Re: ATM Lowest allowable magnification

["Bill T." <twentiethwave@hotmail.com>]

John,

I find it useful to think in "exit pupil" terms, because you can 
characterize a telescope's low power issues with two simple equations.  
We'll assume the basic scope design is ok.

First decide what your dark-adapted eye's pupil size is. 5mm is a realistic 
number for most adults, although some will claim 7mm.

The minimum magnification that will not "waste mirror light outside the 
eye's entrance pupil" at the eyepiece is simply:

1. minimum_mag = mirror_diameter / exit_pupil

That would be 30 power for any 6 inch (150mm), for a 5mm exit pupil.  Or 40 
power for any 8 inch.  Note that's mirror_diameter, not focal_length!

To determine the eyepiece that would yield that magnification:

2. lowest_mag_eyepiece = f_ratio * exit_pupil

or 30mm for any f6 telescope, assuming a 5mm exit pupil.  Of course larger 
apertures will have a higher minimum magnification, per equation 1.

If you have a favorite wide eyepiece, say a 35mm Panoptic, you can calculate 
a "low power optimized f_ratio" for your mirror:

35mm_eyepiece / 5mm_exit_pupil = 7.0_f_ratio

7mm optimists would want to adopt an f5.0 ratio.  Notice that focal_length 
does not directly enter into the above equations, except that at some point 
you do need the focal_length to initially calculate f_ratio.

There's plenty for a purist to pick at here, but these simple equations 
provide practical ball park figures.

Since your 6" and 8" are both f6's, an eyepiece somewhere around 30mm is low 
power optimal for either scope.  That will give you 30 power on the 6", and 
40 power on the 8", while squeezing all of the mirror's reflected light into 
a cone the size of your eye opening.  All shorter focal length eyepieces 
will give more power, and smaller exit pupils.

Bill Tondreau









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