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[ATM] Fun With Airspace



Suppose you start with a "Normal" glass pair; one of the
better ones for Photopic performance is N-BaK1/SF2.  Start
also with a few basic assumptions.  Crown first with the
inner ROCs being equal; the flint with a central thickness
of 7 percent of the aperture; the crown with an edge
thickness of 4 percent of the aperture (after allowing for
some size beyond the aperture itself, say 8 percent). Pick
a nominal aperture (100 mm -- remember this is a "normal"
glass doublet) and field diameter (24 mm -- covers a lot
of 1-1/4 inch diameter eyepieces), but make the analysis
somewhat independent of those choices.


The remaining variables are the focal ratio and the air
space size.  Let v = 1/f#, and A be the aperture; the
desired focal length is F = A/v.  Analysis was done between
about f/6.7 and f/20.

Letting a strong optimizer work on the problem (in my case,
ZEMAX), it turns out the air-space, relative to the aperture,
is not independent of the focal ratio.  I appears that it is
not even monotonic.  A maximal optimal air-space occurs at
about f/11.3.  On the faster side of that peak the optimal space
appears to be quite close to linear in v; however, on the
slower side of that peak, it appears to drop off with a
higher order.  A close description of the points between
f/6.7 and f/20 is given by:

              t2/A = -0.569981 + 31.897051 v - 735.163807 v^2
                               + 9011.632956 v^3 - 61857.249529 v^4
                               + 224966.282382 v^5 - 338175.583967 v^6

(Yes... sixth degree (like a black belt) was the lowest degree
that would closely fit.) A key above is that the constant term
is negative, indicating that for small [positive] v (large f#)
the result is unimplementably negative.  In fact, it is negative
for speeds slower than about f/16.

I set up an approximation:

              t2/A = -0.051440 + 1.960320 v - 24.494773 v^2
                               + 132.439212 v^3 - 265.975044 v^4

by discarding a bunch of the slowest values.  This approximation
is non-negative for the range of about f/5.5 to f/20.  The
fact that it is a poorer approximation to the original near f/20
was deemed acceptable, as, at those speeds, there is more
performance to give away.

When this approximation of air-space is used, the resulting
ROCs are:

              R1/F =  0.707073 - 2.453808 v + 31.611001 v^2
                               - 181.266829 v^3 + 390.441183 v^3
              R2/F = -0.364738 - 0.760965 v + 10.263168 v^2
                               - 62.315806 v^3 + 138.413446 v^3
              R4/F = -1.391462 - 27.355177 v + 343.924817 v^2
                               - 1929.773630 v^3 + 4083.346490 v^3

The back focal length B, and the image ROC RI are
approximated by:

              B/F  =  1.002584 - 0.117757 v - 0.098902 v^2
              RI/F = -0.381584 + 0.212516 v - 1.848904 v^2 + 4.507722 v^3

Funny how such a simple design can appear so complicated;-)
I surmise some of the complexity results from trying to
model the system over the two quite different regions defined
by the primary aberration source.


Although the air-space approximation does not go negative, near the
f/5.5 and f/20 limits it is small and may cause the ghost to get
stronger.  In those cases, a non-spaced model may be desired
(t2/A = 0); of course, you may just want that anyway, as the
optimal gap is never more than 0.6 percent of the clear aperture.

The ROCs are:

               R1/F =  0.684789 - 1.512151 v + 18.064321 v^2
                                - 96.795518 v^3 + 198.884683 v^3
               R2/F = -0.368856 - 0.541451 v + 6.393431 v^2
                                - 35.713438 v^3 + 72.095965 v^3
               R4/F = -1.668378 - 16.139151 v + 190.837960 v^2
                                - 1012.139450 v^3 + 2053.590790 v^4

The back focal length and image ROC are:

               B/F  =  1.000209 - 0.055744 v - 0.338467 v^2
               RI/F = -0.373780 - 0.020781 v - 0.088609 v^2




-- 

Rick S.

http://users.rcn.com/rflrs














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