Damaging Effects of Weapons and Ammunition - Igor A. Balagansky

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Schematic illustration of creation of the law of distribution of the fraction of damaged area for the intermediate values 0 less than u less than umax.

      Source: From Wentzel [2].

      (I.34)upper F left-parenthesis u right-parenthesis equals 1 minus one fourth left-bracket ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis normal beta Subscript u Baseline right-parenthesis minus ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis normal alpha Subscript u Baseline right-parenthesis right-bracket left-bracket ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis normal delta Subscript u Baseline right-parenthesis minus ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis normal gamma Subscript u Baseline right-parenthesis right-bracket comma

Schematic illustration of determination of probability obtaining a given fraction of damaged F(u) using the normalized Laplace function.

      Source: From Wentzel [2].

      Example

      One shot is fired at a target with dimensions Tx = 2; Ty = 1. Damage zone has dimensions Lx = Ly = 3. There is no offset of the aim point. We need to draw on four points the function F(u) of distribution of damage fraction U.

      Solution

StartLayout 1st Row p 0 equals 1 minus ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis 3.5 right-parenthesis dot ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis 4 right-parenthesis equals 0.025 comma 2nd Row p Subscript m Baseline equals ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis 2 right-parenthesis dot ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis 2.5 right-parenthesis equals 0.746 comma 3rd Row upper F left-parenthesis 1 slash 3 u Subscript max Baseline right-parenthesis equals 1 minus ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis 3.33 right-parenthesis dot ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis 3.17 right-parenthesis equals 0.056 comma 4th Row upper F left-parenthesis 2 slash 3 u Subscript max Baseline right-parenthesis equals 1 minus ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis 2.67 right-parenthesis dot ModifyingAbove normal Ф With ampersand c period circ semicolon left-parenthesis 2.83 right-parenthesis equals 0.125 period EndLayout

      Knowing the distribution function F(u) of value U for one shot, it is easy to find its average value (expected value) M = M[U]. For random variables of mixed type, the expected value consists of two parts: sum and integral. The sum is applied to those values that have nonzero probabilities (i.e. where the distribution function makes a jump) and the integral to the area where the distribution function is continuous.

Schematic illustration of the creation of the law of distribution of the fraction of damaged area.

      Source: From Wentzel [2].

Schematic illustration of the function of distribution of damaged area.

      Source: From Wentzel [2].

      (I.35)upper M left-bracket upper U right-bracket equals u Subscript max Baseline p Subscript m Baseline plus integral Subscript 0 Superscript u Subscript max Baseline Baseline u left-parenthesis upper F left-parenthesis u right-parenthesis right-parenthesis Subscript u Superscript prime Baseline italic d u period

      Knowing the function of F(u) distribution of U value for one shot, it is easy to find the probability Ru that for one shot, the damaged fraction U will be not less than the set value u:

upper R Subscript u Baseline equals 1 minus upper F left-parenthesis u right-parenthesis period

      Example

      For the previous example, we need to find the probability that at least 80% of the target area will be damaged.

      Solution

      According to the graph in Figure I.20 at u = 0.8, we have

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