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12th September 2019, 03:56 AM  #1 
Illuminator
Join Date: Jul 2004
Posts: 4,877

Of caps and belts
A belt placed round a sphere has area 2πrh where r is the radius of the sphere and h is the width of the belt.
A cap on a sphere, i.e the intersection of a sphere with a cone with the point at the centre of the sphere, of height h, has to an area of 2πrh. I'd like a method to show the equivalence of these two geometrical areas without resorting to deriving them both separately. Something which is more intuitive for those without a mathematical bent. Anyone any ideas? 
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12th September 2019, 05:30 AM  #2 
Hyperthetical
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For the belt, h is not the width of the belt (which implies some sort of measurement along the surface of the sphere between the edges of the belt), it's the height. That is, it's the perpendicular distance between the two parallel planes whose intersections with the sphere define the edges of the belt.
The cap is simply a special case of the belt, where one of the parallel planes intersects the sphere at a point instead of a circle. 
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12th September 2019, 07:26 AM  #3 
Illuminator
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12th September 2019, 10:22 AM  #4 
Penultimate Amazing
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No, the belt has neither height nor width, it has "percent of the surface of the sphere". It is shperical, but narrow.
So your formula probably needs to calculate the areas of two partial spheres, subtracted from the total area of the whole sphere to give 'belt' area. Pencil that out and the relationships of the two formulas will be obvious? 
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12th September 2019, 10:27 AM  #5 
Penultimate Amazing
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If you think you understand noneuclidean geometry, you have missed something.

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12th September 2019, 10:30 AM  #6 
Hyperthetical
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Hmm, okay. For clarity I'm going to say "cylinder" for what you're describing, and "belt" for the surface of the sphere between two parallel planes that intersect the sphere. I think you might be able to get an intuitive sense of the equivalence, by using the belt as an intermediate case between the cylinder and the cap. Imagine starting with a cylinder around the equator, where h = r/3 (approx.). A few things can be visualized from that: 1. In order to turn the cylinder into a belt of the same h, the rims of the cylinder have to be "pulled in" to be snug with the sphere, which would tend to shrink the area. At the same time, the h of the cylinder gets "wrapped around" the curve of the sphere and therefore has to be extended to reach the latitude of the belt's h, which would tend to increase the area. You can see that these tendencies offset one another, and both tendencies become larger as h increases, though it's not obvious that they offset exactly. 2. When h << r, the difference between the belt and the cylinder becomes small. It's clear intuitively that the belt approaches the cylinder as h approaches 0. ETA: This is clearest, of course, when the belt is around the equator. But you can also look at a cylinder with the same r as the sphere, positioned at other latitudes, where the belt becomes close to a perpendicular slice of a cone when h is small. With a bit of trig you can calculate how much smaller the circumference of the cone slice is at that latitude, and how much wider than h the actual width of the belt along the conical surface is, and see that those factors are inverse. 3. At the opposite extreme, when h = 2*r, the formula for the area of the cylinder becomes the formula for the area of the sphere, which is also the area of the "belt" when the belt goes all the way from pole to pole. This strongly suggests that the two tendencies mentioned above (contraction of longitude and stretching of latitude) do in fact offset exactly, though it's not proof. Once the equivalence of the belt and the cylinder is accepted or assumed, the equivalence of the cap is also established as a special case of the belt. I don't know any intuitive proof that doesn't go through deriving both formulas or using trig functions. It would be nice if there were some way of cutting up a paper cylinder in a way that makes it obvious that the cut pieces would paper the cap perfectly. I don't know of any. 
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12th September 2019, 03:41 PM  #7 
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Thanks Myriad. That solidified the idea for me that it comes down to the two tendencies and that they compensate for each other during the transformation from belt to cylinder.

