Re: [ATM] What is that Image size in Calculations

["Mitchell R" <funnybone101@embarqmail.com>]

I have a few others questions but in the next days ... -:-)

Thanks 

George Nikolidakis

*** Hi, welcome. I typed this up a few years ago and it uses the number for
a 6" mirror, you can replace them where necessary. 

"Secondary Size: A rule of thumb is to have a .50 (half a degree) "fully
illuminated field".
So .0524 is diameter (6) X amount of illuminated field (.5 degrees) divided
by 57.3 (Radians in a degree) So 6X .5/57.3 = .0524 = DF. Then you do DF +
([D-df]/F), which means you take D= diameter (6 inches) minus DF (.0524)
divided by Focal Ratio in inches (48). So 6-.524/48=.114. So then you do X
Lde, take radius of tube (4 inches) from center of mirror to tube side) plus
focuser height (Example: 3.25) equals Lde (7.25*). 

SO, .0524 + .114 X 7.25* = 1.424 , (1.15 for 3 inch focuser)
         DF    D-df/f X Lde

*= need to change to fit different focusers."

Some people say a .75 degree fully illuminated field is better, now really
sure.

Mitch

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