures.<br />
<br />
Note that, like Grubbs, Leslie estimates MR by sampling the mean of radii. This is less efficient than using the Rayleigh estimator on the radii, and then [[Closed_Form_Precision#Mean_Radius_.28MR.29|computing MR based on the sample Rayleigh parameter]]. The latter process is equally and maximally efficient for all invariant measures that are products of the Rayleigh parameter σ.<br />
<br />
= [http://www.geoffrey-kolbe.com/articles/rimfire_accuracy/group_statistics.htm Kolbe, 2010, ''Group Statistics''] =<br />
Attributing the work to [[Media:Sitton 1990.pdf|Sitton et. al., 1990]], [http://www.geoffrey-kolbe.com/articles/rimfire_accuracy/group_statistics.htm Geoffrey Kolbe] walks through applied statistics to show how many groups of how many shots are required to estimate Extreme Spread to within 10% of the true value with 90% confidence. Using the values from [[Prior_Art#Grubbs.2C_1964.2C_Statistical_Measures_of_Accuracy_for_Riflemen_and_Missile_Engineers|Grubb's]] thousand-iteration simulations he determines that 7-shot groups are the most efficient for estimating Extreme Spread.<br />
<br />
Running [[Range_Statistics#Estimation|the same analysis]] with our [[Media:Sigma1RangeStatistics.xls|million-iteration simulation values]] reveals that 6-shot groups are actually optimal, and not significantly more so than 5-shot groups.<br />
<br />
= [http://www.ar15.com/mobile/topic.html?b=3&f=118&t=279218 Molon, 2006, ''The Trouble With 3-Shot Groups''] =<br />
[http://www.ar15.com/mobile/topic.html?b=3&f=118&t=279218 Through an extended forum thread] Molon offers intuitive explanations and illustrations of the problems with Extreme Spread samples.<br />
<br />
''We have not been able to identify the real person behind that now-inactive user account. If anyone knows please contact us so we can give well-deserved credit!''<br />
<br />
= Taylor & Grubbs, 1975, ''Approximate Probability Distributions for the Extreme Spread'' =<br />
[[Media:Approximate Distributions for Extreme Spread - Taylor 1975.pdf|''Approximate Probability Distributions for the Extreme Spread'', Taylor & Grubbs, 1975]]: notes that there is no closed-form expression for [[Describing_Precision#Extreme_Spread|Extreme Spread]]. Uses Monte Carlo simulations of 10,000 iterations to estimate the first four moments of the distribution of extreme spread for shot groups of size 3 to 34. Checks fit against the Chi, LogNormal, and Weibull distributions.<br />
<br />
Note that in [[File:Sigma1RangeStatistics.xls]] we have generated quantiles and the first four moments for the Extreme Spread using 1,000,000 iteration simulations for groups of size 2 to 100.</div>Davidhttp://ballistipedia.com/index.php?title=File:BallisticAccuracyClassification.xlsx&diff=1441File:BallisticAccuracyClassification.xlsx2017-05-30T17:00:05Z<p>David: David uploaded a new version of File:BallisticAccuracyClassification.xlsx</p>
<hr />
<div>Sample calculation of Ballistic Accuracy Classification based on three 10-shot groups.</div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1440Closed Form Precision2017-05-30T16:57:51Z<p>David: /* Confidence Intervals */ Note chi^2 provides a decent estimate of sigma</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} \equiv c_4 = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula.<br />
<br />
=== Multiple Groups ===<br />
<br />
In general we give up two degrees of freedom for every group center we estimate. So if we are aggregating results from ''g'' sample groups then:<br />
:&nbsp; <math>\hat{\sigma} = c_G(2n+1-2g) \sqrt{\frac{\sum r_i^2}{2n-2g}}</math><br />
<br />
[[Rayleigh_sigma_estimate#Known_true_center|When the true center is known]] then this formula is correct with ''g'' = 0.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - ''g'') degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-g)) \leq \chi_L^2) = \alpha/2, \quad Pr(\chi^2(2(n-g)) \leq \chi_U^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2(n-g))</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2(n-g))</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
Note that for large ''n'' the median of this <math>\chi^2</math> distribution is very close to the MLE: <math>\widehat{\sigma} \approx \sqrt{\frac{\sum r^2}{\chi_{50\%}^2}}</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
(In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the ''expected sample median radius'' of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
I.e., the observed sample median tends to be lower than the true median when ''n'' is small.)<br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1439Closed Form Precision2017-05-30T16:40:20Z<p>David: /* Confidence Intervals */ Added term for number of groups</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} \equiv c_4 = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula.<br />
<br />
=== Multiple Groups ===<br />
<br />
In general we give up two degrees of freedom for every group center we estimate. So if we are aggregating results from ''g'' sample groups then:<br />
:&nbsp; <math>\hat{\sigma} = c_G(2n+1-2g) \sqrt{\frac{\sum r_i^2}{2n-2g}}</math><br />
<br />
[[Rayleigh_sigma_estimate#Known_true_center|When the true center is known]] then this formula is correct with ''g'' = 0.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - ''g'') degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-g)) \leq \chi_L^2) = \alpha/2, \quad Pr(\chi^2(2(n-g)) \leq \chi_U^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2(n-g))</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2(n-g))</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
(In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the ''expected sample median radius'' of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
I.e., the observed sample median tends to be lower than the true median when ''n'' is small.)<br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1438Closed Form Precision2017-05-30T16:35:50Z<p>David: /* Rayleigh Estimates */</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} \equiv c_4 = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula.<br />
<br />
=== Multiple Groups ===<br />
<br />
In general we give up two degrees of freedom for every group center we estimate. So if we are aggregating results from ''g'' sample groups then:<br />
:&nbsp; <math>\hat{\sigma} = c_G(2n+1-2g) \sqrt{\frac{\sum r_i^2}{2n-2g}}</math><br />
<br />
[[Rayleigh_sigma_estimate#Known_true_center|When the true center is known]] then this formula is correct with ''g'' = 0.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - 1) degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-1)) \leq \chi_L^2) = \alpha/2, \quad Pr(\chi^2(2(n-1)) \leq \chi_U^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2n-2)</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2n-2)</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
(In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the ''expected sample median radius'' of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
I.e., the observed sample median tends to be lower than the true median when ''n'' is small.)<br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1437Closed Form Precision2017-05-30T16:29:34Z<p>David: /* Rayleigh Estimates */ Give general equation for multiple groups</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} \equiv c_4 = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula.<br />
<br />
=== Multiple Groups ===<br />
<br />
In general we give up two degrees of freedom for every group center we estimate. So if we are aggregating results from ''g'' sample groups then:<br />
:&nbsp; <math>\hat{\sigma} = c_G(2n+1-2g) \sqrt{\frac{\sum r_i^2}{2n-2g}}</math><br />
<br />
[[Rayleigh_sigma_estimate#Known_true_center|When the true center is known]] then this formula is correct with g = 0.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - 1) degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-1)) \leq \chi_L^2) = \alpha/2, \quad Pr(\chi^2(2(n-1)) \leq \chi_U^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2n-2)</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2n-2)</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
(In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the ''expected sample median radius'' of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
I.e., the observed sample median tends to be lower than the true median when ''n'' is small.)<br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1436Closed Form Precision2017-05-30T16:14:14Z<p>David: /* Gaussian correction factor */ Make it clear what conventional c4 refers to</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} \equiv c_4 = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula, as well as a variation for the case in which the true center is known.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - 1) degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-1)) \leq \chi_L^2) = \alpha/2, \quad Pr(\chi^2(2(n-1)) \leq \chi_U^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2n-2)</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2n-2)</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
(In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the ''expected sample median radius'' of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
I.e., the observed sample median tends to be lower than the true median when ''n'' is small.)<br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Rayleigh_sigma_estimate&diff=1435Rayleigh sigma estimate2017-05-30T15:27:21Z<p>David: Section headers</p>
<hr />
<div>= Estimating parameter <math>\sigma</math> for the Rayleigh distribution =<br />
<br />
Let <math>X, Y</math> be uncorrelated, jointly normally distributed cartesian coordinates with means <math>\mu_{X}, \mu_{Y}</math>, and equal variances <math>\sigma_{X}^{2} = \sigma_{Y}^{2}</math>. The radius of an <math>(X,Y)</math>-coordinate pair around the true mean is:<br />
<br />
<math><br />
R := \sqrt{(X-\mu_{X})^{2} + (Y-\mu_{Y})^{2}}<br />
</math><br />
<br />
With <math>N</math> observations of <math>(x,y)</math>-coordinates, <math>\text{SSR}</math> is the sum of squared radii:<br />
<br />
<math><br />
\begin{array}{rcl}<br />
\text{SSR} &:=& \sum\limits_{i=1}^{N} r_{i}^{2} = \sum\limits_{i=1}^{N} ((x_{i}-\mu_{X})^{2} + (y_{i}-\mu_{Y})^{2})\\<br />
&=& \sum\limits_{i=1}^{N} (x_{i}-\mu_{X})^{2} + \sum\limits_{i=1}^{N} (y_{i}-\mu_{Y})^{2}<br />
\end{array}<br />
</math><br />
<br />
The Maximum-Likelihood-estimate of the variance of <math>r</math> is the total variance of all <math>2N</math> separate <math>x</math>- and <math>y</math>-coordinates:<br />
<br />
<math><br />
\widehat{\sigma^{2}_{ML}} = \frac{1}{2N} \cdot \text{SSR}<br />
</math><br />
<br />
== Unknown true center ==<br />
=== Deriving Singh's (1992) <math>C_{2}</math> estimator ===<br />
<br />
When the <math>(\mu_{X}, \mu_{Y})</math>-center is estimated by <math>(\bar{x}, \bar{y})</math>, the [[Closed_Form_Precision#Bessel_correction_factor|Bessel correction]] is required to make the ML-estimate of the variance unbiased:<br />
<br />
<math><br />
\widehat{\sigma^{2}_{ub}} = \frac{N}{N-1} \cdot \frac{1}{2N} \cdot \text{SSR} = \frac{1}{2(N-1)} \cdot \text{SSR}<br />
</math><br />
<br />
However, taking the square root of the unbiased variance estimate makes it biased ([http://en.wikipedia.org/wiki/Jensen%27s_inequality Jensen's inequality]). Specifically, since the square root is concave, the bias is negative and <math>\sqrt{\widehat{\sigma^{2}_{ub}}}</math> underestimates <math>\sigma</math>. Another correction factor is needed to counter this effect. To this end, let <math>Q</math> be a <math>\chi</math>-distributed variable with <math>k</math> degrees of freedom. Then its mean is:<br />
<br />
<math><br />
E(Q) = \sqrt{2} \cdot \frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}\right)}<br />
</math><br />
<br />
When <math>X</math> is a normally distributed random variable with <math>n</math> observations, and <math>s^{2} := \frac{1}{n-1} \sum\limits_{i=1}^{n}(x-\bar{x})^{2}</math> is the Bessel-corrected variance estimate, <math>E(s^{2}) = \sigma^{2}</math>. Then [[Closed_Form_Precision#Gaussian_correction_factor|<math>c_{4}(n)</math> is the correction factor]] such that <math>E(s) = c_{4}(n) \cdot \sigma</math>. With <math>Q</math> as given above, <math>c_{4}(n) = \frac{1}{\sqrt{k}} \cdot E(Q)</math> with <math>k := n-1</math>:<br />
<br />
<math><br />
c_{4}(n) = \sqrt{\frac{2}{n-1}} \cdot \frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}<br />
</math><br />
<br />
We lose 2 degrees of freedom estimating the center. Therefore we really only have 2N - 2 degrees of freedom, so we set <math>n := 2N-1</math> so that <math>c_{4}(n)</math> is the scaled mean of a <math>\chi</math>-distributed variable with <math>k = n-1 = 2N-1-1 = 2N-2</math> degrees of freedom. This gives:<br />
<br />
<math><br />
\begin{array}{rcl}<br />
\widehat{\sigma_{ub}} &=& \frac{1}{c_{4}(2N-1)} \cdot \sqrt{\widehat{\sigma^{2}_{ub}}} \\<br />
&=& \frac{1}{\sqrt{\frac{2}{2N-1-1}} \cdot \frac{\Gamma\left(\frac{2N-1}{2}\right)}{\Gamma\left(\frac{2N-1-1}{2}\right)}} \cdot \sqrt{\frac{N}{N-1} \cdot \frac{1}{2N} \cdot \text{SSR}} \\<br />
&=& \sqrt{\frac{2(N-1)}{2}} \cdot \frac{\Gamma\left(\frac{2(N-1)}{2}\right)}{\Gamma\left(\frac{2N-1}{2}\right)} \cdot \sqrt{\frac{N}{N-1} \cdot \frac{1}{2N} \cdot \text{SSR}} \\ <br />
&=& \sqrt{N} \cdot \frac{\Gamma(N-1)}{\Gamma\left(\frac{2N-1}{2}\right)} \cdot \sqrt{\frac{1}{2N} \cdot \text{SSR}} \\<br />
&=& \frac{\Gamma(N-1)}{\Gamma\left(N - \frac{1}{2}\right)} \cdot \sqrt{\frac{1}{2} \cdot \text{SSR}}<br />
\end{array}<br />
</math><br />
<br />
The penultimate expression is [[CEP_literature#Singh1992|Singh's 1992]] definition 3.3 for estimator <math>C_{2}</math>, also given by [[CEP_literature#Moranda1959|Moranda (1959)]].<br />
<br />
== Known true center ==<br />
=== Deriving Singh's (1992) <math>C_{1}</math> estimator ===<br />
<br />
When the <math>(\mu_{X}, \mu_{Y})</math>-center is known, the Bessel correction is '''not''' required to make the Maximum-Likelihood-estimate of the variance unbiased. In this case we simply set <math>n := 2N+1</math> for the <math>c_{4}(n)</math> correction factor so that <math>c_{4}(n)</math> is the scaled mean of a <math>\chi</math>-distributed variable with <math>k = n-1 = 2N+1-1 = 2N</math> degrees of freedom. This gives:<br />
<br />
<math><br />
\begin{array}{rcl}<br />
\widehat{\sigma_{ub}} &=& \frac{1}{c_{4}(2N+1)} \cdot \sqrt{\widehat{\sigma^{2}_{ub}}} \\<br />
&=& \frac{1}{\sqrt{\frac{2}{2N+1-1}} \cdot \frac{\Gamma\left(\frac{2N+1}{2}\right)}{\Gamma\left(\frac{2N+1-1}{2}\right)}} \cdot \sqrt{\frac{1}{2N} \cdot \text{SSR}} \\<br />
&=& \sqrt{N} \cdot \frac{\Gamma(N)}{\Gamma\left(\frac{2N+1}{2}\right)} \cdot \sqrt{\frac{1}{2N} \cdot \text{SSR}} \\ <br />
&=& \frac{\Gamma(N)}{\Gamma\left(N + \frac{1}{2}\right)} \cdot \sqrt{\frac{1}{2} \cdot \text{SSR}}<br />
\end{array}<br />
</math><br />
<br />
The penultimate expression is [[CEP_literature#Singh1992|Singh's 1992]] definition 2.2 for estimator <math>C_{1}</math>.</div>Davidhttp://ballistipedia.com/index.php?title=File:BallisticAccuracyClassification.xlsx&diff=1434File:BallisticAccuracyClassification.xlsx2017-05-26T18:56:06Z<p>David: David uploaded a new version of File:BallisticAccuracyClassification.xlsx</p>
<hr />
<div>Sample calculation of Ballistic Accuracy Classification based on three 10-shot groups.</div>Davidhttp://ballistipedia.com/index.php?title=Home&diff=1433Home2017-05-18T12:41:21Z<p>David: </p>
<hr />
<div>This site explains and demonstrates statistics for analyzing the precision of projectile weapon systems.<ref name="nuance">Typical examples would be target shooting with a rifle or pistol. Such weapons as shotguns, mortars, and ballistic missiles would have some similar characteristics, but also have factors that are neglected in the discussions and measurements. The wiki will discuss some factors of ballistics, but it is not intended to address all the nuances of internal, external, or terminal ballistics. Rather, the focus is primarily on the analysis of the precision of the whole weapon system which can be observed directly by the relative impact points on a target. Some effort will be made to explore the precision of weapons subsystems.</ref><br />
<br />
High level topics, which are good places to start exploring the site, include:<br />
* [[What is Precision?]]: An important explanation of the difference between precision and accuracy as the terms are used in statistics<br />
* [[Describing Precision]]: Units, terms, and relationships<br />
* [[Precision Models]]: Statistical approaches for efficient estimation and inference of precision <br />
* [[Prior Art]]: Reviews of past efforts to address this question<br />
* [[FAQ]]<br />
* [[Ballistic Accuracy Classification]]: A proposed industry standard for determining and describing precision<br />
<br />
= Synopsis =<br />
<br />
When testing a gun, shooter, and/or ammunition the most popular measure is [[Range Statistics#Extreme Spread|Extreme Spread]] or "group size" of a sample of target shots. However Extreme Spread must be used with care since it is frequently and easily abused. As with all measures, the single best measurement is meaningless in isolation. The proper statistical estimator is an "average" (a.k.a., ''expected value'') of the measurement.<br />
<br />
Another consideration is that some measures, such as Extreme Spread, change value when there are more shots in a group. Measures that have such a dependency will be referred to here as ''variant measures''. There are ''[[Describing_Precision#Invariant_Measures|invariant measures]]'', like [[Circular Error Probable]] or Mean Radius, for which the expected values do not change with the number of shots in a group. Instead [[Precision_Models#How_large_a_sample_do_we_need.3F|having more shots only increases the confidence in the measure's value]]. Of course the experimental error of either type of measurement can also be decreased by increasing the sample size (i.e., shooting more groups). <br />
<br />
Furthermore, by first making assumptions about the inherent shot dispersion, then it is possible to use theoretical models to estimate measurements and their precision. The distributions are of two basic types: If the expected values and the expected precision factor for the measurements depend on distributions which have an [[Closed Form Precision|explicit solution]] then the values can be calculated formulaicly. If the values don't have a distribution with a closed form expression then they can be estimated via Monte Carlo approaches.<br />
<br />
[[:Category:Examples|Examples]] of the application of these [[Precision Models|methods]] and [[Measuring Tools|tools]] include:<br />
* Determining how many sighter shots you should take.<br />
* Determining the likelihood of a hit on a particular target by a zeroed shooting system.<br />
* Comparing the inherent precision of different shooting systems.<br />
* Determining which ammunition shoots better in a particular gun.<br />
<br />
----<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1432Closed Form Precision2017-05-15T18:45:28Z<p>David: /* Circular Error Probable (CEP) */ Clarification that this shows bias in observed median for small n</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula, as well as a variation for the case in which the true center is known.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - 1) degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-1)) \leq \chi_L^2) = \alpha/2, \quad Pr(\chi^2(2(n-1)) \leq \chi_U^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2n-2)</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2n-2)</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
(In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the ''expected sample median radius'' of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
I.e., the observed sample median tends to be lower than the true median when ''n'' is small.)<br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Ballistic_Accuracy_Classification&diff=1431Ballistic Accuracy Classification2017-05-10T15:26:11Z<p>David: /* Classifying a Specimen */ Correction to degrees of freedom for multiple groups</p>
<hr />
<div>== Introduction ==<br />
<br />
Ballistic Accuracy Classification™ is a mathematically rigorous system for describing and understanding the precision of ballistic tools like rifles. It provides information that is:<br />
* Useful and easy for consumers to understand<br />
* Straightforward for enthusiasts and builders to calculate<br />
* Statistically sound enough for experts to validate<br />
''For more background see [[Why BAC]].''<br />
<br />
All firearms and components can be assigned a Ballistic Accuracy Class™ (BAC™), which fully characterizes their accuracy potential. Lower numbers are better. In practice, the lowest possible BAC is 1. (A theoretically perfect rifle system that always puts every shot through the same hole would be BAC 0.)<br />
<br />
Behind the scenes, BAC is defined by a single statistical parameter known as sigma (σ). Using [[Closed_Form_Precision#Symmetric_Bivariate_Normal_Shots_Imply_Rayleigh_Distributed_Distances|the associated statistical model]] (known as the Rayleigh distribution), statisticians can not only determine the BAC for a particular gun or component, but also compute its expected shooting precision.<ref>The proportion of shots expected to fall within radius ''rσ'' of the center of impact is <math>1-e^{-r^2/2}</math>.</ref><br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! Sigma (σ)<br />
! Typical examples<br />
! Shots within<BR/>¼ MOA radius<br />
! Shots within<BR/>½ MOA radius<br />
|-<br />
| Class 1<br />
| < 0.1MOA<br />
| Rail guns<br />
| 96%<br />
| 100%<br />
|-<br />
| Class 2<br />
| < 0.2MOA<br />
| Benchrest guns<br />
| 54%<br />
| 96%<br />
|-<br />
| Class 3<br />
| < 0.3MOA<br />
| Mil-spec for PSR<br />
| 29%<br />
| 75%<br />
|-<br />
| Class 4<br />
| < 0.4MOA<br />
| Competitive auto-loaders<br />
| 18%<br />
| 54%<br />
|-<br />
| Class 5<br />
| < 0.5MOA<br />
| Mil-spec for M110 and M24<br />
| 12%<br />
| 39%<br />
|-<br />
| Class 6<br />
| < 0.6MOA<br />
| Mil-spec for infantry rifles and ammo<br />
| 8%<br />
| 29%<br />
|}<br />
<br />
We can also generate the expected values of more familiar measures, like the extreme spread of a 3- or 5-shot group:<br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! 5-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 5-shot<BR/> Group Spread<br />
! 3-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 3-shot<BR/> Group Spread<br />
|-<br />
| Class 1<br />
| 100%<br />
| 0.3MOA<br />
| 100%<br />
| 0.2MOA<br />
|-<br />
| Class 2<br />
| 98%<br />
| 0.6MOA<br />
| 99%<br />
| 0.5MOA<br />
|-<br />
| Class 3<br />
| 65%<br />
| 0.9MOA<br />
| 85%<br />
| 0.7MOA<br />
|-<br />
| Class 4<br />
| 26%<br />
| 1.2MOA<br />
| 57%<br />
| 0.9MOA<br />
|-<br />
| Class 5<br />
| 9%<br />
| 1.5MOA<br />
| 35%<br />
| 1.2MOA<br />
|-<br />
| Class 6<br />
| 3%<br />
| 1.8MOA<br />
| 21%<br />
| 1.4MOA<br />
|}<br />
<br />
=== Understanding MOA ===<br />
Accuracy is described in [[Angular_Size|angular terms]], most commonly using the unit "Minute of Arc" (MOA). One arc minute spans 1.047" at 100 yards. Rifle shooters often practice on 100 yard targets, and so they often think in terms of how wide their groups are at 100 yards. People often just round it off and think of 1 MOA as "one inch at 100 yards."<br />
<br />
In the absence of an atmosphere, the angular precision measured at one distance would be valid at all other distances. I.e., a 1" group at 100 yards would measure 5" at 500 yards. However, in reality the effects of wind and drag (which, together with gravity, accentuates variations in muzzle velocity) will only increase the angular spread of ballistic groups as distance increases. Therefore, '''in practice one should expect worse-than-advertised accuracy when shooting at longer distances'''. The distance at which atmosphere begins to significantly affect precision depends on a bullet’s muzzle velocity and ballistic coefficient. For high-power rifles this is usually beyond 100 yards. Guns shooting subsonic projectiles can begin to suffer after just 25 yards. <br />
<br />
=== Nomenclature ===<br />
<br />
BAC™ is only meaningful when it conforms to established terms and conventions. BAC must be determined by testing in accordance with [[#Protocol|the BAC Protocol]] described in this document.<br />
<br />
Ballistic Accuracy Classification™ must be supported by the following descriptive parameters:<br />
# Product tested. (Any component, or group of components, associated with accuracy can be tested.)<br />
# Configuration tested. This must include the following details:<br />
## Barrel length, material, profile, rifling.<br/>(E.g., ''20" stainless 1" bull contour with 6-land 1:10"-twist cut rifling''.)<br />
## Receiver, action, and feed mechanism.<br/>(E.g., ''AR-10 magazine-fed semi-automatic''.)<br />
## Ammunition.<br />
### If commercial this must include brand, model, and lot.<br/>(E.g., ''Federal GM308M Lot#214374H077''.)<br />
### If custom, this must list component and load formula.<br/>(E.g., ''Lapua .308 full-sized brass, WLR primer, 168gr SMK, 44gr Varget, 2.800" COAL''.)<br />
# Confidence in BAC. When not conspicuously mentioned, it is assumed that the upper '''90%''' confidence value of sigma is referenced for BAC.<br />
<br />
BAC measures are intentionally kept somewhat coarse. Care should be taken to avoid suggesting more than two significant digits of precision. For example, even after shooting 20 rounds through a gun, [[Closed_Form_Precision#How_large_a_sample_do_we_need.3F|the 80% confidence interval on the precision estimate typically spans 0.9-1.2 times the estimated value]].<br />
<br />
=== Trademarks ===<br />
The following terms are trademarks of Scribe Logistics LLC. They are free to use so long as their use complies with the Nomenclature and Protocol outlined here.<br />
* Ballistic Accuracy Classification™<br />
* Ballistic Accuracy Class™<br />
* BAC™<br />
The trademarks are claimed solely for the purpose of maintaining the integrity of the system and avoiding market confusion.<br />
<br />
== Theory ==<br />
<br />
=== Statistical Model ===<br />
<br />
BAC™ assumes that the impact of ballistic shots on a target are normally distributed with the same variance along any axis. (Empirical data validate this assumption, and it should be true as long as atmospheric effects are negligible.<ref>There are two atmospheric effects that can finally create excess variance in one axis: Variable wind will increase horizontal variance. This is a function of the wind speed and the projectile's time-of-flight to the target. Also, as time-of-flight increases, variations in muzzle velocity will appear as excess vertical variance. Since both of these effects depend on time-of-flight, they are typically negligible for high-velocity rifles before 100 yards, or for subsonic projectiles before 25 yards.</ref>) Therefore, [[Closed_Form_Precision#Symmetric_Bivariate_Normal_Shots_Imply_Rayleigh_Distributed_Distances|we use the Rayleigh distribution to model the radius r]], or dispersion of each shot, from the center of impact. When the coordinates of the shots have independent <math>N(0,\sigma)</math> distributions along orthogonal axes, the radius of each shot is described by the Rayleigh probability density function:<br />
:&nbsp; <math>f(r,\sigma)=\frac{r}{\sigma^2}e^{−r^2/2\sigma^2}</math><br />
The [[Closed_Form_Precision#Rayleigh_Estimates|unbiased estimator for the parameter σ]] comes from <math>\widehat{\sigma^2} = \frac{\sum r_i^2}{2(n−1)}</math>, with confidence <math>\widehat{\sigma^2} \sim \frac{\sum r^2}{\chi_{2n-2}^2}</math>.<br />
<br />
=== Simulation ===<br />
Monte Carlo simulation is adequate for studying and characterizing precision. In fact, many of the results associated with BAC, like the distribution of the extreme spread of a particular number of shots, can only be produced through simulation.<br />
<br />
For simulation purposes random shots should be generated as (x, y) coordinates, where <math>X,Y \sim N(0,\sigma)</math>. It is critical to "forget" the known center when using simulated data. When shooting a real gun we never get to know the true center of impact, and instead have to use the sample center. Likewise, Monte Carlo simulations must not reference the known 0 center, and should instead only reference the sample center of whatever group size is being studied. <br />
<br />
== Protocol ==<br />
The Ballistic Accuracy Classification™ shall be the upper bound of the 90% confidence range on estimated sigma, in units of MOA, multiplied by 10 and rounded to the nearest integer. For example, if the 90% confidence value for sigma on a tested gun is 0.47MOA, then the BAC value is 10 ✕ 0.47 = 4.7, rounded = 5. I.e., in this example we are saying with 90% confidence that the tested gun’s accuracy is no worse than Class 5.<br />
<br />
=== Classifying a Specimen ===<br />
It is not realistic to assign a BAC with fewer than 10 shots. The 90% confidence range with just 10 shots will typically extend to 1.4 times the estimated accuracy value. It will typically take 20 shots to get the outside bound of the 90% confidence interval to within 20% of the estimated value.<ref>The U.S. Army Marksmanship Unit (AMU) has long used a minimum of 3 consecutive 10-shot groups fired from a machine rest to test the accuracy of service rifles.</ref><br />
<br />
You must not discard data points during testing except for an unrelated failure. (E.g., if you are testing a barrel and you encounter a squib load, that shot may be excluded. But "[[fliers]]" should not generally be excluded.)<br />
<br />
'''Data:''' Target distance, and (''x,y'') coordinates of the center of each shot impact. All shots with the same point of aim must be grouped, but multiple groups can be used.<br />
<br />
'''Calculations:''' The formulas to transform these data into a confidence interval for sigma, and the corresponding BAC, are shown in [[Media:BallisticAccuracyClassification.xlsx]]. Given a sample of ''n'' shots, over ''g'' groups, at a distance ''d'':<br />
# For each measurement, convert to units of MOA. For example, if measurements are taken in inches, and the target was shot at a distance of ''d'' yards, then divide each measurement by 0.01047''d''<br />
# For each group ''g'', calculate the center of the group as <math>(\bar{x}_{i \in g},\bar{y}_{i \in g})</math><br />
# For each shot ''i'', find its radius squared relative to the center of the group as <math>r_i^2=(x_i−x_g)^2+(y_i−y_g)^2</math><br />
# Calculate the upper 90% confidence value for sigma as <code>σ<sub>U</sub>=SQRT[SUM(r2)/CHIINV(0.9,2n-2g)]</code><br />
# Calculate the Ballistic Accuracy Class as <code>=ROUND(σ<sub>U</sub>,0)</code><br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Category:Examples&diff=1430Category:Examples2017-05-07T15:08:07Z<p>David: </p>
<hr />
<div>* [[Closed_Form_Precision#How_many_sighter_shots_do_you_need.3F|How many sighter shots do you need]]?<br />
* [[Closed_Form_Precision#The_3-shot_Group|What the 3-shot group says about precision]].<br />
* [[Closed_Form_Precision#Typical_values_of_.CF.83|Typical real-world precision]].<br />
* [[Range_Statistics#Example_1|Expected Extreme Spread]].<br />
* [[Range_Statistics#Example:_Statistical_Inference|Statistical Inference from Extreme Spreads]].<br />
* [http://lstange.github.io/mcgs/ Order statistics for easily calculating 90% hit probability].<br />
* [[Ballistic Accuracy Classification]]<br />
* [http://www.myeley.com/lot-analyzer/919/information#lot-statistics Eley's .22LR Lot Analysis]<br />
<br />
= Sample Targets and Scenarios with Precision Analysis =</div>Davidhttp://ballistipedia.com/index.php?title=Ballistic_Accuracy_Classification&diff=1429Ballistic Accuracy Classification2017-04-23T18:56:15Z<p>David: /* Classifying a Specimen */ Removed incorrect "correction" factor from confidence value</p>
<hr />
<div>== Introduction ==<br />
<br />
Ballistic Accuracy Classification™ is a mathematically rigorous system for describing and understanding the precision of ballistic tools like rifles. It provides information that is:<br />
* Useful and easy for consumers to understand<br />
* Straightforward for enthusiasts and builders to calculate<br />
* Statistically sound enough for experts to validate<br />
''For more background see [[Why BAC]].''<br />
<br />
All firearms and components can be assigned a Ballistic Accuracy Class™ (BAC™), which fully characterizes their accuracy potential. Lower numbers are better. In practice, the lowest possible BAC is 1. (A theoretically perfect rifle system that always puts every shot through the same hole would be BAC 0.)<br />
<br />