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13th September 2019, 10:01 AM  #8 
Hyperthetical
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You're welcome. I'm glad that explanation worked in the end, after a bit of fumbling around on my part.
Putting the argument together a bit more carefully, I get something like this: Imagine you have a paper cylinder of radius r, and a globe also of radius r. The cylinder is of some height h, and it's positioned around the globe with the center of the globe somewhere along the line of the center axis of the cylinder. The cylinder may or may not intersect the globe (that is to say, the intersection of the globe's center with the line of the cylinder's axis may or may not be inside the cylinder). If it does, then the intersection of the globe and the cylinder will be a great circle of the globe, which we can visualize as the equator. Likewise, in any case, we visualize the cylinder's axis line as also being the polar axis of the globe, and the two points where that line intersects the surface of the sphere as the globe's poles. The cylinder itself may not extend beyond either pole. (Thus, h<= 2*r.) Consider the portion of the globe's surface lying between the ends of the cylinder. (If one end of the cylinder is at a pole, and the other end does not extend past the equator, then we can call that portion a cap. The following exercise works the same whether it's a cap or not.) Our task is to use the paper of the cylinder to exactly cover that portion of the globe surface. It appears that an unseemly amount of cutting and rearranging of the paper might be required. But fortunately, the cylinder is made of special paper that can be stretched as needed. We still have to cut the paper as well, but we can do it in a straightforward way. We slice the cylinder, with cuts perpendicular to the axis, into a large number of much shorter cylinders, or rings, all of equal height. If we make a lot of cuts, as we should, the height of each ring will be very small compared to r. (Thus, when talking of a specific ring, it will make more visual sense to talk about the "width" of the ring rather than continuing to say "height.") Where the planes of our cuts intersect the sphere, we draw lines on the sphere similar to latitude lines dividing the surface into bandshaped zones. Each zone has an equal height, measured parallel to the axis. But measured along the globe's surface, zones closer to the equator will be narrower (less range of degrees of latitude) than zones closer to the pole. The key point is now this: each ring, when correctly stretched (preserving its area), will approximately cover the corresponding zone of latitude. How do we have to stretch a ring to make it almost fit? Three ways. One way is we have to "shrink" the radius of the ring (unless the ring's zone is centered on the equator) so that it's the right size to go around its zone. The farther the zone is from the equator, the more we have to shrink the ring to get the best fit for the zone. The amount of radial shrinking needed changes slowly at first moving farther from the equator. It doesn't reach a factor of .5 (decreasing the ring's radius by half) until we're about 87% of the axial distance from the equator to the pole. But if the ring is very near the pole, we will have to shrink it a lot. The second change we have to make is to "tilt" the ring so that it lies flat, or as close to flat as possible, on its zone on the surface of the globe. Tilting in this case means turning the ring, which is a short section of a cylinder, into a short section of a cone instead, while still keeping the same average radius. (The radius of one edge of the ring, after the "tilt", decreases and the radius of the other edge increases by the same amount.) This will also change the area of the ring, but only by a little bit. We don't have to tilt a ring that's going to go around the equator, but the farther the zone is from the equator, the more we have to tilt it to get it to lie flat. The amount of tilting, expressed as an angle, changes slowly at first moving farther from the equator. At about 71% of the axial distance from the equator to the pole the tilt reaches 45 degrees. If the ring is very near the pole, the tilt gets very close to 90 degrees. The third change we have to make is to "widen" the ring. Recall that zones closer to the poles become wider when measured along the globe's surface, while the rings start out as all equally wide. So we have to stretch some or all of the rings to make them wider. The farther the zone is from the equator, the more we have to widen the tilted ring to fill in the full width of the zone. The amount of widening needed changes slowly at first moving away from the equator. It doesn't reach a factor of 2 (doubling ring's width) until we're about 87% of the axial distance from the equator to the pole. But if the ring is very near the pole, we will have to widen it a lot. When we sum up (or more accurately, multiply out) the effects of these changes on the area of each ring, we see that the shrinking of the radius and the widening of the width follow the same pattern, being nearly zero near the equator, increasing slowly at first moving away from the equator, and then increasing quite rapidly near a pole. The tilting doesn't change the area significantly. So it's plausible (and in fact true) that those two factors complement one another, leaving the area of each ring unchanged as it transforms to cover its corresponding zone. (As I mentioned, to actually show this requires basic trigonometry.) You can stop there, but... The other question to consider is that our papered spherical surface, even with our magically stretchy paper, is going to end up imperfect. Does that matter? Even tilted, the rings won't lie perfectly flat on the curved surface of the globe. Furthermore, tilting a ring does change its area a little bit. So can we get from "will approximately cover the corresponding zone..." to "exactly cover"? It turns out we can, because all those irregularities and imperfections decrease when the rings are thinner; that is, when we increase the number of cuts to create a larger number of rings. Given any specification of how close to perfect our papering of the sphere surface has to be, there's some number of cuts/rings we can use that will produce a result that's within that specification. Another way of saying the exact same thing is that the limit of this process, as the number of rings increases without bounds (because there's nothing about the process that puts any upper limit on how thin we can make the slices), is a perfectly papered cap or belt on the sphere. And I'll leave you there, teetering on the edge of calculus. 
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13th September 2019, 11:19 AM  #9 
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Let's take your cap formula
A = 2 pi r h. If h = r, the result is a hemisphere, with the area obviously being 2 pi r^2, right? Now take h > r, and rewrite it as h = r + H Then the area will be 2 pi r (r + H), or 2 pi r^2 + 2 pi r H The first term is the area of the hemisphere from the pole to the equator, and the second term can be considered half of a belt as described in your belt formula. A complete belt will have twice this area, and will be 2 pi r (2H). 2H is the width of the complete belt, or h in your belt equation, so Area = 2 pi r h. 
13th September 2019, 12:58 PM  #10 
الشيطان الأبيض
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For a nifty video illustrating what I think sounds like the same problem in moving 3D, go to the YouTube channel "3 Blue 1 Brown" & search his videos for "sphere" & "cone". ( I can't find & link to the exact one right now.)

13th September 2019, 01:04 PM  #11  
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You beat me to it. I was thinking the exact same thing. The video is this one.


13th September 2019, 02:18 PM  #12  
الشيطان الأبيض
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I had this one in mind, because he puts a cone on a sphere in this one. (The one you linked has no cones.) It turns out that he's actually addressing a different problem in this one too, and putting the cone on the sphere is just part of how he gets at that other question. Still, maybe a couple of 3B1B videos that just seem loosely somewhere near the subject might help indirectly by prompting one to approach this question in a 3B1Bish way.


13th September 2019, 02:37 PM  #13 
Illuminator
Join Date: Jul 2004
Posts: 4,877

Thanks folks,
The 3B1B video fills in Myriad's stretching argument, and basically shows it's necessary to do a bit of precalculus or whatever the young people are calling it these days. Cheers. 
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13th September 2019, 03:14 PM  #14 
Hyperthetical
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TIL if you slice a spherical tomato, such that each slice is equally thick, each slice has the same amount of skin.
The first 3B1B video managed to prove that "stretch * squish = 1" without using any trig functions after all. Similar triangles FTW. That guy's good at this! You do have to know basic SAT level geometry, though. 
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14th September 2019, 10:29 AM  #15 
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In simplest terms, when my pants are too loose and I'm threatened with making a fashion statement that would be unbecoming for one of my generation, I put on a belt capable of maintaining their proper location above my posterior. If it's overtly sunny or cold, I put on a cap best designed to protect my bald head.

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