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Beemer

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The power of geometry [Jun. 17th, 2009|07:59 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 double-check 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 lens-shape 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 small-angle 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 spread-out 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 1-inch-wide cable dangling over a 2-foot-wide 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!
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Comments:
[User Picture]From: pink_halen
2009-06-18 04:50 am (UTC)
I'm happy to say that I didn't have to look up the word penumbra. It's not even the penultimate big word that I know.
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[User Picture]From: goobermunch
2009-06-18 12:15 pm (UTC)
If you were a lawyer, you could have used the word emanation with your penumbra.

--G
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[User Picture]From: annlarimer
2009-06-18 02:47 pm (UTC)
So...you got the same answer that I did ("Dude. It's just a cable."), only...mathier?
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[User Picture]From: dr_tectonic
2009-06-19 03:04 pm (UTC)
Mathier!
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[User Picture]From: annlarimer
2009-06-19 03:05 pm (UTC)
Solid.
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[User Picture]From: zubatac
2009-06-18 04:06 pm (UTC)
But have you considered diffraction at the edges of the object? >=)

[That was a joke, of course. For objects much larger than the wavelength of light, the amount of light affected by diffraction is small compared to the amount of light that's blocked by the object. You'd be better off considering what color your neighbor painted their house.]
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From: detailbear
2009-06-19 04:14 am (UTC)

Or as I learned in my astronomy course...

From http://www.handprint.com/HP/WCL/perspect6.html

the sun has a insignificant angular size (about 1/2° of visual angle) and a relatively small penumbra that grows smaller as the shadow casting object is closer to the shadow receiving surface, which means the sun is effectively a point source that generates parallel rather than divergent rays of light.

for any object at less than astronomical distances.

The difference between the θ/2 (~0.25°) and zero is not measurable by the human eye, and not significant for any man-made structure casting a shadow. (If I've done the math right, the shadow of an object would be about .1 inches wider per 10 feet of distance from the surface, regardless of the width of the object.) On the diagram above, the value of u-x will be approx. 1/1145 of v-y (2 times the inverse of the tangent of 0.25°).
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[User Picture]From: dr_tectonic
2009-06-19 03:03 pm (UTC)

Re: Or as I learned in my astronomy course...

The difference between the θ/2 (~0.25°) and zero is not measurable by the human eye

That's not correct, though. The sun has a visible disc. The moon is the same apparent size as the sun and you can easily make out features much smaller than half its width, right?

For the purposes of a drawing, solar illumination is parallel and point-source-like, sure, especially in comparison to other light sources, but the solar disc is still finite and gives you a penumbra.

It's easy enough to check experimentally: take something small outside and observe how its shadow changes as you move it away from the ground. I used my keyring, which is probably, what, a quarter inch? Sixth of an inch? It's hard to tell exactly where the umbra runs out, but the shadow is definitely completely fuzzed-out by two feet up. If the ratio were 1/1145, it would be at least 11 or 12 feet...
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[User Picture]From: annlarimer
2009-06-19 03:07 pm (UTC)

Re: Or as I learned in my astronomy course...

Wait. Are you saying you've stared at the sun to try to make out individual features?

Because that's really gonna fuck up your math.
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[User Picture]From: dr_tectonic
2009-06-19 03:11 pm (UTC)

Re: Or as I learned in my astronomy course...

OH GOD MY EYES!

THE GOGGLES, THEY DO NOTHING! (Well, aside from look cool.)
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