I want to calculate a light intensity.

How would the light intensity on a plane, as radiated by a somewhat simplified infinitely long fluorescent tube be calculated?

The plane is parallell with the tube, so the light intensity changes along a line perpendicular to the tube. The radius of the tube should be taken into account. No absolute numbers needed, it's ok to assume that the intensity at the closest point is 1.0.




Note:
For line sources (tube radius = 0), I did it like this:

E(h, x) = h / (h^2 + x^2);

where h = height from plane to tube at closest spot, and x is displacement from the closest spot, along the line perpendicular to the tube.

If you have Maple, you can plot 6 0-radius tubes like this:

E := (h, x) -> h / (h^2 + x^2);
Es := (a, b, c, d, e, f, g, h, i, j, k, l) -> E(a, g-t) + E(b, h-t) + E(c, i-t) + E(d, j-t) + E(e, k-t) + E(f, l-t);
plot([Es(5, 5, 5, 5, 9, 9, -4, 4, 12, 20, 4, 12), 0], t=0..16);

[quote author=Adron link=board=36;threadid=7218;start=0#msg64841 date=1087058772]
E(h, x) = h / (h^2 + x^2);
[/quote]
Hmm, is that right? I would expect E(h, 0) = 1.
Maybe a sqrt is needed in the denominator?

Anyway, studying history for tomorrow's final now. I'll think about it tomorrow if nobody else solves it.

[quote author=Yoni link=board=36;threadid=7218;start=0#msg64852 date=1087061018]
[quote author=Adron link=board=36;threadid=7218;start=0#msg64841 date=1087058772]
E(h, x) = h / (h^2 + x^2);
[/quote]
Hmm, is that right? I would expect E(h, 0) = 1.
Maybe a sqrt is needed in the denominator?
[/quote]

Well, it isn't normalized. When I was doing this, I was a bit lazy, and just scaled the plot of the function.

The light drop-off with distance should become 1/distance for the infinite length tube. That's 1 / sqrt(h^2 + x^2). In addition to that, there's a reduction because the rays will be hitting the plane at an angle. That reduction is cos(angle of incidence) or h / sqrt(h^2 + x^2). To normalize, a multiplication by h should do the thing, making the whole equation h^2 / (h^2 + x^2) for a single tube.

However, when I start adding more lights, the normalization has to be the same for all, and so the expression for the normalization constant would become rather complicated.

I discussed the problem with Adron (on MSN) and found a nice solution, for the sake of communism I will share it here.

My calculations:
http://yoni.valhallalegends.com/stuff/Light1.jpg (215kb)
http://yoni.valhallalegends.com/stuff/Light2.jpg (106kb)

(The result in Light2.jpg fixes a mistake in Light1.jpg.)

Here is the Mathematica 5 processing of the last formula, showing a graph of the light intensity as a function of the distance from the center, for R = 0.8 and h = 4.5, with one light source, two light sources, and three light sources.

http://yoni.valhallalegends.com/stuff/Light.png (27kb)

Good luck with the construction :)

[quote]C = some constant, I guess[/quote]

Har har har.

[quote author=j0k3r link=board=36;threadid=7218;start=0#msg65182 date=1087219028]
[quote]C = some constant, I guess[/quote]

Har har har.
[/quote]I was serious, it's the maximum light intensity constant.