Behind the scenes, BAC is defined by a single statistical parameter known as sigma (σ). Using [[Closed_Form_Precision#Symmetric_Bivariate_Normal_Shots_Imply_Rayleigh_Distributed_Distances|the associated statistical model]] (known as the Rayleigh distribution), statisticians can not only determine the BAC for a particular gun or component, but also compute its expected shooting precision.<ref>The proportion of shots expected to fall within radius ''rσ'' of the center of impact is <math>1-e^{-r^2/2}</math>.</ref><br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! Sigma (σ)<br />
! Typical examples<br />
! Shots within<BR/>¼ MOA radius<br />
! Shots within<BR/>½ MOA radius<br />
|-<br />
| Class 1<br />
| < 0.1MOA<br />
| Rail guns<br />
| 96%<br />
| 100%<br />
|-<br />
| Class 2<br />
| < 0.2MOA<br />
| Benchrest guns<br />
| 54%<br />
| 96%<br />
|-<br />
| Class 3<br />
| < 0.3MOA<br />
| Mil-spec for PSR<br />
| 29%<br />
| 75%<br />
|-<br />
| Class 4<br />
| < 0.4MOA<br />
| Competitive auto-loaders<br />
| 18%<br />
| 54%<br />
|-<br />
| Class 5<br />
| < 0.5MOA<br />
| Mil-spec for M110 and M24<br />
| 12%<br />
| 39%<br />
|-<br />
| Class 6<br />
| < 0.6MOA<br />
| Mil-spec for infantry rifles and ammo<br />
| 8%<br />
| 29%<br />
|}<br />
<br />
We can also generate the expected values of more familiar measures, like the extreme spread of a 3- or 5-shot group:<br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! 5-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 5-shot<BR/> Group Spread<br />
! 3-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 3-shot<BR/> Group Spread<br />
|-<br />
| Class 1<br />
| 100%<br />
| 0.3MOA<br />
| 100%<br />
| 0.2MOA<br />
|-<br />
| Class 2<br />
| 98%<br />
| 0.6MOA<br />
| 99%<br />
| 0.5MOA<br />
|-<br />
| Class 3<br />
| 65%<br />
| 0.9MOA<br />
| 85%<br />
| 0.7MOA<br />
|-<br />
| Class 4<br />
| 26%<br />
| 1.2MOA<br />
| 57%<br />
| 0.9MOA<br />
|-<br />
| Class 5<br />
| 9%<br />
| 1.5MOA<br />
| 35%<br />
| 1.2MOA<br />
|-<br />
| Class 6<br />
| 3%<br />
| 1.8MOA<br />
| 21%<br />
| 1.4MOA<br />
|}<br />
<br />
=== Understanding MOA ===<br />
Accuracy is described in [[Angular_Size|angular terms]], most commonly using the unit "Minute of Arc" (MOA). One arc minute spans 1.047" at 100 yards. Rifle shooters often practice on 100 yard targets, and so they often think in terms of how wide their groups are at 100 yards. People often just round it off and think of 1 MOA as "one inch at 100 yards."<br />
<br />
In the absence of an atmosphere, the angular precision measured at one distance would be valid at all other distances. I.e., a 1" group at 100 yards would measure 5" at 500 yards. However, in reality the effects of wind and drag (which, together with gravity, accentuates variations in muzzle velocity) will only increase the angular spread of ballistic groups as distance increases. Therefore, '''in practice one should expect worse-than-advertised accuracy when shooting at longer distances'''. The distance at which atmosphere begins to significantly affect precision depends on a bullet’s muzzle velocity and ballistic coefficient. For high-power rifles this is usually beyond 100 yards. Guns shooting subsonic projectiles can begin to suffer after just 25 yards. <br />
<br />
=== Nomenclature ===<br />
<br />
BAC™ is only meaningful when it conforms to established terms and conventions. BAC must be determined by testing in accordance with [[#Protocol|the BAC Protocol]] described in this document.<br />
<br />
Ballistic Accuracy Classification™ must be supported by the following descriptive parameters:<br />
# Product tested. (Any component, or group of components, associated with accuracy can be tested.)<br />
# Configuration tested. This must include the following details:<br />
## Barrel length, material, profile, rifling.<br/>(E.g., ''20" stainless 1" bull contour with 6-land 1:10"-twist cut rifling''.)<br />
## Receiver, action, and feed mechanism.<br/>(E.g., ''AR-10 magazine-fed semi-automatic''.)<br />
## Ammunition.<br />
### If commercial this must include brand, model, and lot.<br/>(E.g., ''Federal GM308M Lot#214374H077''.)<br />
### If custom, this must list component and load formula.<br/>(E.g., ''Lapua .308 full-sized brass, WLR primer, 168gr SMK, 44gr Varget, 2.800" COAL''.)<br />
# Confidence in BAC. When not conspicuously mentioned, it is assumed that the upper '''90%''' confidence value of sigma is referenced for BAC.<br />
<br />
BAC measures are intentionally kept somewhat coarse. Care should be taken to avoid suggesting more than two significant digits of precision. For example, even after shooting 20 rounds through a gun, [[Closed_Form_Precision#How_large_a_sample_do_we_need.3F|the 80% confidence interval on the precision estimate typically spans 0.9-1.2 times the estimated value]].<br />
<br />
=== Trademarks ===<br />
The following terms are trademarks of Scribe Logistics LLC. They are free to use so long as their use complies with the Nomenclature and Protocol outlined here.<br />
* Ballistic Accuracy Classification™<br />
* Ballistic Accuracy Class™<br />
* BAC™<br />
The trademarks are claimed solely for the purpose of maintaining the integrity of the system and avoiding market confusion.<br />
<br />
== Theory ==<br />
<br />
=== Statistical Model ===<br />
<br />
BAC™ assumes that the impact of ballistic shots on a target are normally distributed with the same variance along any axis. (Empirical data validate this assumption, and it should be true as long as atmospheric effects are negligible.<ref>There are two atmospheric effects that can finally create excess variance in one axis: Variable wind will increase horizontal variance. This is a function of the wind speed and the projectile's time-of-flight to the target. Also, as time-of-flight increases, variations in muzzle velocity will appear as excess vertical variance. Since both of these effects depend on time-of-flight, they are typically negligible for high-velocity rifles before 100 yards, or for subsonic projectiles before 25 yards.</ref>) Therefore, [[Closed_Form_Precision#Symmetric_Bivariate_Normal_Shots_Imply_Rayleigh_Distributed_Distances|we use the Rayleigh distribution to model the radius r]], or dispersion of each shot, from the center of impact. When the coordinates of the shots have independent <math>N(0,\sigma)</math> distributions along orthogonal axes, the radius of each shot is described by the Rayleigh probability density function:<br />
:&nbsp; <math>f(r,\sigma)=\frac{r}{\sigma^2}e^{−r^2/2\sigma^2}</math><br />
The [[Closed_Form_Precision#Rayleigh_Estimates|unbiased estimator for the parameter σ]] comes from <math>\widehat{\sigma^2} = \frac{\sum r_i^2}{2(n−1)}</math>, with confidence <math>\widehat{\sigma^2} \sim \frac{\sum r^2}{\chi_{2n-2}^2}</math>.<br />
<br />
=== Simulation ===<br />
Monte Carlo simulation is adequate for studying and characterizing precision. In fact, many of the results associated with BAC, like the distribution of the extreme spread of a particular number of shots, can only be produced through simulation.<br />
<br />
For simulation purposes random shots should be generated as (x, y) coordinates, where <math>X,Y \sim N(0,\sigma)</math>. It is critical to "forget" the known center when using simulated data. When shooting a real gun we never get to know the true center of impact, and instead have to use the sample center. Likewise, Monte Carlo simulations must not reference the known 0 center, and should instead only reference the sample center of whatever group size is being studied. <br />
<br />
== Protocol ==<br />
The Ballistic Accuracy Classification™ shall be the upper bound of the 90% confidence range on estimated sigma, in units of MOA, multiplied by 10 and rounded to the nearest integer. For example, if the 90% confidence value for sigma on a tested gun is 0.47MOA, then the BAC value is 10 ✕ 0.47 = 4.7, rounded = 5. I.e., in this example we are saying with 90% confidence that the tested gun’s accuracy is no worse than Class 5.<br />
<br />
=== Classifying a Specimen ===<br />
It is not realistic to assign a BAC with fewer than 10 shots. The 90% confidence range with just 10 shots will typically extend to 1.4 times the estimated accuracy value. It will typically take 20 shots to get the outside bound of the 90% confidence interval to within 20% of the estimated value.<ref>The U.S. Army Marksmanship Unit (AMU) has long used a minimum of 3 consecutive 10-shot groups fired from a machine rest to test the accuracy of service rifles.</ref><br />
<br />
You must not discard data points during testing except for an unrelated failure. (E.g., if you are testing a barrel and you encounter a squib load, that shot may be excluded. But "[[fliers]]" should not generally be excluded.)<br />
<br />
'''Data:''' Target distance, and (''x,y'') coordinates of the center of each shot impact. All shots with the same point of aim must be grouped, but multiple groups can be used.<br />
<br />
'''Calculations:''' The formulas to transform these data into a confidence interval for sigma, and the corresponding BAC, are shown in [[Media:BallisticAccuracyClassification.xlsx]]. Given a sample of ''n'' shots, over ''g'' groups, at a distance ''d'':<br />
# For each measurement, convert to units of MOA. For example, if measurements are taken in inches, and the target was shot at a distance of ''d'' yards, then divide each measurement by 0.01047''d''<br />
# For each group ''g'', calculate the center of the group as <math>(\bar{x}_{i \in g},\bar{y}_{i \in g})</math><br />
# For each shot ''i'', find its radius squared relative to the center of the group as <math>r_i^2=(x_i−x_g)^2+(y_i−y_g)^2</math><br />
# Calculate the upper 90% confidence value for sigma as <code>σ<sub>U</sub>=SQRT[SUM(r2)/CHIINV(0.9,2n-2)]</code><br />
# Calculate the Ballistic Accuracy Class as <code>=ROUND(σ<sub>U</sub>,0)</code><br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1428Closed Form Precision2017-04-23T18:54:13Z<p>David: /* Confidence Intervals */ Switched to less confusing subscripts for chi^2 values</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula, as well as a variation for the case in which the true center is known.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - 1) degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_L^2, \ \chi_U^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-1)) \leq \chi_L^2) = \alpha/2, \quad Pr(\chi^2(2(n-1)) \leq \chi_U^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_L^2</math> = <code>CHIINV(α/2, 2n-2)</code>,<math>\quad \chi_U^2</math> = <code>CHIINV((1-α/2), 2n-2)</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_L^2}, \ \frac{2(n-1) s^2}{\chi_U^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_L^2}, \ \frac{\sum r^2}{\chi_U^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_L^2}}, \ \sqrt{\frac{\sum r^2}{\chi_U^2}} \right]</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 10% (i.e., 90% confidence interval)<br />
:&nbsp; <math>\chi_L^2(4) \approx 9.49, \quad \chi_U^2(4) \approx 0.711</math>. Therefore,<br />
:&nbsp; <math>0.03 \approx \frac{1}{4 \chi_L^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_U^2} \approx 0.35</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.59</math><br />
so with 90% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.6MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the expected sample median radius of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1427Closed Form Precision2017-04-23T18:38:17Z<p>David: /* Confidence Intervals */ Convexity cancels out; no need for correction factor in confidence interval ratio! Verified in simulation.</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula, as well as a variation for the case in which the true center is known.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - 1) degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_1^2, \ \chi_2^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-1)) \leq \chi_1^2) = \alpha/2, \quad Pr(\chi^2(2(n-1)) \leq \chi_2^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_1^2</math> = <code>CHIINV(α/2, 2n-2)</code>,<math>\quad \chi_2^2</math> = <code>CHIINV((1-α/2), 2n-2)</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_2^2}, \ \frac{2(n-1) s^2}{\chi_1^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_2^2}, \ \frac{\sum r^2}{\chi_1^2} \right]</math><br />
<br />
Since convexity in the numerator and denominator cancel out, no correction factors are needed to compute the confidence interval on the Rayleigh parameter itself:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ \sqrt{\frac{\sum r^2}{\chi_2^2}}, \ \sqrt{\frac{\sum r^2}{\chi_1^2}} \right]</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 5% (i.e., 95% confidence intervals)<br />
:&nbsp; <math>\chi_1^2(4) \approx 0.484, \quad \chi_2^2(4) \approx 11.14</math>. Therefore,<br />
:&nbsp; <math>0.02 \approx \frac{1}{4 \chi_2^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_1^2} \approx 0.52</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.76</math><br />
so with 95% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.8MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the expected sample median radius of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Why_BAC&diff=1426Why BAC2017-03-26T16:48:12Z<p>David: /* The Problem */</p>
<hr />
<div>== The Pursuit of Accuracy (or ''Precision''<ref>Most of the time, [[What_is_Precision%3F#Precision_vs_Accuracy|when people talk about "accuracy" they're actually talking about "precision."]]</ref>) ==<br />
'''''Accuracy:''''' Most shooters and manufacturers care about it, but they don't have a proper way to evaluate or describe it.<br />
<br />
Present customs are to make unqualified references to "MOA" performance and group size. Manufacturers give [[Closed_Form_Precision#The_3-shot_Group|"MOA" guarantees]], and shooters show off small shot groups, but neither of those are particularly meaningful. They don't inform essential questions like:<br />
<br />
# Is my gun holding me back in competition?<br />
# Are my gun and ammo accurate enough to take a particular shot on an animal?<br />
# Does my gun even have the capability to hit a given target with a high probability?<br />
<br />
[[Ballistic_Accuracy_Classification|BAC™]] provides an efficient and standardized system for characterizing and talking about ballistic accuracy.<br />
<br />
What follows is an exposition of current customs and their shortcoming, which are mitigated by the [[Ballistic Accuracy Classification]]™ system.<br />
<br />
== The Problem ==<br />
''It's easy to fool yourself with Extreme Spread, and it’s even easier to fool others.''<br />
<br />
First note that even the most precise rail gun will occasionally print a "[[Fliers|flier]]." If 99 out of 100 shots nestle into a dime group, but one breaks away by an inch, would you characterize the accuracy of that gun as shooting one-inch groups?<br />
<br />
Now pick up a rifle, pop in a beta mag, and empty it from the hip. There’s a chance that 3 of those shots will be touching at 100 yards. Would you cut out those three shots and declare that you have a half-MOA gun?<br />
<br />
Maybe your manufacturer gave you a "1-MOA guarantee," by which they mean that your gun will shoot a 3-shot group into 1” at 100 yards with quality ammo. So you hit the range, tighten your scope, and fire three shots at the same point. Bingo: You don’t need calipers to see that they’re within an inch of each other. You’ve really got a sub-MOA rifle!<br />
<br />
Well you’ve got a full box of ammo, so why not knock the center out of that pretty little group? You take another shot and, damnit, it goes wide. (You’ll have to crop that group pretty tightly to show it off now!) But everyone knows you can call fliers, so you take a breath and try some more. Before you’re through the box you will notice something annoying: ''The more shots you take, the wider your group tends to get.''<br />
<br />
Now something doesn’t add up here. The manufacturer guaranteed your rifle would shoot three shots within 1 MOA. But neither they, nor you, nor your gun could predict which order the shots in that group would appear. If your gun is really sub-MOA you should be able to pick ''any three'' as your "group" and it should measure under 1 MOA. You’ve just discovered one of the industry’s inside jokes: Accuracy guarantees expressed in terms of group sizes are either impossible or meaningless.<br />
<br />
You’ve also discovered one of the problems with the Extreme Spread: It depends on the number of shots you take. Worse yet, it doesn’t differentiate between a target where most of the shots are in a tight group and there’s a lone "flier," and one with the same extreme spread but with every shot scattered about the same distance from the center.<br />
<br />
If you kept your sight zeroed and marked every shot taken with a given rifle and lot of ammunition, after 1000 shots<ref>This is at a "point-blank" target, by which we mean one that is close enough that wind variation and muzzle velocity variation are insignificant. For high-velocity bullets these conditions hold out to at least 100 yards. For subsonic bullets they are valid to at least 25 yards.</ref> your aggregated target would look something like this: <br />
<br />
[[File:1000Shots.png|frame|center|1000 shot simulation. Dots represent the center of each hit, not the size of the holes cut by the bullets.]]<br />
<br />
It doesn’t matter how accurate or inaccurate your gun: its shot distribution is the same as the one that produced this picture. The only thing that varies with accuracy is how large or small this cluster is.<ref>The red circle is drawn around the center of the distribution and contains exactly half of the shots in the picture. The radius of that red circle is called the [[Describing_Precision#Circular_Error_Probable_.28CEP.29|Circular Error Probable (CEP)]], and that single value is sufficient to characterize the accuracy of a gun. Some real-world CEP values:<br />
* The U.S. Precision Sniper Rifle contract called for CEP better than 0.3MOA<br />
* The M24 and M110 acceptance standards require CEP better than 0.6MOA<br />
* XM193 ammunition from a test barrel has to shoot tighter than 0.7MOA CEP<br />
I.e., from a fixed barrel XM193 ammunition should put at least half its shots inside a 1.5” diameter circle at 100 yards.<br />
</ref><br />
<br />
What does this mean for accuracy? You don’t get to pick the order in which those shots are fired. Any individual shot is essentially selected at random from that distribution of shots. If you pick 3 shots at random from that distribution you could end up with three holes practically touching. They might be near the center of impact, or they might be far away. Conversely, you could end up with three shots quite far from each other. Three shots don’t tell you very much!<br />
<br />
== The Solution ==<br />
Fortunately, there are statistically rigorous ways of sampling and describing the accuracy of a gun. And a shooter doesn't have to understand the statistics to understand what those statistics imply for performance.<br />
<br />
The [[Ballistic Accuracy Classification]] (BAC™) system allows anyone – a shooter or a manufacturer – to determine how accurate a gun is using a single number. And that number lets you derive all sorts of useful and statistically meaningful information. For example:<br />
<br />
* What proportion of shots will hit within a 1MOA circle? How about a half MOA?<br />
* How many 3- or 5-shot groups should measure under an inch at 100 yards?<br />
<br />
The [[Ballistic Accuracy Classification]] white paper explains exactly how BAC works. It provides a step-by-step guide to what information you need and how to actually calculate BAC. Everything can be done in a simple spreadsheet.<br />
<br />
For the curious who want to read up on the statistical model that BAC uses for making inferences on expected future performance of a gun, the white paper links to pages on Ballistipedia that provide all background information. Every formula is laid out and backed up with relevant literature sources. However, these technical details are not required to simply use BAC for your own purposes. You can enjoy driving a car without being an engineer, and you can benefit from BAC without being a statistician.<br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Why_BAC&diff=1425Why BAC2017-03-26T16:46:32Z<p>David: /* The Solution */</p>
<hr />
<div>== The Pursuit of Accuracy (or ''Precision''<ref>Most of the time, [[What_is_Precision%3F#Precision_vs_Accuracy|when people talk about "accuracy" they're actually talking about "precision."]]</ref>) ==<br />
'''''Accuracy:''''' Most shooters and manufacturers care about it, but they don't have a proper way to evaluate or describe it.<br />
<br />
Present customs are to make unqualified references to "MOA" performance and group size. Manufacturers give [[Closed_Form_Precision#The_3-shot_Group|"MOA" guarantees]], and shooters show off small shot groups, but neither of those are particularly meaningful. They don't inform essential questions like:<br />
<br />
# Is my gun holding me back in competition?<br />
# Are my gun and ammo accurate enough to take a particular shot on an animal?<br />
# Does my gun even have the capability to hit a given target with a high probability?<br />
<br />
[[Ballistic_Accuracy_Classification|BAC™]] provides an efficient and standardized system for characterizing and talking about ballistic accuracy.<br />
<br />
What follows is an exposition of current customs and their shortcoming, which are mitigated by the [[Ballistic Accuracy Classification]]™ system.<br />
<br />
== The Problem ==<br />
''It's easy to fool yourself with Extreme Spread, and it’s even easier to fool others.''<br />
<br />
First note that even the most precise rail gun will occasionally print a "[[Fliers|flier]]." If 99 out of 100 shots nestle into a dime group, but one breaks away by an inch, would you characterize the accuracy of that gun as shooting one-inch groups?<br />
<br />
Now pick up a rifle, pop in a beta mag, and empty it from the hip. There’s a chance that 3 of those shots will be touching at 100 yards. Would you cut out those three shots and declare that you have a half-MOA gun?<br />
<br />
Maybe your manufacturer gave you a "1-MOA guarantee," by which they mean that your gun will shoot a 3-shot group into 1” at 100 yards with quality ammo. So you hit the range, tighten your scope, and fire three shots at the same point. Bingo: You don’t need calipers to see that they’re within an inch of each other. You’ve really got a sub-MOA rifle!<br />
<br />
Well you’ve got a full box of ammo, so why not knock the center out of that pretty little group? You take another shot and, damnit, it goes wide. (You’ll have to crop that group pretty tightly to show it off now!) But everyone knows you can call fliers, so you take a breath and try some more. Before you’re through the box you will notice something annoying: ''The more shots you take, the wider your group tends to get.''<br />
<br />
Now something doesn’t add up here. The manufacturer guaranteed your rifle would shoot three shots within 1 MOA. But neither they, nor you, nor your gun could predict which order the shots in that group would appear. If your gun is really sub-MOA you should be able to pick ''any three'' as your "group" and it should measure under 1 MOA. You’ve just discovered one of the industry’s inside jokes: Accuracy guarantees expressed in terms of group sizes are either impossible or meaningless.<br />
<br />
You’ve also discovered one of the problems with the Extreme Spread: It depends on the number of shots you take. Worse yet, it doesn’t differentiate between a target where most of the shots are in a tight group and there’s a lone "flier," and one with the same extreme spread but with every shot scattered about the same distance from the center.<br />
<br />
If you kept your sight zeroed and marked every shot taken with a given rifle and lot of ammunition, after 1000 shots<ref>This is at a "point-blank" target, by which we mean one that is close enough that wind variation and muzzle velocity variation are insignificant. For high-velocity bullets these conditions hold out to at least 100 yards. For subsonic bullets they are valid to at least 25 yards.</ref> your aggregated target would look something like this: <br />
<br />
[[File:1000Shots.png|frame|center|1000 shot simulation. Dots represent the center of each hit, not the size of the holes cut by the bullets.]]<br />
<br />
It doesn’t matter how accurate or inaccurate your gun: its shot distribution is the same as the one that produced this picture. The only thing that varies with accuracy is how large or small this cluster is.<ref>The red circle is drawn around the center of the distribution and contains exactly half of the shots in the picture. The radius of that red circle is called the Circular Error Probable (CEP), and that single value is sufficient to characterize the accuracy of a gun. Some real-world CEP values:<br />
* The U.S. Precision Sniper Rifle contract called for CEP better than 0.3MOA<br />
* The M24 and M110 acceptance standards require CEP better than 0.6MOA<br />
* XM193 ammunition from a test barrel has to shoot tighter than 0.7MOA CEP<br />
I.e., from a fixed barrel XM193 ammunition should put at least half its shots inside a 1.5” diameter circle at 100 yards.<br />
</ref><br />
<br />
What does this mean for accuracy? You don’t get to pick the order in which those shots are fired. Any individual shot is essentially selected at random from that distribution of shots. If you pick 3 shots at random from that distribution you could end up with three holes practically touching. They might be near the center of impact, or they might be far away. Conversely, you could end up with three shots quite far from each other. Three shots don’t tell you very much!<br />
<br />
== The Solution ==<br />
Fortunately, there are statistically rigorous ways of sampling and describing the accuracy of a gun. And a shooter doesn't have to understand the statistics to understand what those statistics imply for performance.<br />
<br />
The [[Ballistic Accuracy Classification]] (BAC™) system allows anyone – a shooter or a manufacturer – to determine how accurate a gun is using a single number. And that number lets you derive all sorts of useful and statistically meaningful information. For example:<br />
<br />
* What proportion of shots will hit within a 1MOA circle? How about a half MOA?<br />
* How many 3- or 5-shot groups should measure under an inch at 100 yards?<br />
<br />
The [[Ballistic Accuracy Classification]] white paper explains exactly how BAC works. It provides a step-by-step guide to what information you need and how to actually calculate BAC. Everything can be done in a simple spreadsheet.<br />
<br />
For the curious who want to read up on the statistical model that BAC uses for making inferences on expected future performance of a gun, the white paper links to pages on Ballistipedia that provide all background information. Every formula is laid out and backed up with relevant literature sources. However, these technical details are not required to simply use BAC for your own purposes. You can enjoy driving a car without being an engineer, and you can benefit from BAC without being a statistician.<br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Why_BAC&diff=1424Why BAC2017-03-26T16:41:33Z<p>David: /* The Problem */</p>
<hr />
<div>== The Pursuit of Accuracy (or ''Precision''<ref>Most of the time, [[What_is_Precision%3F#Precision_vs_Accuracy|when people talk about "accuracy" they're actually talking about "precision."]]</ref>) ==<br />
'''''Accuracy:''''' Most shooters and manufacturers care about it, but they don't have a proper way to evaluate or describe it.<br />
<br />
Present customs are to make unqualified references to "MOA" performance and group size. Manufacturers give [[Closed_Form_Precision#The_3-shot_Group|"MOA" guarantees]], and shooters show off small shot groups, but neither of those are particularly meaningful. They don't inform essential questions like:<br />
<br />
# Is my gun holding me back in competition?<br />
# Are my gun and ammo accurate enough to take a particular shot on an animal?<br />
# Does my gun even have the capability to hit a given target with a high probability?<br />
<br />
[[Ballistic_Accuracy_Classification|BAC™]] provides an efficient and standardized system for characterizing and talking about ballistic accuracy.<br />
<br />
What follows is an exposition of current customs and their shortcoming, which are mitigated by the [[Ballistic Accuracy Classification]]™ system.<br />
<br />
== The Problem ==<br />
''It's easy to fool yourself with Extreme Spread, and it’s even easier to fool others.''<br />
<br />
First note that even the most precise rail gun will occasionally print a "[[Fliers|flier]]." If 99 out of 100 shots nestle into a dime group, but one breaks away by an inch, would you characterize the accuracy of that gun as shooting one-inch groups?<br />
<br />
Now pick up a rifle, pop in a beta mag, and empty it from the hip. There’s a chance that 3 of those shots will be touching at 100 yards. Would you cut out those three shots and declare that you have a half-MOA gun?<br />
<br />
Maybe your manufacturer gave you a "1-MOA guarantee," by which they mean that your gun will shoot a 3-shot group into 1” at 100 yards with quality ammo. So you hit the range, tighten your scope, and fire three shots at the same point. Bingo: You don’t need calipers to see that they’re within an inch of each other. You’ve really got a sub-MOA rifle!<br />
<br />
Well you’ve got a full box of ammo, so why not knock the center out of that pretty little group? You take another shot and, damnit, it goes wide. (You’ll have to crop that group pretty tightly to show it off now!) But everyone knows you can call fliers, so you take a breath and try some more. Before you’re through the box you will notice something annoying: ''The more shots you take, the wider your group tends to get.''<br />
<br />
Now something doesn’t add up here. The manufacturer guaranteed your rifle would shoot three shots within 1 MOA. But neither they, nor you, nor your gun could predict which order the shots in that group would appear. If your gun is really sub-MOA you should be able to pick ''any three'' as your "group" and it should measure under 1 MOA. You’ve just discovered one of the industry’s inside jokes: Accuracy guarantees expressed in terms of group sizes are either impossible or meaningless.<br />
<br />
You’ve also discovered one of the problems with the Extreme Spread: It depends on the number of shots you take. Worse yet, it doesn’t differentiate between a target where most of the shots are in a tight group and there’s a lone "flier," and one with the same extreme spread but with every shot scattered about the same distance from the center.<br />
<br />
If you kept your sight zeroed and marked every shot taken with a given rifle and lot of ammunition, after 1000 shots<ref>This is at a "point-blank" target, by which we mean one that is close enough that wind variation and muzzle velocity variation are insignificant. For high-velocity bullets these conditions hold out to at least 100 yards. For subsonic bullets they are valid to at least 25 yards.</ref> your aggregated target would look something like this: <br />
<br />
[[File:1000Shots.png|frame|center|1000 shot simulation. Dots represent the center of each hit, not the size of the holes cut by the bullets.]]<br />
<br />
It doesn’t matter how accurate or inaccurate your gun: its shot distribution is the same as the one that produced this picture. The only thing that varies with accuracy is how large or small this cluster is.<ref>The red circle is drawn around the center of the distribution and contains exactly half of the shots in the picture. The radius of that red circle is called the Circular Error Probable (CEP), and that single value is sufficient to characterize the accuracy of a gun. Some real-world CEP values:<br />
* The U.S. Precision Sniper Rifle contract called for CEP better than 0.3MOA<br />
* The M24 and M110 acceptance standards require CEP better than 0.6MOA<br />
* XM193 ammunition from a test barrel has to shoot tighter than 0.7MOA CEP<br />
I.e., from a fixed barrel XM193 ammunition should put at least half its shots inside a 1.5” diameter circle at 100 yards.<br />
</ref><br />
<br />
What does this mean for accuracy? You don’t get to pick the order in which those shots are fired. Any individual shot is essentially selected at random from that distribution of shots. If you pick 3 shots at random from that distribution you could end up with three holes practically touching. They might be near the center of impact, or they might be far away. Conversely, you could end up with three shots quite far from each other. Three shots don’t tell you very much!<br />
