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Re: [ATM] Baffles
Anthony Stillman wrote:
>>>I uploaded a optically simplified drawing of a 6 inch f/12 system at
<http://www.atmlist.net/contrib/atmer-at-flash-dot-net/6f12.jpg>.
The fully illuminated field is 0.7 inches and the secondary (identified by
the arrowed lines), which
functions as a baffle, is six inches from the focal plane. The top of the
drawing shows baffling for a 10 inch tube, the bottom for an 8 inch tube.>>>
The 6" f/12 is an extreme example, which is good to illustrate the point.
But how much of the tube needs to be baffled? Calculating back, I estimate
your secondary to be 1.2 inch dia, and let's consider a 8in. inner dia.
tube.
If so: The top part of the tube visible from the FOV I calculate as 44 in.
from the primary. What is the lowest part of the tube to be illuminated at
all via reflection in the primary? assuming that the tube is 1 f.l. long,
the ray passing by the tube edge hitting the opposite edge of the mirror has
an angle arctan 7/72 = 5.55 deg to the tube axis. The normal to the mirror
here is arctan 3/144 =1.19 deg, and the reflected ray has an angle 3.17 deg
and hits the tube wall 18 in. above the primary.
Thus, there is a potentially 26 in. long illuminated part of the tube from
18 in. to 44 in. that needs baffling, shading the outer part from
illumination via the primary and the lower part from sight from the FOV. If
we do it with one baffle, where will it be placed and will it be reasonably
low?
Assume it is x high, and 18+y from the mirror. The worst case ray to hit
the wall 44 in. up still comes from the edge of the primary, so the first
condition gives x/(26-y) = 1/44;
and the second x/y=4.35/54; solving gives x=0.46, y=5.7. Thus, if my
calculations are correct, a 0.46 in. wide ring baffle (opening dia. = 7.08
in.placed 23.7 in. from the primary would give complete baffling from
grazing reflections via the primary.
I have used your original to outline this baffling scheme:
http://www.atmlist.net/contrib/nilsolof-dot-carlin-at-telia-dot-com/6in-f12.gif
Nils Olof
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