Postscript: a wrinkle concerning planar orientation. Four of the eight cuboid corner-slice planes comprising one tetrahedron form the two pairs of base planes required to specify each pencil correspondence. If these are oriented consistently (positive constant component 1), the correspondence must map s E + F -> s G - H to generate the required one-sheet hyperboloid. Ellipsoids are generated by reguli of lines in complex space: setting z -> z \i, s,t -> u +/- v \i in the hyperboloid gives ellipsoid x^2/a^2 + y^2/b^2 + z^2/c^2 = w^2 parameterised by Q(u, v) = [w,x,y,z] = [u^2 + v^2 + 1, 2*a*u, b*(u^2 + v^2 - 1), 2*c*v] . Fred Lunnon On 11/25/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Implementing the method I sketched earlier for a general one-sheet hyperboloid in canonical position x^2/a^2 + y^2/b^2 - z^2/c^2 = w^2 makes it easy to locate lines missing from reguli using symmetry. Four lines from each regulus are face diagonals of the cuboid with corners [+/- a, +/- b, +/- c], and base planes for the correspondence between pencils through two of them can be selected from corner-slices +/- x/a +/- y/b +/- z/c = 1 .
The result is gratifyingly elegant: a general point P(s, t) on the surface has homogeneous coordinate vector
[w,x,y,z] = [s*t+1, -a*(s+t), b*(s*t-1), c*(-s+t)] .
It is trivial to check that P satisfies the (homogeneous) quadric equation; and being linear in s and t separately, P describes a line when either parameter is fixed. A missing subset of measure zero is reached by letting s -> oo or t -> oo .
Showing their line intersection sweeping out the surface as the two moving planes rotate about their pencil axes in synchrony might make a worthwhile demonstration?
Fred Lunnon
On 11/23/11, James Propp <jamespropp@gmail.com> wrote:
Thanks!
It occurs to me to ask: Is there a parametrization of the one-sheeted hyperboloid that makes it manifest that it's a ruled surface?
And: Is there a parametrization that makes it manifest that it's a doubly-ruled surface?
Jim