<br />
== The Solution ==<br />
Fortunately, there are statistically rigorous ways of sampling and describing the accuracy of a gun. And a shooter doesn't have to understand the statistics to understand what those statistics imply for performance.<br />
<br />
The [[Ballistic Accuracy Classification]] (BAC™) system allows anyone – a shooter or a manufacturer – to determine how accurate a gun is using a single number. And that number lets you derive all sorts of useful and statistically meaningful information. For example:<br />
<br />
* What proportion of shots will hit within a 1MOA circle? How about a half MOA?<br />
* How many 3- or 5-shot groups should measure under an inch at 100 yards?<br />
<br />
The [[Ballistic Accuracy Classification]] white paper explains how exactly BAC™ works. It provides a step-by-step guide on what information you need to calculate BAC™, and what formulas are used. A simple spreadsheet is everything you need.<br />
<br />
For the curious who want to read up on the statistical model that BAC™ uses for making inferences on expected future performance of a gun, the white paper links to pages on Ballistipedia that provide all background information. Every formula is laid out and backed up with relevant literature sources. However, becoming acquainted with these technical details is not required to simply use BAC™ for your own purposes. You can enjoy driving a car without being an engineer, and you can benefit from BAC™ without being a statistician.<br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Why_BAC&diff=1421Why BAC2017-03-22T17:39:50Z<p>David: /* The Pursuit of Accuracy */</p>
<hr />
<div>== The Pursuit of Accuracy (or ''Precision''<ref>Most of the time, [[What_is_Precision%3F#Precision_vs_Accuracy|when people talk about "accuracy" they're actually talking about "precision."]]</ref>) ==<br />
'''''Accuracy:''''' Most shooters and manufacturers care about it, but they don't have a proper way to evaluate or describe it.<br />
<br />
Present customs are to make unqualified references to "MOA" performance and group size. Manufacturers give [[Closed_Form_Precision#The_3-shot_Group|"MOA" guarantees]], and shooters show off small shot groups, but neither of those are particularly meaningful. They don't inform essential questions like:<br />
<br />
# Is my gun holding me back in competition?<br />
# Are my gun and ammo accurate enough to take a particular shot on an animal?<br />
# Does my gun even have the capability to hit a given target with a high probability?<br />
<br />
[[Ballistic_Accuracy_Classification|BAC™]] provides an efficient and standardized system for characterizing and talking about ballistic accuracy.<br />
<br />
What follows is an exposition of current customs and their shortcoming, which are mitigated by the [[Ballistic Accuracy Classification]]™ system.<br />
<br />
== The Problem ==<br />
''It's easy to fool yourself with Extreme Spread, and it’s even easier to fool others.''<br />
<br />
First note that even the most precise rail gun will occasionally print a "flier." If 99 out of 100 shots nestle into a dime group, but one breaks away by an inch, would you characterize the accuracy of that gun as shooting one-inch groups?<br />
<br />
Now pick up a rifle, pop in a beta mag, and empty it from the hip. There’s a chance that 3 of those shots will be touching at 100 yards. Would you cut out those three shots and declare that you have a half-MOA gun?<br />
<br />
Maybe your manufacturer gave you a "1-MOA guarantee," by which they mean that your gun will shoot a 3-shot group into 1” at 100 yards with quality ammo. So you hit the range, tighten your scope, and fire three shots at the same point. Bingo: You don’t need calipers to see that they’re within an inch of each other. You’ve really got a sub-MOA rifle!<br />
<br />
Well you’ve got a full box of ammo, so why not knock the center out of that pretty little group? You take another shot and, damnit, it goes wide. (You’ll have to crop that group pretty tightly to show it off now!) But everyone knows you can call fliers, so you take a breath and try some more. Before you’re through the box you will notice something annoying: ''The more shots you take, the wider your group tends to get.''<br />
<br />
Now something doesn’t add up here. The manufacturer guaranteed your rifle would shoot three shots within 1 MOA. But neither they, nor you, nor your gun could predict which order the shots in that group would appear. If your gun is really sub-MOA you should be able to pick ''any three'' as your "group" and it should measure under 1 MOA. You’ve just discovered one of the industry’s inside jokes: Accuracy guarantees expressed in terms of group sizes are either impossible or meaningless.<br />
<br />
You’ve also discovered one of the problems with the Extreme Spread: It depends on the number of shots you take. Worse yet, it doesn’t differentiate between a target where most of the shots are in a tight group and there’s a lone "flier," and one with the same extreme spread but with every shot scattered about the same distance from the center.<br />
<br />
If you kept your sight zeroed and marked every shot taken with a given rifle and lot of ammunition, after 1000 shots<ref>This is at a "point-blank" target, by which we mean one that is close enough that wind variation and muzzle velocity variation are insignificant. For high-velocity bullets these conditions hold out to at least 100 yards. For subsonic bullets they are valid to at least 25 yards.</ref> your aggregated target would look something like this: <br />
<br />
[[File:1000Shots.png|frame|center|1000 shot simulation. Dots represent the center of each hit, not the size of the holes cut by the bullets.]]<br />
<br />
It doesn’t matter how accurate or inaccurate your gun: its shot distribution is the same as the one that produced this picture. The only thing that varies with accuracy is how large or small this cluster is.<ref>The red circle is drawn around the center of the distribution and contains exactly half of the shots in the picture. The radius of that red circle is called the Circular Error Probable (CEP), and that single value is sufficient to characterize the accuracy of a gun. Some real-world CEP values:<br />
* The U.S. Precision Sniper Rifle contract called for CEP better than 0.3MOA<br />
* The M24 and M110 acceptance standards require CEP better than 0.6MOA<br />
* XM193 ammunition from a test barrel has to shoot tighter than 0.7MOA CEP<br />
I.e., from a fixed barrel XM193 ammunition should put at least half its shots inside a 1.5” diameter circle at 100 yards.<br />
</ref><br />
<br />
What does this mean for accuracy? You don’t get to pick the order in which those shots are fired. Any individual shot is essentially selected at random from that distribution of shots. If you pick 3 shots at random from that distribution you could end up with three holes practically touching. They might be near the center of impact, or they might be far away. Conversely, you could end up with three shots quite far from each other. Three shots don’t tell you very much!<br />
<br />
== The Solution ==<br />
Fortunately, there are statistically rigorous ways of sampling and describing the accuracy of a gun. And a shooter doesn't have to understand the statistics to understand what those statistics imply for performance.<br />
<br />
The [[Ballistic Accuracy Classification]] system allows anyone – a shooter or a manufacturer – to determine how accurate a gun is using a single number. And that number determines all sorts of useful and statistically meaningful information. For example:<br />
<br />
* What proportion of shots will hit within a 1MOA circle? How about a half MOA?<br />
* How many 3- or 5-shot groups should measure under an inch at 100 yards?<br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Why_BAC&diff=1420Why BAC2017-03-22T17:38:40Z<p>David: /* The Pursuit of Accuracy */</p>
<hr />
<div>== The Pursuit of Accuracy ==<br />
'''''Accuracy:'''''<ref>Most of the time, [[What_is_Precision%3F#Precision_vs_Accuracy|when people talk about "accuracy" they're actually talking about "precision."]]</ref>More precisely, Most shooters and manufacturers care about it, but they don't have a proper way to evaluate or describe it.<br />
<br />
Present customs are to make unqualified references to "MOA" performance and group size. Manufacturers give [[Closed_Form_Precision#The_3-shot_Group|"MOA" guarantees]], and shooters show off small shot groups, but neither of those are particularly meaningful. They don't inform essential questions like:<br />
<br />
# Is my gun holding me back in competition?<br />
# Are my gun and ammo accurate enough to take a particular shot on an animal?<br />
# Does my gun even have the capability to hit a given target with a high probability?<br />
<br />
[[Ballistic_Accuracy_Classification|BAC™]] provides an efficient and standardized system for characterizing and talking about ballistic accuracy.<br />
<br />
What follows is an exposition of current customs and their shortcoming, which are mitigated by the [[Ballistic Accuracy Classification]]™ system.<br />
<br />
== The Problem ==<br />
''It's easy to fool yourself with Extreme Spread, and it’s even easier to fool others.''<br />
<br />
First note that even the most precise rail gun will occasionally print a "flier." If 99 out of 100 shots nestle into a dime group, but one breaks away by an inch, would you characterize the accuracy of that gun as shooting one-inch groups?<br />
<br />
Now pick up a rifle, pop in a beta mag, and empty it from the hip. There’s a chance that 3 of those shots will be touching at 100 yards. Would you cut out those three shots and declare that you have a half-MOA gun?<br />
<br />
Maybe your manufacturer gave you a "1-MOA guarantee," by which they mean that your gun will shoot a 3-shot group into 1” at 100 yards with quality ammo. So you hit the range, tighten your scope, and fire three shots at the same point. Bingo: You don’t need calipers to see that they’re within an inch of each other. You’ve really got a sub-MOA rifle!<br />
<br />
Well you’ve got a full box of ammo, so why not knock the center out of that pretty little group? You take another shot and, damnit, it goes wide. (You’ll have to crop that group pretty tightly to show it off now!) But everyone knows you can call fliers, so you take a breath and try some more. Before you’re through the box you will notice something annoying: ''The more shots you take, the wider your group tends to get.''<br />
<br />
Now something doesn’t add up here. The manufacturer guaranteed your rifle would shoot three shots within 1 MOA. But neither they, nor you, nor your gun could predict which order the shots in that group would appear. If your gun is really sub-MOA you should be able to pick ''any three'' as your "group" and it should measure under 1 MOA. You’ve just discovered one of the industry’s inside jokes: Accuracy guarantees expressed in terms of group sizes are either impossible or meaningless.<br />
<br />
You’ve also discovered one of the problems with the Extreme Spread: It depends on the number of shots you take. Worse yet, it doesn’t differentiate between a target where most of the shots are in a tight group and there’s a lone "flier," and one with the same extreme spread but with every shot scattered about the same distance from the center.<br />
<br />
If you kept your sight zeroed and marked every shot taken with a given rifle and lot of ammunition, after 1000 shots<ref>This is at a "point-blank" target, by which we mean one that is close enough that wind variation and muzzle velocity variation are insignificant. For high-velocity bullets these conditions hold out to at least 100 yards. For subsonic bullets they are valid to at least 25 yards.</ref> your aggregated target would look something like this: <br />
<br />
[[File:1000Shots.png|frame|center|1000 shot simulation. Dots represent the center of each hit, not the size of the holes cut by the bullets.]]<br />
<br />
It doesn’t matter how accurate or inaccurate your gun: its shot distribution is the same as the one that produced this picture. The only thing that varies with accuracy is how large or small this cluster is.<ref>The red circle is drawn around the center of the distribution and contains exactly half of the shots in the picture. The radius of that red circle is called the Circular Error Probable (CEP), and that single value is sufficient to characterize the accuracy of a gun. Some real-world CEP values:<br />
* The U.S. Precision Sniper Rifle contract called for CEP better than 0.3MOA<br />
* The M24 and M110 acceptance standards require CEP better than 0.6MOA<br />
* XM193 ammunition from a test barrel has to shoot tighter than 0.7MOA CEP<br />
I.e., from a fixed barrel XM193 ammunition should put at least half its shots inside a 1.5” diameter circle at 100 yards.<br />
</ref><br />
<br />
What does this mean for accuracy? You don’t get to pick the order in which those shots are fired. Any individual shot is essentially selected at random from that distribution of shots. If you pick 3 shots at random from that distribution you could end up with three holes practically touching. They might be near the center of impact, or they might be far away. Conversely, you could end up with three shots quite far from each other. Three shots don’t tell you very much!<br />
<br />
== The Solution ==<br />
Fortunately, there are statistically rigorous ways of sampling and describing the accuracy of a gun. And a shooter doesn't have to understand the statistics to understand what those statistics imply for performance.<br />
<br />
The [[Ballistic Accuracy Classification]] system allows anyone – a shooter or a manufacturer – to determine how accurate a gun is using a single number. And that number determines all sorts of useful and statistically meaningful information. For example:<br />
<br />
* What proportion of shots will hit within a 1MOA circle? How about a half MOA?<br />
* How many 3- or 5-shot groups should measure under an inch at 100 yards?<br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Why_BAC&diff=1419Why BAC2017-03-21T18:00:03Z<p>David: Created page with "== The Pursuit of Accuracy == '''''Accuracy:''''' Most shooters and manufacturers care about it, but they don't have a proper way to evaluate or describe it. Present customs..."</p>
<hr />
<div>== The Pursuit of Accuracy ==<br />
'''''Accuracy:''''' Most shooters and manufacturers care about it, but they don't have a proper way to evaluate or describe it.<br />
<br />
Present customs are to make unqualified references to "MOA" performance and group size. Manufacturers give [[Closed_Form_Precision#The_3-shot_Group|"MOA" guarantees]], and shooters show off small shot groups, but neither of those are particularly meaningful. They don't inform essential questions like:<br />
<br />
# Is my gun holding me back in competition?<br />
# Are my gun and ammo accurate enough to take a particular shot on an animal?<br />
# Does my gun even have the capability to hit a given target with a high probability?<br />
<br />
[[Ballistic_Accuracy_Classification|BAC™]] provides an efficient and standardized system for characterizing and talking about ballistic accuracy.<br />
<br />
What follows is an exposition of current customs and their shortcoming, which are mitigated by the [[Ballistic Accuracy Classification]]™ system.<br />
<br />
== The Problem ==<br />
''It's easy to fool yourself with Extreme Spread, and it’s even easier to fool others.''<br />
<br />
First note that even the most precise rail gun will occasionally print a "flier." If 99 out of 100 shots nestle into a dime group, but one breaks away by an inch, would you characterize the accuracy of that gun as shooting one-inch groups?<br />
<br />
Now pick up a rifle, pop in a beta mag, and empty it from the hip. There’s a chance that 3 of those shots will be touching at 100 yards. Would you cut out those three shots and declare that you have a half-MOA gun?<br />
<br />
Maybe your manufacturer gave you a "1-MOA guarantee," by which they mean that your gun will shoot a 3-shot group into 1” at 100 yards with quality ammo. So you hit the range, tighten your scope, and fire three shots at the same point. Bingo: You don’t need calipers to see that they’re within an inch of each other. You’ve really got a sub-MOA rifle!<br />
<br />
Well you’ve got a full box of ammo, so why not knock the center out of that pretty little group? You take another shot and, damnit, it goes wide. (You’ll have to crop that group pretty tightly to show it off now!) But everyone knows you can call fliers, so you take a breath and try some more. Before you’re through the box you will notice something annoying: ''The more shots you take, the wider your group tends to get.''<br />
<br />
Now something doesn’t add up here. The manufacturer guaranteed your rifle would shoot three shots within 1 MOA. But neither they, nor you, nor your gun could predict which order the shots in that group would appear. If your gun is really sub-MOA you should be able to pick ''any three'' as your "group" and it should measure under 1 MOA. You’ve just discovered one of the industry’s inside jokes: Accuracy guarantees expressed in terms of group sizes are either impossible or meaningless.<br />
<br />
You’ve also discovered one of the problems with the Extreme Spread: It depends on the number of shots you take. Worse yet, it doesn’t differentiate between a target where most of the shots are in a tight group and there’s a lone "flier," and one with the same extreme spread but with every shot scattered about the same distance from the center.<br />
<br />
If you kept your sight zeroed and marked every shot taken with a given rifle and lot of ammunition, after 1000 shots<ref>This is at a "point-blank" target, by which we mean one that is close enough that wind variation and muzzle velocity variation are insignificant. For high-velocity bullets these conditions hold out to at least 100 yards. For subsonic bullets they are valid to at least 25 yards.</ref> your aggregated target would look something like this: <br />
<br />
[[File:1000Shots.png|frame|center|1000 shot simulation. Dots represent the center of each hit, not the size of the holes cut by the bullets.]]<br />
<br />
It doesn’t matter how accurate or inaccurate your gun: its shot distribution is the same as the one that produced this picture. The only thing that varies with accuracy is how large or small this cluster is.<ref>The red circle is drawn around the center of the distribution and contains exactly half of the shots in the picture. The radius of that red circle is called the Circular Error Probable (CEP), and that single value is sufficient to characterize the accuracy of a gun. Some real-world CEP values:<br />
* The U.S. Precision Sniper Rifle contract called for CEP better than 0.3MOA<br />
* The M24 and M110 acceptance standards require CEP better than 0.6MOA<br />
* XM193 ammunition from a test barrel has to shoot tighter than 0.7MOA CEP<br />
I.e., from a fixed barrel XM193 ammunition should put at least half its shots inside a 1.5” diameter circle at 100 yards.<br />
</ref><br />
<br />
What does this mean for accuracy? You don’t get to pick the order in which those shots are fired. Any individual shot is essentially selected at random from that distribution of shots. If you pick 3 shots at random from that distribution you could end up with three holes practically touching. They might be near the center of impact, or they might be far away. Conversely, you could end up with three shots quite far from each other. Three shots don’t tell you very much!<br />
<br />
== The Solution ==<br />
Fortunately, there are statistically rigorous ways of sampling and describing the accuracy of a gun. And a shooter doesn't have to understand the statistics to understand what those statistics imply for performance.<br />
<br />
The [[Ballistic Accuracy Classification]] system allows anyone – a shooter or a manufacturer – to determine how accurate a gun is using a single number. And that number determines all sorts of useful and statistically meaningful information. For example:<br />
<br />
* What proportion of shots will hit within a 1MOA circle? How about a half MOA?<br />
* How many 3- or 5-shot groups should measure under an inch at 100 yards?<br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=File:1000Shots.png&diff=1418File:1000Shots.png2017-03-21T17:57:02Z<p>David: David uploaded a new version of File:1000Shots.png</p>
<hr />
<div>1000 symmetric bivariate normal shots</div>Davidhttp://ballistipedia.com/index.php?title=File:1000Shots.png&diff=1417File:1000Shots.png2017-03-21T01:13:32Z<p>David: 1000 symmetric bivariate normal shots</p>
<hr />
<div>1000 symmetric bivariate normal shots</div>Davidhttp://ballistipedia.com/index.php?title=Ballistic_Accuracy_Classification&diff=1416Ballistic Accuracy Classification2017-03-20T18:44:00Z<p>David: Added internal links</p>
<hr />
<div>== Introduction ==<br />
<br />
Ballistic Accuracy Classification™ is a mathematically rigorous system for describing and understanding the precision of ballistic tools like rifles. It provides information that is:<br />
* Useful and easy for consumers to understand<br />
* Straightforward for enthusiasts and builders to calculate<br />
* Statistically sound enough for experts to validate<br />
''For more background see [[Why BAC]].''<br />
<br />
All firearms and components can be assigned a Ballistic Accuracy Class™ (BAC™), which fully characterizes their accuracy potential. Lower numbers are better. In practice, the lowest possible BAC is 1. (A theoretically perfect rifle system that always puts every shot through the same hole would be BAC 0.)<br />
<br />
Behind the scenes, BAC is defined by a single statistical parameter known as sigma (σ). Using [[Closed_Form_Precision#Symmetric_Bivariate_Normal_Shots_Imply_Rayleigh_Distributed_Distances|the associated statistical model]] (known as the Rayleigh distribution), statisticians can not only determine the BAC for a particular gun or component, but also compute its expected shooting precision.<ref>The proportion of shots expected to fall within radius ''rσ'' of the center of impact is <math>1-e^{-r^2/2}</math>.</ref><br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! Sigma (σ)<br />
! Typical examples<br />
! Shots within<BR/>¼ MOA radius<br />
! Shots within<BR/>½ MOA radius<br />
|-<br />
| Class 1<br />
| < 0.1MOA<br />
| Rail guns<br />
| 96%<br />
| 100%<br />
|-<br />
| Class 2<br />
| < 0.2MOA<br />
| Benchrest guns<br />
| 54%<br />
| 96%<br />
|-<br />
| Class 3<br />
| < 0.3MOA<br />
| Mil-spec for PSR<br />
| 29%<br />
| 75%<br />
|-<br />
| Class 4<br />
| < 0.4MOA<br />
| Competitive auto-loaders<br />
| 18%<br />
| 54%<br />
|-<br />
| Class 5<br />
| < 0.5MOA<br />
| Mil-spec for M110 and M24<br />
| 12%<br />
| 39%<br />
|-<br />
| Class 6<br />
| < 0.6MOA<br />
| Mil-spec for infantry rifles and ammo<br />
| 8%<br />
| 29%<br />
|}<br />
<br />
We can also generate the expected values of more familiar measures, like the extreme spread of a 3- or 5-shot group:<br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! 5-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 5-shot<BR/> Group Spread<br />
! 3-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 3-shot<BR/> Group Spread<br />
|-<br />
| Class 1<br />
| 100%<br />
| 0.3MOA<br />
| 100%<br />
| 0.2MOA<br />
|-<br />
| Class 2<br />
| 98%<br />
| 0.6MOA<br />
| 99%<br />
| 0.5MOA<br />
|-<br />
| Class 3<br />
| 65%<br />
| 0.9MOA<br />
| 85%<br />
| 0.7MOA<br />
|-<br />
| Class 4<br />
| 26%<br />
| 1.2MOA<br />
| 57%<br />
| 0.9MOA<br />
|-<br />
| Class 5<br />
| 9%<br />
| 1.5MOA<br />
| 35%<br />
| 1.2MOA<br />
|-<br />
| Class 6<br />
| 3%<br />
| 1.8MOA<br />
| 21%<br />
| 1.4MOA<br />
|}<br />
<br />
=== Understanding MOA ===<br />
Accuracy is described in [[Angular_Size|angular terms]], most commonly using the unit "Minute of Arc" (MOA). One arc minute spans 1.047" at 100 yards. Rifle shooters often practice on 100 yard targets, and so they often think in terms of how wide their groups are at 100 yards. People often just round it off and think of 1 MOA as "one inch at 100 yards."<br />
<br />
In the absence of an atmosphere, the angular precision measured at one distance would be valid at all other distances. I.e., a 1" group at 100 yards would measure 5" at 500 yards. However, in reality the effects of wind and drag (which, together with gravity, accentuates variations in muzzle velocity) will only increase the angular spread of ballistic groups as distance increases. Therefore, '''in practice one should expect worse-than-advertised accuracy when shooting at longer distances'''. The distance at which atmosphere begins to significantly affect precision depends on a bullet’s muzzle velocity and ballistic coefficient. For high-power rifles this is usually beyond 100 yards. Guns shooting subsonic projectiles can begin to suffer after just 25 yards. <br />
<br />
=== Nomenclature ===<br />
<br />
BAC™ is only meaningful when it conforms to established terms and conventions. BAC must be determined by testing in accordance with [[#Protocol|the BAC Protocol]] described in this document.<br />
<br />
Ballistic Accuracy Classification™ must be supported by the following descriptive parameters:<br />
# Product tested. (Any component, or group of components, associated with accuracy can be tested.)<br />
# Configuration tested. This must include the following details:<br />
## Barrel length, material, profile, rifling.<br/>(E.g., ''20" stainless 1" bull contour with 6-land 1:10"-twist cut rifling''.)<br />
## Receiver, action, and feed mechanism.<br/>(E.g., ''AR-10 magazine-fed semi-automatic''.)<br />
## Ammunition.<br />
### If commercial this must include brand, model, and lot.<br/>(E.g., ''Federal GM308M Lot#214374H077''.)<br />
### If custom, this must list component and load formula.<br/>(E.g., ''Lapua .308 full-sized brass, WLR primer, 168gr SMK, 44gr Varget, 2.800" COAL''.)<br />
# Confidence in BAC. When not conspicuously mentioned, it is assumed that the upper '''90%''' confidence value of sigma is referenced for BAC.<br />
<br />
BAC measures are intentionally kept somewhat coarse. Care should be taken to avoid suggesting more than two significant digits of precision. For example, even after shooting 20 rounds through a gun, [[Closed_Form_Precision#How_large_a_sample_do_we_need.3F|the 80% confidence interval on the precision estimate typically spans 0.9-1.2 times the estimated value]].<br />
<br />
=== Trademarks ===<br />
The following terms are trademarks of Scribe Logistics LLC. They are free to use so long as their use complies with the Nomenclature and Protocol outlined here.<br />
* Ballistic Accuracy Classification™<br />
* Ballistic Accuracy Class™<br />
* BAC™<br />
The trademarks are claimed solely for the purpose of maintaining the integrity of the system and avoiding market confusion.<br />
<br />
== Theory ==<br />
<br />
=== Statistical Model ===<br />
<br />
BAC™ assumes that the impact of ballistic shots on a target are normally distributed with the same variance along any axis. (Empirical data validate this assumption, and it should be true as long as atmospheric effects are negligible.<ref>There are two atmospheric effects that can finally create excess variance in one axis: Variable wind will increase horizontal variance. This is a function of the wind speed and the projectile's time-of-flight to the target. Also, as time-of-flight increases, variations in muzzle velocity will appear as excess vertical variance. Since both of these effects depend on time-of-flight, they are typically negligible for high-velocity rifles before 100 yards, or for subsonic projectiles before 25 yards.</ref>) Therefore, [[Closed_Form_Precision#Symmetric_Bivariate_Normal_Shots_Imply_Rayleigh_Distributed_Distances|we use the Rayleigh distribution to model the radius r]], or dispersion of each shot, from the center of impact. When the coordinates of the shots have independent <math>N(0,\sigma)</math> distributions along orthogonal axes, the radius of each shot is described by the Rayleigh probability density function:<br />
:&nbsp; <math>f(r,\sigma)=\frac{r}{\sigma^2}e^{−r^2/2\sigma^2}</math><br />
The [[Closed_Form_Precision#Rayleigh_Estimates|unbiased estimator for the parameter σ]] comes from <math>\widehat{\sigma^2} = \frac{\sum r_i^2}{2(n−1)}</math>, with confidence <math>\widehat{\sigma^2} \sim \frac{\sum r^2}{\chi_{2n-2}^2}</math>.<br />
<br />
=== Simulation ===<br />
Monte Carlo simulation is adequate for studying and characterizing precision. In fact, many of the results associated with BAC, like the distribution of the extreme spread of a particular number of shots, can only be produced through simulation.<br />
<br />
For simulation purposes random shots should be generated as (x, y) coordinates, where <math>X,Y \sim N(0,\sigma)</math>. It is critical to "forget" the known center when using simulated data. When shooting a real gun we never get to know the true center of impact, and instead have to use the sample center. Likewise, Monte Carlo simulations must not reference the known 0 center, and should instead only reference the sample center of whatever group size is being studied. <br />
<br />
== Protocol ==<br />
The Ballistic Accuracy Classification™ shall be the upper bound of the 90% confidence range on estimated sigma, in units of MOA, multiplied by 10 and rounded to the nearest integer. For example, if the 90% confidence value for sigma on a tested gun is 0.47MOA, then the BAC value is 10 ✕ 0.47 = 4.7, rounded = 5. I.e., in this example we are saying with 90% confidence that the tested gun’s accuracy is no worse than Class 5.<br />
<br />
=== Classifying a Specimen ===<br />
It is not realistic to assign a BAC with fewer than 10 shots. The 90% confidence range with just 10 shots will typically extend to 1.4 times the estimated accuracy value. It will typically take 20 shots to get the outside bound of the 90% confidence interval to within 20% of the estimated value.<ref>The U.S. Army Marksmanship Unit (AMU) has long used a minimum of 3 consecutive 10-shot groups fired from a machine rest to test the accuracy of service rifles.</ref><br />
<br />
You must not discard data points during testing except for an unrelated failure. (E.g., if you are testing a barrel and you encounter a squib load, that shot may be excluded. But "[[fliers]]" should not generally be excluded.)<br />
<br />
'''Data:''' Target distance, and (''x,y'') coordinates of the center of each shot impact. All shots with the same point of aim must be grouped, but multiple groups can be used.<br />
<br />
'''Calculations:''' The formulas to transform these data into a confidence interval for sigma, and the corresponding BAC, are shown in [[Media:BallisticAccuracyClassification.xlsx]]. Given a sample of ''n'' shots, over ''g'' groups, at a distance ''d'':<br />
# For each measurement, convert to units of MOA. For example, if measurements are taken in inches, and the target was shot at a distance of ''d'' yards, then divide each measurement by 0.01047''d''<br />
# For each group ''g'', calculate the center of the group as <math>(\bar{x}_{i \in g},\bar{y}_{i \in g})</math><br />
# For each shot ''i'', find its radius squared relative to the center of the group as <math>r_i^2=(x_i−x_g)^2+(y_i−y_g)^2</math><br />
# Calculate the Gaussian correction factor <code>c<sub>G</sub>=1/EXP(LN(SQRT(2/(2n-2))) + GAMMALN((2n-1)/2) - GAMMALN((2n-2)/2))</code><br />
# Calculate the upper 90% confidence value for sigma as <code>σ<sub>U</sub>=cG*SQRT[SUM(r2)/CHIINV(0.9,2n-2)]</code><br />
# Calculate the Ballistic Accuracy Class as <code>=ROUND(σ<sub>U</sub>,0)</code><br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Ballistic_Accuracy_Classification&diff=1415Ballistic Accuracy Classification2017-03-19T18:00:37Z<p>David: Reworded to reference 90% confidence value, since we only care about upper bound (single-tail)</p>
<hr />
<div>== Introduction ==<br />
<br />
Ballistic Accuracy Classification™ is a mathematically rigorous system for describing and understanding the precision of ballistic tools like rifles. It provides information that is:<br />
* Useful and easy for consumers to understand<br />
* Straightforward for enthusiasts and builders to calculate<br />
* Statistically sound enough for experts to validate<br />
<br />
All firearms and components can be assigned a Ballistic Accuracy Class™ (BAC™), which fully characterizes their accuracy potential. Lower numbers are better. In practice, the lowest possible BAC is 1. (A theoretically perfect rifle system that always puts every shot through the same hole would be BAC 0.)<br />
<br />
