Making a homemade focal reducer
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I want to make my own focal reducer for my 11" SCT. Can anyone tell me
how to figure
the actual focal reduction based on the f.l. of the achromat lens?
There are several
48mm achromats available online with focal lengths from 208mm to 360mm
Any help is appreciated. Thanks.
The formula for reduction is:
Where 'Rf', is the reduction/magnification factor (the same formula works
for Barlows as well), 'f' is the focal length of the lens assembly
concerned, and 's' is the seperation between the optical centre of the
lens, and the focal 'plane'
So if you have a 208mm lens, and space it 69mm from the focal plane, you
Rf=(208-69)/208 = 0.668*
However there is a large 'caveat' with what you are talking about. The
actual focal field of an SCT,is significantly curved. On your 11" unit,
probably with a radius of about 12" (it depends on the focal lengths of
the two mirrors). When you apply a focal reducer, this curvature is made
worse, reducing the field diameter at the CCD/film etc., that can be used
before the defocus produced by field curvature, becomes unacceptable. This
is why the commercial SCT reducers, are 'reducer/correctors'. They
normally have plano convex lenses, in at least one element, to produce a
field curvature (the other way)themseles, which at least partially
corrects for this problem.
Many people have used lenses from binoculars or similar sources to make
reducers like you describe, and for small fields (a small CCD etc.), the
results can be acceptable. However one of the big advantages of the
commercial reducers, is the field flattening effect, which is why the view
through (for example), the Celestron/Meade F*.63 unit, combined with
perhaps a 26mm eyepiece, can in some cases be more useable than a 2" 40mm
eyepiece. Though the human eye accomodates quite well for the field
curvature, the effect of seeing well focussed stars across the larger
field, can be impressive.