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Topic: DISCO balls (Read 563 times) |
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Noke Lieu
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DISCO balls
« on: Jul 26th, 2004, 11:28pm » |
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Haven't had time to polish this up, doesn't look like I am going to get time either. so here it is. Rough and ready.
Disco balls. Glitter balls. Call them what you like. You can buy cheap ones. expanded polystyrene balls with mirrors glued on them.
Lets say a 10 cm ball. Mirrors are 1 cm square. Bevelled so that the squares touch perfectly, and remain squares (or treat them as zero thickness)
Obviously, they are a bit gappy- but next time you see a cheap one, you'll see that's the way it happens.
Soooooo....
How many mirrors can you fit on? What is the difference between having one mirror at the poles and having three or four nestled around the pole?
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Eigenray
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Re: DISCO balls
« Reply #1 on: Jul 27th, 2004, 8:00am » |
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Let R be the ratio of the radius of the sphere to half the side length of the mirror.
Around a great circle of the sphere, one mirror takes up an angle
[alpha] = 2tan-1(1/R).
Thus we can get floor(2[pi]/[alpha]) mirrors around the equator.
Now we add circles of mirrors centered at a latitude k[alpha]. The distance from the center of such a mirror to the axis of the sphere is Rcos(k[alpha]). However, the mirror tilts towards the axis, so the effective radius is
rk = Rcos(k[alpha]) - sin(k[alpha]).
The tops of the mirrors form a polygon with inscribed radius rk. Each one takes up an angle
[alpha]k = 2tan-1(1/rk),
and thus the number we can fit on the k-th circle is
nk = floor(2[pi]/[alpha]k).
We can form a circle as long as
rk = Rcos(k[alpha]) - sin(k[alpha]) [ge] 0
k [le] floor(tan-1(R)/[alpha]) = M
The total number of tiles with this scheme is:
N = n0 + 2[sum]k=1M nk.
So when R = 10 (5cm radius), this gives 281. If you mean a 10cm radius, then R = 20, which gives 1190 (compared to a surface area of ~1257cm2).
Of course there's no reason to think this is optimal.
Taking limits R[to][infty] to check this:
M ~ tan-1(R)/(2tan-1(1/R)) ~ [pi]R/4
nk ~ [pi]/tan-1(1/r(u)) ~ [pi]Rcos u,
where r(u) = Rcos(u) - sin(u) ~ Rcos u,
u = 2ktan-1(1/R) ~ 2k/R.
N ~ 2[int]0[pi]R/4 n(k) dk
~ R[int]0[pi]/2 n(u) du.
~ R[int]0[pi]/2 [pi]Rcos u du = [pi]R2.
Yay!
Of course, I brought a limit inside the integral, but blah blah uniform convergence blah blah or something.
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| « Last Edit: Jul 27th, 2004, 8:12am by Eigenray » |
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Noke Lieu
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Re: DISCO balls
« Reply #2 on: Jul 27th, 2004, 10:37pm » |
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Beauty. Was considering 5cm radius. Haven't had opportunity to tinker with it, so I don't know if changing what happens at the poles is significant.....
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| « Last Edit: Jul 27th, 2004, 10:39pm by Noke Lieu » |
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Nigel_Parsons
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Re: DISCO balls
« Reply #3 on: Sep 25th, 2004, 12:01pm » |
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To the sub-question " What is the difference between having one mirror at the poles and having three or four nestled around the pole?"
There is no difference. The poles (assuming you mean top/bottom, or North/South) are arbitrary points. You can start at any point on the shere with a single mirror, and work away from that. All other arrangements are just Reflections (groan)
Of course, if by 'pole' you mean the hanging support, then the size of the pole & whether it could be mirrored will affect the answer.
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