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Monday, January 31, 2011
A trivial math question
I am not a mathematician, but I occasionally dabble in recreational mathematics. One thing I want to occasionally do with this blog is to throw out some observations or questions that occur to a layman like myself, someone who asks questions that are probably simple enough but to which I don't know the answer. These aren't meant as "brain teasers" (though sometimes they serve as such for me or for some of the kids I work with); I'm just leaving them out in case someone wants to point me to an obvious solution (or at least a method). I'm painfully aware that in asking this I make clear my own mathematical limitations, but there it is.
Everyone knows that the ratio of a Euclidean circle's circumference to its diameter is π:1. And these days, we also are familiar with the fact that Euclidean geometry does not transfer automatically to other surfaces. Thus, on a the surface of a regular 3-D sphere, the ratio of circumference to diameter changes depending on how big the circle is. You can visualize this most easily by imagining a globe. The "great circle" drawn on the globe by the equator has a diameter that is exactly half of its circumference; imagine the diameter drawing a line from one point on the equator, up through the north pole, and back to a spot on the equator 180° removed from the starting point. This diameter is demonstrably half of the circumference; you could slide it down the face of the globe till it coincided with half of the original circle. Thus, on a sphere's surface, this circle's ratio of circumference to diameter is 2:1.
But not every circle will have a diameter of half its circumference; only a circle as big as the equator (a "great circle") will have this. Any smaller circle (and no circle on a sphere's surface can be larger, of course) will have a circumference whose ratio to the diameter is larger than this. The smaller the circle, the larger this ratio. You can visualize this if you imagine shrinking the circle, starting at the equator and edging up closer towards the pole. The closer you get to the pole, the more you approximate the Euclidean plane, and so the closer the circumference:diameter ratio gets to π:1. (This is so for the same reason that the Earth looks flat if you are standing on it. A sufficiently small circle can leave the curvature of the sphere's surface out of account.)
This is what occasions my question. If this ratio starts (at the equator, or any great circle) at 2, and reaches π at the pole (or at any geometrical point), then there is a particular size of circle on the surface of a sphere where the ratio of circumference:diameter is precisely 3:1. Where is this circle? I.e., how big is it in comparison to a great circle on the same sphere? (Or, say, to the diameter of the sphere itself?)
What is interesting about this (to me) is that one can see, in the case of a great circle, why the ratio has to be 2:1. And you can just manage to stipulate that, since the point is where the sphere becomes a "virtual plane," so to speak, Euclidean values become valid there...well, because they do. (Just why Euclidean constants are as they are is a different question.) But in the case of the circle I've got in mind, there's not really an obvious reason why the integer 3 should obtain in that ratio just there.
I've no idea how many SCT readers are qualified to venture advice here for an amateur, but this is obviously not one of the really hard questions; it's just an interesting one that happens to have me slightly out of my mathematical depth. If anyone wants to pipe up, I'll be more than happy to hear from them.
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On a sphere the circumference C is C=pi*D*sin(d/D), where pi=3.14..., d is the diameter of the circle on the sphere, and D is the diameter of the sphere. To see if this formula reproduces the great circle C/d=2 result, let d=pi*D/2. Now C=pi*D*sin(pi/2) which is equal to pi*D, or 2*d.
ReplyDeleteIf you want to find the size of a circle so that C/d=3 (reading too much Old Testament?), then you have to solve the equation pi*D*sin(d/D)=3*d, or rearranging it: pi*sin(x)=3*x, where x=d/D. This, unfortunately, is a transcendental equation that isn't amenable to the usual methods of solving, but if you simply graph the lines y=3*x and y=pi*sin(x) and look where they intersect, you'll find that x=pi/6=0.52359... Thus, C/d=3 when d=pi*D/6 (i.e. Solomon must have been richer and more powerful than we thought!).
Is general relativity next on the agenda?
Grad,
ReplyDeleteThanks very much for this. I'm going to try to work it out on paper. If I bump into a problem, I'll mention it.
Clever of you to notice the Solomon connection... and funny you should ask. It is possible that a year or so ago when the puzzle first occurred to me, this was in my head (it's the sort of thing I might well have been thinking of, God knows), but when I wrote the post, that question had long since been forgotten. However, today as I was wondering whether anyone would take me up on this, it suddenly (re?)occurred to me that this would make a nice answer to the conundrum. (For those reading along at home, the issue at hand is that in 1 Kings 7:23, the "sea of cast bronze" in the Jerusalem Temple is said to be 30 cubits around and 10 cubits in diameter. This has given no end of merriment to that curious form of Biblical literalism which endeavors to "prove" that the Biblical God is a bungler.) Unfortunately, I don't think a curved diameter of 10 cubits and a circumference of 30 cubits makes a very deep "sea"-- certainly not "five cubits high." For all I know, there could be something to this solution, but I am not overly concerned to salvage the literal text of the Tanakh. I leave this to people who know it better than I and have a better sense of why it should be kept letter-for-letter. (Alas, of course, it isn't always just these who make it their business to "defend" it.)
General Relativity? Nah... I'm after the Riemann Hypothesis!
The Riemann hypothesis! Sounds like fun. Just to further comment on the original problem, I think it's interesting that when you choose an integer ratio of C:d, you get such a clean number as d/D=pi/6 as you do when C/d=2.
ReplyDeleteAlso, for some reason when I wrote my first comment, I only considered the possibility that the Sea of bronze's diameter was measured over the curved surface of Earth, not the idea that the curved surface was that of the Sea of bronze itself. Perhaps Solomon wasn't that wealthy and powerful after all ;)
(For related amusement, check out the Indiana Pi bill of 1897.)
"I think it's interesting that when you choose an integer ratio of C:d, you get such a clean number as d/D=pi/6 as you do when C/d=2."
ReplyDeleteI agree, and it sort of confirms my intuition that there was bound to be something worth attention in this particular case.
Thanks again for this. Trigonometry was exactly where I bottomed out, and I knew the solution was going to involve radians.