Extreme Spread

Experimental Summary

Given	set of n shots {\( (h_1, v_1), (h_2, v_2), ..., (h_n, v_n) \)} All of the (h,v) positions do not need to be known so a ragged hole will suffice.
Assumptions	Ideally any shot \(i\) would follow a Rayleigh Distribution \(h_i \sim \mathcal{N}(\bar{h},\sigma_h^2), v_i \sim \mathcal{N}(\bar{v},\sigma_v^2)\) Horizontal and vertical dispersion are independent. \(\sigma_h = \sigma_v\) (realistically \(\sigma_h \approx \sigma_v\)) \(\rho = 0\) \(PDF_{r_i}(r) = \frac{r}{\sigma^2}e^{-r^2/2\sigma^2}\) Note: It is not necessary to fit \(\sigma\) to calculate the Extreme Spread. No Fliers
Data transformation	Identify two holes, \(i, j\) which are the farthest apart and measure ES  \(ES = \sqrt{(x_i - x_j)^2 + (y_i - y_j)^2}\)
Experimental Measure	ES

Given

The requirements for this test are very basic. Just a target with \(n\) shots, and some measuring device. Assuming an Extreme spread of under 6 inches then a vernier caliper is used. A measurement is possible to a few thousandths of an inch which is vastly more precision than is usually required. From longer distance a ruler, or perhaps a tape measure.

Assumptions

None are needed to make measurement. However making assumptions about the dispersion will enable theoretical predictions about the measurement.

Data transformation

The data transformation for a human has simple requirements, just the ability to locate the holes which are the furthest apart and measure the distance between them. If the target has a ragged hole it can be a bit tricky, but the edges of the hole should have enough curvature to make shot location possible.

If measuring on the range, then the center of the hole is difficult to locate. Typically a vernier caliper (cheap is fine!) would be used to measure the distance from the outside edges of the holes, then the bullet caliber subtracted to get a c-t-c measurement.

A cheap ($10-$20) vernier caliper works fine. There is no need for a $2,000 one that measures to 1/10,000th of an inch and has National Bureau of Standards calibration. The vernier caliper is nice for the c-t-c measurement because the knife edges will be parallel and won't obscure the edges of the bullet hole. Thus it is easy to accurately place both of the knife edges on a tangent to the curved bullet holes.

If using a computer then the center location would be a matter programming. For example a mouse might be used simply to point out the holes, or to drop a dot at the center of the hole, or to drag a circle over the hole. The computer would then make the c-t-c measurement.

Theoretical \(ES\) Distribution

Dispersion Follows Rayleigh Distribution

Assuming that the shots are Rayleigh distributed allows us to make some theoretical estimates.

Theoretical \(ES\) Distribution

Parameters Needed
\(PDF(r; \sigma)\)	no deterministic solution, must be simulated via Monte Carlo
\(CDF(r; \sigma)\)	no deterministic solution, must be simulated via Monte Carlo
Mode of PDF)	depends on \(n\), Mode increases as number of shots increases.
Median of PDF	depends on \(n\), Median increases as number of shots increases.
Mean of PDF	depends on \(n\), Median increases as number of shots increases
Variance	no deterministic solution, must be simulated via Monte Carlo
Variance Distribution
(h,v) for all points?	yes for simulation.
Symmetric about Mean?	No, skewed to larger values. More symmetric about mean as the number of shots increases.

Mode, Median, Mean, Standard Deviation, %RSD

Since the distribution is positively skewed:

Mean > Median > Mode

Table columns for "ES Values from Monte Carlo Simulation" Table

number of shots	the number of shots used per target
Mode	The peak of the distribution
Median	The 50th percentile
Mean	the average value over the whole distrubtion
"Normality Error"	As sort of a crude indication of normality let's use the value: "Normality Error" = \( \frac{\frac{CDF(5) - CDF(95)}{2} - Mean}{Mean} {\cdot 100}\) So we measure half the distance between the 5th percentile and the 95th percentile to determine where the Mean should be if the distribution was symmetrical, and determine the % error based on the actual value of the mean. + value means positively skewed, - value means negatively skewed.
Standard Deviation
%RSD	The relative standard deviation which is a ratio of the standard deviation to the mean as a percentage.

The point of the "Normality Error" is to give the reader a quinsy feeling about using Student's T-Test for groups with few shots, or the average of a small number of targets.

Theoretical ES Values from Monte Carlo Simulation Distribution

number of shots	Mode	Median	Mean	"Normality Error"	Standard Deviation	%RSD
2			1.772		0.932	52.6%
3			2.406	4.95%	0.887	36.9%
4			2.787		0.856	30.7%
5			3.066	3.06%	0.828	27.0%
6			3.277		0.806
7			3.443		0.783
9			3.710		0.754
10			3.813		0.745
20
30			4.788	1.63%	0.745	15.6%

The tabular values can be used in a number of ways:

Estimate a 95% confidence Interval for Given 2-shot groups based on one ES measuremnt

So a 2-shot group has been measured. If the measured value is accepted as the true value, what would the standard deviation of multiple 2-shot groups be?

This is another example to warn the reader. Just because you can calculate a standard deviation doesn't mean that a Student's T Test will work. A typical 95% confidence Interval for an individual ES measurement is \(\pm 1.96 \sigma\) and for a 2-shot group that is:

 \(\pm 1.96 \cdot 52.6\% = \pm 103.1\%\) of the measurement

so the lower confidence interval would be at -3.1% !!! The nonsensical result is because the distribution is skewed. A negative extreme spread measurement is impossible. It isn't the standard deviation that is wrong, it is the assumption that the confidence interval would be \(\pm 1.96 \sigma\) that is the problem. Since the distribution is skewed, the low side of the confidence interval at the 2.5 percentile is at ?? and the high side of the confidence interval at the 97.5 percentile is at +??.

At 5 shots the T-test is reasonable, and at 10 shots pretty good.

Given ES of one 5-shot group is 1.53 inches

• Estimate ES values for different group sizes.

A 3-shot group would be given by measured size times ratios of the Means from the table

 \(1.53 \frac{2.406}{3.066} = 1.20\) inches

A 10-shot group would be given by measured size times ratios of Means from the table

 \(1.53 \frac{3.813}{3.066} = 1.90\) inches

• Estimate the expected standard deviation from the measured ES value

The %RSD value for 5-shots is 27.0% so:

 \(\hat{s} = 1.53 \text{ inches} \cdot 0.270 = 0.413 \text{ inches} \)

• Estimate the expected Standard Deviation of the average of 4 targets

 \(\text{SD}_{ES\ 4 \ Targets}\ = \frac{27.0\%}{\sqrt{4}} = 13.5\% \)

See Also

Dispersion Assumptions - A discussion of the different cases for shot dispersion

Other measurements practical for range use are:

• Covering Circle Radius - about same precision as Extreme Spread if Rayleigh distributed
• Diagonal - somewhat better precision than Extreme Spread if Rayleigh distributed
• Figure of Merit - somewhat better precision than Extreme Spread if Rayleigh distributed