Sighting a Weapon

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Revision as of 18:43, 2 June 2015 by Herb (talk | contribs) (Sighting Process)
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"Sighting" a weapon involves trying to align its sighting device so that the Point Of Aim (POA) coincides with the Center Of Impact (COI). The COI is from a sample of the infinite number of possible shots, the population of shots. The statistical challenge with sighting a weapon is that given any shot dispersion, and every weapon has some, then we can only estimate the POI. In principle we only know that increasing the number of shots (samples) will improve our estimate.

Sighting Process

As Einstein said, "Everything is relative." So for the purpose of this wiki we need to develop some frame of reference around which we can analyze situations. The sighting process nails down the theoretical perspective.

Assuming that "perfect" horizontal and vertical sight adjustments can be made for the fixed target distance used for sighting, the process is to:

  • Take \(n\) shots at target
  • Measure the horizontal and vertical distance from each shot to the center of the target.
  • Calculate the mean horizontal, \(\bar{h}\), and vertical, \(\bar{v}\), distances from the sample COI to the target center
  • Adjust the Horizontal and vertical sights perfectly to correct for the error.

It is also assumed that the weapon can maintain this "zero" until the weapon is sighted again. So we're ignoring the real world problems of the weapon getting banged around, undergoing temperature changes and a zillion other real world details that effect the ballistics of the weapon.

The only thing that we can measure is the COI for the shot pattern, and hence this is generally our point of reference.

Sighting Bias

Breaking the sighting error into parts assuming each error source is independent and normally distributed we have:

Sighting Analysis for Rayleigh Shot Distribution

To analyze the sighting bias we continue to use the symmetric bivariate normal model of shot impacts.

First using Cartesian coordinates, we assume that both the measurements along horizontal and vertical axis have a Normal Distribution with the same variance of \(\sigma^2\) and that the distributions are uncorrelated. Thus:

    \(h \sim \mathcal{N}(\mu_h, \sigma^2)\) and \(v \sim \mathcal{N}(\mu_v, \sigma^2)\)

From the \(n\) "sighter" shots and the mean horizontal, \(\bar{h}\), and vertical, \(\bar{v}\), distances have been calculated.

Derivation of the Theoretical Distribution of Weapon Zero

For those not satisfied with just assigning the Rayleigh Distribution to the shot pattern, the following proof is offered thanks to Alecos Papadopoulos. (Blame the editors of this wiki page for any errors or lack of clarity!)

For our n shots let the distance from the COI to the population center be:

$$r_n = \sqrt{(\mu_h - \overline{h_c})^2 + (\mu_v - \overline{v_c})^2}$$

Given the preceding definitions, the sample means are also mean-zero normal random variables with $$\bar X, \bar Y \sim N(0,\sigma^2/n)$$

Define random variables $$Z_x = \left(\frac{\bar X}{\sigma/\sqrt n}\right)^2 = \frac n{\sigma^2}\bar X^2 \sim \chi^2(1),\;\;Z_y = \left(\frac{\bar Y}{\sigma/\sqrt n}\right)^2 =\frac n{\sigma^2}\bar Y^2\sim \chi^2(1),$$

Then the random variable $$W = Z_x + Z_y =\frac n{\sigma^2}\left(\bar X^2+\bar Y^2\right)\sim \chi^2(2)$$

By the properties of a chi-square random variable, we have $$W_n=\frac {\sigma^2}nW \sim \text {Gamma}(k=1, \theta = 2\sigma^2/n) = \text{Exp}(2\sigma^2/n)$$ i.e. $$f_{W_n}(w_n) = \frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} w_n\Big\}$$

Define \(R_n = \sqrt {W_n}\). By the change-of-variable formula we have

$$W_n = R_n^2 \Rightarrow \frac {dW_n}{dR_n} = 2R_n$$ and so

$$f_{R_n}(r_n) = 2r_n\frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} r_n^2\Big\} = \frac {r_n}{\alpha^2} \exp\Big \{-\frac {r_n^2}{2\alpha^2} \Big\},\;\;\alpha \equiv \sigma/\sqrt n$$

So \(R_n\) follows a Rayleigh distribution with parameter \(\alpha = \sigma / \sqrt{n}\).

Experimental Estimation of Population Center From Sighting