Behind the scenes, BAC is defined by a single statistical parameter known as sigma (σ). Using the associated statistical model (known as the Rayleigh distribution), statisticians can not only determine the BAC for a particular gun or component, but also compute its expected shooting precision.<ref>The proportion of shots expected to fall within radius ''rσ'' of the center of impact is <math>1-e^{-r^2/2}</math>.</ref><br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! Sigma (σ)<br />
! Typical examples<br />
! Shots within<BR/>¼ MOA radius<br />
! Shots within<BR/>½ MOA radius<br />
|-<br />
| Class 1<br />
| < 0.1MOA<br />
| Rail guns<br />
| 96%<br />
| 100%<br />
|-<br />
| Class 2<br />
| < 0.2MOA<br />
| Benchrest guns<br />
| 54%<br />
| 96%<br />
|-<br />
| Class 3<br />
| < 0.3MOA<br />
| Mil-spec for PSR<br />
| 29%<br />
| 75%<br />
|-<br />
| Class 4<br />
| < 0.4MOA<br />
| Competitive auto-loaders<br />
| 18%<br />
| 54%<br />
|-<br />
| Class 5<br />
| < 0.5MOA<br />
| Mil-spec for M110 and M24<br />
| 12%<br />
| 39%<br />
|-<br />
| Class 6<br />
| < 0.6MOA<br />
| Mil-spec for infantry rifles and ammo<br />
| 8%<br />
| 29%<br />
|}<br />
<br />
We can also generate the expected values of more familiar measures, like the extreme spread of a 3- or 5-shot group:<br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! 5-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 5-shot<BR/> Group Spread<br />
! 3-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 3-shot<BR/> Group Spread<br />
|-<br />
| Class 1<br />
| 100%<br />
| 0.3MOA<br />
| 100%<br />
| 0.2MOA<br />
|-<br />
| Class 2<br />
| 98%<br />
| 0.6MOA<br />
| 99%<br />
| 0.5MOA<br />
|-<br />
| Class 3<br />
| 65%<br />
| 0.9MOA<br />
| 85%<br />
| 0.7MOA<br />
|-<br />
| Class 4<br />
| 26%<br />
| 1.2MOA<br />
| 57%<br />
| 0.9MOA<br />
|-<br />
| Class 5<br />
| 9%<br />
| 1.5MOA<br />
| 35%<br />
| 1.2MOA<br />
|-<br />
| Class 6<br />
| 3%<br />
| 1.8MOA<br />
| 21%<br />
| 1.4MOA<br />
|}<br />
<br />
=== Understanding MOA ===<br />
Accuracy is described in angular terms, most commonly using the unit "Minute of Arc" (MOA). One arc minute spans 1.047" at 100 yards. Rifle shooters often practice on 100 yard targets, and so they often think in terms of how wide their groups are at 100 yards. People often just round it off and think of 1 MOA as one inch at 100 yards.<br />
<br />
In the absence of an atmosphere, the angular precision measured at one distance would be valid at all other distances. I.e., a 1" group at 100 yards would measure 5" at 500 yards. However, in reality the effects of wind and drag (which, together with gravity, accentuates variations in muzzle velocity) will only increase the angular spread of ballistic groups as distance increases. Therefore, '''in practice one should expect worse-than-advertised accuracy when shooting at longer distances'''. The distance at which atmosphere begins to significantly affect precision depends on a bullet’s muzzle velocity and ballistic coefficient. For high-power rifles this is usually beyond 100 yards. Guns shooting subsonic projectiles can begin to suffer after just 25 yards. <br />
<br />
=== Nomenclature ===<br />
<br />
BAC™ is only meaningful when it conforms to established terms and conventions. BAC must be determined by testing in accordance with the BAC Protocol described in this document.<br />
<br />
Ballistic Accuracy Classification™ must be supported by the following descriptive parameters:<br />
# Product tested. (Any component, or group of components, associated with accuracy can be tested.)<br />
# Configuration tested. This must include the following details:<br />
## Barrel length, material, profile, rifling.<br/>(E.g., ''20" stainless 1" bull contour with 6-land 1:10"-twist cut rifling''.)<br />
## Receiver, action, and feed mechanism.<br/>(E.g., ''AR-10 magazine-fed semi-automatic''.)<br />
## Ammunition.<br />
### If commercial this must include brand, model, and lot.<br/>(E.g., ''Federal GM308M Lot#214374H077''.)<br />
### If custom, this must list component and load formula.<br/>(E.g., ''Lapua .308 full-sized brass, WLR primer, 168gr SMK, 44gr Varget, 2.800" COAL''.)<br />
# Confidence in BAC. When not conspicuously mentioned, it is assumed that the higher '''90%''' confidence value of sigma is referenced for BAC.<br />
<br />
BAC measures are intentionally kept somewhat coarse. Care should be taken to avoid suggesting more than two significant digits of precision. For example, even after shooting 20 rounds through a gun, the 80% confidence interval on the precision estimate is still from 0.9-1.2 times the estimated value.<br />
<br />
=== Trademarks ===<br />
The following terms are trademarks of Scribe Logistics LLC. They are free to use so long as their use complies with the Nomenclature and Protocol outlined here.<br />
* Ballistic Accuracy Classification™<br />
* Ballistic Accuracy Class™<br />
* BAC™<br />
The trademarks are claimed solely for the purpose of maintaining the integrity of the system and avoiding market confusion.<br />
<br />
== Theory ==<br />
<br />
=== Statistical Model ===<br />
<br />
BAC™ assumes that the impact of ballistic shots on a target are normally distributed along any axis. (Empirical data validate this assumption, and it should be true as long as atmospheric effects are negligible.<ref>There are two atmospheric effects that can finally create excess variance in one axis: Variable wind will increase horizontal variance. This is a function of the wind speed and the projectile's time-of-flight to the target. Also, as time-of-flight increases, variations in muzzle velocity will appear as excess vertical variance. Since both of these effects depend on time-of-flight, they are typically negligible for high-velocity rifles before 100 yards, or for subsonic projectiles before 25 yards.</ref>) Therefore, we use the Rayleigh distribution to model the radius r, or dispersion of each shot, from the center of impact. When the coordinates of the shots have independent <math>N(0,\sigma)</math> distributions along orthogonal axes, the radius of each shot is described by the Rayleigh probability density function:<br />
:&nbsp; <math>f(r,\sigma)=\frac{r}{\sigma^2}e^{−r^2/2\sigma^2}</math><br />
The unbiased estimator for the parameter σ comes from <math>\widehat{\sigma^2} = \frac{\sum r_i^2}{2(n−1)}</math>, with confidence <math>\widehat{\sigma^2} \sim \frac{\sum r^2}{\chi_{2n-2}^2}</math>.<br />
<br />
=== Simulation ===<br />
Monte Carlo simulation is adequate for studying and characterizing precision. In fact, many of the results associated with BAC, like the distribution of the extreme spread of a particular number of shots, can only be produced through simulation.<br />
<br />
For simulation purposes random shots should be generated as (x, y) coordinates, where <math>X,Y \sim N(0,\sigma)</math>. It is critical to "forget" the known center when using simulated data. When shooting a real gun we never get to know the true center of impact, and instead have to use the sample center. Likewise, Monte Carlo simulations must not reference the known 0 center, and should instead only reference the sample center of whatever group size is being studied. <br />
<br />
== Protocol ==<br />
The Ballistic Accuracy Classification™ shall be the upper bound of the 90% confidence range on estimated sigma, in units of MOA, multiplied by 10 and rounded to the nearest integer. For example, if the 90% confidence value for sigma on a tested gun is 0.47MOA, then the BAC value is 10 ✕ 0.47 = 4.7, rounded = 5. I.e., in this example we are saying with 90% confidence that the tested gun’s accuracy is no worse than Class 5.<br />
<br />
=== Classifying a Specimen ===<br />
It is not realistic to assign a BAC with fewer than 10 shots. The 90% confidence range with just 10 shots will typically extend to 1.4 times the estimated accuracy value. It will typically take 20 shots to get the outside bound of the 90% confidence interval to within 20% of the estimated value.<ref>The U.S. Army Marksmanship Unit (AMU) has long used a minimum of 3 consecutive 10-shot groups fired from a machine rest to test the accuracy of service rifles.</ref><br />
<br />
You must not discard data points during testing except for an unrelated failure. (E.g., if you are testing a barrel and you encounter a squib load, that shot may be excluded. But "[[fliers]]" should not generally be excluded.)<br />
<br />
'''Data:''' Target distance, and (''x,y'') coordinates of the center of each shot impact. All shots with the same point of aim must be grouped, but multiple groups can be used.<br />
<br />
'''Calculations:''' The formulas to transform these data into a confidence interval for sigma, and the corresponding BAC, are shown in [[Media:BallisticAccuracyClassification.xlsx]]. Given a sample of ''n'' shots, over ''g'' groups, at a distance ''d'':<br />
# For each measurement, convert to units of MOA. For example, if measurements are taken in inches, and the target was shot at a distance of ''d'' yards, then divide each measurement by 0.01047''d''<br />
# For each group ''g'', calculate the center of the group as <math>(\bar{x}_{i \in g},\bar{y}_{i \in g})</math><br />
# For each shot ''i'', find its radius squared relative to the center of the group as <math>r_i^2=(x_i−x_g)^2+(y_i−y_g)^2</math><br />
# Calculate the Gaussian correction factor <code>c<sub>G</sub>=1/EXP(LN(SQRT(2/(2n-2))) + GAMMALN((2n-1)/2) - GAMMALN((2n-2)/2))</code><br />
# Calculate the upper 90% confidence value for sigma as <code>σ<sub>U</sub>=cG*SQRT[SUM(r2)/CHIINV(0.9,2n-2)]</code><br />
# Calculate the Ballistic Accuracy Class as <code>=ROUND(σ<sub>U</sub>,0)</code><br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Ballistic_Accuracy_Classification&diff=1414Ballistic Accuracy Classification2017-03-18T16:49:40Z<p>David: /* Introduction */</p>
<hr />
<div>== Introduction ==<br />
<br />
Ballistic Accuracy Classification™ is a mathematically rigorous system for describing and understanding the precision of ballistic tools like rifles. It provides information that is:<br />
* Useful and easy for consumers to understand<br />
* Straightforward for enthusiasts and builders to calculate<br />
* Statistically sound enough for experts to validate<br />
<br />
All firearms and components can be assigned a Ballistic Accuracy Class™ (BAC™), which fully characterizes their accuracy potential. Lower numbers are better. In practice, the lowest possible BAC is 1. (A theoretically perfect rifle system that always puts every shot through the same hole would be BAC 0.)<br />
<br />
Behind the scenes, BAC is defined by a single statistical parameter known as sigma (σ). Using the associated statistical model (known as the Rayleigh distribution), statisticians can not only determine the BAC for a particular gun or component, but also compute its expected shooting precision.<ref>The proportion of shots expected to fall within radius ''rσ'' of the center of impact is <math>1-e^{-r^2/2}</math>.</ref><br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! Sigma (σ)<br />
! Typical examples<br />
! Shots within<BR/>¼ MOA radius<br />
! Shots within<BR/>½ MOA radius<br />
|-<br />
| Class 1<br />
| < 0.1MOA<br />
| Rail guns<br />
| 96%<br />
| 100%<br />
|-<br />
| Class 2<br />
| < 0.2MOA<br />
| Benchrest guns<br />
| 54%<br />
| 96%<br />
|-<br />
| Class 3<br />
| < 0.3MOA<br />
| Mil-spec for PSR<br />
| 29%<br />
| 75%<br />
|-<br />
| Class 4<br />
| < 0.4MOA<br />
| Competitive auto-loaders<br />
| 18%<br />
| 54%<br />
|-<br />
| Class 5<br />
| < 0.5MOA<br />
| Mil-spec for M110 and M24<br />
| 12%<br />
| 39%<br />
|-<br />
| Class 6<br />
| < 0.6MOA<br />
| Mil-spec for infantry rifles and ammo<br />
| 8%<br />
| 29%<br />
|}<br />
<br />
We can also generate the expected values of more familiar measures, like the extreme spread of a 3- or 5-shot group:<br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! 5-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 5-shot<BR/> Group Spread<br />
! 3-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 3-shot<BR/> Group Spread<br />
|-<br />
| Class 1<br />
| 100%<br />
| 0.3MOA<br />
| 100%<br />
| 0.2MOA<br />
|-<br />
| Class 2<br />
| 98%<br />
| 0.6MOA<br />
| 99%<br />
| 0.5MOA<br />
|-<br />
| Class 3<br />
| 65%<br />
| 0.9MOA<br />
| 85%<br />
| 0.7MOA<br />
|-<br />
| Class 4<br />
| 26%<br />
| 1.2MOA<br />
| 57%<br />
| 0.9MOA<br />
|-<br />
| Class 5<br />
| 9%<br />
| 1.5MOA<br />
| 35%<br />
| 1.2MOA<br />
|-<br />
| Class 6<br />
| 3%<br />
| 1.8MOA<br />
| 21%<br />
| 1.4MOA<br />
|}<br />
<br />
=== Understanding MOA ===<br />
Accuracy is described in angular terms, most commonly using the unit "Minute of Arc" (MOA). One arc minute spans 1.047" at 100 yards. Rifle shooters often practice on 100 yard targets, and so they often think in terms of how wide their groups are at 100 yards. People often just round it off and think of 1 MOA as one inch at 100 yards.<br />
<br />
In the absence of an atmosphere, the angular precision measured at one distance would be valid at all other distances. I.e., a 1" group at 100 yards would measure 5" at 500 yards. However, in reality the effects of wind and drag (which, together with gravity, accentuates variations in muzzle velocity) will only increase the angular spread of ballistic groups as distance increases. Therefore, '''in practice one should expect worse-than-advertised accuracy when shooting at longer distances'''. The distance at which atmosphere begins to significantly affect precision depends on a bullet’s muzzle velocity and ballistic coefficient. For high-power rifles this is usually beyond 100 yards. Guns shooting subsonic projectiles can begin to suffer after just 25 yards. <br />
<br />
=== Nomenclature ===<br />
<br />
BAC™ is only meaningful when it conforms to established terms and conventions. BAC must be determined by testing in accordance with the BAC Protocol described in this document.<br />
<br />
Ballistic Accuracy Classification™ must be supported by the following descriptive parameters:<br />
# Product tested. (Any component, or group of components, associated with accuracy can be tested.)<br />
# Configuration tested. This must include the following details:<br />
## Barrel length, material, profile, rifling.<br/>(E.g., ''20" stainless 1" bull contour with 6-land 1:10"-twist cut rifling''.)<br />
## Receiver, action, and feed mechanism.<br/>(E.g., ''AR-10 magazine-fed semi-automatic''.)<br />
## Ammunition.<br />
### If commercial this must include brand, model, and lot.<br/>(E.g., ''Federal GM308M Lot#214374H077''.)<br />
### If custom, this must list component and load formula.<br/>(E.g., ''Lapua .308 full-sized brass, WLR primer, 168gr SMK, 44gr Varget, 2.800" COAL''.)<br />
# Confidence interval on BAC. When not conspicuously mentioned, it is assumed that the upper bound of the 80% confidence interval is referenced for BAC.<br />
<br />
BAC measures are intentionally kept somewhat coarse. Care should be taken to avoid suggesting more than two significant digits of precision. For example, even after shooting 20 rounds through a gun, the 80% confidence interval on the precision estimate is still from 0.9-1.2 times the estimated value.<br />
<br />
=== Trademarks ===<br />
The following terms are trademarks of Scribe Logistics LLC. They are free to use so long as their use complies with the Nomenclature and Protocol outlined here.<br />
* Ballistic Accuracy Classification™<br />
* Ballistic Accuracy Class™<br />
* BAC™<br />
The trademarks are claimed solely for the purpose of maintaining the integrity of the system and avoiding market confusion.<br />
<br />
== Theory ==<br />
<br />
=== Statistical Model ===<br />
<br />
BAC™ assumes that the impact of ballistic shots on a target are normally distributed along any axis. (Empirical data validate this assumption, and it should be true as long as atmospheric effects are negligible.<ref>There are two atmospheric effects that can finally create excess variance in one axis: Variable wind will increase horizontal variance. This is a function of the wind speed and the projectile's time-of-flight to the target. Also, as time-of-flight increases, variations in muzzle velocity will appear as excess vertical variance. Since both of these effects depend on time-of-flight, they are typically negligible for high-velocity rifles before 100 yards, or for subsonic projectiles before 25 yards.</ref>) Therefore, we use the Rayleigh distribution to model the radius r, or dispersion of each shot, from the center of impact. When the coordinates of the shots have independent <math>N(0,\sigma)</math> distributions along orthogonal axes, the radius of each shot is described by the Rayleigh probability density function:<br />
:&nbsp; <math>f(r,\sigma)=\frac{r}{\sigma^2}e^{−r^2/2\sigma^2}</math><br />
The unbiased estimator for the parameter σ comes from <math>\widehat{\sigma^2} = \frac{\sum r_i^2}{2(n−1)}</math>, with confidence interval <math>\widehat{\sigma^2} \in \left[ \frac{\sum r^2}{\chi_2^2}, \ \frac{\sum r^2}{\chi_1^2} \right]</math>.<br />
<br />
=== Simulation ===<br />
Monte Carlo simulation is adequate for studying and characterizing precision. In fact, many of the results associated with BAC, like the distribution of the extreme spread of a particular number of shots, can only be produced through simulation.<br />
<br />
For simulation purposes random shots should be generated as (x, y) coordinates, where <math>X,Y \sim N(0,\sigma)</math>. It is critical to "forget" the known center when using simulated data. When shooting a real gun we never get to know the true center of impact, and instead have to use the sample center. Likewise, Monte Carlo simulations must not reference the known 0 center, and should instead only reference the sample center of whatever group size is being studied. <br />
<br />
== Protocol ==<br />
The Ballistic Accuracy Classification™ shall be the upper bound of the 80% confidence range on estimated sigma, in units of MOA, multiplied by 10 and rounded to the nearest integer. For example, if the 80% confidence interval for sigma on a tested gun is [0.33MOA, 0.47MOA], then the BAC value is 10 ✕ 0.47 = 4.7, rounded = 5. I.e., in this example we are saying with 80% confidence that the tested gun’s accuracy is no worse than Class 5.<br />
<br />
=== Classifying a Specimen ===<br />
It is not realistic to assign a BAC with fewer than 10 shots. The 80% confidence range with just 10 shots will typically extend to 1.4 times the estimated accuracy value. It will typically take 20 shots to get the outside bound of the 80% confidence interval to within 20% of the estimated value.<ref>The U.S. Army Marksmanship Unit (AMU) has long used a minimum of 3 consecutive 10-shot groups fired from a machine rest to test the accuracy of service rifles.</ref><br />
<br />
You must not discard data points during testing except for an unrelated failure. (E.g., if you are testing a barrel and you encounter a squib load, that shot may be excluded. But "[[fliers]]" should not generally be excluded.)<br />
<br />
'''Data:''' Target distance, and (''x,y'') coordinates of the center of each shot impact. All shots with the same point of aim must be grouped, but multiple groups can be used.<br />
<br />
'''Calculations:''' The formulas to transform these data into an 80% confidence interval for sigma, and the corresponding BAC, are shown in [[Media:BallisticAccuracyClassification.xlsx]]. Given a sample of ''n'' shots, over ''g'' groups, at a distance ''d'':<br />
# For each measurement, convert to units of MOA. For example, if measurements are taken in inches, and the target was shot at a distance of ''d'' yards, then divide each measurement by 0.01047''d''<br />
# For each group ''g'', calculate the center of the group as <math>(\bar{x}_{i \in g},\bar{y}_{i \in g})</math><br />
# For each shot ''i'', find its radius squared relative to the center of the group as <math>r_i^2=(x_i−x_g)^2+(y_i−y_g)^2</math><br />
# Calculate the Gaussian correction factor <code>c<sub>G</sub>=1/EXP(LN(SQRT(2/(2n-2))) + GAMMALN((2n-1)/2) - GAMMALN((2n-2)/2))</code><br />
# Calculate the upper bound on the 80% confidence interval for sigma as <code>σ<sub>U</sub>=cG*SQRT[SUM(r2)/CHIINV(0.9,2n-2)]</code><br />
# Calculate the Ballistic Accuracy Class as <code>=ROUND(σ<sub>U</sub>,0)</code><br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=300BLK_Test&diff=1413300BLK Test2017-03-15T23:00:18Z<p>David: /* hBN plating */ Added micrographs</p>
<hr />
<div>= Objective =<br />
Improve the accuracy of 300BLK subsonic ammunition.<br />
<br />
The standard subsonic 300BLK load exhibits significant variation in muzzle velocity, which results in [http://www.300blktalk.com/forum/viewtopic.php?f=140&t=86927 excessive vertical dispersion].<br />
<br />
We will test various products and hypotheses to determine whether improving internal ballistics can statistically increase the precision of subsonic 300BLK.<br />
<br />
= Background =<br />
The objective of subsonic ammunition is to avoid the sonic crack that occurs as bullet velocities approach the speed of sound, which is typically about 1100fps. Given this upper bound on velocity the only way to improve external and terminal ballistics is to increase the mass of the projectile. The upper limit on mass of standard .30" jacketed lead-core bullets that fit the AR-15 platform with a standard bolt and magazine is 225gr, which in the BTHP profile is a bullet 1.57" long, with a bearing length of 0.70-0.75".<br />
<br />
The standard Remington factory subsonic load, designed to cycle reliably in the greatest variety of guns, consists of:<br />
* 220gr BTHP bullet<br />
* 10.4gr A1680<br />
* 2.12" COAL<br />
QuickLOAD estimates that this generates MAP of 33kpsi and only 81% propellant burnt within a 16" barrel.<br />
<br />
= Hypotheses =<br />
# Reducing friction between the bullet and bore should reduce the variation of muzzle velocity.<br />
# Increasing peak muzzle pressure and/or burning efficiency should reduce the variation of interior ballistics, and therefore also reduce variation of muzzle velocity.<br />
<br />
== Pressure and Accuracy ==<br />
The standard subsonic 300BLK load puts a bullet optimized for supersonic flight down a bore at little more than half the peak pressure for which modern ballistic components are designed. This may result in higher interior ballistic variances, which would also result in higher muzzle velocity variance.<br />
<br />
We will test loads that produce higher peak pressures than the standard 33kpsi and higher interior burn ratios than the standard 80% through a 16" barrel.<br />
<br />
We will also test Lapua's 200gr subsonic bullets. These have a reduced bearing surface, which should reduce bore friction and any variance associated with that. They also have a profile optimized for subsonic flight, which may improve the exterior ballistic precision.<br />
<br />
== Friction ==<br />
Variables:<br />
# Bullet coating<br />
# Barrel alloy<br />
# Barrel condition<br />
# [[300BLK_Test#Bore_Cross-section_Correlation_to_Friction|Bore/rifling dimensions?]]<br />
<br />
hBN has a lower coefficient of friction and higher stability than established but messier friction-proofing compounds like MoS2 ("moly") or WS2 ("Danzac"). However it is unclear whether common uses of hBN actually reduce friction. In general we expect reduced bore friction to result in markedly lower muzzle velocities &mdash; a well-known phenomenon with MoS2. But no hBN users consulted to date have observed significantly reduced muzzle velocities. Rather, the most frequently cited benefit of the proper use of hBN is reduced bore fouling. Experts like David Tubb and [http://theswissriflesdotcommessageboard.yuku.com/topic/9031/hBN-again-and-this-time-Ill-leave-it#.U6BsPPldXEV Swiss Products] claim that shooting only hBN-plated bullets and bores result in bores that can be cleaned with a single dry patch. The fact that copper is not being deposited on the bore does sound consistent with reduced friction. But our final objective is to reduce muzzle velocity variance; if a friction-proofed system fails to achieve this then we have not met our objective.<br />
<br />
Here we will test four variations of hBN:<br />
# Tubb et. al. advocate impact-plating bullets with HCPL-grade hBN, which averages 10 microns and is among the most lubricious formulations. For our tests we use a vibratory tumbler and steel shot to produce what we will call '''''plated bullets'''''.<br />
# Tubb et. al. also advocate coating a completely clean bore with sub-micron hBN (we use AC6111-grade) suspended in isopropyl alcohol, and then firing a plated bullet through it to fix the hBN. We refer to this as a '''''coated bore'''''.<br />
# [http://rydol.com/ Rydol] addresses a common concern with plated bullets which is that the impact-plating process results in a non-uniform and potentially excessive build-up of hBN. Rydol advocates bullet lubrication via ultrasonic embedment of sub-micron hBN and PTFE. We will test bullets treated by their method as '''''Polysonic bullets'''''.<br />
# Rydol also sells a bore conditioner designed to embed sub-micron hBN and PTFE particles in bore irregularities, creating a smoother and more lubricious surface. We will test this as a '''''Rydol bore'''''.<br />
<br />
The consensus is that hBN is easily stripped from both ''coated'' and ''Rydol'' bores by most cleaning chemicals. Therefore we will be able to "reset" test barrels when necessary via regular cleaning.<br />
<br />
We already performed one test of plated bullets: [http://www.300blktalk.com/forum/viewtopic.php?p=817836#p817836 We compared 50 rounds handloaded with the standard specifications to 50 rounds using the same specifications but with hBN-plated bullets]. From a 16" stainless barrel there was no significant difference in muzzle velocity variance between the plated and unplated bullets:<br />
* Uncoated average = 923fps, stdev = 25fps<br />
* hBN coated average = 941fps, stdev = 27fps<br />
<br />
Subsequently we have found suggestions that hBN does not embed or interact with stainless steel as effectively as it does with other standard barrel alloys. Therefore this test will be repeated with 4140 chrome-moly nitride barrels.<br />
<br />
While studying [[300BLK_Test#Guns|the barrels assembled for this test]] we found one more potential variable: the bore and rifling dimensions. Although all three use 5-land rifling the lands on the Noveske barrel are far wider than on the AAC barrel. ''We need to slug the barrels and determine whether cross section varies enough to explain variations in friction.'' See [[300BLK_Test#Bore_Cross-section_Correlation_to_Friction]] in Ancillary Tests and Questions below.<br />
<br />
=== Pre-test on hBN coating ===<br />
After running a vibratory coating cycle for 3 hours in airtight containers I noticed a significant ammonia odor on opening them. Subsequent testing confirmed the presence of substantial ammonia, which suggests contamination and/or chemical degradation of the hBN. After consulting with an engineer at Momentive Performance Materials, the source of the hBN, they agreed to analyze my samples to try to determine what's going wrong. Their current theory is that the mechanical action of the impact-plating with steel media is causing a degradation of the hBN. But in order to rule out contamination they are sending a new batch of HCPL-grade hBN. I will conduct controlled tests with different media and different run-times, as well as peak temperatures during operation, to see if there is any relationship to ammonia production.<br />
<br />
= Test outline =<br />
<br />
Our objective is to look for statistically significant improvements over the standard load and untreated barrels in terms of:<br />
# Reduced variation of muzzle velocity.<br />
# Reduced mean radius (i.e., increased on-target precision).<br />
# Reduced velocities or bore temperatures as evidence that any variable has reduced bullet-bore friction.<br />
<br />
Therefore we will run the following test strings:<br />
<br />
== Friction Sequence ==<br />
Using standard load:<br />
<br />
Load Batch #0:<br />
# 16" bbl:<br />
## Normal bullet x50: μ = 923fps; σ = 25fps. MR = 1.37MOA, 90% = (1.23, 1.56)<br />
## Plated bullet x50: μ = 941fps; σ = 27fps. MR = 1.46MOA, 90% = (1.30, 1.65) <br />
<br />
Load Batch #1: This was loaded with 13.4gr A1680<br />
# 16" bbl:<br />
## Normal bullet x20: μ = 1286fps; σ = 16fps. MR = 1.75MOA, 90% = (1.46, 2.19)<br />
## Polysonic bullet x30: μ = 1231fps; σ = 24fps. MR = 1.06MOA, 90% = (.92, 1.26)<br />
# 12" coated bbl<br />
## Plated bullet x40: μ = 1264fps; σ = 22fps. MR = 2.33MOA, 90% = (1.96, 2.87)<br />
# 12" clean bbl:<br />
## Normal bullet x20: μ = 1250fps; σ = 20fps. MR = 1.85MOA, 90% = (1.56, 2.29)<br />
## Polysonic bullet x30: μ = 1230fps; σ = 21fps. MR = 1.55MOA, 90% = (1.35, 1.83)<br />
# 12" Rydol bbl:<br />
## Normal bullet x30: μ = 1274fps; σ = 17fps. MR = 1.50MOA, 90% = (1.30, 1.77)<br />
## Polysonic bullet x30: μ = 1260fps; σ = 26fps. MR = 1.11MOA, 90% = (0.97, 1.32)<br />
<br />
Load Batch #2:<br />
# 12" clean bbl<br />
## Normal bullet x25: μ = 929fps; σ = 16fps. <br />
## B-coated bullet x25: μ = 901fps; σ = 33fps. <br />
# 16" clean bbl<br />
## B-coated bullet x25: μ = 954fps; σ = 24fps (17fps excluding first two as "foulers"). <br />
## Normal bullet x25: μ = 981fps; σ = 24fps. <br />
# Clean and B-coat 12" bbl<br />
## Normal bullet x25: μ = 948fps; σ = 26fps. <br />
## B-coated bullet x25: μ = 935fps; σ = 22fps. <br />
# Clean and B-coat 16" bbl<br />
## B-coated bullet x25: μ = 984fps; σ = 25fps<br />
## Normal bullet x25: μ = 1010fps; σ = 21fps<br />
<br />
== Pressure/Accuracy Sequence ==<br />
Note that bullets are seated to max magazine length to minimize jump to lands.<br />
# 220gr load with MAP > 45kpsi and > 99% interior burn: 10.1gr VV N110, 2.20" COAL.<br />
## 16" clean bbl <br />
### Unsuppressed x20: μ = 1147fps; σ = 12fps. MR = 0.84MOA, 90% = (0.71, 1.04)<br />
### Suppressed x20: μ = 1151fps; σ = 11fps. MR = 0.62MOA, 90% = (0.52, 0.78)<br />
## 12" Rydol bbl<br />
### Unsuppressed x10 (fails to cycle): μ = 1166fps; σ = 15fps.<br />
### Suppressed x20: μ = 1170fps; σ = 13fps. MR = 0.70MOA, 90% = (0.59, 0.86)<br />
# 200gr Lapua .308" load with MAP > 45kpsi and > 99% interior burn: 10.2gr VV N110, 2.10" COAL.<br />
## 16" clean bbl<br />
### Unsuppressed x20: μ = 1170fps; σ = 11fps. MR = 0.86MOA, 90% = (0.72, 1.06)<br />
### Suppressed x20: μ = 1203fps; σ = 12fps. MR = 0.79MOA, 90% = (0.67, 0.98)<br />
## 12" Rydol bbl with suppressor x20: μ = 1218fps; σ = 19fps. MR = 1.11MOA, 90% = (0.94, 1.37)<br />
<br />
= Equipment =<br />
<br />
== Guns ==<br />
All guns have pistol-length gas systems and will run the same NP3-coated BCG.<br />
<br />
All barrels have 5-land rifling (''TBD: rifling method and cross section via slugging''):<br />
# Noveske 16" 1:7 stainless. Chambers 220gr SMKs up to 2.27" COAL and 200gr Lapua subsonics up to 2.14" COAL.<br />
# AAC 12.5" 1:7 4140 chorme-moly nitride. Chambers 220gr SMKs up to 2.31" COAL. [http://i.stack.imgur.com/xFGC3.jpg Muzzle photo]<br />
# CMMG 8" 1:7 4140 chorme-moly nitride. Chambers 220gr SMKs up to 2.27" COAL and 200gr Lapua subsonics up to 2.14" COAL.<br />
<br />
== Loads ==<br />
All cases are fired Remington 300BLK brass. Cases are full-length sized, then case mouth is chamferred, deburred, brushed; then primer pocket is cleaned.<br />
<br />
Primers are Remington #7.5.<br />
<br />
Powder is dispensed by volume after calibration. Bullets in a subsequence are alternated so that any drift in powder dispenser is evenly distributed between the strings.<br />
<br />
A single lot of A1680 and VV N110 will be used for all loads.<br />
<br />
Bullets are seated with a Forster Benchrest Seater die.<br />
<br />
524 Bullets required:<br />
* 185 220gr SMKs<br />
* 93 Polysonic coated 220gr SMKs<br />
* 82 Plated 220gr SMKs<br />
* 82 Lapua 200gr .308"<br />
* 82 Lapua 200gr .310"<br />
<br />
== Measurements ==<br />
Two chronographs will be positioned in line starting 15 feet from muzzle.<br />
# [https://www.competitionelectronics.com/index.php?page=shop.product_details&flypage=flypage.tpl&product_id=20&category_id=7&vmcchk=1&option=com_virtuemart&Itemid=79 Competition Electronics ProChono Digital]<br />
# [http://www.btibrands.com/product/ballistic-precision-chronograph/ Caldwell Ballistic Precision Chronograph]<br />
<br />
Thermocouple to check barrel temperature secured with electrical tape to top of barrel 2.5" forward of gas block.<br />
<br />
= Procedures =<br />
All guns will be shot without suppressors, [http://emptormaven.com/2007/03/freebore-boost-effects/ which increase dispersion of muzzle velocity].<br />
<br />
Muzzles will be shot with bare threads, to avoid the risk that thread protectors or any other device might loosen or affect harmonics.<br />
<br />
One sample of every bullet/bore condition will be fired into a water tank so bullet engraving can be examined.<br />
<br />
