Update on Enneper-sphere intersections: For the nice tennis-ball curve with parallel osculating planes at its tips, the exact sphere radius is the root of (9r)^4 + 14(9r)^2 - 79 , ie. r ~ 0.2307718797455473 --- rather less than 1/4 . At the tips [x,y,z] ~ [0, +/- 0.1849108100, +/- 0.1380711874] , so the cuboid boxing the curve is somewhat flattened. For the extreme waisted curve with tacnodes at its tips, dividing the sphere into four teardrop regions, radius 1 turns out to be exact. Note that the usual parameterisation for Enneper's surface scales all coordinates involved up by a factor 3 . Magma script and Maple graphic are available on request. Fred Lunnon On 11/4/13, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I'd always casually assumed that the tennis-ball / baseball curve was probably some simple space quartic that everybody but me knew about. But since it seems this is not the case, I'll put in my two-penn'orth.
The classic minimal Enneper's surface intersects concentric spheres in a family of such curves. With the parameterisation in the Wikipedia page, a sphere of small radius meets it in an approximate circle; a sphere of radius (approx?) 1/4 meets it in a typical tennis-ball curve, with expected symmetry and arcs parallel at the extremities; a sphere of radius (approx?) 1 meets it in a curve with touching arcs; for larger radius the curve has 4 self-intersections.
See http://www.indiana.edu/~minimal/maze/enneper.html http://en.wikipedia.org/wiki/Enneper_surface
The degree of Enneper's surface equals 9, so presumably these curves have degree 18.
Fred Lunnon
On 11/3/13, Henry Baker <hbaker1@pipeline.com> wrote:
Is a tennis ball seam the same shape as a baseball seam?
http://math.arizona.edu/~rbt/baseball.PDF
"Designing a Baseball Cover"
Richard B. Thompson
College Mathematics Journal, Jan. 1998.
At 09:09 AM 11/3/2013, rkg wrote:
Dear funsters, A tennis-ball appears to be made from 2 congruent pieces of material, seamed together in a curve. Are possible equations to the curve known? I'd like a smooth algebraic equation, probably of degree 4, and preferably with a maximum number of rational points on it. A first approximation might be to take a sphere of radius root(3) and centre at (0,0,0) and take the 8 great circle arcs (1,1,1) to (-1,1,1) to (-1,-1,1), (1,-1,1), (1,-1,-1), (-1,-1,-1), (-1,1,-1), (1,1,-1) and back to (1,1,1). However, this isn't smooth at the 8 corners of the cube, and I think that it doesn't even partition the sphere into two congruent pieces.
Is this well-known to those who well know it? What do the tennis-ball manufacturers do? R.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun