Closed Form Precision

The Precision Parameter

Measuring Precision showed how a single parameter σ characterizes the precision of a shooting system.

Rayleigh distribution of shots given σ

This σ is the parameter for the Rayleigh distribution with probability density function \(\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}\). The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:

  \(Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma}\)

Therefore, we expect 39% of shots to fall within a circle of radius σ, 86% within 2σ, and 99% within 3σ.

Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful precision measures:

Radial Standard Deviation (RSD)

Radial Standard Deviation \(RSD = \sigma \sqrt{2}\). The expected sample RSD of a group of size n is

  \(RSD_n = \sigma \sqrt{\frac{2}{c_{G}(n)}} \approx \sigma \sqrt{2 - \frac{1}{2n} - \frac{7}{16n^2} - \frac{19}{64n^3}}\)

Mean Radius (MR)

Mean Radius \(MR = \sigma \sqrt{\frac{\pi}{2}}\). The expected sample MR of a group of size n is

  \(MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}\)

Circular Error Probable (CEP)

Circular Error Probable \(CEP = \sigma \sqrt{\ln(4)}\). The expected sample CEP of a group of size n is

  \(CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}\)

Spread Measures

Expected and standard deviation of values for size statistics when σ = 1

The three measures that vary with n (Extreme Spread, Diagonal, and Figure of Merit) are range statistics that lack convenient functional forms. However both the mean and standard deviation of their expected value scales with σ, so it is sufficient to calculate them once for σ = 1 and multiply the resulting values by the desired σ. Media:Sigma1ShotStatistics.ods contains those values for n up to 100.

Example

The extreme spread value from the table for 5 shots is 3.06σ. I've determined my rifle has precision σ = ½MOA. If I take five shots at 100 yards I would expect an extreme spread of 3.06/2 = 1.53MOA \(\approx\) 1.6". And since the standard deviation of that value is 0.83/2 = 0.415σ I expect that 68% of the time my five shot groups will have an extreme spread in the range (1.12, 1.94) MOA — i.e., between 1.2" and 2.0".