[help-3dldf] Fwd: Re: [metafont] Re: button-hole problem

[Laurence Finston]

------ Forwarded message -------


From: Dan Luecking <address@hidden>
To: address@hidden
Date: Tue, 26 Apr 2005 14:23:44 -0500

At 02:17 PM 4/25/2005, you wrote:
>However, I'm still a bit at sea about the problem of finding equations
>(implicit and/or parametric) for the projections

You have not made it clear what the input data is (i.e., how a particular
conic section is to be described). Perhaps you can get something out of the 
following:

A conic section is a plane figure, so I will assume you have such figure
in some plane P_0. Now you want to project it through a focal point F onto
another plane P_1. One can rotate, translate and scale the 3d coordinates
so that the initial plane P_0 is taken to the xy-plane and the focal point F
is taken to (0,0,1). This transforms the conic section in P_0 to one in the
xy-plane with some formula Q(x,y) = 0, where Q is quadratic function in x
and y. (And the target plane P_1 has been transformed to some new plane P_2.)

Projection involves taking all possible lines containing (0,0,1) that pass
through points of the conic section and finding where they intersect the
transformed target plane P_2. The set of all such lines form a surface of
which the equation is
   (1-z)^2 Q(x/(1-z), y/(1-z)) = 0.
This is a quardatic function in x, y and z. If Q(x,y) can be explicitly
found, so can this surface. The projected conic section is the intesection
of this surface with P_2. This could be found using the equation for P_2
to replace one of the variables in the above surface equation. Or one could
transform coordinates again so that P_2 coincides with the xy-plane and
the above surface formula becomes some new formula R(x,y,z) = 0. The new
conic section has equation R(x,y,0) = 0. Now reverse all previous
transformations to get the new conic.

For example, suppose the original focus is already (0,0,1) and the initial
plane P_0 is the xy-plane. Suppose the conic is the circle center at (1,0)
with radius 1:
   x^2 - 2x + y^2 = 0
Suppose we want to project this to the plane P_1 which is perpendicular to
the x-axis at x=2. The projection surface is then
   x^2 - 2x(1 - z) + y^2 = 0
We can get the intersection of this surface with P_1 by putting using the
equation of P_1 directly, I.e., put x=2 in the equation of the above surface
to get
   4 - 4(1 - z) + y^2 = 0
i.e.,
    z = -y^2/4.
This (along with the equation of the plane x=2) specifies the projected
conic section. Notice that this projection of a circle is a parabola.

>and finding the intersections
>of conic sections.

I guess here you mean solving two simultaneous equations, each of which is
a quadratic in x and y. I'm sure there are well know numerical recipes for
that sort of thing. I don't believe there are any directly usable formulas
for the intersection points.


Dan


Daniel H. Luecking
Department of Mathematical Sciences
University of Arkansas
"Be kind. Every person you meet is fighting a hard battle." - Anon.