Re: [ATM] Ellipses, etc.

["Bob May" <rmay@nethere.com>]

Jim, THANKS!!!
Although I hate the equations where the one side of the equation
is equal to 0, I'll get this one figured out.  It's like saying
1+1-2=0.  Somewhere in an old Analog magazine, I've got the proof
that 1+1=2.
Bob May

rmay at nethere.com
http: slash /nav.to slash bobmay
http: slash /bobmay dot astronomy.net

----- Original Message -----
From: Jim Burrows <burrjaw@earthlink.net>
To: ATM List <atm@atmlist.net>
Sent: Monday, April 28, 2008 4:52 PM
Subject: [ATM] Ellipses, etc.


In Smith's "Modern Optical Engineering", p. 392,
he says a conic section through the origin satisfies the equation

         y² - 2Rx + (b+1)x² = 0

with y axis along the optical axis, x
perpendicular, radius of curvature R, and Tex's
deformation constant (conic constant) b.  The conic-b
relationship is

         ellipse b>0
         circle  b=0
         ellipse -1<b<0
         parabola        b=-1
         hyperbola       b<-1

Now with e = eccentricity,

         circle  e=0
         ellipse 0<e<1
         parabola        e=1
         hyperbola       e>1

and b = -e², which doesn't work for the b>0
ellipse case.  The problem is, in that case, the
optical axis is along the minor axis of the
ellipse.  Essentially, the eccentricity is
describing the shape of the ellipse rather than its optical
properties.

         -- Jim Burrows
         -- http://home.earthlink.net/~burrjaw
         -- mailto:burrjaw@earthlink.net
         -- Seattle N47.4723 W122.3662 (WGS84)

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