First, thanks Darren and Thorsten. No sooner had I posted the message saying
how weird it was then the problem resolved of its own accord. Go figure.

Le_Forgeron <lef### [at] freefr> wrote:
> Le 04/10/2010 15:05, Graham a écrit :
>
> The problem is solving the equation of a hyperbolic paraboloid with ABCD.
> Notice that the points A, B, C & D are not enough (4 non-coplanar points
> do not fix an hyperbolic paraboloid), but you can add that each points
> in AB, BC, CD & DA must also be on the surface.
So, if three non-collinear points determine a plane, then how many for a HP?
Surely this isn't new ground.
>
> >>
> >> Hint: transition from #2 to #3 seems easy. But How do you evolve from #1
> >> to #3 ?
> > I'm guessing the best answer would involve Mesh2(to preserve CSG, and given
> > what I'm already thinking about the surface it would be relatively simple to
> > generate a number of points), but is there a better non-Mesh solution?
> I was not speaking from the POV point of view, but from the mathematical
> surface point of view.
> The case #3 to #2 is a degeneration of the HP into a plane. no real
> discontinuity, as both are conic with 1 fold.
> But I believe you have to sacrifice a fold when coming from #1.
>
> Have you look at
>
> http://mathworld.wolfram.com/SkewQuadrilateral.html
>
Yes. I mocked up this based on the alternative z=xy,
where (-1,-1)<=(x,y)<=(1,1)
http://www.youtube.com/watch?v=ik6Kty1HSUw
The protrusions of the tips through the two triangular sides should
give an indication as to how this SQ was constructed.

I enjoy math, and from memory my maths of a surface got as far as grad,
div and curl, but I have no experience with Abelian Integrals. That was
half a lifetime ago...

I'm prepared to learn, but I thought you should be aware of my starting point.

Post a reply to this message