Re: ATM Secondary size

[Dwight Elvey <elvey@hal.com>]

"Richard Blackburn" <rblackburn@mindspring.com> wrote:
> 
> Putting together the parts before I assemble my telescope.  I am building a
> 6" f/4, and was wondering what the best secondary size to use would be?
> 
> Thanks!
> 
> Richard

Hi Richard
 If you are not interested in the exact formula, that
includes offset effects ( quite small ) you can do
some simple triangles. First draw out on a piece of paper
the rough scale of the size of your mirror. Now draw a center
line down the axis of the mirror. On this line put your
focal distance ( rough scale, we are only doing this for
understanding. Now the hard part. You need to select a
100% illuminate field of view. For visual, the smaller the
secondary the better. For photography, you'll want a
100% illuminated field for the size of at least the
minor axis of the film if not the full corner to corner.
Let us assume that it is to be a visual. You have selected
tha magic number of 1.5 degrees ( because your most likely
a wide field type ) of the sky. 2 * SIN( 1.5/2 ) * 24 = 0.63
inches. This is the size of the spot at focus that this angle
goes to. We do the 2* and /2 because the trig is only correct
on right triangles and not equal lateral, so we split it
into 2 and add them back to gether afterwards. The 24
is of course, your focal length. Now back to the drawing.
Draw a .63 inch line centered on the focal point and
perpendicular to the axis line. Draw two lines from the
edges of your mirror and the end points of the focal line,
we just made. This will make a larger triangle that has its
end point on the axis farther out than the focal point.
let us call the distance between the focal point and the
end of the triangle, distance "A". From simple geometry,
we know that the triangle .63/A is equal to the
triangle 6/(24+A). Or .63/A=6/(24+A). Solving for A
we get A = 2.82 inches. Now we know the total size of
our triangle, it is 24 + 2.82 = 26.82 inches long.
Now measure back from the focal point, the distance
that you'll need from the center of the tube to the
top of your focuser ( I'll guess here, not having your
dimmensions ). Tube size = 8 diameter and focuser height
is 2.5 inches ( your mileage may vary ). 8/2 + 2.5 = 6.5 inches.
Draw a line perpendicular to axis to the edges of the triangle.
This is the size of the secondary you'll need ( minus
a small amount that you'll need for offset. Like I say,
since diagonals only come in certain sizes, I don't need
an exact number yet ). Now we go back to similar triangles
again. 6/(24 + 2.82) = D/(6.5 + 2.83) where D is the minor
diameter of the secondary. Solving D = 2.09. Looking in
the catalog, I see that 2.14 and 1.83 are standard sizes.
If I was demanding on 100% illuminated field of view,
such as for photography, I'd take the 2.14. For visual,
the 1.83 will most likely work. I'd still need to see
that I had some field of view at the focal point but
I think, I'll leave that triangle drawing to you.
 The reason I went through this is to show that with
simple math, anyone can figure this. You don't have
to be some kind of math wiz, you just need to draw things
out and think about them a little. If I gave you an equation,
you'd never understand why it was done. I think the
the best part about making telescopes is there should be
nothing that you don't have at least a feel for why
it is so.
 I also didn't just give you a number because the requirements
of your telescope are a personal decission that only
you can deside. Use your desired values and fixed
mechanical sizes and find the diagonal for your dream
telescope.
Dwight