Re: ATM 100% illumination and secondary sizing

["John Lynch" <john_p_lynch@hotmail.com>]

David,

I was going to send you an E-mail regarding the question about
the use of Newt and your 12.5" f/5 reflector.  I ran the numbers,
out of curiosity, because I've used this software extensively to
design and optimize my 6" f/6 and an idea for a future 8" f/6.25.  I've also 
used Texereau for this effort, and your E-mail has
spawned yet another question for the ATM listserv.

Here goes:  Using 12.5" for mirror diameter, focal ratio of f/5,
2" I.D. focuser with minimum height of 1.6" (like a Crayford),
and 1" of extra travel for visual use only, a 14.5" I.D. tube with
0.375" tube thickness and a 2.6" secondary minor axis, I get
0.66" as 100% illumination and 1.7" as 75% illumination.  The 0.66
is good for full moon, which Texereau calculates as 12.5x5x0.99=
0.56" as a minimum.  If you use the 2.14" diagonal, the 100%
illumination would be 0.11" and 75% illumination would be 1.2".
Neither case introduces vignetting, according to Newt.  However,
the smaller diagonal doesn't meet the minimum criteria for the
full moon of 0.56".

Now, here's the question:  In Texereau's book, 2nd Ed. on page 110 (Table 
VI.  Permissible Field Diameters), he computes the value to
be 0.95 inches for a 12" f/5 reflector.  In cases where I've used
Newt and this table for 6" f/6 and 8" f/6.25, I've always been able
to choose a diagonal minor axis to get a good enough 100% illumination value 
(above f.l. X 0.009) and a 75% illumination below the value in this table 
(which I then consider an optimal design/tradeoff).  Hope I haven't made 
this too confusing, but I'd like someone to set me straight if I'm 
misinterpreting the data.  And what implication does this have for David's 
design?

Thanks.

John
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