The power of geometry 
[Jun. 17th, 200907:59 pm]
Beemer

So like I said the other day, my friend Peter presented me with a problem when we had lunch. At first I thought it was some kind of complicated optics problem, but as I thought about it, I realized it's just geometry. And fairly simple geometry, at that. I thought I'd share, because the answer is a kind of neat "hooray for science!" puzzle. (Plus, well, I wanna let other people doublecheck my thinking  just to make sure I'm not being dumb...)
Also, I will get to use the word 'penumbra'.
Anyway, Peter works in solar, and when you're putting solar panels up, you want them to be in the sun, not in the shade. Now, an object that's far away is going to block less light than one that's close up, right? Eventually it gets so small it doesn't block the sun at all. So the question is, what's the equation that describes how much clearance you need to minimize your energy loss?
And the answer is: it doesn't matter, actually.
Here's a diagram. It is very, very not to scale. Sun's to the left, solar panel is to the right. The black lensshape is our occluder. (Shape is irrelevant; I just wanted something with sharp edges for clarity.) The sun symbol shows how the rays incoming from the two opposite edges of the solar disc are at different angles, although, again, very not to scale.
The purple region is the umbra, where the sun is totally blocked; the green region is the penumbra, where the occluder only partially blocks the sun; and the blue region is the antumbra, where the occluder is in front of the sun, but far enough away that it's too small to block it completely.
Now, with a little bit of trigonometry, we can figure out what the length of the shadow cone (the umbral region) is and work out how far point p is from the occluder. There are a bunch of parallel lines on the diagram, so all those marked angles are congruent to one another. More importantly, we know what the angle θ is: it's the angular diameter of the sun. That turns out to be about half a degree, or 1/108 radians.
That's small, so we can use the smallangle approximation: cos(θ)≈1 and tan(θ)≈sin(θ)≈θ . Thus, for the right triangle from p to the occluder, tan(θ/2) = (x/2)/y = (1/2)/108 , so y = 108 * x .
So if our occluder is something like a power cable with a width of 1 inch, the shadow cone is going to be 108 inches long. Give the solar panel a clearance of at least 108 inches, and it'll be entirely outside the umbra. And then you could do some calculus to figure out the incoming intensity at various points in the penumbra and the antumbra and... hang on, hang on. Something's not right.
Turn the problem around for a minute. Have a look at the occluder. How much sun does it block? No matter how close or far the solar panel is, the occluder is always going to be absorbing the same amount of light. So all that happens by changing the clearance is that you've changed how concentrated or spreadout the shadow is; it's always blocking the same amount of incoming solar power.
So assuming that the solar panel's efficiency is constant with the intensity of light (which is a big assumption that I will have to check), it doesn't matter how close the solar panel is. The only way to decrease the amount of light that the occluder is blocking is to have it so far away that the shadow is bigger than the solar panel, and part of it falls off the edge.
Back to the diagram, that means we're now looking at the u/v triangle, instead of the x/y triangle. Again, all those angles are congruent and you have the same width/height ratio, so the minimum distance for the shadow to be as big as the solar panel is 108 times the width of the solar panel (minus the length of the shadow cone).
So for a 1inchwide cable dangling over a 2footwide solar panel, you need to move it more than (216 feet minus 108 inches =) 207 feet away in order to have any effect at all. Which is so much more clearance than you'll ever actually have that it Just. Doesn't. Matter.
Isn't that interesting? Hooray for math! 

