Error Propagation
Contents
Error Propagation
Basic Assumptions
In general shooting has a number of different error sources. The "basic" assumption is that the error sources are normally distributed and independent. Thus for example the vertical error component would be something like:
\(\sigma_v^2 = \sigma_{v,Weapon}^2 + \sigma_{v,Ammunition}^2 + \sigma_{v,shooter}^2\)
and there would similar equations for \(\sigma_h^2\) and \(\sigma_{RSD}^2\).
Notes:
- The error sources can't be measured independently. Only the total error is observable. This is a very important consideration since the total error thus limits the precision which any of the individual error factors can be determined.
- If any one of the error sources is more than about three times the sum of the other two, then that error source essentially controls the overall error of the measurement.
- These error sources could be decomposed further into "finer" errors. For instance consider the multitude of variables that a handloader controls when loading a cartridge.
- The fact that a very small amount of data is typically collected by a shooter greatly limits the overall precision of the mathematical analysis. In statistics this is known as the "small sample" problem. For example to get "good" estimates for the normal distribution parameters \(\bar{x}\) and \(\sigma\), 30 measurements are desired.
- In general for a given sample size of n measurements the mean, \(\bar{x}\), is determined relatively (i.e as a percentage) much more precisely than the standard deviation, \(\sigma\) (or other dispersion factor).
Error Handling Examples
Examples Using System Errors Of The Populations
Let's assume the following for an Extreme Spread measurements for the population values:
\(\sigma_{ES}^2 = \sigma_{ES,Weapon}^2 + \sigma_{ES,Ammunition}^2 + \sigma_{ES,shooter}^2\)
\(\sigma_{ES,Weapon}^2 = 0.30\) inches
\(\sigma_{ES,Ammunition(1)}^2 = 0.50\) inches
\(\sigma_{ES,Ammunition(2)}^2 = 0.70\) inches
\(\sigma_{ES,Shooter}^2 = 0.40\) inches
What is the expected \(\sigma_{ES(1)}\) for type 1 ammo?
Substituting for the values into the general equation:
\(\sigma_{ES(1)}^2 = 0.30^2 + 0.50^2 + 0.40^2 = 0.50\)
thus \(\sigma_{ES(1)} = \sqrt{0.5} = 0.707 \) inches
What is the expected \(\sigma_{ES(2)}\) for type 2 ammo?
Substituting for the values into the general equation:
\(\sigma_{ES(2)}^2 = 0.30^2 + 0.70^2 + 0.40^2 = 0.740 \)
thus \(\sigma_{ES(2)} = \sqrt{0.5} = 0.860 \) inches
From the population values, what is the expected difference between Type(1) and Type(2) ammo?
Examples Using Experimental System Errors Of The Populations
Given:
\(\sigma_{ES}^2 = \sigma_{ES,Weapon}^2 + \sigma_{ES,Ammunition}^2 + \sigma_{ES,shooter}^2\)
\(\sigma_{ES(1)} = 0.707 \) inches
\(\sigma_{ES(2)} = 0.860 \) inches
From the experimental values, what is the precision of the difference measurement between Type(1) and Type(2) ammo?
There is a nasty surprise in this.
The precision with which this measurement can be made is very poor.
\(\sigma_{diff}^2 = \sigma_{ES(1)}^2 + \sigma_{ES(2)}^2<br/> inches<br/>
or:<br/>
<math>\sigma_{diff} = \sqrt{0.707^2 + 0.860^2} = 1.113\) inches
The nasty surprise is that while variances can be subtracted to get differences, the standard deviation of the difference is equal to the sum of the variances. Thus the trend is that since both measurements have some error, that the errors don't negate each other but combine.
If (which we can't do!!) we could measure just the isolated ammunition factor,
Error Dominance
Large Ammunition Error
Let's again assume the basic equation for the Extreme Spread measurement, but in addition let's assume that:
\(\sigma_{ES,Ammunition} = 3\sqrt{\sigma_{ES,Weapon}^2 + \sigma_{ES,shooter}^2}\).
What is the effect of this situation?
Substituting into the basic equation we can see that:
\(\sigma_{ES}^2 = \sigma_{ES,Ammunition}^2 + \frac{1}{9}\sigma_{ES,Ammunition}^2 = \frac{10}{9}\sigma_{ES,Ammunition}^2\)
Thus in this case the ammunition error dominates the total error of the measurement. Experiments using different kinds of ammunition would tend to readily show differences between ammunition types. Shooting a few groups of each type should be sufficient.
Small Ammunition Error
As before let's assume the basic equation for the Extreme Spread measurement, but in let's flip the assumption about ammunition controlling the experimental error around. Now let's assume that:
\(\sigma_{ES,Ammunition} = \frac{1}{3}\sqrt{\sigma_{ES,Weapon}^2 + \sigma_{ES,shooter}^2}\).
What is the effect of this situation?
Now we can see that the ammunition is only 1/10th of the total error. In order to precisely measure differences between ammunition types a lot of measurements are going to be needed. Thus a large number of groups of each ammunition types will be needed.
