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






From: "Geogre Nikolidakis" <geoniko@otenet.gr>


In my calculation about the size of secondary mirror one of the parameters
is the image size.

D=( L(D-I)) / F  + I

Where I = image size , L = the distance from the Secondary to focal plane

D= Primary Size , F= focal lenght

I am not sure what it means, for example I like to full illuminate a 2"
eyepiece.

Is that means the image size it will be 2"?

Also L = the distance from the Secondary to focal plane .

The point for the focal plane it will be the base, I mean the low end of the
eyepiece ?

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

Thanks

George Nikolidakis

The choice of a secondary size for a visual scope is easily made much more complicated than it needs to be. You are trying to accomplish two things: 1. provide a reasonable level of illumination (not necessarily "full illumination") for the edges of the field of view and, 2. minimize diffraction effects from the secondary. The simple rule of thumb to reach these somewhat contradictory goals is to choose an elliptical secondary with a minor axis equal to 20% to 25% of the primary. For a 20" (508mm) primary, that would be a minor axis between 4" and 5" (102mm to 127mm). If you are going to use wide field eyepieces, choose a secondary near the large end of the range. If your eyepieces, like mine, have a more ordinary apparent field of 50 degrees or less, choose a secondary closer to 100mm to minimize diffraction.

I am building a 20" scope but it will be used almost exclusively for CCD photography. In my case, a small (18%) secondary of 92mm minor axis will provide ample illumination to cover a small CCD chip.

The best location for the focal plane depends on the design of your focuser and the eyepieces you plan to use. Generally, 50mm beyond the outer surface of the tube should work.

Jim Sturtevant

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