Test firing shall be at 100-yard targets from a shaded bunker. If accuracy fixture is operational at time of test guns will be locked in fixture for firing. Otherwise firing will be from sandbags with a benchrest scope in LaRue QD mount. Trigger is Timney 3.5# single-stage.<br />
<br />
Every attempt will be made to avoid firing when wind gusts appear to exceed 5mph.<br />
<br />
Every sample string will start with a cold bore and we will attempt to fire them at the same rate so that the final barrel temperatures can be fairly compared.<br />
<br />
= Data =<br />
For each string:<br />
# Load<br />
# Upper<br />
# Bore condition<br />
<br />
== Temperature ==<br />
# String<br />
# Start time, barrel temp, and ambient temp & humidity<br />
# End time, barrel temp, and ambient temp & humidity<br />
<br />
== Chronograph ==<br />
# String<br />
# Shot #<br />
# Chronograph #1 FPS<br />
# Chronograph #2 FPS<br />
<br />
== Precision ==<br />
# String<br />
# Mean Radius with 90% confidence interval<br />
<br />
== Failures ==<br />
Failures to feed, fire or cycle will be noted but corrected as quickly as possible to avoid interrupting the string.<br />
<br />
= Ancillary Tests and Questions =<br />
<br />
== Atomic Analysis ==<br />
Fired and unfired samples of bullets, both coated and uncoated, were sent to an imaging specialist with access to scanning electron and atomic force microscopes, as well as energy-dispersive x-ray (EDX) equipment. We hope to see:<br />
* What are the coefficients of friction on the bearing surfaces of the three unloaded rounds?<br />
* Can you determine how thick and uniform the hBN coating is? If so,<br />
* Can you determine how much, if any, hBN survives the firing process – especially in the engraved portions of the bullets where the friction is most substantial?<br />
<br />
== Bore Cross-section Correlation to Friction ==<br />
''Does the bore cross-section vary enough between barrels to affect bullet friction?''<br />
<br />
The bore cross-section is specified by SAAMI, but that dimension is often only loosely referenced by machinists.<br />
<br />
Therefore we will carefully slug the barrels to see if we can detect variations in cross-section. If that is insufficiently precise we will measure the bore volume (using water-soluble cutting fluid).<br />
<br />
= Conclusions =<br />
<br />
Our objective was to increase the precision of subsonic 300BLK rifles by looking at modifications to the ammunition, with a special emphasis on trying to reduce the variance of muzzle velocity.<br />
<br />
Neither bullet coatings nor barrel coatings could significantly reduce the variance of muzzle velocity using A1680 powder. However two of the three bullet coatings tested ''did'' show evidence of friction-proofing.<br />
<br />
# '''Bullet Coatings:'''<br />
## hBN impact-plated bullets always resulted in wider groups, higher muzzle velocities, and no improvement in the variance of velocity.<br />
## Rydol Polysonic-treated bullets always resulted in significantly tighter groups and slightly lower muzzle velocities than the same load with regular untreated bullets.<br />
## Company ''B'' coated bullets always resulted in lower muzzle velocities (about 2.5% lower), indicating consistent friction proofing.<br />
# '''Barrel Coatings:'''<br />
## Test strings 1.3 and 1.4 show that the Rydol conditioned bore produced higher muzzle velocities than the control (i.e., clean bore before treatment) -- about 2.4% higher. In general I do not find this to be a good thing, since it's the opposite of what good lubrication is supposed to do. However, the ultimate objective here was to look at tightening groups, and it does appear to have delivered on that front: In fact, I show better than 90% confidence that the Polysonic bullets were more precise in the coated bore than the uncoated.<br />
## Test strings 2.3 and 2.4 show that the Company ''B'' conditioned bores also produced higher muzzle velocities -- almost 3% higher than the controls. However I may have applied too thick a coating, constricting the bore cross-section and thereby producing the higher pressures indicated by the velocity increase. This test was run at lower velocities than the first, making it impractical to gauge relative precision at 100 yards. I did note that absolutely no copper fouling was evident in the coated bores after the test, so the coating is performing its intended function.<br />
# '''Loads:''' I suspected that the MAP (about 33kpsi) and incomplete interior burn of the standard load is a significant contributor to the high standard deviation of muzzle velocity – above 20fps – observed in all tests using that load. I tested a load using VV N110, which QuickLOAD estimates achieves MAP of 45kpsi and almost complete interior burn. While marginally supersonic (1150fps) and unable to cycle the AAC upper without a suppressor, this '''did appear to validate the theory since in every test standard deviation of muzzle velocity was about 12fps, and groups were better than half the mean radius of the control groups!'''<br />
<br />
Therefore, if you want to produce ammunition that should give better accuracy in ''every'' gun I would have Rydol treat the bullets. Company ''B'''s bullet coating is probably more expensive but also more reliable in its effect.<br />
<br />
I would also spend more time looking for more efficient loads with higher MAPs. If you can’t do this without compromising compatibility I would produce “special-purpose” ammunition that is advertised as having higher accuracy, but that may require a suppressor or wider gas port to cycle reliably.<br />
<br />
One other observation concerns barrels: The 12” 4140 barrel had a leade that was about .04” longer than the leade in the 16” stainless barrel. The stainless barrel was always more accurate, and I suspect the reduced bullet jump was a significant contributor to that accuracy. <br />
<br />
== hBN plating ==<br />
[[File:UncoatedFired.jpg|320px|thumb|left|Micrograph of the beginning of rifled groove engraved on a .30" copper-jacketed bullet.]] <br />
[[File:HBNfired.jpg|320px|thumb|right|Micrograph of rifled engraving on a .30" copper-jacketed bullet that was impact-plated with hBN (which appears in black).]]<br />
Testing the impact-plated bullets in both stainless and 4140 barrels, coated or uncoated, shows the bullet coating actually ''adversely impacts performance'': '''muzzle velocity is slightly higher, velocity dispersion is the same, and accuracy on target is reduced'''! Electron micrographs of a plated bullet show that absolutely no hBN is left in the bullet grooves, suggesting it is scraped away by the rifling during engraving. Which means that all the impact plating process is doing is dinging up the bullets on a microscopic level and leaving (as the imaging technician described it) puddles of hBN in the resulting jacket scars.<br />
<br />
A number of variations were attempted on impact plating, including adding heat, varying run-time, and varying media. None created any notable difference. I even sent a batch to WMD to attempt spin-coating. On their own initiative they even bead-blasted some bullets to try to get the hBN to adhere better and they concluded it would not. Even with a batch of HCPL hBN direct from the manufacturer degradation into ammonia always resulted from aggressive and/or extended run times. I’m convinced impact-plating hBN is a dead end.<br />
<br />
== Polysonic coating ==<br />
Show reduced muzzle velocity, which is the surest sign of friction proofing I know. Specifically, in a 12" barrel I measured muzzle velocity falling from 1250fps to 1230fps, and in a 16" barrel it falls from 1286fps to 1231fps. (The fact that we measure the same muzzle velocity out of two barrels with a length difference of 4" is not surprising: Remember this cartridge is intentionally loaded with a fast powder that doesn't exploit barrel length to build velocity, and also that the rifling and composition of the two barrels are different.)<br />
<br />
The tests involved shooting 20 rounds of the untreated "control" bullets followed by 30 rounds of Polysonic bullets, in each barrel.<br />
<br />
Polysonic bullets also show statistically significant increases in accuracy on target vs regular bullets. From the 16" barrel the target statistics show Polysonic to be more accurate at the 98% confidence level!<br />
<br />
== Rydol bore coating ==<br />
The tests show this coating produces a further increase in precision, but it's unclear to me why that would happen when muzzle velocities are increasing.<br />
<br />
Unfortunately I don't have a good protocol to measure the other benefits Rydol should provide, like reduced fouling and easier cleaning.<br />
<br />
It's also possible I didn't properly condition the barrel, so there's a potential user/tester error on this product that doesn't exist with the factory-coated Polysonic bullets.<br />
<br />
During this test I was only able to evaluate this product on one "virgin barrel." Ideally I would do a much more extended test involving multiple barrels as well as a full-rifle caliber (e.g., .223 or .308) to better understand the performance of the bore conditioner.</div>Davidhttp://ballistipedia.com/index.php?title=File:HBNfired.jpg&diff=1412File:HBNfired.jpg2017-03-15T22:47:32Z<p>David: Micrograph of a fired copper-jacketed bullet that was impact-plated with hBN (which appears in black). Shown is the beginning of one groove at the bullet's ogive.</p>
<hr />
<div>Micrograph of a fired copper-jacketed bullet that was impact-plated with hBN (which appears in black). Shown is the beginning of one groove at the bullet's ogive.</div>Davidhttp://ballistipedia.com/index.php?title=File:UncoatedFired.jpg&diff=1411File:UncoatedFired.jpg2017-03-15T22:46:25Z<p>David: Micrograph of a fired copper-jacketed bullet. Shown is the beginning of one groove at the bullet's ogive.</p>
<hr />
<div>Micrograph of a fired copper-jacketed bullet. Shown is the beginning of one groove at the bullet's ogive.</div>Davidhttp://ballistipedia.com/index.php?title=ShotStat:About&diff=1410ShotStat:About2017-03-14T14:51:01Z<p>David: </p>
<hr />
<div>This project began as a collaboration between David Bookstaber, Charles McMillan, and Daniel Wollschlaeger.<br />
<br />
If you would like to contribute please [mailto:bookstaber@hotmail.com Email David] and he will give you an account.</div>Davidhttp://ballistipedia.com/index.php?title=Ballistic_Accuracy_Classification&diff=1409Ballistic Accuracy Classification2017-03-13T18:51:23Z<p>David: Created page with "== Introduction == Ballistic Accuracy Classification™ is a mathematically rigorous system for describing and understanding the precision of ballistic tools like rifles. It..."</p>
<hr />
<div>== Introduction ==<br />
<br />
Ballistic Accuracy Classification™ is a mathematically rigorous system for describing and understanding the precision of ballistic tools like rifles. It provides information that is:<br />
Useful and easy for consumers to understand<br />
Straightforward for enthusiasts and builders to calculate<br />
Statistically sound enough for experts to validate<br />
<br />
All firearms and components can be assigned a Ballistic Accuracy Class™ (BAC™), which fully characterizes their accuracy potential. Lower numbers are better. In practice, the lowest possible BAC™ is 1. (A theoretically perfect rifle system that always puts every shot through the same hole would be BAC™ 0.)<br />
<br />
Behind the scenes, BAC™ is defined by a single statistical parameter known as sigma (σ). Using the associated statistical model (known as the Rayleigh distribution), statisticians can not only determine the BAC™ for a particular gun or component, but also compute its expected shooting precision.<ref>The proportion of shots expected to fall within radius ''rσ'' of the center of impact is <math>1-e^{-r^2/2}</math>.</ref><br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! Sigma (σ)<br />
! Typical examples<br />
! Shots within<BR/>¼ MOA radius<br />
! Shots within<BR/>½ MOA radius<br />
|-<br />
| Class 1<br />
| < 0.1MOA<br />
| Rail guns<br />
| 96%<br />
| 100%<br />
|-<br />
| Class 2<br />
| < 0.2MOA<br />
| Benchrest guns<br />
| 54%<br />
| 96%<br />
|-<br />
| Class 3<br />
| < 0.3MOA<br />
| Mil-spec for PSR<br />
| 29%<br />
| 75%<br />
|-<br />
| Class 4<br />
| < 0.4MOA<br />
| Competitive auto-loaders<br />
| 18%<br />
| 54%<br />
|-<br />
| Class 5<br />
| < 0.5MOA<br />
| Mil-spec for M110 and M24<br />
| 12%<br />
| 39%<br />
|-<br />
| Class 6<br />
| < 0.6MOA<br />
| Mil-spec for infantry rifles and ammo<br />
| 8%<br />
| 29%<br />
|}<br />
<br />
We can also generate the expected values of more familiar measures, like the extreme spread of a 3- or 5-shot group:<br />
<br />
{| class="wikitable"<br />
! BAC™<br />
! 5-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 5-shot<BR/> Group Spread<br />
! 3-shot Groups<BR/> ⌀ < 1MOA<br />
! Median 3-shot<BR/> Group Spread<br />
|-<br />
| Class 1<br />
| 100%<br />
| 0.3MOA<br />
| 100%<br />
| 0.2MOA<br />
|-<br />
| Class 2<br />
| 98%<br />
| 0.6MOA<br />
| 99%<br />
| 0.5MOA<br />
|-<br />
| Class 3<br />
| 65%<br />
| 0.9MOA<br />
| 85%<br />
| 0.7MOA<br />
|-<br />
| Class 4<br />
| 26%<br />
| 1.2MOA<br />
| 57%<br />
| 0.9MOA<br />
|-<br />
| Class 5<br />
| 9%<br />
| 1.5MOA<br />
| 35%<br />
| 1.2MOA<br />
|-<br />
| Class 6<br />
| 3%<br />
| 1.8MOA<br />
| 21%<br />
| 1.4MOA<br />
|}<br />
<br />
=== Understanding MOA ===<br />
Accuracy is described in angular terms, most commonly using the unit "Minute of Arc" (MOA). One arc minute spans 1.047" at 100 yards. Rifle shooters often practice on 100 yard targets, and so they often think in terms of how wide their groups are at 100 yards. It is sufficient to think of 1 MOA as one inch at 100 yards.<br />
<br />
In the absence of an atmosphere, the angular precision measured at one distance would be valid at all other distances. I.e., a 1" group at 100 yards would measure 5" at 500 yards. However, in reality the effects of wind and drag (which, together with gravity, accentuates variations in muzzle velocity) will only increase the angular spread of ballistic groups as distance increases. Therefore, '''in practice one should expect worse-than-advertised accuracy when shooting at longer distances'''. The distance at which atmosphere begins to significantly affect precision depends on a bullet’s muzzle velocity and ballistic coefficient. For high-power rifles this is usually beyond 100 yards. Guns shooting subsonic projectiles can begin to suffer after just 25 yards. <br />
<br />
=== Nomenclature ===<br />
<br />
BAC™ is only meaningful when it conforms to established terms and conventions. BAC must be determined by testing in accordance with the BAC Protocol described in this document.<br />
<br />
Ballistic Accuracy Classification™ must be supported by the following descriptive parameters:<br />
# Product tested. (Any component, or group of components, associated with accuracy can be tested.)<br />
# Configuration tested. This must include the following details:<br />
## Barrel length, material, profile, rifling.<br/>(E.g., ''20" stainless 1" bull contour with 6-land 1:10"-twist cut rifling''.)<br />
## Receiver, action, and feed mechanism.<br/>(E.g., ''AR-10 magazine-fed semi-automatic''.)<br />
## Ammunition.<br />
### If commercial this must include brand, model, and lot.<br/>(E.g., ''Federal GM308M Lot#214374H077''.)<br />
### If custom, this must list component and load formula.<br/>(E.g., ''Lapua .308 full-sized brass, WLR primer, 168gr SMK, 44gr Varget, 2.800" COAL''.)<br />
# Confidence interval on BAC. When not conspicuously mentioned, it is assumed that the upper bound of the 80% confidence interval is less than the specified BAC threshold.<br />
<br />
BAC measures are intentionally kept somewhat coarse. Care should be taken to avoid suggesting more than one significant digit of precision. For example, even after shooting 20 rounds through a gun, the 80% confidence interval on the precision estimate is still from 0.9-1.2 times the estimated value.<br />
<br />
=== Trademarks ===<br />
The following terms are trademarks of Scribe Logistics LLC. They are free to use so long as their use complies with the Nomenclature and Protocol outlined here.<br />
* Ballistic Accuracy Classification™<br />
* Ballistic Accuracy Class™<br />
* BAC™<br />
The trademarks are claimed solely for the purpose of maintaining the integrity of the system and avoiding market confusion.<br />
<br />
== Theory ==<br />
<br />
=== Statistical Model ===<br />
<br />
BAC™ assumes that the impact of ballistic shots on a target are normally distributed along any axis. (Empirical data validate this assumption, and it should be true as long as atmospheric effects are negligible.<ref>There are two atmospheric effects that can finally create excess variance in one axis: Variable wind will increase horizontal variance. This is a function of the wind speed and the projectile's time-of-flight to the target. Also, as time-of-flight increases, variations in muzzle velocity will appear as excess vertical variance. Since both of these effects depend on time-of-flight, they are typically negligible for high-velocity rifles before 100 yards, or for subsonic projectiles before 25 yards.</ref>) Therefore, we use the Rayleigh distribution to model the radius r, or dispersion of each shot, from the center of impact. When the coordinates of the shots have independent <math>N(0,\sigma)</math> distributions along orthogonal axes, the radius of each shot is described by the Rayleigh probability density function:<br />
:&nbsp; <math>f(r,\sigma)=\frac{r}{\sigma^2}e^{−r^2/2\sigma^2}</math><br />
The unbiased estimator for the parameter σ comes from <math>\widehat{\sigma^2} = \frac{\sum r_i^2}{2(n−1)}</math>, with confidence interval <math>\widehat{\sigma^2} \in \left[ \frac{\sum r^2}{\chi_2^2}, \ \frac{\sum r^2}{\chi_1^2} \right]</math>.<br />
<br />
=== Simulation ===<br />
Monte Carlo simulation is adequate for studying and characterizing precision. In fact, many of the results associated with BAC, like the distribution of the extreme spread of a particular number of shots, can only be produced through simulation.<br />
<br />
For simulation purposes random shots should be generated as (x, y) coordinates, where <math>X,Y \sim N(0,\sigma)</math>. It is critical to "forget" the known center when using simulated data. When shooting a real gun we never get to know the true center of impact, and instead have to use the sample center. Likewise, Monte Carlo simulations must not reference the known 0 center, and should instead only reference the sample center of whatever group size is being studied. <br />
<br />
== Protocol ==<br />
The Ballistic Accuracy Classification™ shall be the upper bound of the 80% confidence range on estimated sigma, in units of MOA, multiplied by 10 and rounded to the nearest integer. For example, if the 80% confidence interval for sigma on a tested gun is [0.33MOA, 0.47MOA], then the BAC value is 10 ✕ 0.47 = 4.7, rounded = 5. I.e., in this example we are saying with 80% confidence that the tested gun’s accuracy is no worse than Class 5.<br />
<br />
=== Classifying a Specimen ===<br />
It is not realistic to assign a BAC with fewer than 10 shots. The 80% confidence range with just 10 shots will typically extend to 1.4 times the estimated accuracy value. It will typically take 20 shots to get the outside bound of the 80% confidence interval to within 20% of the estimated value.<ref>The U.S. Army Marksmanship Unit (AMU) has long used a minimum of 3 consecutive 10-shot groups fired from a machine rest to test the accuracy of service rifles.</ref><br />
<br />
You must not discard data points during testing except for an unrelated failure. (E.g., if you are testing a barrel and you encounter a squib load, that shot may be excluded. But "[[fliers]]" should not generally be excluded.)<br />
<br />
'''Data:''' Target distance, and (''x,y'') coordinates of the center of each shot impact. All shots with the same point of aim must be grouped, but multiple groups can be used.<br />
<br />
'''Calculations:''' The formulas to transform these data into an 80% confidence interval for sigma, and the corresponding BAC, are shown in [[Media:BallisticAccuracyClassification.xlsx]]. Given a sample of ''n'' shots, over ''g'' groups, at a distance ''d'':<br />
# For each measurement, convert to units of MOA. For example, if measurements are taken in inches, and the target was shot at a distance of ''d'' yards, then divide each measurement by 0.01047''d''<br />
# For each group ''g'', calculate the center of the group as <math>(\bar{x}_{i \in g},\bar{y}_{i \in g})</math><br />
# For each shot ''i'', find its radius squared relative to the center of the group as <math>r_i^2=(x_i−x_g)^2+(y_i−y_g)^2</math><br />
# Calculate the Gaussian correction factor <code>c<sub>G</sub>=1/EXP(LN(SQRT(2/(2n-2))) + GAMMALN((2n-1)/2) - GAMMALN((2n-2)/2))</code><br />
# Calculate the upper bound on the 80% confidence interval for sigma as <code>σ<sub>U</sub>=cG*SQRT[SUM(r2)/CHIINV(0.9,2n-2)]</code><br />
# Calculate the Ballistic Accuracy Class as <code>=ROUND(σ<sub>U</sub>,0)</code><br />
<br />
== Notes ==<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=File:BallisticAccuracyClassification.xlsx&diff=1408File:BallisticAccuracyClassification.xlsx2017-03-13T18:43:43Z<p>David: Sample calculation of Ballistic Accuracy Classification based on three 10-shot groups.</p>
<hr />
<div>Sample calculation of Ballistic Accuracy Classification based on three 10-shot groups.</div>Davidhttp://ballistipedia.com/index.php?title=Closed_Form_Precision&diff=1407Closed Form Precision2017-03-13T17:19:26Z<p>David: Fixed reference to Hogema</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Precision Models]]</p><br />
<br />
= Symmetric Bivariate Normal Shots Imply Rayleigh Distributed Distances =<br />
[[File:Bivariate.png|400px|thumb|right|Distribution of samples from a symmetric bivariate normal distribution. Axis units are multiples of σ.]]<br />
After factoring out the known sources of asymmetry in the bivariate normal model we might conclude that shot groups are sufficiently symmetric that we can assume <math>\sigma_x = \sigma_y</math>. In the case of the symmetric bivariate normal distribution, the distance of each shot from the center of impact (COI) follows the Rayleigh distribution with parameter ''σ''.<ref>[[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|''Shot group statistics'', Jeroen Hogema, 2005]]</ref><br />
<br />
NB: It is common to describe normal distributions using variance, or <math>\sigma^2</math>, because variances have some convenient linear characteristics that are lost when we take the square root. For similar reasons many prefer to describe the Rayleigh distribution using a parameter <math>\gamma = \sigma^2</math>. To clarify our parameterization '''the ''σ'' we will be describing is the standard deviation of the bivariate normal distribution, and the parameter that produces the following pdf for the Rayleigh distribution''':<br />
:&nbsp; <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math><br />
<br />
Where the bivariate normal distribution describes the coordinates (''x'', ''y'') of shots on target, the Rayleigh distribution describes the distance, or radius, <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math> of those shots from the center point of impact.<br />
<br />
= Estimating ''σ'' =<br />
The Rayleigh distribution provides closed form expressions for precision. However, when estimating ''σ'' from sample sets we will most often use methods associated with the normal distribution for one essential reason: ''We never observe the true center of the distribution''. When we calculate the center of a group on a target it will almost certainly be some distance from the true center, and thus underestimate the true distance of the sample shots to the distribution center. (Average distance from sample center to true center is listed in the second column of [[Media:Sigma1ShotStatistics.ods]].) The Rayleigh model describes the distribution of shots from the (unobservable) true center. When the center is unknown we have to use the sample center, and we fall back on characteristics of the normal distribution with unknown mean.<br />
<br />
== Correction Factors ==<br />
The following three correction factors will be used throughout this statistical inference and deduction. <br />
<br />
Note that all of these correction factors are > 1, are significant for very small ''n'', and converge towards 1 as <math>n \to \infty</math>. Their values are listed for ''n'' up to 100 in [[Media:Sigma1ShotStatistics.ods]]. [[File:SymmetricBivariate.c]] uses Monte Carlo simulation to confirm that their application produces valid corrected estimates.<br />
<br />
=== [http://en.wikipedia.org/wiki/Bessel%27s_correction Bessel correction factor] ===<br />
The Bessel correction removes bias in sample variance.<br />
:&nbsp; <math>c_{B}(n) = \frac{n}{n-1}</math><br />
<br />
=== [http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation#Results_for_the_normal_distribution Gaussian correction factor] ===<br />
The Gaussian correction (sometimes called <math>c_4</math>) removes bias introduced by taking the square root of variance.<br />
:&nbsp; <math>\frac{1}{c_{G}(n)} = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})</math><br />
<br />
The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:&nbsp; <math>c_{G}(n)</math> <code>=1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))</code><br />
<br />
=== Rayleigh correction factor ===<br />
The unbiased estimator for the Rayleigh distribution is also for <math>\sigma^2</math>. The following corrects for the concavity introduced by taking the square root to get ''σ''.<br />
:&nbsp; <math>c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}</math> <ref>[[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|''Statistical Inference for Rayleigh Distributions'', M. M. Siddiqui, 1964, p.1007]]</ref><br />
<br />
To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula: <code>=EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))</code><br />
<br />
== Data ==<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
(One easy way to compile these data is to process an image of the target through a program like [http://ontargetshooting.com/features.html OnTarget Precision Calculator].)<br />
<br />
== Variance Estimates ==<br />
For a single axis the [http://en.wikipedia.org/wiki/Bessel's_correction#Formula unbiased estimate of variance] for a normal distribution is <math>s_x^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} </math>, from which the unbiased estimate of standard deviation is <math>\widehat{\sigma_x} = c_G(n) \sqrt{(s_x^2)}</math>.<br />
<br />
Since we are assuming that the shot dispersion is jointly independent and identically distributed along the ''x'' and ''y'' axes we improve our estimate by aggregating the data from both dimensions. I.e., we look at the average sample variance <math>s^2 = (s_x^2 + s_y^2)/2</math>, and <math>\hat{\sigma} = c_G(2n-1) \sqrt{s^2}</math>. This turns out to be identical to the Rayleigh estimator.<br />
<br />
== Rayleigh Estimates ==<br />
The Rayleigh distribution describes the random variable ''R'' defined as the distance of each shot from the center of the distribution. Again, we never get to observe the true center, so we begin by calculating the sample center <math>(\bar{x}, \bar{y})</math>. Then for each shot we can compute the sample radius <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math>.<br />
<br />
The [http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation unbiased Rayleigh estimator] is <math>\widehat{\sigma_R^2} = c_B(n) \frac{\sum r_i^2}{2n} = \frac{c_B(n)}{2} \overline{r^2}</math>, which is literally a restatement of the combined variance estimate <math>s^2</math>. Hence the unbiased parameter estimate is once again <math>\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}</math>.<br />
<br />
[[Rayleigh sigma estimate]] provides a derivation of this formula, as well as a variation for the case in which the true center is known.<br />
<br />
== Confidence Intervals ==<br />
Siddiqui<ref>[[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|''Some Problems Connected With Rayleigh Distributions'', M. M. Siddiqui, 1961, p.169]]</ref> shows that the confidence intervals are given by the <math>\chi^2</math> distribution with 2''n'' degrees of freedom. However this assumes we know the true center of the distribution. We lose two degrees of freedom (one in each dimension) by using the sample center, so we actually have only 2(''n'' - 1) degrees of freedom. (Here again we will get the same equations if we instead follow the derivation of confidence intervals for the combined variance <math>s^2</math>.)<br />
<br />
To find the (1 - ''α'') confidence interval, first find <math>\chi_1^2, \ \chi_2^2</math> where:<br />
:&nbsp; <math>Pr(\chi^2(2(n-1)) \leq \chi_1^2) = \alpha/2, \quad Pr(\chi^2(2(n-1)) \leq \chi_2^2) = 1 - \alpha/2</math><br />
For example, using spreadsheet functions we have <math>\chi_1^2</math> = <code>CHIINV(α/2, 2n-2)</code>,<math>\quad \chi_2^2</math> = <code>CHIINV((1-α/2), 2n-2)</code>.<br />
<br />
Now the confidence intervals are given by the following:<br />
:&nbsp; <math>s^2 \in \left[ \frac{2(n-1) s^2}{\chi_2^2}, \ \frac{2(n-1) s^2}{\chi_1^2} \right]</math>, or in equivalent Rayleigh terms <math>\widehat{\sigma_R^2} \in \left[ \frac{\sum r^2}{\chi_2^2}, \ \frac{\sum r^2}{\chi_1^2} \right]</math><br />
<br />
Using the more convenient Rayleigh expression the confidence interval for the precision parameter is:<br />
:&nbsp; <math>\widehat{\sigma} \in \left[ c_G(2n-1) \sqrt{\frac{\sum r^2}{\chi_2^2}}, \ c_G(2n-1) \sqrt{\frac{\sum r^2}{\chi_1^2}} \right]</math><br />
<br />
=== How large a sample do we need? ===<br />
[[File:ConfidenceIntervals.png|450px|thumb|right]]<br />
Note that confidence intervals are a function of both the sample size and the average radius in the sample. If we hold the mean sample radius constant we can see how the confidence interval tightens with sample size. The adjacent chart shows the 95% confidence intervals for σ when the estimate is 1.0 and the mean sample radius is held constant at <math>\overline{r}^2 = 2</math>. (NB: This is an extraordinarily skewed scenario, since typically each sample radius varies from the average.)<br />
<br />
With a sample of 10 shots our 95% confidence interval is 77% as large as the parameter σ itself. At 20 it's just under 50%. It takes a group of 66 shots to get it under 25% and 100 to get it to 20% of the estimated σ.<br />
<br />
<br clear=all><br />
<br />
=== The 3-shot Group ===<br />
[[File:3ShotSample.png|210px|thumb|right|Sample 3-shot group|Sample 3-shot group with 1/2" extreme spread. Sample center is in red. Each shot has ''r'' = .29".]]<br />
A rifle builder sends you a [[FAQ#How_meaningful_is_a_3-shot_precision_guarantee.3F|3-shot group]] measuring ½" between each of three centers to prove how accurate your rifle is. ''What does that really say about the gun's accuracy?''<br />
In the ''best'' case &mdash; i.e.:<br />
# The group was actually fired from your gun<br />
# The group was actually fired at the distance indicated (in this case 100 yards)<br />
# The group was not cherry-picked from a larger sample &mdash; e.g., the best of an unknown number of test 3-shot groups<br />
# The group was not clipped from a larger group (in the style of [http://www.ar15.com/forums/t_3_118/500913_.html the "Texas Sharpshooter"])<br />
&mdash; if all of these conditions are satisfied, then we have a statistically valid sample. In this case our group is an equilateral triangle with ½" sides. A little geometry shows the distance from each point to sample center is <math>r_i = \frac{1}{2 \sqrt{3}} \approx .29"</math>.<br />
<br />
The Rayleigh estimator <math>\widehat{\sigma_R^2} = c_B(3) \frac{\sum r_i^2}{6} = \frac{3}{2} \frac{1}{24} = \frac{1}{16}</math>. So <math>\hat{\sigma} = c_G(2n - 1) \sqrt{1/16} = (\frac{4}{3}\sqrt{\frac{2}{\pi}})\frac{1}{4} \approx .25MOA</math>. Not bad! But not very significant. Let's check the confidence intervals: For ''α'' = 5% (i.e., 95% confidence intervals)<br />
:&nbsp; <math>\chi_1^2(4) \approx 0.484, \quad \chi_2^2(4) \approx 11.14</math>. Therefore,<br />
:&nbsp; <math>0.02 \approx \frac{1}{4 \chi_2^2} \leq \widehat{\sigma_R^2} \leq \frac{1}{4 \chi_1^2} \approx 0.52</math>, and<br />
:&nbsp; <math>0.16 \leq \hat{\sigma} \leq 0.76</math><br />
so with 95% certainty we can only say that the gun's true precision ''σ'' is somewhere in the range from approximately 0.2MOA to 0.8MOA.<br />
<br />
= Using ''σ'' =<br />
[[File:RayleighProcess.png|250px|thumb|right|Rayleigh Probabilities|Rayleigh distribution of shots given ''σ'']]<br />
