How do I calculate a specific variation for a known value of the normal distribution function? I am writing a Gaussian blur filter in graphics shader code.  I want to make the blur parameterized by radius from the users perspective.  The best method I can figure to do this is to pick a suitable stopping point for y, say .001, and solve for the variance to plug into the normal distribution function that will achieve that value of y.
Unfortunately I cannot for the life of me solve this equation for v...
$x = 10$ (blur radius)
$$.001 = \frac{1}{2 \pi v^2}e^{-\frac{x^2}{2v^2}}$$.
 A: To expand on the suggestion to use the Lambert function, I'll show how it arises in your equation of interest.
Starting with
$y=\frac1{2\pi v^2}\exp\left(-\frac{x^2}{2v^2}\right)$
we multiply both sides by $-\pi x^2$ to give
$-\pi x^2 y=-\frac{x^2}{2v^2}\exp\left(-\frac{x^2}{2v^2}\right)$
which can now be inverted to the Lambert function (recall that the Lambert function $W(z)$ is the inverse function of $z\exp(z)$, $W(z)\exp(W(z))=z$):
$-\frac{x^2}{2v^2}=W(-\pi x^2 y)$
which we can now solve for $v$
$v=\frac{x}{\sqrt{-2W(-\pi x^2 y)}}$
The choice of sign for the square root is motivated by the fact that variances are conventionally taken to be positive.
A: I don't think the radius has to involve the variance. You could just use the radius to scale the x-axis. 
So you could sample the distribution from -1 to 1, always. If the radius is 4, would sample 9 regions. For 10 you sample 21 times between -1 and 1. The variance can stay the same. 
You might want to try this question here too https://gamedev.stackexchange.com/. 
