Re: [ATM] Cassegrain vs. Gregorian (secondary testing)

["Richard F.L.R. Snashall" <rflrs@rcn.com>]

Bob May wrote:
  > Decent start for a page.  Now lets have the formulas and so forth so
that we
  > can see what is needed for the two doublets of the test lenses.

However, I'm going to hypothesize a few things and see if they seem to
at least start to pass muster.

If M is the Gaussian transformation matrix from the point source to the
secondary, the final vector for initial vector

               [  y  ]
           V = [     ]
               [  u  ]

is:

                   [  1      0 ]
          M^(-1) * [           ] * M . V  + M^(-1) . D
                   [ -2/R   -1 ]

where R is the secondary RoC,  D is reflected angular difference
associated with the secondary deviation from a sphere and
is approximated by:

          D = transpose(( 0, (1-k) * (ym/R)^3 ))

for the zone ym and conic k (k = 1 is a sphere, k = 0 is a paraboloid).

The first term in the final vector above has a zero lateral component.
The second term has nearly zero u component.  The longitudinal error is
then  -yf/uf, where yf is the lateral component of the second term
and uf is the angular component of the first term.

If the lens set is a pair with the lens nearest the mirror having
a focal length close to the mirror RoC, then I hypothesize that
there is little change in lateral value across the lens set and,
for the zone ym, the return angle is simply ym/t where t is the
distance of the focus from the lens set.  I also think
the matrix M^(-1) has little impact on the nature of D so that
the lateral value at the focus is simply the angle translated
through the distance.  The longitudinal error would then be:


        LA =   - (( (1-k) * (ym/R)^3 ) * ( -t )) / ( ym/t )

or:

        LA =   (1-k) * ym^2 * t^2 / R^3

Assuming I made this at all comprehendable, is it reasonable?

-- 

Rick S.

http://users.rcn.com/rflrs



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