The ''σ'' we have carefully sampled and estimated is the parameter for the Rayleigh distribution with probability density function <math>\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}</math>. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:<br />
:&nbsp; <math>Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}</math><br />
Therefore, we expect 39% of shots to fall within a circle of radius ''σ'', 86% within ''2σ'', and 99% within ''3σ''.<br />
<br />
Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful [[Describing_Precision#Measures|precision measures]]:<br />
<br />
== Mean Radius (MR) ==<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
<math>1 - e^{-\frac{\pi}{4}} \approx 54\%</math> of shots should fall within the mean radius. 96% of shots should fall within the Mean Diameter (MD = 2 MR).<br />
<br />
''Given σ'', the expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
(This sample size adjustment doesn't use the Gaussian correction factor because the mean radius is not an estimator for ''σ'', even though in the limit the true value of one is a constant multiple of the other.)<br />
<br />
== Circular Error Probable (CEP) ==<br />
For the Rayleigh distribution, the 50%-Circular Error Probable is <math>CEP(0.5) = \sigma \sqrt{\ln(4)} \ \approx 1.18 \ \sigma</math>. 50% of shots should fall within a circle with this radius around the point-of-aim. See [[Circular Error Probable]] for a more detailed discussion.<br />
<br />
In theory CEP is the median radius, but especially for small ''n'' '''the sample median is a very bad estimator for the true median'''. Given ''σ'', the following is a good estimate of the expected sample median radius of a group of size ''n'':<br />
:&nbsp; <math>CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}</math><br />
<br />
== Summary Probabilities ==<br />
From the Rayleigh quantile function we can compute the radius expected to cover proportion ''F'' of shots as <math>CEP(F) = \sigma \sqrt{-2 \ln(1-F)}</math>. E.g.,<br />
{| class="wikitable"<br />
|-<br />
! Name !! Multiple ''x'' of ''σ'' !! Shots Covered by Circle of Radius ''x σ''<br />
|-<br />
| || 1 || 39%<br />
|-<br />
| CEP || 1.18 || 50%<br />
|-<br />
| MR || 1.25 || 54%<br />
|-<br />
| || 2 || 86%<br />
|-<br />
| MD || 2.5 || 96%<br />
|-<br />
| || 3 || 99%<br />
|}<br />
<br />
== Typical values of ''σ'' ==<br />
A lower bound on ''σ'' is probably that displayed by rail guns in 100-yard competition. On average they can place 10 rounds into a quarter-inch group, which [[Predicting_Precision#Spread_Measures|as we will see shortly]] suggests ''σ'' = 0.070MOA, or under 0.025mil.<br />
<br />
The U.S. Precision Sniper Rifle specification requires a statistically significant number of 10-round groups fall under 1MOA. This means ''σ'' = 0.28MOA, or under 0.1mil.<br />
<br />
The specification for the M110 semi-automatic sniper rifle (MIL-PRF-32316) as well as the M24 sniper rifle (MIL-R-71126) requires MR below 0.65SMOA, which means ''σ'' = 0.5MOA. The latter spec indicates that an M24 barrel is not considered worn out until MR exceeds 1.2MOA, or ''σ'' = 1MOA!<br />
<br />
XM193 ammunition specifications require 10-round groups to fall under 2MOA. This means ''σ'' = 0.6MOA or 0.2mil, and it is a good minimum precision standard for light rifles.<br />
<br />
== How many sighter shots do you need? ==<br />
[[File:3ShotSighterError.png|265px|thumb|right|99% shooting errors expected from 3-shot sighting groups|99% shooting errors expected from 3-shot sighting groups, which on average impact .7σ from the Point of Aim.]]<br />
How many shots do you need to zero your scope? As detailed in [[Sighter Distribution]] we know that the distance from the true center of a "sighting group" of ''n'' shots has a Rayleigh distribution with parameter <math>\sigma / \sqrt{n}</math>. Following is a table showing the mean distance of a sighting group from the true zero for groups of different sizes, in terms of ''σ''. To illustrate the implications for a typical precision gun we convert this to inches of error at 100 yards for ''σ'' = 0.5MOA.<br />
<center><br />
{| class="wikitable"<br />
|-<br />
! Sighter<br>Group Size !! Average Distance<br>from True Zero !! Error at 100 yards<br>for ''σ'' = 0.5MOA !! Shots Lost to<br>Sighting Error<br>on 50% Target !! Shots Lost to<br>Sighting Error<br>on 96% Target<br />
|-<br />
| 3 || 0.7 ''σ'' || 0.4" || 8% || 4%<br />
|-<br />
| 5 || 0.6 ''σ'' || 0.3" || 6% || 3%<br />
|-<br />
| 10 || 0.4 ''σ'' || 0.2" || 3% || 1%<br />
|-<br />
| 20 || 0.3 ''σ'' || 0.15" || 2% || <1%<br />
|}<br />
</center><br />
<br />
The [http://en.wikipedia.org/wiki/Rice_distribution Rice distribution] gives the expected hit probabilities when incorporating a sighting error ''ε σ''. The Rice CDF is hard to calculate so here we used [http://www.wolframalpha.com/ Wolfram Alpha] to compute CDF values as <math>F(x|ε, \sigma) = P(X \leq x)</math> <code>= MarcumQ[1, ε, 0, x]</code>, which gives the probability of a hit within distance ''x σ'' given a sighting error of ''ε σ''.<br />
<br />
We define a '''''t''% target''' as a target large enough that ''t''% of shots fired would be expected to hit it, if the gun were perfectly sighted in. (This value is given by the Rayleigh distribution.)<br />
<br />
"Shots Lost to Sighting Error" on a ''t''% target is the difference between the proportion of shots that would hit if perfectly sighted and the proportion expected to hit with a sighting error of ''ε'': i.e., <math>F(t|0, \sigma) - F(t|ε, \sigma)</math>.<br />
<br />
There are probably better ways to characterize the importance and impact of the sighting error depending on the application.<br />
<br />
= References =<br />
<references /><br />
<br />
<BR/><br />
<HR/><br />
<p style="text-align:right"><B>Next:</B> [[:Category:Examples|Examples]]</p></div>Davidhttp://ballistipedia.com/index.php?title=Category:Examples&diff=1406Category:Examples2017-02-23T03:53:20Z<p>David: </p>
<hr />
<div>* [[Closed_Form_Precision#How_many_sighter_shots_do_you_need.3F|How many sighter shots do you need]]?<br />
* [[Closed_Form_Precision#The_3-shot_Group|What the 3-shot group says about precision]].<br />
* [[Closed_Form_Precision#Typical_values_of_.CF.83|Typical real-world precision]].<br />
* [[Range_Statistics#Example_1|Expected Extreme Spread]].<br />
* [[Range_Statistics#Example:_Statistical_Inference|Statistical Inference from Extreme Spreads]].<br />
* [http://lstange.github.io/mcgs/ Order statistics for easily calculating 90% hit probability].<br />
* [[Ballistic Accuracy Classification]]<br />
<br />
= Sample Targets and Scenarios with Precision Analysis =</div>Davidhttp://ballistipedia.com/index.php?title=Talk:Closed_Form_Precision&diff=1405Talk:Closed Form Precision2017-02-14T15:40:09Z<p>David: /* How large a sample do we need? */</p>
<hr />
<div>For the most part the discussion of the Bessel, Gaussian and Rayleigh correction factors is off the deep end. All elementary text books would use (n-1) as the divisor when calculating the standard deviation, but few would name it. The Gaussian correction is small compared to the overall size of the confidence interval with small samples. For measures like group size it is also unnecessary. For small samples the distribution is significantly skewed so you should use tables based on Monte Carlo results which would have the correction factor "built in." By the time you get enough data for group size measurements to be really be normally distributed the Gaussian correction factor is then totally negligible. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
<br />
: '''Perhaps the correction factors could be buried deeper or on another page. But we can't just wave them away and we shouldn't pull mysterious correction factors out of thin air. This is, after all, the closed form analysis. Yes, you can get the same results via Monte Carlo and use that approach instead, but for those who want to see the pure math these terms are part of it. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
<br />
The fact that most of the measures are really skewed distributions deserves more serious discussion. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
<br />
: '''Agreed. Higher moments are noted in a few of the spreadsheets, like [[Media:Sigma1RangeStatistics.xls]], but a discourse would be a worthwhile addition. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
<br />
I must admit a bit of curiosity here. I don't know if the Student's T test has the Gaussian correction built into the tables or not. For small samples (2-5 degrees of freedom) the confidence interval is so large that it wouldn't make a lot of difference, and when doing various error combining to get an "effective" number for the degrees of freedom then having factor built in might mess things up. I looked and can't find the answer readily available. I'll try to remember to ask the question on the math stats forum. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
<br />
== σ ==<br />
<br />
The paragraph under heading "Estimating σ" misses the salient points entirely. First there are so many different uses of <math>\sigma</math> that it is confusing when hopping around on wiki. <math>\sigma_{RSD}</math> should be used consistently for the radial standard deviation of the population of the Rayleigh distribution. Second <math>s_{RSD}</math> can't be calculated directly from the experimental data like ''s'' can for the normal distribution. You have to calculate <math>s_r</math> from the data, then use a theoretical proportionality constant to get <math>s_{RSD}</math>. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
<br />
: '''As noted at the top of the page this use of σ is ''not'' confusing because it turns out to be the same σ used in the parameterization of the normal distribution used as the model, which is the same as the standard deviation. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
<br />
The discussion about the ''unbiased Rayleigh estimator'' is just wrong. I'll admit another limit on my knowledge here. You could calculate <math>s_{RSD}</math> from either the experimental <math>\bar{r}</math> or <math>s_r</math>. It seems some combination of the two would give the "best" result. I don't have a clue how to do that error combination. I do know that in general the measurement of <math>\bar{r}</math> would have a greater relative precision (as % error) than <math>s_r</math>. In fact the ratio of the two estimates could be used as a test for outliers. The standard deviation is more sensitive to an outlier than the mean. <br />
<br />
: '''I don't understand what's wrong: the math is right there in [[Closed Form Precision#Variance_Estimates]] and [[Closed Form Precision#Rayleigh_Estimates]]. Expand the equations and they are identical. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
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:: The point that I am trying to make here, and above in my criticism of the paragraph under heading "Estimating σ", is that there are two different uses of <math>\sigma</math> connected to the Rayleigh distribution, and the two are not equal.<br /><math>\sigma_r</math> is the standard deviation of the mean radius measurement<br /><math>\sigma_{RSD}</math> is the shape factor used in the PDF<br /> Also the wikipedia section http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation is a bit confusing, but it essentially refers to N dimensions, not N data points in the sample. In our case N=2.<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 22:08, 24 May 2015 (EDT)<br />
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::: !@#$%^ For the wikipedia link, N=1 in our case since the Rayleigh distribution converts to polar coordinates. r is Rayleigh distribued, <math>\theta</math> is not. If we used (h,v) coordinates, then N=2.< br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 01:59, 25 May 2015 (EDT) <br />
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:: read this: http://ballistipedia.com/index.php?title=User_talk:Herb#Radial_Standard_Deviation_of_the_Rayleigh_Distribution<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 12:39, 25 May 2015 (EDT)<br />
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: '''In order to avoid this confusion [[Describing_Precision#Radial_Standard_Deviation_.28RSD.29|I make a point here]] of suggesting we avoid talking about "radial standard deviation" (i.e., the standard deviation of radii). If we do that then the only σ we're talking about is the one used in the Rayleigh parameterization which (via transformation to polar coordinates) ''is'' the same as the σ parameter of the normal distribution associated with this model. If you see any other σ with another meaning please point it out so we can either remove it or clarify if it can't be removed. [[User:David|David]] ([[User talk:David|talk]]) 12:31, 25 May 2015 (EDT)'''<br />
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:: Making the simplification doesn't avoid the confusion it creates it. Wikipedia doesn't create unique σ's for every function and use by labeling them with subscripts since there are thousands of cases. We should since the σ in the Rayleigh distribution has other relationships to the distribution that σ in the normal distribution doesn't have. <br />
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:: As I noted above read this: http://ballistipedia.com/index.php?title=User_talk:Herb#Radial_Standard_Deviation_of_the_Rayleigh_Distribution<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 13:06, 25 May 2015 (EDT)<br />
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::: '''"Radial Standard Deviation" is the problem. If by RSD you mean "the standard deviation of radii" then it is ''not'' the Rayleigh scale parameter. But if you introduce the term then people ''will'' confuse it with the Rayleigh parameter ... when they're not confusing it with earlier definitions like <math>\sqrt{\sigma_h^2 + \sigma_v^2}</math>. I haven't found a reason to even refer to the standard deviation of radii: it is not used in any calculation or estimation, and it is not a helpful measure given the others in use.<br />
::: '''As for the unlabelled σ: When used as the Rayleigh parameter it is done in the context of the symmetric bivariate normal distribution, in which case every σ ''is'' identical. Elsewhere when we talk about non-symmetric distributions, or specific axes, we give it a subscript. Where else should it be labelled? [[User:David|David]] ([[User talk:David|talk]]) 14:03, 25 May 2015 (EDT)'''<br />
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Your calculation for σ in the Rayleigh equation is correct. When you peel back the math, it is simply the pooled average <math>(s_h + s_v)/2</math>. That is no doubt the best way to calculate the parameter. <br />
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[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 20:52, 25 May 2015 (EDT)<br />
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== Mean Diameter ==<br />
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In the section on ''Mean Radius'' the use of the phrase ''Mean Diameter'' seems out of whack. Using a "diameter" as a radius is just confusing for no good reason. Seems like you should just use 2<math>\bar{r}</math>. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
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: '''''Mean Diameter'' has come up as an attractive standard, since it corresponds to the 96% CEP which is closer to the measures people are used to than 50% CEP or the mean radius, and rolls off the tongue more easily than "two mean radii" or any other variation I've heard of. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
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:: Yes, it is cutesy phrase, but I find it confusing. My perception is that this wiki is aimed shooter who is a neophyte statistician. I'd vote for clarity over cutesy. <br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 22:08, 24 May 2015 (EDT)<br />
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== Sighter Shots ==<br />
A number of things are stuffed on this wiki page because they fit some erroneous notion of "Closed form". For example the discussion of sighter shots. That should be a a separate wiki page. The whole thing about the errors for sighter shots also depends on the assumptions for the dispersion for which there are 4 general cases. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
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: '''Yes, we should probably break that onto a separate page. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
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== Student's T & Gaussian Correction factor ==<br />
<br />
Asked question on Stack Exchange about if Student's T table has correction included. I'll update this if the question is answered. <br />
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http://stats.stackexchange.com/questions/153978/should-the-standard-deviation-be-corrected-in-a-students-t-test<br />
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: Good question, and good answer! [[User:David|David]] ([[User talk:David|talk]]) 09:37, 26 May 2015 (EDT)<br />
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:: All in all I think mentioning the correction factors are overkill. So I'd recommend just using (n-1) when calculating variance and stop at that. I've read dozens of statistics books where the Student's T test was described. I never saw anything about needing to correct the bias of the sample standard deviation due to sample size. The gist is that if I want to use the extreme spread measure for a 3-shot group to characterize my rifle, then the statistics are abysmally bad. Thus making the "Gaussian correction" won't magically fix that problem. Shooters typically use small sample statistics without realizing how bad that limits analysis. In other words... Buba says "but I measured the 3-shot group to a thousandth of an inch with a $200 pair of vernier calipers. It is 1.756 inches ever time I measure it." LOL <br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 13:11, 26 May 2015 (EDT)<br />
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== Mean Radius (MR) Section ==<br />
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Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
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'''**** that is correct. ****'''<br />
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The expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
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'''**** This is wrong. This is where "correcting" a sample standard deviation gets you into trouble. The mean radius doesn't depend on sample size. The precision with which the mean radius is measured does depend on sample size. It would be far better to discuss the dependency of <math>\sigma_{MR}</math> as a function of sample size. ****'''<br />
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: This is a key point that perhaps bears elaboration, and which is why the correction factors get so much attention here: The expected value of a sample ''does'' require a correction for the sample size. If you Monte Carlo this (see, e.g., [[File:SymmetricBivariateSigma1.xls]]) you will see that right away, and if we don't include the correction factors then the numbers produced by the formulas are just not right &ndash; especially for the relatively small ''n'' values shooters typically use. Add the correction factors and they match for all ''n''. I think [[Closed_Form_Precision#Estimating_.CF.83|the reason is easy to follow in the case of sample shot groups]]: Nobody gets to measure their groups from the "true" center. The center itself is estimated from the sample, and it is always closer to the sample shots than the true center.<br />
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: So yes, the true Mean Radius does not depend on sampling. But, given a Mean Radius, the expected value of a sample does depend on the number of shots sampled. [[User:David|David]] ([[User talk:David|talk]])<br />
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::&nbsp;&nbsp;&nbsp;No, no, no, no.... This is why I think labeling the various <math>\sigma</math>'s is important. First, the <math>\sigma</math> that you're using is presumably from the Rayleigh parameter fitting. For a sample size of N shots, there are (2N-2) degrees of freedom. <br />
::&nbsp;&nbsp;&nbsp;You're correct on Monte Carlo point. You can easily simulate millions of shots which is really impossible at a range. So discuss and make such corrections only in that context. In "everyday" use the corrections just aren't large enough to be significant. <br />
::&nbsp;&nbsp;&nbsp;In the everyday use of the mean radius measurements you'd actually calculate all the <math>r_i</math> values and average them. The correct way to discuss the variability of that measurement then would be to use <math>\sigma_{\bar{MR}}</math>, the standard deviation of the mean which does depend on sample size. But the value of the mean won't change, just the precision with which it is measured. In other words the "corrected" value for the mean would be well within any "reasonable" confidence interval for the mean. <br />
::&nbsp;&nbsp;&nbsp;The Gaussian correction isn't magic that makes small sample statistics work. Think of it as lipstick on a bulldog. Doesn't make it pretty.<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 15:47, 26 May 2015 (EDT)<br />
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::: '''Maybe it's not significant for a lot of real-world situations. Although I think it's important that people know that, for example, on average their 3-shot groups will show a Mean Radius that's only 80% of their gun's true MR. We can tuck these details out of the way where they're distracting, but anyone who wants to check the math has to see them. [[User:David|David]] ([[User talk:David|talk]]) 17:39, 26 May 2015 (EDT)'''<br />
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::::&nbsp;&nbsp;&nbsp;You have to realize that a particular mean radius for a 3-shot sample may be low or high. Ideally you like the measurement to be normally distributed so that there was a 50% chance of high and a 50% change of low - ''on average''. But if the ratio gets skewed a bit so that 40% are high and 60% are low, then the skewness is virtually undetectable in real world experiments. There are some subtleties here that perhaps you're missing. <br />
::::&nbsp;&nbsp;&nbsp;(1) Did you notice that we may be using two different degrees of freedom here? If a sample is used to calculate the <math>\sigma</math> of the Raleigh distribution that would be one use. If we later try to predict the mean radius for a different 3-shot group sample, then that would be a 2nd sample. We have to consider degrees of freedom from both. <br />
::::&nbsp;&nbsp;&nbsp;(2) The mean radius of a 3-shot group isn't normally distributed. It has a Chi-Squared distribution which is skewed. (Chi-Sqaured uses variance, not standard deviation). You'd need about 10 shots in a group to get a decently normally distributed mean radius. So at about 10-shots per group there would be a "smooth" transition going from the Chi-Squared distribution to the normal distribution approximation. At that point the correction is exceedingly small, and hence rather pointless.<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 19:26, 26 May 2015 (EDT)<br />
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::::: '''There are a lot of correct statements in here. The complexities of statistics should be spread out for those who want to be as precise and pedantic as possible. And for those who just want a simple tool they can use, or a straightforward answer to a straightforward question, we should endeavor to encapsulate as many of the details as possible.<br />
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::::: '''With regard to the second statement at the top of this thread: This is the simplest answer to one of the first questions I had when I began looking at this problem, which was, "If I somehow know the precision of my gun then what sorts of groups should I expect to see?" The answer is important for small ''n'' because we don't always have the luxury of shooting groups with large ''n''. One can certainly delve deeper, but that is the formula for the first sample moment. If someone wants to work out the formulas for higher sample moments as a function of ''n'' that would be great.<br />
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::::: '''Looking at it now I am thinking that perhaps it would avoid confusion if I prefaced it by saying, "''Given σ'', the expected sample MR of a group of size ''n'' is..."?<br />
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::::: '''In any case, everything that I've written is in response to a question that I had or was asked. If something is incorrect then we should correct it. If it's not clear what question it answers or what purpose it serves then whoever wrote it should be ready to illuminate that. Hopefully I have at least confirmed that the original statement is in fact correct, and also provided reasonable justification for why it is worthy of note? [[User:David|David]] ([[User talk:David|talk]]) 21:12, 26 May 2015 (EDT)'''<br />
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1 Symmetric Bivariate Normal = Rayleigh Distribution<br />
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... being more repetitive: ''"Symmetric Bivariate Normal = Rayleigh Distribution''" This title is misleading as the (univariate) Rayleigh distribution is not the same as a circular bivariate normal distribution. If the shots (in the sense of (h,v) coordinates) follow a circular bivariate normal distribution, then their distances to the true COI follows a Rayleigh distribution.<br />
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''"In this case the dispersion of shots is modeled by a symmetric bivariate normal, which is equivalent[1] to the Rayleigh distribution"'' - see above.<br />
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[[User:Armadillo|Armadillo]]<br />
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: Just put in a fix for this. [[User:David|David]] ([[User talk:David|talk]]) 09:57, 24 June 2015 (EDT)<br />
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== How large a sample do we need? ==<br />
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[[Closed_Form_Precision#How_large_a_sample_do_we_need.3F|This section]] could probably use improvement. Here's a rewrite from scratch that attempts to explain it more clearly:<br />
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One of the big questions people have is, “How many shots do I have to take to know how accurate something is?”<br />
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First, we need to realize, “You can never '''know for sure'''.” All you can do is increase your confidence that what you have observed (your sample) reflects what you would continue to observe if you kept shooting (drawing more samples). In order to do this with parametric statistics:<br />
# We establish a model (in this case the symmetric bivariate normal, which is equivalent to the Rayleigh model)<br />
# We figure out how to estimate the model parameter based on a sample that (we assume) is drawn from the model distribution<br />
# We figure out how “confident” we are in a given estimate. (Confidence has a formal definition: Pick an interval about your estimate. Now, if say that we are x% confident in that interval, then we expect that if we repeat the experiment of shooting the same number of shots and estimating the model parameter, then x% of those experiments will result in a parameter estimate within the interval.)<br />
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So we’ve gone through this tedious exercise, and we can correctly calculate all these things, but people still want to know essentially, “How confident should I be that a group of n shots is indicative of my accuracy?”<br />
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[[File:RadiusBounds.png|300px|thumb|right]]<br />
The answer is that it depends on two things: The number of shots, ''and'' the particular way those shots land. Perhaps I should construct examples of two groups that give the same parameter estimate but different confidence estimates – something analogous to the one I did to the right for Extreme Spread?<br />
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Well, outside of contrived examples in which the sample significantly alters the confidence, people are more concerned with ''how the number of shots affects confidence''. So in order to illustrate that, I hold the distribution of the shots constant. The constant I pick is <math>\overline{r}^2 = 2</math> because in that case the estimated parameter σ = 1, and the chart in this section is then easy to understand because the confidence interval is a range about the estimated value (y = 1). Now people can get a feel for how confidence increases with sample size. For example, looking at that graph, hopefully people will say, “Dang, fewer than ten shots leave a lot of doubt in the estimate!”<br />
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They can also see from the chart that the confidence interval is skewed to the outside of the estimate. I.e., especially with small groups, it’s as likely that the “true” value is ''much larger'' than the sample estimate as it is likely that it’s ''just a little smaller'' than what the sample indicated. Or, put another way (if you click through to the spreadsheet behind that chart): If you shoot a 3-shot group that suggests your accuracy is σ = 1, all you can say with 95% confidence is that subsequent 3-shot groups will show σ in the range (0.78, 3.74). I.e., you should expect to see a lot larger groups, but not much smaller groups.<br />
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[[User:David|David]] ([[User talk:David|talk]]) 11:38, 7 February 2017 (EST)<br />
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: David, I'd like to split hairs w/r/t bullet point 3: The technical meaning of a confidence interval to level 95% is the following: If you repeat the experiment many times and each time calculate the confidence interval, then 95% of these intervals will contain the true value. Note that all confidence intervals will be (slightly) different since they are calculated based on different samples. The same is true for the point estimates. This implies that the confidence is technically not in one particular interval, but in the method used to construct the interval. Daniel</div>Davidhttp://ballistipedia.com/index.php?title=Talk:Closed_Form_Precision&diff=1402Talk:Closed Form Precision2017-02-07T16:38:06Z<p>David: /* How large a sample do we need? */ new section</p>
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<div>For the most part the discussion of the Bessel, Gaussian and Rayleigh correction factors is off the deep end. All elementary text books would use (n-1) as the divisor when calculating the standard deviation, but few would name it. The Gaussian correction is small compared to the overall size of the confidence interval with small samples. For measures like group size it is also unnecessary. For small samples the distribution is significantly skewed so you should use tables based on Monte Carlo results which would have the correction factor "built in." By the time you get enough data for group size measurements to be really be normally distributed the Gaussian correction factor is then totally negligible. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
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: '''Perhaps the correction factors could be buried deeper or on another page. But we can't just wave them away and we shouldn't pull mysterious correction factors out of thin air. This is, after all, the closed form analysis. Yes, you can get the same results via Monte Carlo and use that approach instead, but for those who want to see the pure math these terms are part of it. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
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The fact that most of the measures are really skewed distributions deserves more serious discussion. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
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: '''Agreed. Higher moments are noted in a few of the spreadsheets, like [[Media:Sigma1RangeStatistics.xls]], but a discourse would be a worthwhile addition. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
