Why do we need to keep error types separate?

We saw in the previous article how it is important to categorize a specific error type as either occasion-to-occasion or round-to-round since they are treated differently in the effectiveness calculations that require accuracy data. To illustrate this, assume that we provide accuracy data to a Monte-Carlo effectiveness method. The correct form of such a simulation is shown diagrammatically in Figure 1, and shows how each Monte-Carlo iteration corresponds to a single occasion while for each occasion a precision error is generated for each weapon delivered in that occasion.

Figure 1  Using accuracy measures in effectiveness calculations

Multiple occasion-to-occasion error sources may be combined as can multiple round-to-round errors, but only in the case of single weapon/munition delivery can occasion-to-occasion and round-to-round errors be combined directly together. To illustrate the effect of treating these errors separately, two examples will be considered

Example 1

Consider the scenario shown below. Here we have a target represented by a 20’x20’ rectangular vulnerable area and fired on by a weapon with a 3 round burst. Given the aiming error has a standard deviation of 5’ in range and deflection, and the precision error is 2’, what is the expected probability of killing the target? To simulate this we implement the program illustrated functionally in Figure 1, insert the given error data, run for 10,000 occasions and get the result P_K = 0.105.

Figure 2 Monte Carlo simulation example

If now however we assume (incorrectly) that the two error sources are independent and root sum square (RSS) them together we get a combined error standard deviation of 5.385’. The P_K then becomes

               (1)

Clearly the difference in results, 0.105 vs 0.875, shows the effect of assuming independence of error sources introduces considerable error into the calculation of P_K.

Example 2

A GPS/INS weapon has round-to-round accuracy of σ_x=100’ and an occasion-to-occasion error of σ_μ=210’ and a lethal area of radius 220’. How many rounds are needed to give a P_K of 0.7 against a unitary (point) target?

This may be interpreted as: how many rounds are needed to obtain a 0.7 probability of hitting a 220’ radius target? Given R_T, σ_x, σ_μ, we can actually estimate the correct answer by using the rather complicated salvo equation

           (2)

Solving this equation repetitively for increasing number of weapons (n) until a value of P_h=0.7 is found, results in n=7. Suppose we had incorrectly RSS the two dispersions and used the Rayleigh distribution we would obtain the probability of one round hitting the target is  

           (3)

This has to be powered up to get a probability of 0.7 giving

                 (4)

Again we see the danger of RSS the error sources since if we do the result suggests that 3 weapons will suffice whereas the correct answer is 7. The reason there is a difference may be explained by considering a hypothetical case of using 50 weapons against a unitary target. These weapons are employed in two ways

1.     Fifty independent weapons aimed and fired one at a time

2.     Ten independent volleys (occasions) of 5 weapons each

For the case of shooting one weapon at a time, each weapon may or may not kill the target but whether it did or did not, the next weapon sees a fresh target which it too may or may not kill irrespective of what the first weapon did

For the salvo case where 5 weapons are fired rapidly against the target, suppose that the first weapon to arrive does kill the target. If the next weapon to arrive would have killed the target if the first weapon did, this second kill is not counted since the target is already killed and cannot be killed again. This second kill would have counted if the weapon had been fired independently, but in salvo fire it is not. This comparison is shown in Figure 3.

Figure 3 Comparison of single round and salvo fire

It is expected therefore that salvo fire will produce a lower probability of kill compared to the same number of weapons used independently, as confirmed by the numerical example discussed earlier

An interesting point arises if we are concerned only in accuracy metrics such as CEP, as is typically the case in weapon testing. Whether we deliver the weapon in 50 independent attacks or 10 volleys of 5 weapons each, the resultant CEP is the same. The different delivery methods may only be discriminated by the resulting effectiveness calculations, not by impact data alone.