Trying to "create" an equation. And failing miserably.

I've been working on a science fiction story for a ridiculously long amount of time, and I'm trying to nail down the properties of a particular piece of fictional technology. I'd like to do this as accurately as I can, to the point where an equation or two can describe it.

I'm trying to show that as the distance from a central point increases, the probability of accurately targeting a second point decreases. Naturally I looked at the inverse-square law. Unfortunately, I'm not too great with math beyond basic algebra and geometry.

What I want is for the probability to be 100% (or 1.00) when the target point is the same as the central point (radius = 0), then decrease from there, somehow according to the inverse-square law. I also want to be able to show that some other factor can multiply this probability (increasing the range at which the second point can be reliably targeted).

I'm not asking anyone to do the work for me, but I described it so that someone can hopefully point me in the right direction. Can one of you resident geniuses give me some tips, or tell me what I should be studying so I can figure this out?

I don't think you are looking for a 1/r^2, but a 1/r^3 relationship.

Let R be the distance between source and target, and let r be the distance of a point from the target. Let's assume that the probability distribution F of your targetting mechanism deploying a charge at distance r from the target is radially symmetric, for example with a Gaussian profile:

[math]

F = F(r) = \alpha e^{-\frac{r^2}{2\sigma^2}}

[/math]

The normalisation factor, alpha, is just there to ensure

[math]

\int_0^\infty \int_0^{2 \pi} \int_0^\pi F r^2 \sin\theta d\phi d\theta dr = 1

[/math]

Radial symmetry is a sensible assumption. The width of the profile, \sigma, we assume to be a linear function of R. For example, you may be able to launch your charge with a given angular precision, so the lateral accuracy will be a linear function of R. Similar argument for radial precision.

The exact shape of the probability distribution isn't important, so we can make life easier by assuming a simpler profile:

[math]

F(r) = \left\lbrace \begin{array}{ll}

\alpha & \mathrm{if}\; r < \sigma\\

0 & \mathrm{otherwise}

\end{array}\right.

[/math]

i.e. a ball with constant density. Since \alpha must be chosen so that the integral over the ball sums up to 1, it should be clear that density F goes down with the cube of the ball radius, i.e.

[math]

\alpha \sim \sigma^{-3}

[/math]

Let's assume that in order to score a "hit" you have to end up within a given radius r_0 of the target. Therefore, the hit probability P is the integral of F over a ball of radius r_0. since F is constant, it follows that

[math]

P = \frac{4\pi}{3}r_0^3 F \sim \frac{4\pi}{3} \frac{r_0^3}{\sigma^3} \sim \frac{4\pi}{3} \frac{r_0^3}{R^3}

[/math]

The same is true for any other radially symmetric profile F. Your assumption of a 1/r^2 would only make sense if for example your lateral precision was a linear function of distance, but the radial targetting were always spot on, i.e. your charges would always end up on the surface of a sphere with radius R.