A string is wound around a cone with base radius 1 cm and slant height 10 cm. The string is wrapped in a helical manner, as if a strand of lights decorating a Christmas tree, except that it is pulled taut around the cone so as to have no local "wiggles", thus making what is known mathematically as a
geodesic curve on the surface of the cone.
What is the maximum length of such a spiral around the cone before it starts to intersect itself?
Since the cone is developable, its surface can be laid out flat, and the geodesic will be mapped as a straight line. As it continues more than just once around the cone, multiple copied of the cone, meeting at their mapped vertices and attached edge to edge can be placed so that the full spiral can be mapped out, as a straight line segment.
The cone's base has a circumference of 2*pi. As its slant height is 10, when laid out flat it has a radius of 10. Its full circle therefore has a circumference of 20*pi, and the arc of one occurrence of a map of the cone occupies 1/10 of the full circle, or 36°.
The line will start at the mapped base, on the circumference of a 1cmradius circle, and head off at some angle to the vertex. When that line segment grows to a point where, together with the radius from the starting point on the circumference to the vertex, it forms one leg of a right triangle with that radius as a hypotenuse, that will comprise the complete upward portion of the spiral. From then on, as the segment extends further, its leading end point gets farther from the vertex, and is therefore headed downward on the original cone that is being mapped here.
When the endpoint has subtended a further angle at the vertex equal to half the total arc of one laidout (flattened) cone, the endpoint will be at the point of self intersection due to the symmetry of one full loop around the cone about the highest point on the spiral. In this case, that's 18° farther.
Therefore the portion of the spiral leading up to the highest point on it has length 10*cos(A), where A is the angle the spiral path makes with the radius of the mapping circle at the starting point. But the line continues beyond this, to subtend another 18° as seen by the vertex. The leg from the vertex to the point already reached (the high point of the spiral) is 10*sin(A), and this must be multiplied by tan(18°) to get the length of the extension to the point of selfintersection.
The following program originally had a greater span of angles (x here represents angle A mentioned above), but that has been narrowed down to near the sought maximum:
DEFDBL AZ
PRINT
pi = ATN(1) * 4
dr = pi / 180
FOR x = 17.99# TO 18.01# STEP .001#
PRINT x, 10 * (COS(x * dr) + SIN(x * dr) * TAN(18 * dr))
NEXT
PRINT
PRINT 1 / COS(18 * dr)
The significant portion of the output is
17.99600000000001 10.51462221675918
17.99700000000001 10.51462222796946
17.99800000000001 10.5146222359768
17.99900000000001 10.5146222407812
18.00000000000001 10.51462224238267
18.00100000000001 10.5146222407812
18.00200000000001 10.5146222359768
18.00300000000001 10.51462222796946
which seems to indicate A = 18° reaches the maximum, of 10.51462224238267, which is therefore
10 (cos(18) + sin(18)tan(18))
using degree measure.

Posted by Charlie
on 20090806 17:38:33 