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I must admit a bit of curiosity here. I don't know if the Student's T test has the Gaussian correction built into the tables or not. For small samples (2-5 degrees of freedom) the confidence interval is so large that it wouldn't make a lot of difference, and when doing various error combining to get an "effective" number for the degrees of freedom then having factor built in might mess things up. I looked and can't find the answer readily available. I'll try to remember to ask the question on the math stats forum. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
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== σ ==<br />
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The paragraph under heading "Estimating σ" misses the salient points entirely. First there are so many different uses of <math>\sigma</math> that it is confusing when hopping around on wiki. <math>\sigma_{RSD}</math> should be used consistently for the radial standard deviation of the population of the Rayleigh distribution. Second <math>s_{RSD}</math> can't be calculated directly from the experimental data like ''s'' can for the normal distribution. You have to calculate <math>s_r</math> from the data, then use a theoretical proportionality constant to get <math>s_{RSD}</math>. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
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: '''As noted at the top of the page this use of σ is ''not'' confusing because it turns out to be the same σ used in the parameterization of the normal distribution used as the model, which is the same as the standard deviation. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
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The discussion about the ''unbiased Rayleigh estimator'' is just wrong. I'll admit another limit on my knowledge here. You could calculate <math>s_{RSD}</math> from either the experimental <math>\bar{r}</math> or <math>s_r</math>. It seems some combination of the two would give the "best" result. I don't have a clue how to do that error combination. I do know that in general the measurement of <math>\bar{r}</math> would have a greater relative precision (as % error) than <math>s_r</math>. In fact the ratio of the two estimates could be used as a test for outliers. The standard deviation is more sensitive to an outlier than the mean. <br />
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: '''I don't understand what's wrong: the math is right there in [[Closed Form Precision#Variance_Estimates]] and [[Closed Form Precision#Rayleigh_Estimates]]. Expand the equations and they are identical. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
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:: The point that I am trying to make here, and above in my criticism of the paragraph under heading "Estimating σ", is that there are two different uses of <math>\sigma</math> connected to the Rayleigh distribution, and the two are not equal.<br /><math>\sigma_r</math> is the standard deviation of the mean radius measurement<br /><math>\sigma_{RSD}</math> is the shape factor used in the PDF<br /> Also the wikipedia section http://en.wikipedia.org/wiki/Rayleigh_distribution#Parameter_estimation is a bit confusing, but it essentially refers to N dimensions, not N data points in the sample. In our case N=2.<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 22:08, 24 May 2015 (EDT)<br />
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::: !@#$%^ For the wikipedia link, N=1 in our case since the Rayleigh distribution converts to polar coordinates. r is Rayleigh distribued, <math>\theta</math> is not. If we used (h,v) coordinates, then N=2.< br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 01:59, 25 May 2015 (EDT) <br />
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:: read this: http://ballistipedia.com/index.php?title=User_talk:Herb#Radial_Standard_Deviation_of_the_Rayleigh_Distribution<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 12:39, 25 May 2015 (EDT)<br />
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: '''In order to avoid this confusion [[Describing_Precision#Radial_Standard_Deviation_.28RSD.29|I make a point here]] of suggesting we avoid talking about "radial standard deviation" (i.e., the standard deviation of radii). If we do that then the only σ we're talking about is the one used in the Rayleigh parameterization which (via transformation to polar coordinates) ''is'' the same as the σ parameter of the normal distribution associated with this model. If you see any other σ with another meaning please point it out so we can either remove it or clarify if it can't be removed. [[User:David|David]] ([[User talk:David|talk]]) 12:31, 25 May 2015 (EDT)'''<br />
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:: Making the simplification doesn't avoid the confusion it creates it. Wikipedia doesn't create unique σ's for every function and use by labeling them with subscripts since there are thousands of cases. We should since the σ in the Rayleigh distribution has other relationships to the distribution that σ in the normal distribution doesn't have. <br />
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:: As I noted above read this: http://ballistipedia.com/index.php?title=User_talk:Herb#Radial_Standard_Deviation_of_the_Rayleigh_Distribution<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 13:06, 25 May 2015 (EDT)<br />
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::: '''"Radial Standard Deviation" is the problem. If by RSD you mean "the standard deviation of radii" then it is ''not'' the Rayleigh scale parameter. But if you introduce the term then people ''will'' confuse it with the Rayleigh parameter ... when they're not confusing it with earlier definitions like <math>\sqrt{\sigma_h^2 + \sigma_v^2}</math>. I haven't found a reason to even refer to the standard deviation of radii: it is not used in any calculation or estimation, and it is not a helpful measure given the others in use.<br />
::: '''As for the unlabelled σ: When used as the Rayleigh parameter it is done in the context of the symmetric bivariate normal distribution, in which case every σ ''is'' identical. Elsewhere when we talk about non-symmetric distributions, or specific axes, we give it a subscript. Where else should it be labelled? [[User:David|David]] ([[User talk:David|talk]]) 14:03, 25 May 2015 (EDT)'''<br />
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***** Eating Crow *****<br />
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Your calculation for σ in the Rayleigh equation is correct. When you peel back the math, it is simply the pooled average <math>(s_h + s_v)/2</math>. That is no doubt the best way to calculate the parameter. <br />
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[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 20:52, 25 May 2015 (EDT)<br />
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== Mean Diameter ==<br />
<br />
In the section on ''Mean Radius'' the use of the phrase ''Mean Diameter'' seems out of whack. Using a "diameter" as a radius is just confusing for no good reason. Seems like you should just use 2<math>\bar{r}</math>. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
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: '''''Mean Diameter'' has come up as an attractive standard, since it corresponds to the 96% CEP which is closer to the measures people are used to than 50% CEP or the mean radius, and rolls off the tongue more easily than "two mean radii" or any other variation I've heard of. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
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:: Yes, it is cutesy phrase, but I find it confusing. My perception is that this wiki is aimed shooter who is a neophyte statistician. I'd vote for clarity over cutesy. <br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 22:08, 24 May 2015 (EDT)<br />
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== Sighter Shots ==<br />
A number of things are stuffed on this wiki page because they fit some erroneous notion of "Closed form". For example the discussion of sighter shots. That should be a a separate wiki page. The whole thing about the errors for sighter shots also depends on the assumptions for the dispersion for which there are 4 general cases. [[User:Herb|Herb]] ([[User talk:Herb|talk]]) 16:36, 24 May 2015 (EDT)<br />
<br />
: '''Yes, we should probably break that onto a separate page. [[User:David|David]] ([[User talk:David|talk]]) 17:56, 24 May 2015 (EDT)'''<br />
<br />
== Student's T & Gaussian Correction factor ==<br />
<br />
Asked question on Stack Exchange about if Student's T table has correction included. I'll update this if the question is answered. <br />
<br />
http://stats.stackexchange.com/questions/153978/should-the-standard-deviation-be-corrected-in-a-students-t-test<br />
<br />
: Good question, and good answer! [[User:David|David]] ([[User talk:David|talk]]) 09:37, 26 May 2015 (EDT)<br />
<br />
:: All in all I think mentioning the correction factors are overkill. So I'd recommend just using (n-1) when calculating variance and stop at that. I've read dozens of statistics books where the Student's T test was described. I never saw anything about needing to correct the bias of the sample standard deviation due to sample size. The gist is that if I want to use the extreme spread measure for a 3-shot group to characterize my rifle, then the statistics are abysmally bad. Thus making the "Gaussian correction" won't magically fix that problem. Shooters typically use small sample statistics without realizing how bad that limits analysis. In other words... Buba says "but I measured the 3-shot group to a thousandth of an inch with a $200 pair of vernier calipers. It is 1.756 inches ever time I measure it." LOL <br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 13:11, 26 May 2015 (EDT)<br />
<br />
== Mean Radius (MR) Section ==<br />
<br />
Mean Radius <math>MR = \sigma \sqrt{\frac{\pi}{2}} \ \approx 1.25 \ \sigma</math>.<br />
<br />
'''**** that is correct. ****'''<br />
<br />
The expected sample MR of a group of size ''n'' is<br />
:&nbsp; <math>MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}</math><br />
<br />
'''**** This is wrong. This is where "correcting" a sample standard deviation gets you into trouble. The mean radius doesn't depend on sample size. The precision with which the mean radius is measured does depend on sample size. It would be far better to discuss the dependency of <math>\sigma_{MR}</math> as a function of sample size. ****'''<br />
<br />
: This is a key point that perhaps bears elaboration, and which is why the correction factors get so much attention here: The expected value of a sample ''does'' require a correction for the sample size. If you Monte Carlo this (see, e.g., [[File:SymmetricBivariateSigma1.xls]]) you will see that right away, and if we don't include the correction factors then the numbers produced by the formulas are just not right &ndash; especially for the relatively small ''n'' values shooters typically use. Add the correction factors and they match for all ''n''. I think [[Closed_Form_Precision#Estimating_.CF.83|the reason is easy to follow in the case of sample shot groups]]: Nobody gets to measure their groups from the "true" center. The center itself is estimated from the sample, and it is always closer to the sample shots than the true center.<br />
<br />
: So yes, the true Mean Radius does not depend on sampling. But, given a Mean Radius, the expected value of a sample does depend on the number of shots sampled. [[User:David|David]] ([[User talk:David|talk]])<br />
<br />
::&nbsp;&nbsp;&nbsp;No, no, no, no.... This is why I think labeling the various <math>\sigma</math>'s is important. First, the <math>\sigma</math> that you're using is presumably from the Rayleigh parameter fitting. For a sample size of N shots, there are (2N-2) degrees of freedom. <br />
::&nbsp;&nbsp;&nbsp;You're correct on Monte Carlo point. You can easily simulate millions of shots which is really impossible at a range. So discuss and make such corrections only in that context. In "everyday" use the corrections just aren't large enough to be significant. <br />
::&nbsp;&nbsp;&nbsp;In the everyday use of the mean radius measurements you'd actually calculate all the <math>r_i</math> values and average them. The correct way to discuss the variability of that measurement then would be to use <math>\sigma_{\bar{MR}}</math>, the standard deviation of the mean which does depend on sample size. But the value of the mean won't change, just the precision with which it is measured. In other words the "corrected" value for the mean would be well within any "reasonable" confidence interval for the mean. <br />
::&nbsp;&nbsp;&nbsp;The Gaussian correction isn't magic that makes small sample statistics work. Think of it as lipstick on a bulldog. Doesn't make it pretty.<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 15:47, 26 May 2015 (EDT)<br />
<br />
::: '''Maybe it's not significant for a lot of real-world situations. Although I think it's important that people know that, for example, on average their 3-shot groups will show a Mean Radius that's only 80% of their gun's true MR. We can tuck these details out of the way where they're distracting, but anyone who wants to check the math has to see them. [[User:David|David]] ([[User talk:David|talk]]) 17:39, 26 May 2015 (EDT)'''<br />
<br />
::::&nbsp;&nbsp;&nbsp;You have to realize that a particular mean radius for a 3-shot sample may be low or high. Ideally you like the measurement to be normally distributed so that there was a 50% chance of high and a 50% change of low - ''on average''. But if the ratio gets skewed a bit so that 40% are high and 60% are low, then the skewness is virtually undetectable in real world experiments. There are some subtleties here that perhaps you're missing. <br />
::::&nbsp;&nbsp;&nbsp;(1) Did you notice that we may be using two different degrees of freedom here? If a sample is used to calculate the <math>\sigma</math> of the Raleigh distribution that would be one use. If we later try to predict the mean radius for a different 3-shot group sample, then that would be a 2nd sample. We have to consider degrees of freedom from both. <br />
::::&nbsp;&nbsp;&nbsp;(2) The mean radius of a 3-shot group isn't normally distributed. It has a Chi-Squared distribution which is skewed. (Chi-Sqaured uses variance, not standard deviation). You'd need about 10 shots in a group to get a decently normally distributed mean radius. So at about 10-shots per group there would be a "smooth" transition going from the Chi-Squared distribution to the normal distribution approximation. At that point the correction is exceedingly small, and hence rather pointless.<br />[[User:Herb|Herb]] ([[User talk:Herb|talk]]) 19:26, 26 May 2015 (EDT)<br />
<br />
::::: '''There are a lot of correct statements in here. The complexities of statistics should be spread out for those who want to be as precise and pedantic as possible. And for those who just want a simple tool they can use, or a straightforward answer to a straightforward question, we should endeavor to encapsulate as many of the details as possible.<br />
<br />
::::: '''With regard to the second statement at the top of this thread: This is the simplest answer to one of the first questions I had when I began looking at this problem, which was, "If I somehow know the precision of my gun then what sorts of groups should I expect to see?" The answer is important for small ''n'' because we don't always have the luxury of shooting groups with large ''n''. One can certainly delve deeper, but that is the formula for the first sample moment. If someone wants to work out the formulas for higher sample moments as a function of ''n'' that would be great.<br />
<br />
::::: '''Looking at it now I am thinking that perhaps it would avoid confusion if I prefaced it by saying, "''Given σ'', the expected sample MR of a group of size ''n'' is..."?<br />
<br />
::::: '''In any case, everything that I've written is in response to a question that I had or was asked. If something is incorrect then we should correct it. If it's not clear what question it answers or what purpose it serves then whoever wrote it should be ready to illuminate that. Hopefully I have at least confirmed that the original statement is in fact correct, and also provided reasonable justification for why it is worthy of note? [[User:David|David]] ([[User talk:David|talk]]) 21:12, 26 May 2015 (EDT)'''<br />
<br />
----<br />
<br />
1 Symmetric Bivariate Normal = Rayleigh Distribution<br />
<br />
... being more repetitive: ''"Symmetric Bivariate Normal = Rayleigh Distribution''" This title is misleading as the (univariate) Rayleigh distribution is not the same as a circular bivariate normal distribution. If the shots (in the sense of (h,v) coordinates) follow a circular bivariate normal distribution, then their distances to the true COI follows a Rayleigh distribution.<br />
<br />
''"In this case the dispersion of shots is modeled by a symmetric bivariate normal, which is equivalent[1] to the Rayleigh distribution"'' - see above.<br />
<br />
[[User:Armadillo|Armadillo]]<br />
<br />
: Just put in a fix for this. [[User:David|David]] ([[User talk:David|talk]]) 09:57, 24 June 2015 (EDT)<br />
<br />
== How large a sample do we need? ==<br />
<br />
[[Closed_Form_Precision#How_large_a_sample_do_we_need.3F|This section]] could probably use improvement. Here's a rewrite from scratch that attempts to explain it more clearly:<br />
<br />
One of the big questions people have is, “How many shots do I have to take to know how accurate something is?”<br />
<br />
First, we need to realize, “You can never '''know for sure'''.” All you can do is increase your confidence that what you have observed (your sample) reflects what you would continue to observe if you kept shooting (drawing more samples). In order to do this with parametric statistics:<br />
# We establish a model (in this case the symmetric bivariate normal, which is equivalent to the Rayleigh model)<br />
# We figure out how to estimate the model parameter based on a sample that (we assume) is drawn from the model distribution<br />
# We figure out how “confident” we are in a given estimate. (Confidence has a formal definition: Pick an interval about your estimate. Now, if say that we are x% confident in that interval, then we expect that if we repeat the experiment of shooting the same number of shots and estimating the model parameter, then x% of those experiments will result in a parameter estimate within the interval.)<br />
<br />
So we’ve gone through this tedious exercise, and we can correctly calculate all these things, but people still want to know essentially, “How confident should I be that a group of n shots is indicative of my accuracy?”<br />
<br />
[[File:RadiusBounds.png|300px|thumb|right]]<br />
The answer is that it depends on two things: The number of shots, ''and'' the particular way those shots land. Perhaps I should construct examples of two groups that give the same parameter estimate but different confidence estimates – something analogous to the one I did to the right for Extreme Spread?<br />
<br />
Well, outside of contrived examples in which the sample significantly alters the confidence, people are more concerned with ''how the number of shots affects confidence''. So in order to illustrate that, I hold the distribution of the shots constant. The constant I pick is <math>\overline{r^2} = 2</math> because in that case the estimated parameter σ = 1, and the chart in this section is then easy to understand because the confidence interval is a range about the estimated value (y = 1). Now people can get a feel for how confidence increases with sample size. For example, looking at that graph, hopefully people will say, “Dang, fewer than ten shots leave a lot of doubt in the estimate!”<br />
<br />
They can also see from the chart that the confidence interval is skewed to the outside of the estimate. I.e., especially with small groups, it’s as likely that the “true” value is ''much larger'' than the sample estimate as it is likely that it’s ''just a little smaller'' than what the sample indicated. Or, put another way (if you click through to the spreadsheet behind that chart): If you shoot a 3-shot group that suggests your accuracy is σ = 1, all you can say with 95% confidence is that subsequent 3-shot groups will show σ in the range (0.78, 3.74). I.e., you should expect to see a lot larger groups, but not much smaller groups.<br />
<br />
[[User:David|David]] ([[User talk:David|talk]]) 11:38, 7 February 2017 (EST)</div>Davidhttp://ballistipedia.com/index.php?title=Prior_Art&diff=1401Prior Art2017-01-05T01:59:55Z<p>David: /* Hogema, 2006, Measuring Precision */ Partial archive; better than nothing...</p>
<hr />
<div>= [http://www.public.iastate.edu/~jessie/PPB/Stats/Testing%20loads.htm Danielson, 2005, ''Testing loads''] =<br />
[http://www.public.iastate.edu/~jessie/PPB/Stats/Testing%20loads.htm Brent J. Danielson] suggests that a practical way to assess and compare precision is to shoot many 2-shot groups and measure the extreme spread of each.<br />
* When one simply wants to assess whether one sample is more precise than another that probability is given by the one-tailed T-test for two samples with unequal variance -- i.e., the spreadsheet function <code>=1-TTEST({Sample1},{Sample2},1,3)</code>.<br />
* Using the same data it is possible to determine the precision parameter. It is also noteworthy that the square of the 2-tailed T-Test gives the exact confidence range for which the precision parameters of two samples do not overlap -- i.e., the probability that two samples have different precision parameters is <code>=POWER(1-TTEST({Sample1},{Sample2},2,3), 2)</code>. This is all illustrated using Danielson's own data set in [[Media:DanielsonExample.xlsx]].<br />
<br />
= Grubbs, 1964, ''Statistical Measures of Accuracy for Riflemen and Missile Engineers'' =<br />
[[Media:Statistical Measures for Riflemen and Missile Engineers - Grubbs 1964.pdf|''Statistical Measures of Accuracy for Riflemen and Missile Engineers'', Frank E. Grubbs, 1964]].<br />
<br />
= [http://web.archive.org/web/20131105232146/http://home.kpn.nl/jhhogema1966/skeetn/ballist/sgs/sgs.htm Hogema, 2005, ''Shot group statistics''] =<br />
[http://web.archive.org/web/20131105232146/http://home.kpn.nl/jhhogema1966/skeetn/ballist/sgs/sgs.htm Jeroen Hogema] provides an accessible proof of the equivalence between the symmetric bivariate normal and Rayleigh distributions. He provides extensive examples, simulations, and applications to scoring and load selection, and begins to address the problem of estimating the Rayleigh parameter.<br />
<br />
= [http://web.archive.org/web/20131105221113/http://home.kpn.nl/jhhogema1966/skeetn/ballist/precision/Measuring_precision.htm Hogema, 2006, ''Measuring Precision''] =<br />
[http://web.archive.org/web/20131105221113/http://home.kpn.nl/jhhogema1966/skeetn/ballist/precision/Measuring_precision.htm ''Picking the most precise ammo, probably.''] Jeroen Hogema:<br />
* Reproduces Leslie’s 1993 results.<br />
* Confirms that for radius measures it is preferable to incorporate all data at once, not to break them into separate groups.<br />
* For FOM and ES it is best to generate many groups so as to preserve more data points.<br />
* Looks at T-tests for significance and shows very large groups are needed to detect statistically meaningful differences.<br />
<br />
= Leslie, 1993, ''Is "Group Size" the Best Measure of Accuracy?'' =<br />
[[Media:Is_Group_Size_the_Best_Measure_of_Accuracy_by_J.E._Leslie_III.pdf|''Is "Group Size" the Best Measure of Accuracy?'', John "Jack" E. Leslie III, 1993]]. Notes:<br />
* Extreme Spread: Maximum distance between any two shots in group. Note that this effectively only uses two data points.<br />
* Figure of Merit (FoM): Average of the maximum horizontal group spread and the maximum vertical group spread. This uses only 2-4 data points depending on the group. Like Diagonal, FoM becomes more efficient than Extreme Spread for larger group sizes.<br />
* Mean Radius: Average distance to center of group for all shots.<br />
* Radial Standard Deviation: Sqrt (Horizontal Variance + Vertical Variance).<br />
* Found military using RSD and Mean Radius as early a 1918.<br />
<br />
His Monte Carlo analysis shows sample RSD to be most efficient predictor of precision, followed closely by Mean Radius. I.e., they can distinguish between loads of different inherent precision more accurately and using fewer sample shots than the other measures.<br />
<br />
Note that, like Grubbs, Leslie estimates MR by sampling the mean of radii. This is less efficient than using the Rayleigh estimator on the radii, and then [[Closed_Form_Precision#Mean_Radius_.28MR.29|computing MR based on the sample Rayleigh parameter]]. The latter process is equally and maximally efficient for all invariant measures that are products of the Rayleigh parameter σ.<br />
<br />
= [http://www.geoffrey-kolbe.com/articles/rimfire_accuracy/group_statistics.htm Kolbe, 2010, ''Group Statistics''] =<br />
Attributing the work to [[Media:Sitton 1990.pdf|Sitton et. al., 1990]], [http://www.geoffrey-kolbe.com/articles/rimfire_accuracy/group_statistics.htm Geoffrey Kolbe] walks through applied statistics to show how many groups of how many shots are required to estimate Extreme Spread to within 10% of the true value with 90% confidence. Using the values from [[Prior_Art#Grubbs.2C_1964.2C_Statistical_Measures_of_Accuracy_for_Riflemen_and_Missile_Engineers|Grubb's]] thousand-iteration simulations he determines that 7-shot groups are the most efficient for estimating Extreme Spread.<br />
<br />
Running [[Range_Statistics#Estimation|the same analysis]] with our [[Media:Sigma1RangeStatistics.xls|million-iteration simulation values]] reveals that 6-shot groups are actually optimal, and not significantly more so than 5-shot groups.<br />
<br />
= [http://www.ar15.com/mobile/topic.html?b=3&f=118&t=279218 Molon, 2006, ''The Trouble With 3-Shot Groups''] =<br />
[http://www.ar15.com/mobile/topic.html?b=3&f=118&t=279218 Through an extended forum thread] Molon offers intuitive explanations and illustrations of the problems with Extreme Spread samples.<br />
<br />
''We have not been able to identify the real person behind that now-inactive user account. If anyone knows please contact us so we can give well-deserved credit!''<br />
<br />
= Taylor & Grubbs, 1975, ''Approximate Probability Distributions for the Extreme Spread'' =<br />
[[Media:Approximate Distributions for Extreme Spread - Taylor 1975.pdf|''Approximate Probability Distributions for the Extreme Spread'', Taylor & Grubbs, 1975]]: notes that there is no closed-form expression for [[Describing_Precision#Extreme_Spread|Extreme Spread]]. Uses Monte Carlo simulations of 10,000 iterations to estimate the first four moments of the distribution of extreme spread for shot groups of size 3 to 34. Checks fit against the Chi, LogNormal, and Weibull distributions.<br />
<br />
Note that in [[File:Sigma1RangeStatistics.xls]] we have generated quantiles and the first four moments for the Extreme Spread using 1,000,000 iteration simulations for groups of size 2 to 100.</div>Davidhttp://ballistipedia.com/index.php?title=Prior_Art&diff=1400Prior Art2017-01-05T01:58:40Z<p>David: /* Hogema, 2005, Shot group statistics */ Found archived version</p>
<hr />
<div>= [http://www.public.iastate.edu/~jessie/PPB/Stats/Testing%20loads.htm Danielson, 2005, ''Testing loads''] =<br />
[http://www.public.iastate.edu/~jessie/PPB/Stats/Testing%20loads.htm Brent J. Danielson] suggests that a practical way to assess and compare precision is to shoot many 2-shot groups and measure the extreme spread of each.<br />
* When one simply wants to assess whether one sample is more precise than another that probability is given by the one-tailed T-test for two samples with unequal variance -- i.e., the spreadsheet function <code>=1-TTEST({Sample1},{Sample2},1,3)</code>.<br />
* Using the same data it is possible to determine the precision parameter. It is also noteworthy that the square of the 2-tailed T-Test gives the exact confidence range for which the precision parameters of two samples do not overlap -- i.e., the probability that two samples have different precision parameters is <code>=POWER(1-TTEST({Sample1},{Sample2},2,3), 2)</code>. This is all illustrated using Danielson's own data set in [[Media:DanielsonExample.xlsx]].<br />
<br />
= Grubbs, 1964, ''Statistical Measures of Accuracy for Riflemen and Missile Engineers'' =<br />
[[Media:Statistical Measures for Riflemen and Missile Engineers - Grubbs 1964.pdf|''Statistical Measures of Accuracy for Riflemen and Missile Engineers'', Frank E. Grubbs, 1964]].<br />
<br />
= [http://web.archive.org/web/20131105232146/http://home.kpn.nl/jhhogema1966/skeetn/ballist/sgs/sgs.htm Hogema, 2005, ''Shot group statistics''] =<br />
[http://web.archive.org/web/20131105232146/http://home.kpn.nl/jhhogema1966/skeetn/ballist/sgs/sgs.htm Jeroen Hogema] provides an accessible proof of the equivalence between the symmetric bivariate normal and Rayleigh distributions. He provides extensive examples, simulations, and applications to scoring and load selection, and begins to address the problem of estimating the Rayleigh parameter.<br />
<br />
= [http://home.kpn.nl/jhhogema1966/skeetn/ballist/precision/Measuring_precision.htm Hogema, 2006, ''Measuring Precision''] =<br />
[http://home.kpn.nl/jhhogema1966/skeetn/ballist/precision/Measuring_precision.htm ''Picking the most precise ammo, probably.''] Jeroen Hogema:<br />
* Reproduces Leslie’s 1993 results.<br />
* Confirms that for radius measures it is preferable to incorporate all data at once, not to break them into separate groups.<br />
* For FOM and ES it is best to generate many groups so as to preserve more data points.<br />
* Looks at T-tests for significance and shows very large groups are needed to detect statistically meaningful differences.<br />
<br />
= Leslie, 1993, ''Is "Group Size" the Best Measure of Accuracy?'' =<br />
[[Media:Is_Group_Size_the_Best_Measure_of_Accuracy_by_J.E._Leslie_III.pdf|''Is "Group Size" the Best Measure of Accuracy?'', John "Jack" E. Leslie III, 1993]]. Notes:<br />
* Extreme Spread: Maximum distance between any two shots in group. Note that this effectively only uses two data points.<br />
* Figure of Merit (FoM): Average of the maximum horizontal group spread and the maximum vertical group spread. This uses only 2-4 data points depending on the group. Like Diagonal, FoM becomes more efficient than Extreme Spread for larger group sizes.<br />
* Mean Radius: Average distance to center of group for all shots.<br />
* Radial Standard Deviation: Sqrt (Horizontal Variance + Vertical Variance).<br />
* Found military using RSD and Mean Radius as early a 1918.<br />
<br />
His Monte Carlo analysis shows sample RSD to be most efficient predictor of precision, followed closely by Mean Radius. I.e., they can distinguish between loads of different inherent precision more accurately and using fewer sample shots than the other measures.<br />
<br />
Note that, like Grubbs, Leslie estimates MR by sampling the mean of radii. This is less efficient than using the Rayleigh estimator on the radii, and then [[Closed_Form_Precision#Mean_Radius_.28MR.29|computing MR based on the sample Rayleigh parameter]]. The latter process is equally and maximally efficient for all invariant measures that are products of the Rayleigh parameter σ.<br />
<br />
= [http://www.geoffrey-kolbe.com/articles/rimfire_accuracy/group_statistics.htm Kolbe, 2010, ''Group Statistics''] =<br />
Attributing the work to [[Media:Sitton 1990.pdf|Sitton et. al., 1990]], [http://www.geoffrey-kolbe.com/articles/rimfire_accuracy/group_statistics.htm Geoffrey Kolbe] walks through applied statistics to show how many groups of how many shots are required to estimate Extreme Spread to within 10% of the true value with 90% confidence. Using the values from [[Prior_Art#Grubbs.2C_1964.2C_Statistical_Measures_of_Accuracy_for_Riflemen_and_Missile_Engineers|Grubb's]] thousand-iteration simulations he determines that 7-shot groups are the most efficient for estimating Extreme Spread.<br />
<br />
Running [[Range_Statistics#Estimation|the same analysis]] with our [[Media:Sigma1RangeStatistics.xls|million-iteration simulation values]] reveals that 6-shot groups are actually optimal, and not significantly more so than 5-shot groups.<br />
<br />
= [http://www.ar15.com/mobile/topic.html?b=3&f=118&t=279218 Molon, 2006, ''The Trouble With 3-Shot Groups''] =<br />
[http://www.ar15.com/mobile/topic.html?b=3&f=118&t=279218 Through an extended forum thread] Molon offers intuitive explanations and illustrations of the problems with Extreme Spread samples.<br />
<br />
''We have not been able to identify the real person behind that now-inactive user account. If anyone knows please contact us so we can give well-deserved credit!''<br />
