Inverse square falloff is the same for any scale - units don't affect the relationship. So no need for concern on that.

Want proof?

Letting I be the measured intensity at some pre-defined distance d expressed in some arbitrary unit u, the resulting intensity taking inverse square falloff into account is:

f(x) = I(du/xu)^2

It should immediately be apparent that u/u cancels out, so the equation is:

f(x) = I(d/x)^2

See? Units don't matter as long as the reference distance, d, and the rendering distance, x, are in the same units.

Now let's examine the relationship (ratio) between the intensity at some arbitrary distance a, versus some arbitrary distance b:

f(a) / f(b) = (I(d/a)^2) / (I(d/b)^2)

It should be apparent that I / I cancels out. This tells us that the ratio is not affected by the actual intensity of the light - the ratio will be based on the distances a and b alone.

f(a) / f(b) = ((d/a)^2) / ((d/b)^2)

Further rearranging:

(since these are not imaginary numbers, it is safe to factor out the exponent)

f(a) / f(b) = ((d/a) / (d/b))^2

(exchange division with multiplication by the inverse)

f(a) / f(b) = ((d / a) * (b / d))^2

(multiply the fractions)

f(a) / f(b) = (db / da)^2

(d / d cancels out - this means we don't even care what the reference distance is)

f(a) / f(b) = (b / a)^2

Voila - the ratio of light intensity at two distances is equal to the inverse ratio of those distances, squared - i.e. inverse square with no interest in the original intensity or what the distance was that you measured that intensity

 

 

 

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Renderosity forum reply notifications are wonky. If I read a follow-up in a thread, but I don't myself reply, then notifications no longer happen AT ALL on that thread. So if I seem to be ignoring a question, that's why. (Updated September 23, 2019)