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Conic Section of an Elliptical Cone

Date: 08/01/2007 at 22:25:41
From: Tom
Subject: conic section of an elliptical cone

If a flashlight with an elliptical beam is shining on the wall at a
slant, is the spot elliptical and, if so, how can the parameters be
computed?

This problem involves a lot of parameters.  There is the semi-major
and semi-minor axes of the ellipse, the orientation of the semi-major
axis, and the angle of incidence.

I solved this numerically by curve-fitting the boundary of the spot to
an ellipse.  I used a hundred points.  It worked OK, but I'm not too
happy about it.  There ought to be a formula for it.



Date: 08/02/2007 at 06:18:47
From: Doctor George
Subject: Re: conic section of an elliptical cone

Hi Tom,

Thanks for writing to Doctor Math.  The intersection is only an 
ellipse if the center of the beam intersects the wall.  Otherwise you
get a different conic section.

I think we need to model the light beam as a cone.  I'll assume that
the cone can be elliptical for more generality.  Start by putting a
coordinate system at the origin of the cone so that the cone is in
standard position.

Now write the equation of the plane in that coordinate system as follows.

  a1 x + b1 y + c1 z + d = 0

We can also write it like this.
             _   _
             | x |
  [a1 b1 c1] | y | + d = 0                            (1)
             | z |
             -   -

Now let
  _   _     _   _
  | x |     | u |
  | y | = Q | v |                                     (2)  
  | z |     | w |
  -   -     -   -

where Q is an orthogonal matrix such that

  [a1 b1 c1] Q = [0 0 m]                              (3)

where m is the length of [a1 b1 c1].  The strategy is to simplify the
problem by rotating the cone and the plane so that the plane is
horizontal in the coordinate system.

Substituting (2) into (1) and utilizing (3) gives us

  w = -d / m

Now if the cone is

  x^2/(a2)^2 + y^2/(b2)^2 = z^2/(c2)^2

we can write it like this.
          _                         _ _   _
          | 1/(a2)^2   0      0     | | x |
  [x y z] |   0    1/(b2)^2   0     | | y | = 0       (4)
          |   0        0  -1/(c2^)2 | | z |
          -                         - -   -

Substituting (2) into (4) and utilizing (3) gives us
                 _                         _   _       _
                 | 1/(a2)^2   0      0     |   |   u   |
  [u v -d/m] Q^T |   0    1/(b2)^2   0     | Q |   v   | = 0
                 |   0        0  -1/(c2)^2 |   | -d/m  |
                 -                         -   -       -

Carrying out the multiplication gives us a general quadratic in u and
v that is the equation of the intersection.  If the axis of the cone
intersects the plane at a steep enough angle the result will be an
ellipse.  An shallower angle will lead to other conic types.  [The
breakdown of the cases is more complex that what I indicated in a
previous message.  The complement of the half angle of the cone
appears to be the critical angle value.]

A rotation and translation in the (u,v) plane will lead to the 
standard form of the intersection in (u',v') coordinates.  Once you
find all of the points and vectors of interest in (u',v') coordinates
and then (u,v,w) coordinates, you can multiply by Q to find them in
(x,y,z) coordinates.

All that remains is to find Q.  There are a number of ways to find
such a matrix. That can be left for a separate discussion if need be.

Does that make sense?  Write again if you need more help.

- Doctor George, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 08/03/2007 at 07:28:24
From: Tom
Subject: Thank you (conic section of an elliptical cone)

That is a nice, simple solution.  Thank you very much.
Associated Topics:
College Conic Sections/Circles

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