<br />
= Taylor & Grubbs, 1975, ''Approximate Probability Distributions for the Extreme Spread'' =<br />
[[Media:Approximate Distributions for Extreme Spread - Taylor 1975.pdf|''Approximate Probability Distributions for the Extreme Spread'', Taylor & Grubbs, 1975]]: notes that there is no closed-form expression for [[Describing_Precision#Extreme_Spread|Extreme Spread]]. Uses Monte Carlo simulations of 10,000 iterations to estimate the first four moments of the distribution of extreme spread for shot groups of size 3 to 34. Checks fit against the Chi, LogNormal, and Weibull distributions.<br />
<br />
Note that in [[File:Sigma1RangeStatistics.xls]] we have generated quantiles and the first four moments for the Extreme Spread using 1,000,000 iteration simulations for groups of size 2 to 100.</div>Davidhttp://ballistipedia.com/index.php?title=References&diff=1399References2017-01-05T01:06:25Z<p>David: Added clarification on Siddiqui's parameterization.</p>
<hr />
<div>'''[[Prior Art]]''' details previous work on the problem of estimating shooting statistics.<br />
<br />
'''[[CEP literature]]''' focuses on the broader body of work related to characterizing Circular Error Probable, which is applicable not only to ballistics but also to fields like navigation and signal processing.<br />
<br />
Following is a complete list of References and Prior Art:<br />
<br />
* Bookstaber, David (2014). [http://www.thetruthaboutguns.com/2014/12/daniel-zimmerman/understanding-rifle-precision/ '''Understanding Rifle Precision'''].<br />
<br />
* Danielson, Brent J. (2005). [[Prior_Art#Danielson.2C_2005.2C_Testing_loads|'''Testing Loads''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Grubbs, Frank E. (1964). [[Prior_Art#Grubbs.2C_1964.2C_Statistical_Measures_of_Accuracy_for_Riflemen_and_Missile_Engineers|'''Statistical Measures of Accuracy for Riflemen and Missile Engineers''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Hogema, Jeroen (2005). [[Prior_Art#Hogema.2C_2005.2C_Shot_group_statistics|'''Shot group statistics''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Hogema, Jeroen (2006). [[Prior_Art#Hogema.2C_2006.2C_Measuring_Precision|'''Measuring Precision''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Kolbe, Geoffrey (2010). [[Prior_Art#Kolbe.2C_2010.2C_Group_Statistics|'''Group Statistics''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Leslia, John E. III (1993). [[Prior_Art#Leslie.2C_1993.2C_Is_.22Group_Size.22_the_Best_Measure_of_Accuracy.3F|'''Is "Group Size" the Best Measure of Accuracy?''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Molon (2006). [[Prior_Art#Molon.2C_2006.2C_The_Trouble_With_3-Shot_Groups|'''The Trouble With 3-Shot Groups''' &ndash; ''detailed in Prior Art'']].<br />
<br />
* Rifleslinger (2014). [http://artoftherifleblog.com/on-zeroing/2014/02/on-zeroing.html '''On Zeroing'''].<br />
<br />
* Siddiqui, M. M. (1961). [[Media:Some Problems Connected With Rayleigh Distributions - Siddiqui 1961.pdf|'''Some Problems Connected With Rayleigh Distributions''']]. The Journal of Research of the National Bureau of Standards, Sec. D: Radio Science, Vol. 68D, No. 9.<br />
<br />
* Siddiqui, M. M. (1964). [[Media:Statistical Inference for Rayleigh Distributions - Siddiqui, 1964.pdf|'''Statistical Inference for Rayleigh Distributions''']]. The Journal of Research of the National Bureau of Standards, Sec. D: Radio Propagation, Vol. 66D, No. 2. (''Summarizes and extends Siddiqui, 1961.'')<br />
<br />
'''''Important Note on Siddiqui''': Siddiqui parameterizes the Rayleigh distribution with <math>\frac{\sigma}{\sqrt{2}}</math>. Therefore, should you endeavor to relate Siddiqui's work to that referenced here and in more modern usage, remember that <math>\sigma_{modern} = \sqrt{2} \sigma_{Siddiqui}</math>.''<br />
<br />
* Taylor, M. S. & Grubbs, Frank E. (1975). [[Prior_Art#Taylor_.26_Grubbs.2C_1975.2C_Approximate_Probability_Distributions_for_the_Extreme_Spread|'''Approximate Probability Distributions for the Extreme Spread''' &ndash; ''detailed in Prior Art'']].<br />
<br />
= Reference Data =<br />
<br />
* [[File:Confidence Interval Convergence.xlsx]]: Shows how precision confidence intervals shrink as sample size increases.<br />
<br />
* [[File:Sigma1RangeStatistics.xls]]: Simulated median, 50%, 80%, and 95% quantiles, plus first four sample moments, for shot groups containing 2 to 100 shots, of: Extreme Spread, Diagonal, Figure of Merit.<br />
<br />
* [[File:SymmetricBivariateSigma1.xls]]: Monte Carlo simulation results validating the [[Closed Form Precision]] math.</div>Davidhttp://ballistipedia.com/index.php?title=Category:Examples&diff=1386Category:Examples2016-06-05T21:12:50Z<p>David: </p>
<hr />
<div>* [[Closed_Form_Precision#How_many_sighter_shots_do_you_need.3F|How many sighter shots do you need]]?<br />
* [[Closed_Form_Precision#The_3-shot_Group|What the 3-shot group says about precision]].<br />
* [[Closed_Form_Precision#Typical_values_of_.CF.83|Typical real-world precision]].<br />
* [[Range_Statistics#Example_1|Expected Extreme Spread]].<br />
* [[Range_Statistics#Example:_Statistical_Inference|Statistical Inference from Extreme Spreads]].<br />
* [http://lstange.github.io/mcgs/ Order statistics for easily calculating 90% hit probability].<br />
<br />
= Sample Targets and Scenarios with Precision Analysis =</div>Davidhttp://ballistipedia.com/index.php?title=Order_Statistics&diff=1385Order Statistics2016-06-05T21:03:24Z<p>David: /* tl;dr */ Added link to http://lstange.github.io/mcgs/</p>
<hr />
<div>==tl;dr==<br />
<br />
A typical hunter is interested in the 90% CEP, which we'll call here <math>R_{90}</math>: the radius around point of aim expected to contain 90% of shots.<br />
<br />
Using order statistics, we will derive "field estimates" that are easy to measure and compute. ([http://lstange.github.io/mcgs/ Monte Carlo analysis of robustness and comparison with other statistics available here.])<br />
<br />
Let <math>R_{m:n}</math> be the ''m<sup>th</sup>'' smallest radius (i.e., ''(n-m+1)<sup>th</sup>'' "worst") of ''n'' sample shots on a target.<br />
<br />
* If you can only take one shot then <math>\hat{R_{90}} = 3 R_{1:1}</math><br />
* If you can only take three shots then <math>\hat{R_{90}} = \sqrt{2} R_{3:3}</math><br />
* If you take five shots and throw away the worst, <math>\hat{R_{90}} \approx 1.6 R_{4:5}</math><br />
* With ten shots, <math>\hat{R_{90}} \approx 0.7 (R_{6:10} + R_{9:10})</math><br />
<br />
==Motivational Example==<br />
<br />
You borrow a rifle for the hunt. The owner says it is zeroed and you trust him. But you don’t know how well you can shoot this rifle, or how your ammo performs in it. You want to determine how big a target you can hit, or equivalently how far you can hit a target of a given size. For simplicity assume <s>spherical horse in vacuum</s> that the target is round and the distance is small enough that you don’t have to worry about holdovers and can ignore wind.<br />
<br />
By the way, this is what professional hunters do when they bring their client to the range the day before the hunt to “zero the rifle”. They know the rifle is zeroed. They want to know how close they need to get this particular client to the game animal for the hunt to be successful.<br />
<br />
==Metric Choice==<br />
<br />
One metric particularly useful in practice is <math>R_{90}</math>: the radius of a circle that is expected to contain 90% of all impacts.<br />
<br />
This choice is a compromise. <math>R_{100}</math> is much easier to estimate, but knowing the trivial fact that <math>R_{100}=\infty</math> does not help much. Given available number of shots, there would not be enough tail events to determine <math>R_{99}</math> or even <math>R_{95}</math> with sufficient accuracy. From the other side, <math>R_{50}</math> aka CEP might be reasonable for military applications, but is too low for civilian use.<br />
<br />
In theory it’s easy to convert between different metrics, but in practice it’s better to have a way to get the metric you want directly with the least amount of math to be done in the field.<br />
<br />
One nice side effect of choosing <math>R_{90}</math> is that it allows to sidestep the tricky [[Fliers_vs._Outliers]] question. We simply decide to not care about the few worst shots, regardless of the reason. And if fliers are more frequent, they probably aren’t really fliers.<br />
<br />
So you go to the range, set up some paper targets at typical engagement range, shoot at them, and look at the results.<br />
<br />
==One Shot==<br />
<br />
===Summary===<br />
<br />
Even one shot is better than nothing. If miss radius (the distance between target center and impact) is <math>R_{1:1}</math>, then estimated <math>R_{90}</math> is<br />
<br />
:<math>\hat{R_{90}} = 3 R_{1:1}</math><br />
<br />
In examples below red circle has radius <math>R_{1:1}</math> and blue circle has radius <math>\hat{R_{90}}</math>.<br />
<br />
[[File:OrderStatisticsFig1.png|800px]]<br />
<br />
All impact coordinates were pulled from the same standard bivariate normal distribution, but the circles look very different in size, which means that the variance of the estimate is quite high.<br />
<br />
===Details===<br />
<br />
Here’s the math behind the magic number 3. Assume miss radiuses of individual shots follow Rayleigh distribution with <math>\sigma = 1</math>. Its probability density function is<br />
<br />
:<math>f(x)=x e^{-\frac{x^2}{2}}</math><br />
<br />
and cumulative distribution function is<br />
<br />
:<math>F(x)=1-e^{-\frac{x^2}{2}}</math><br />
<br />
<math>x</math> here is miss radius rather than horizontal coordinate. <math>R_{90}</math> and <math>x</math> are measured in the same units and there are no other factors involved, so <math>R_{90}</math> should be proportional to <math>x</math> with some yet unknown dimensionless coefficient <math>k</math>. Probability that miss radius of the next shot is greater than <math>kx</math> (complementary cumulative distribution function) is<br />
<br />
:<math>1-F(kx) = e^{-\frac{k^2 x^2}{2}}</math><br />
<br />
Integrating over all values of <math>x</math> gives total probability of the next miss radius exceeding <math>R_{90}</math>, which should be equal to 10%:<br />
<br />
:<math>\int_{0}^{\infty}f(x)\left(1-F(kx)\right) dx = \int_{0}^{\infty}x e^{-\frac{x^2}{2}} e^{-\frac{k^2 x^2}{2}} dx = (1 - 90\%)</math><br />
<br />
This equation is easy to solve on paper, but it will soon get more interesitng so why not warm [http://maxima.sourceforge.net Maxima] up now to find <math>k</math>.<br />
<br />
find_root(integrate(x*exp(-x^2/2)*exp(-k^2*x^2/2), x, 0, inf)=0.1, k, 1, 10);<br />
3.0<br />
<br />
==Three Shots==<br />
<br />
===Summary===<br />
<br />
If worst miss radius in the three-shot group is <math>R_{3:3}</math>, then<br />
<math>\hat{R_{90}} = \sqrt{2} R_{3:3} \approx 1.4 R_{3:3}</math><br />
<br />
In examples below red circle has radius <math>R_{3:3}</math> and blue circle has radius <math>\hat{R_{90}}</math>.<br />
<br />
[[File:OrderStatisticsFig3.png|800px]]<br />
<br />
===Details===<br />
<br />
Here’s where <math>\sqrt{2}</math> comes from. Probability density of <math>m</math>th miss radius in a group of <math>n</math> shots is<br />
<br />
:<math>f_{m:n}(x)=\frac{n!}{(m-1)!(n-m)!}\left(F(x)\right)^{m-1}\left(1-F(x)\right)^{n-m}f(x)</math><br />
<br />
For <math>m=3</math> and <math>n=3</math><br />
<br />
:<math>f_{3:3}(x)=\frac{3!}{(3-1)!(3-3)!}\left(F(x)\right)^{3-1}\left(1-F(x)\right)^{3-3}f(x)=3\left(F(x)\right)^{2}f(x)=3\left(1-e^{-\frac{x^2}{2}}\right)^{2} x e^{-\frac{x^2}{2}}</math><br />
<br />
Integrating over all values of <math>x</math><br />
<br />
:<math>\int_{0}^{\infty}f_{3:3}(x) (1-F(kx)) dx = \int_{0}^{\infty}3\left(1-e^{-\frac{x^2}{2}}\right)^{2} x e^{-\frac{x^2}{2}} e^{-\frac{k^2 x^2}{2}} dx = (1 - 90\%)</math><br />
<br />
Solving for <math>k</math>, we get<br />
<br />
find_root(integrate(3*(1-exp(-x^2/2))^2*x*exp(-x^2/2)*exp(-k^2*x^2/2), x, 0, inf)=0.1, k, 1, 10);<br />
1.414213562373095<br />
<br />
==Five Shots==<br />
<br />
===Summary===<br />
<br />
So far we have ignored the issue of outliers (fliers), but with five or more shots we can finally do something about them: look at the <em>second worst</em> miss radius instead of the worst miss radius.<br />
<br />
:<math>\hat{R_{90}} \approx 1.6 R_{4:5}</math><br />
<br />
In examples below red circle has radius <math>R_{4:5}</math> (4<sup>th</sup>-smallest miss radius in a 5-shot group), and blue circle has radius <math>\hat{R_{90}}</math>.<br />
<br />
[[File:OrderStatisticsFig5.png|800px]]<br />
<br />
===Details===<br />
<br />
According to formula 3.13 in M. M. Siddiqui 1964 paper in [[References]], optimum unbiased estimator from single order statistic <math>R_{m:n}</math> is<br />
<br />
:<math>m \approx 0.79681(n+1)-0.39841+\frac{1.16312}{n+1}</math><br />
<br />
For <math>n=5</math> optimum estimator is <math>m=5</math>, but <math>m=4</math> is not far behind and is less sensitive to outliers.<br />
<br />
Probability density of <math>m</math>th miss radius in a group of <math>n</math> shots is<br />
<br />
:<math>f_{m:n}(x)=\frac{n!}{(m-1)!(n-m)!}\left(F(x)\right)^{m-1}\left(1-F(x)\right)^{n-m}f(x)</math><br />
<br />
For <math>m=4</math> and <math>n=5</math><br />
<br />
:<math>f_{4:5}(x) = \frac{5!}{(4-1)!(5-4)!}\left(F(x)\right)^{4-1}\left(1-F(x)\right)^{5-4}f(x) = 20\left(F(x)\right)^{3}\left(1-F(x)\right)f(x) = 20\left(1-e^{-\frac{x^2}{2}}\right)^{3} e^{-\frac{x^2}{2}} x e^{-\frac{x^2}{2}}</math><br />
<br />
Integrating over all values of <math>x</math><br />
<br />
:<math>\int_{0}^{\infty}f_{4:5}(x) (1-F(kx)) dx = \int_{0}^{\infty} 20\left(1-e^{-\frac{x^2}{2}}\right)^{3} e^{-\frac{x^2}{2}} x e^{-\frac{x^2}{2}} x e^{-\frac{k^2 x^2}{2}} dx = (1 - 90\%)</math><br />
<br />
Solving for <math>k</math><br />
<br />
f45(x):=20*(1-exp(-x^2/2))^3*x*exp(-x^2);<br />
find_root(integrate(f45(u)*exp(-k^2*u^2/2),u,0,inf)=0.1, k, 1, 2);<br />
1.578643529936508<br />
<br />
==Ten Shots==<br />
<br />
===Summary===<br />
<br />
We can simply split ten shots into two five-shot groups and take the average, but adding 6th and 9th miss radiuses works [http://lstange.github.io/mcgs even better]:<br />
<br />
:<math>\hat{R_{90}} \approx 0.7 (R_{6:10} + R_{9:10})</math><br />
<br />
In examples below the two red circles have radiuses <math>R_{6:10}</math> and <math>R_{9:10}</math>, and blue circle has radius <math>\hat{R_{90}}</math>.<br />
<br />
[[File:OrderStatisticsFig10.png|800px]]<br />
<br />
===Details===<br />
<br />
According to M. M. Siddiqui 1964 paper in [[References]], the two optimal order statistics are <math>0.639(n+1)</math> and <math>0.927(n+1)</math>. For <math>n=10</math> it’s 7 and 10, but 6 and 9 work [http://lstange.github.io/mcgs nearly as well]. Not using the worst miss radius means that R<sub>6:10</sub> + R<sub>9:10</sub> is a more robust estimator.<br />
<br />
Joint probability distribution of 6<sup>th</sup> miss radius <math>x=R_{6:10}</math> and 9<sup>th</sup> miss radius <math>y=R_{9:10}</math> is<br />
<br />
:<math>f(x,y)=\frac{n!}{(j-1)!(k-j-1)!(n-k)!}\left[F(x)\right]^{j-1}\left(F(y)-F(x)\right)^{k-j-1}\left(1-F(y)\right)^{n-k}f(x)f(y)</math><br />
:<math>=15120 \left[1-e^{-\frac{x^2}{2}}\right]^5 \left(e^{-\frac{x^2}{2}}-e^{-\frac{y^2}{2}}\right)^2 e^{-\frac{y^2}{2}} x e^{-\frac{x^2}{2}} y e^{-\frac{y^2}{2}}, x \le y</math><br />
<br />
Since we only care about <math>x+y</math>, we can rotate axes by 45° and integrate over <math>x-y</math>:<br />
<br />
:<math>p(u)=\frac{1}{2}\int_{0}^{u}f\left(\frac{u-v}{2},\frac{u+v}{2}\right)dv</math><br />
<br />
Integrating over all values of <math>u</math><br />
<br />
:<math>\int_{0}^{\infty}p(u)e^{-\frac{k^2 u^2}{2}}du = (1 - 90\%)</math><br />
<br />
This seems to be too hard even for Maxima to solve analytically, so we’ll have to resort to numeric integration using [https://en.wikipedia.org/wiki/Romberg%27s_method Romberg’s method]. Upper limit of 10 instead of <math>\infty</math> in second integration prevents underflow in the tail.<br />
<br />
f(x,y):=15120*(1-exp(-x^2/2))^5*(exp(-x^2/2)-exp(-y^2/2))^2*exp(-y^2/2)*x*exp(-x^2/2)*y*exp(-y^2/2);<br />
p(u):=romberg(f((u-v)/2,(u+v)/2)/2,v,0,u);<br />
find_root(romberg(p(u)*exp(-k^2*u^2/2),u,0,10)=0.1,k,0.6,0.8);<br />
0.70766872521906<br />
<br />
==Multiple Groups==<br />
<br />
===Summary===<br />
<br />
With more than ten shots, it’s easier to split them into five- or ten-shot groups and take the average. The factors will be somewhat smaller:<br />
<br />
:<math>\hat{R_{90}} \approx 1.4 R_{4:5} \approx 0.67 (R_{6:10} + R_{9:10})</math><br />
<br />
===Details===<br />
<br />
With very large number of groups, the factors are just ratios of expected values:<br />
<br />
:<math>\frac{E(R_{90})}{E(R_{4:5})}</math><br />
<br />
f45(x):=20*(1-exp(-x^2/2))^3*x*exp(-x^2);<br />
float(sqrt(2*log(10))/integrate(x*f45(x), x, 0, inf));<br />
1.38619009633813<br />
<br />
:<math>\frac{E(R_{90})}{E(R_{6:10})+E(R_{9:10})}</math><br />
<br />
f610(x):=1260*(1-exp(-x^2/2))^5*exp(-2*x^2)*x*exp(-x^2/2);<br />
f910(x):=90*(1-exp(-x^2/2))^8*x*exp(-x^2);<br />
float(sqrt(2*log(10))/(integrate(x*f610(x),x,0,inf)+integrate(x*f910(x),x,0,inf)));<br />
0.6702446399176036</div>Davidhttp://ballistipedia.com/index.php?title=Precision_Models&diff=1384Precision Models2016-06-05T20:59:43Z<p>David: /* Statistical Analysis of Dispersion */</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[Describing Precision]]</p><br />
<br />
= Models of Dispersion =<br />
<br />
We present four options for measuring and analyzing precision:<br />
# [[Closed Form Precision]]<br />
# [[Circular Error Probable]]<br />
# [[Elliptic Error Probable]]<br />
# [[Range Statistics]]<br />
# [[Order Statistics]]<br />
<br />
Before selecting one consider the following background:<br />
<br />
== General Bivariate Normal ==<br />
[http://en.wikipedia.org/wiki/Normal_distribution The normal, a.k.a. Gaussian, distribution] is the broadly accepted model of a random variable like the dispersion of a physical gunshot from its center point. The normal distribution is parameterized by its mean and standard deviation, or <math>(\mu, \sigma)</math>. As explained in ''[[What is Precision?]]'' we are only interested in the dispersion component, since the center point of impact is controlled by [[FAQ#How_many_shots_do_I_need_to_sight_in.3F|sighting in the gun]] (i.e., adjusting its aiming device). Therefore we will assume that a gunner can dial <math>\mu \approx 0</math> and leave that parameter out of the question in what follows.<br />
<br />
Since we are interested in shot dispersion on a two-dimensional target we will look at a [http://en.wikipedia.org/wiki/Bivariate_normal_distribution bivariate normal distribution], which has separate parameters for the standard deviation in each dimension, <math>\sigma_x, \sigma_y</math>, as well as a correlation parameter ''ρ''.<br />
<br />
== Uncorrelated Bivariate Normal ==<br />
We don't have any compelling evidence that in general there is, or should be, correlation between the horizontal and vertical dispersion of gunshots. Therefore, for most of our analysis we will assume ''ρ'' = 0.<br />
<br />
We do know that targets can often exhibit vertical or horizontal stringing, and therefore <math>\sigma_x \neq \sigma_y</math>. To the extent these parameters are not equal they produce elliptical instead of circular shot groups.<br />
<br />
However, we know some of the significant sources of stringing and can potentially factor them out:<br />
<br />
# The primary source of x-specific variance is crosswind. If we measure the wind while shooting we can bound and remove a “wind variance” term from that axis. E.g., "Suppose the orthogonal component of wind is ranging at random from 0-10mph during the shooting. Given lag-time ''t'' this will expand the no-wind horizontal dispersion at the target by <math>\sigma_w</math>."<ref>Wind deflection is a function of the ballistic curve and distance, but can be expressed as a simple product of the cross-wind velocity and lag time. For more information on the "lag rule" see Bryan Litz, ''Applied Ballistics for Long Range Shooting, 2<sup>nd</sup> Edition'' (2011) A4; or Robert McCoy, ''Modern Exterior Ballistics, 2<sup>nd</sup> Edition'' (2012) 7.27.</ref> Since variances are additive we could adjust <math>\sigma_x</math> via the equation <math>{\sigma'}_x^2 = \sigma_x^2 - \sigma_w^2</math>.<br />
# The primary source of y-specific variance is muzzle velocity, which we can actually measure with a chronograph (or assert) and then remove from that axis. E.g., "If standard deviation of muzzle velocity is <math>\sigma_{mv}</math> then, given the bullet's ballistic model for the given target distance, the vertical spread attributable to that is some <math>\sigma_v</math>. Here too we can remove this known source of dispersion from our samples via the equation <math>{\sigma'}_y^2 = \sigma_y^2 - \sigma_v^2</math>. This adjustment is shown in several of the examples:<br />
#* [[22LR CCI 40gr HV 40-shot 100-yard Example]]<br />
#* [[300BLK Subsonic 20-shot 100-yard Example]]<br />
<br />
= Statistical Analysis of Dispersion =<br />
<br />
In view of the preceding:<br />
# The [[Closed Form Precision]] model requires that we assume the shot group is, or can be normalized to be, a fairly symmetric bivariate Gaussian process. This assumption is the most amenable to statistical analysis.<br />
# [[Order Statistics]] are slightly less efficient and amenable to abstract analysis, but are both more robust and easier to apply "in the field." <br />
# [[Circular Error Probable]] disregards any ellipticity in the actual shot process in order to characterize precision using a single parameter. Since most of precision estimation is for the purposes of comparing loads, rifles, and shooters, we need a single number and we don't care if the dispersion is elliptic: tighter is always better.<br />
# [[Elliptic Error Probable]] allows for a full characterization of the General Bivariate Normal model. For some applications &ndash; e.g., computing hit probabilities on non-circular targets &ndash; we want to preserve statistically significant ellipticity. And for a few &ndash; e.g., harmonic dampening, and perhaps shooter technique correction &ndash; the orientation of the ellipse produced by non-zero ''ρ'' could be helpful if it can be estimated.<br />
# Extreme Spread and the other [[Range Statistics]], which increase with number of shots per group ''n'', do not have any useful functional forms. The characteristics of these measures have to be derived from Monte Carlo simulation. They are the least efficient statistics but are also the most commonly used because they are so easy to measure in the field and so familiar to shooters.<br />
<br />
One practical question that many shooters raise is what to do with outliers, known in the sport as "fliers." We address [[Fliers|fliers here]].<br />
<br />
= Tools =<br />
See [[Measuring Tools]] for convenient ways of measuring and analyzing precision.<br />
<br />
= References =<br />
<references /></div>Davidhttp://ballistipedia.com/index.php?title=Describing_Precision&diff=1383Describing Precision2016-06-05T20:47:31Z<p>David: /* Which Measure is Best? */</p>
<hr />
<div><p style="text-align:right"><B>Previous:</B> [[What is Precision?]]</p><br />
<br />
= [[Angular Size|Units]] =<br />
<br />
When we talk about shooting precision we are referring to the amount of dispersion we expect to see of each shot about a center point (which shooters try to adjust to match the point of aim). Precision is like a cone of error that projects out from the muzzle of the gun. I.e., double the distance and the dispersion also doubles. We can describe this error by referring to dispersion at a specific distance. For example, it is common to quote precision in inches of extreme spread at 100 yards, or "inches per hundred yards" (IPHY).<br />
<br />
It is more common, however, to describe [[Angular Size]] &ndash; i.e., the angle of the cone at its tip &ndash; since this is independent of the distance at which a target is shot. The higher the precision, the tighter the cone and the smaller the angle at its tip.<br />
<br />
One of two popular angular units used by shooters is '''MOA''', though there is some ambiguity in this term. MOA was initially short for Minute of Arc, or arc minute, which is one sixtieth of one degree. '''At 100 yards (3600 inches) one MOA is 3600" tan (1/60 degrees) = 1.047"'''. At some point shooters began to expand the acronym as Minute of Angle. They also and rounded its correct value to 1” at 100 yards, though for clarity the latter unit is properly called "Shooters MOA," or '''SMOA'''.<br />
<br />
The other common unit is the '''mil''', which simply means thousandth. For example, '''at 100 yards a mil is 100 yards / 1000 = 3.6"'''. Some more benign confusion also persists around this term, with some assuming "mil" is short for milliradian, which is another angular unit. Fortunately, a milliradian &mdash; 3600" tan (1/1000 radians) ≈ 3.600001" inches at 100 yards &mdash; is almost exactly equal to a mil so there’s no harm interchanging ''mil'', ''mrad'', ''milrad'', and ''milliradian''.<br />
<br />
'''NB: 1 mil = 3.6 SMOA <math>\approx</math> 3.44 MOA'''. See [[Angular Size]] for detailed illustrations and conversion formulas.<br />
<!--<br />
Note also: Even '''mil''' is encumbered by some [http://en.wikipedia.org/wiki/Angular_mil#Definitions_of_the_angular_mil historical ambiguity]. For example, western militaries going back at least a century used an angular unit for artillery calculations that divided the circle into 6400 "mils," which persists the "NATO mil."--><br />
<br />
= Examples =<br />
<br />
One of the important questions addressed here is ''what'' to measure in order to determine the intrinsic precision of a shooting system, and what sample size is sufficient to achieve any degree of statistical significance.<br />
<br />
Following are common measurements used in the industry:<br />
* [[Range Statistics|Extreme Spread]] of a 3-shot group, usually at 100 yards. This is statistically almost meaningless, especially when there is no reference to how many 3-shot groups were sampled. ([http://www.ar15.com/mobile/topic.html?b=3&f=118&t=279218 An extended practical, and amusing, critique of the 3-shot group is archived here].) <br />
* Extreme Spread of a 5-shot group, sometimes excluding the worst shot. Hardly any better.<br />
* Average, Max, and Min Extreme Spread of five 5-shot groups. ([[Range_Statistics#Example:_NRA.27s_Test_Protocol|This is the protocol used by the NRA's magazines and is actually rather efficient]].)<br />
* The US Army Marksmanship Unit at Ft. Benning, GA uses a minimum of 3 consecutive 10-shot groups fired with the rifle in a machine rest when testing service rifles. Armed forces also often explicitly uses the more statistically powerful Mean Radius and Circular Error Probable measures.<br />
<br />
= Measures =<br />
[[File:SCAR17 150gr 100yd.png|365px|thumb|right|Precision Measures diagrammed on a 10-shot 100-yard group. Data in [[Media:SCAR17_150gr_100yd.xls]]]]<br />
Eight different measures have been used to characterize the dispersion of bullet holes in a sample target. Some are easier to calculate than others.<br />
<br />
In the following formulas assume that we are looking at a target reflecting ''n'' shots and that we are able to determine the center coordinates ''x'' and ''y'' for each shot.<br />
<br />
As is customary, the mean value of a set of values <math>\lbrace x \rbrace</math> is called "x-bar" and defined as <math>\bar{x} \equiv \sum_{}^n x_i / n</math>.<br />
<br />
(There are some methods for dealing with the case of targets with [[ragged holes]] in which we can't determine the precise coordinates of each shot.)<br />
<br />
== Mean Radius (MR) ==<br />
:&nbsp; <math>\bar{R} = \sum_{}^n r_i / n</math> where <math>r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}</math><br />
<br />
As we will see in [[Closed Form Precision]], the Mean Radius is typically only 6% larger than the Circular Error Probable. Since this is within the margin of error of most real-world usage the terms MR and CEP may be casually interchanged. <br />
<br />
== [[Circular Error Probable]] (CEP) ==<br />
CEP(''p''), for <math>p \in [0, 1)</math>, is the expected radius of the smallest circle that covers proportion ''p'' of the shot group. When ''p'' is not indicated it is assumed to be 50%, which is the true '''median shot radius'''.<br />
<br />
== Horizontal and Vertical Variance ==<br />
:&nbsp; <math>\sigma_h^2 = \frac{\sum^{n}(x_i - \bar{x})^2}{n - 1}, \quad \sigma_v^2 = \frac{\sum^{n}(y_i - \bar{y})^2}{n - 1}</math><br />
(Often these will be given as standard deviations, which is just the square root of variance.)<br />
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== Radial Standard Deviation (RSD) ==<br />
RSD is typically defined in the [[Prior Art]] as <math>\sqrt{\sigma_x^2 + \sigma_y^2}</math>. This does not express the standard deviation of shot radii. However it turns out to be <math>\sqrt{2}</math> times the biased estimator of the [[Closed Form Precision]] model and has therefore served as a useful reference.<br />
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In order to avoid confusion with this measure that is both biased and misnamed we will try to avoid any further reference to RSD.<br />
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Note that in general the actual standard deviation of radii <math>r_i</math> is not easy to calculate: When <math>\sigma_x \neq \sigma_y</math> the variance of radii is <math>\frac{\sigma_x^2}{\pi} (\pi - 2 K^2(1 - \frac{\sigma_y^2}{\sigma_x^2})) + \sigma_y^2</math> where ''K'' is the complete elliptic integral.<br />
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In the special case that <math>\sigma_x = \sigma_y</math> then, per the Rayleigh model described in [[Closed Form Precision]], we get the following simple formula:<br />
:&nbsp; <math>RSD = \sigma \sqrt{\frac{4 - \pi}{2}}</math><br />
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== Diagonal ==<br />
Let <math>\hat{X} = x_\max - x_\min, \quad \hat{Y} = y_\max - y_\min</math> &mdash; i.e., the ranges of ''x'' and ''y'' values.<br />
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The diagonal <math>D = \sqrt{\hat{X}^2 + \hat{Y}^2}</math> is the length of the diagonal line through the smallest rectangle covering the sample group.<br />
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== Figure of Merit ==<br />
FoM = <math>(\hat{X} + \hat{Y}) / 2</math> is the average extreme width and height of the group.<br />
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== Covering Circle Radius ==<br />
CCR is the radius of the smallest circle containing all shot centers. This will either pass through the two extreme points &ndash; in which case CCR = (Extreme Spread) / 2 &ndash; or else it will pass through three outside points.<br />
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== Extreme Spread ==<br />
The Extreme Spread <math>ES = \max \sqrt{(x_i - x_j)^2 - (y_i - y_j)^2)}</math> is the largest distance between any two points, ''i'' and ''j'', in the group. Statistical literature has also used the term ''bivariant range''. Shooters typically called this measure ''group size''.<br />
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= Which Measure is Best? =<br />
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[[Precision Models]] will detail the mathematical relationships between many of these measures.<br />
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== Invariant Measures ==<br />
* Mean Radius<br />
* Circular Error Probable<br />
* Variance<br />
* Standard Deviation<br />
It is worth noting that these first four measures do not vary with group size. I.e., taking more shots tightens their confidence interval but doesn't change their expected value.<br />
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In fact we will see that when <math>\sigma_h = \sigma_v</math> each of these four measures is just a scalar function of ''σ'', so they all convey the same underlying information.<br />
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== [[Range Statistics]] ==<br />
* Extreme Spread<br />
* Diagonal<br />
* Figure of Merit<br />
* Covering Circle Radius<br />
These measures increase with group size. They are more commonly used because they are easier to calculate. But they are statistically far weaker because they virtually ignore inner data points.<br />
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== [[Order Statistics]] ==<br />
This relatively new approach combines the computational ease of Range Statistics with more statistical power and robustness.<br />
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<p style="text-align:right"><B>Next:</B> [[Precision Models]]</p></div>David