Re: ATM The offset bugbear.

["James W. Burrows" <burrjaw@halcyon.com>]

On Sat, 14 Jun 1997 andydtg@phoenix.net wrote:

> BTW the solution for the exact offset between the center of the ellipse
> and the optical axis intersection, given the other parameters of focal
> length, field size, and primary-secondary spacing is not trivial & looks
> like the solution to a 4th order 'quadratic' equation. I plugged this
> into MathCad & it spit out a set of equations that works but is a page
> full of algebraic expressions. I managed to stuff it into a spreadsheet
> & verified the empirical solution given by drawing the system in
> AutoCAD, drawing it is *much* easier. 

It's just some trig.  Let A = arctan(1/(2*f/)), E = diagonal-focus
distance (along the optical axis), then, from the law of sines for oblique
plane triangles (remember your high school trig?)

        offset = EsinA( 1/sin(45-A) - 1/sin(135-A) )/2

You saw it first on the ATM list.

     -- Jim Burrows:     (206)244-2933, fax -0294
     -- E-mail:     burrjaw@halcyon.com
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