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Puzzles Spiral Maths A cylinder is 12 inches high and has a circular base with radius 2 inches. A string is wrapped around this cylinder in a spiral fashion, as in the diagram at right.
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It begins at a point on the upper edge and ends at the point directly below this on the lower edge, and it circles the cylinder exactly three times.
Your puzzle is this: what is the shortest length of string which can make such a spiral? | |
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Comments | terveloc |  09:57:13, 18 Nov 07 |  Newbie
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| This can be solved by the integral int( sqrt( r ^2 + (dr/dTheta)^2 + (dz/dTheta)^2 ), Theta = 0 to Theta_final). Since r = 2, and the thing circles 3 times to bring us to z=12, this becomes int(sqrt(4+(2/Pi)^2), Theta = 0 to 6*Pi), which simplifies further.
Another method is to just realize that each full rotation can be broken down into a right triangle with base 4Pi and height 4. So, by Pythagorean Theorem, each curve rotation has the hypotenuse length 4*sqrt(1+Pi^2). There are three such triangles, giving a total length of 12*sqrt(1+Pi^2).
Both methods work equally well, but with trade-offs. The integral method is a simple plug and chug integral, but requires advanced tools (e.g. calculus). The Pythagorean method is mechanically easier, but requires you to be a bit more clever in setting it up.
| | | | terveloc | 09:48:17, 18 Nov 07 | Newbie
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| 12*sqrt(1+Pi^2)
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