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Finding the Axes of an Ellipse from a Known Cone

Date: 01/26/2001 at 15:22:14
From: Ron Mion
Subject: Finding the axes of an ellipse from a known cone

I'm trying to derive a formula and can't find anything related to the 
specific situation I'm trying to solve regarding lighting when viewed 
as an oblique circular cone.

I have already determined how to find the projected size of a 
spotlight on a wall assuming the beam is perpendicular to the wall and 
forms a right circular cone. Using trig, if I know the angle of the 
beam spread, I can determine the radius of the spot of light. However, 
if the light shines on the wall at other than a right angle and forms 
an ellipse, then I only know one axis of the resulting ellipse, the 
short axis. I would like to be able to also calculate the long axis to 
determine how the long axis increases in relation to the degrees away 
from perpendicular.

The ultimate formula would allow me to find the long axis of the spot 
(ellipse) when I know the angle of the beam spread (steepness of the 
cone), the angle at which the center beam is hitting the wall (angle 
of the plane), and the distance (height) to the beam's vertex.  
Ideally, some of the variables could be backed out to produce a simple 
formula that calculates the ratio of the two axes based on the angle 
of the intersecting plane (wall) relative to the vertex (light 

From this I could back into the proper angle for the light based on 
the aspect ratio of the subject (e.g., for a painting that is 2' wide 
and 4' high, the best lighting angle would be x).

I hope you can help me apply your geometry knowledge to this real 
world problem confronting me. Thanks.

Date: 01/26/2001 at 16:45:26
From: Doctor Rob
Subject: Re: Finding the axes of an ellipse from a known cone

Thanks for writing to Ask Dr. Math, Ron.

You have posed a very interesting question, which I have enjoyed
investigating. I drew a diagram that looks like this:

                                _,-'   \
                            _,-'        \
                        _,-'             \ x
                    _,-'                  \
                _,-'                       \
            _,-'alpha                  beta \
 Source S o----------------------------------o B
            `-. alpha         d               \
                `-._                           \
                    `-._                        \
                        `-._                     \
                            `-._                  \
                                `-._               \ y
                                    `-._            \
                                        `-._         \
                                            `-._      \
                                                `-._   \

Using this diagram, the fact that the sum of the angles of a triangle
makes two right angles, and the law of sines, I found that

   <SAB = 180 - alpha - beta,
   <SBC = 180 - beta,
   <SCB = beta - alpha,
   x/sin(alpha) = d/sin(180-alpha-beta) = d/sin(beta+alpha),
   y/sin(alpha) = d/sin(beta-alpha).


   x = d*sin(alpha)/sin(beta+alpha),
   y = d*sin(alpha)/sin(beta-alpha).

Then AC, the length of the major axis of the cone, is given by

   2*a = x + y,
       = d*sin(alpha)*(1/sin[beta+alpha]+1/sin[beta-alpha]),
       = d*sin(2*alpha)*sin(beta)/[sin(beta+alpha)*sin(beta-alpha)].

That's the easy part. The length of the minor axis of the cone is
somewhat more involved. Because y > x, point B does not lie on that 
minor axis, and so the length of the minor axis is actually more than 
you would expect, which is 2*d/tan(alpha). The way to find the minor 
axis is to look at a point B' that is the midpoint of AC, and find the 
width of the cone at a plane perpendicular to SB and passing through 
B'.  This creates a diagram like this:

                                      A _,-'   |
                                    _,o'       |
                                _,-'   \       |
                            _,-'        \      |b
                        _,-'             \ x   |
                    _,-'                  \    |
                _,-'                       \   |
            _,-'alpha                  beta \B |B"
 Source S o----------------------------------o-o---------
            `-. alpha         d       (y-x)/2 \|w
                `-._                           o B'
                    `-._                       |\
                        `-._                   | \
                            `-._           b-w |  \
                                `-._           |   \ (y+x)/2
                                    `-._       |    \
                                        `-._   |     \
                                            `-.|      \
                                              C'`-._   \
This gives the additional facts that

   <CC'B' = 90 + alpha,
   <CB'C' = <B"B'B = 90 - beta,
   <B'BB" = beta,
   z = B'B" = B"C',
   w = B'B".

Now the Law of Sines again gives us

   sin(beta-alpha)/(b-w) = sin(90+alpha)/([y+x]/2),
   sin(beta)/w = 2/(y-x),


   w = (y-x)*sin(beta)/2,
   b = (y+x)*sin(beta-alpha)/[2*cos(alpha)] + (y-x)*sin(beta)/2.

Substituting into this equation for y and x, which are given above
as functions of d, alpha, and beta, you get

   b = d*sin(2*alpha)*sin^2(beta)/[2*sin(beta+alpha)*sin(beta-alpha)],

so the minor axis is

   2*sqrt(b^2-w^2) = 2*d*sin(alpha)*sin(beta)/sqrt[sin(beta+alpha)*sin

The area can also be computed using the formula

   A = Pi*a*b,

if that is desired.

I hope this is all the data you need.  If not, write again.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Practical Geometry

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