In principle, an answer is that the lines of a regulus have Pluecker coordinate 6-vectors which are linear combinations of any 3 distinct fixed members. Since these coordinates are projective / homogeneous, and a line must also satisfy the (quadratic) Grassmann constraint L ^ L = 0, there is only one degree of freedom left over to generate the real regulus. Given a pair of conjugate reguli, the points can be now generated as intersections of pairs of lines, one from each regulus. Each point is specified by two pairs of (inhomogenous) parameters, each pair subject to a quadratic constraint: I think it must be straightforward to parameterise the solutions, though I don't know the answer offhand. 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
On Tue, Nov 22, 2011 at 5:11 PM, Joshua Zucker <joshua.zucker@gmail.com>wrote:
The Exploratorium has this exhibit but I was only able to find still photos of it in my searches. They call it "Hyperbolic Slot". Then I found http://www.youtube.com/watch?v=OJGkdO9gcdY which isn't gerat but should suffice.
--Joshua Zucker
On Tue, Nov 22, 2011 at 1:58 PM, James Propp <jamespropp@gmail.com> wrote:
Is there a video on the web of a line segment passing through a hyperbola-shaped slit, as in the Eames Mathematica exhibit (and probably other math museum exhibits as well)?
I want to show this to my calculus students tomorrow, and was astonished that I couldn't find it on the web even after several minutes of searching ("... Eames ruled surface hyperbola ...".
Thanks,
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