Otherwise the it cannot be solved in terms of elementary
(high-school) functions.

Best regards.

Put your known variables into Excel in cells A2:A5 (F,Y,D,V in that order – just because that’s the order they cropped up in the formula), you’ll have to make something up for D, then in A1 type the formula in:
=(3.07*A2*A3*(1+1.4*(A4/A5)*EXP(-2*A4/A5)))/(A4^2)

With A1 highlighted, go to Tools > Goal Seek. Set “to value” to what you want Q to be, and “by changing cell” to A4, and it will give you an approximation to about 3 or 4 dp.

“People are, in this case, correct. Variables in a linear term and an exponential term are rarely solvable algebraically.“

Jason’s code iterates until two consecutive values of the sequence (x0 & x1) are close enough (10^-6 in his code) to be considered a solution X. But X=-2*D/V (see comment #25) so D=-X*V/2

Daniel and Jason are both relying on a magic trick, the “iterative solution”: find an alchemy turning a good guess into a better guess. If you find such an alchemy and invoke it on a reasonable guess many, many times, you can find a result as close to the true answer as you have patience and bits.

Here is a working python function:

import math
def distance_for_luminance(Q, F, Y, V):
# set up some stuff
c = 3.07 * F * Y / Q
a = 1.04 / V
f = math.exp(-2.0/V)
# start with a terrible guess
D = math.sqrt(a)
# A whole bunch of times ("more than enough"),
for i in xrange(10000):
# Turn the good guess into a better guess
D_new = (c / D) + a*c*(f**D)
# And adopt the better guess as our new value
D = D_new
return D

Perhaps by iteration, or perhaps from a whisper by the goddess Namagiri into your ear as you slept, just plug all values into both sides and see how far off it is.

@Meyers:
Regarding the iteration, think of it as making a guess (x_0), then using that guess to make a *better* guess (e.g. going from x_n to x_{n+1}).

So:
var x0 = -Math.sqrt(a);
var x1 = -Math.sqrt(a*(1 – 0.7 * x0 * Math.exp(x0)));
while (Math.abs(x1-x0) > 1e-6) {
x0 = x1;
x1 = -Math.sqrt(a*(1 – 0.7 * x0 * Math.exp(x0)));
}
alert(x1);

That would be some JS to quickly calculate my re-parameterized value of x. (Of course, x=-2*D/V, so you are actually interested in D=-x1*V/2.)

[Fished this version of the comment out of the spam filters and restored it. Sorry about that, Jason! -E.]

And Daniel’s i is just to express that repeating the assignation below 1000 times you’ll get an accurate enough solution

Like, at the end of your comment, you do a “for i in” operation with no actual i in the equation, which has the solve-for term D on both sides of the equation. At which point my brain shorts out. Sorry.

I think maybe all the brain cells that would normally go toward understanding that kind of thing are currently devoted to holding a functional copy of the CSS line layout model.

You can’t avoid iteration. But the recurrence I already suggested is way simpler than the bisection method. Here is the core of it:

c = 3.07/1.4*F*Y*V
d = -Q*V/1.4
f = math.exp(-2.0/V)
D = 0.0
for i in xrange(1000) :
D = -c/(d*D+f**D)

Or try Jason’s recurrence, it’s mostly the same.

When D = 0, Q is undefined.
When D > 0, Q is positive.
When D < 0, Q is negative.

I don’t know where you got your equation, but the derivative of the equation at the top of the post is undefined at D = 0. Therefore, the function is not continuous.

The function may be solvable for only positive values of D, but I have no idea how to do that.

Incidentally, for those who are curious:

Q = radiant energy (cal/cm^2)
F = thermal partition (unitless)
Y = yield (KT)
V = visibility (miles)
D = distance (miles)

So it’s easy (relatively speaking) to calculate the radiant energy at a given distance. I had assumed that the inverse, finding the distance at which one find a given radiant energy, was a matter of reshuffling. Kind of frustrating to find that’s not so.

Thanks for all the effort and input, folks!

QV = (3.07FYV / D^2) + ((4.298FY / D) * (e^-2)^(D / V))

or if you prefer

Q = (3.07FY / D^2) + ((4.298FY / D^2) * (D / V) * (e^-2)^(D/V))

The exponentials are getting to me :(