I would like some comments/alternatives to calculating true f-ratios
when using focal reducers. My approach is outlined below.

I was playing around with some formulae I found to figure out the true
f-ratio when using a focal reducer. The 80mm focal reducer in my
Lumicon Giant Easy Guider (GEG) can be moved forward or back to vary the
focal ratio. However, the real f-ratio of a 12" LX200 is something
greater than f/10 to start, and the length of the GEG and resulting back
focus means that the true f-ratios need to be calculated.

With the GEG in place and the focal reducer in the rear position in the
GEG, I placed the Meade 12.4mm Series 4000 eyepiece where the camera
(35mm or CCD) would go using a 1-1/4" adapter. The *apparent* field of
view for that eyepiece is 52 degrees. I timed how long a star (near the
meridian and 0 degrees dec) took to drift across the field of view and
converted to degrees (time * 15) to get the *true* field of view.

Resulting magnification (M) = apparent field of view(AF)/true field of
view(TF)

In this case M=0.28/52 or 184x

Given the focal length (Fe) of the eyepiece is 12.4mm:

Now the effective telescope focal length (Ft) is given by M * Fe

Ft = 184 * 12.4 = 2276mm

And given the aperature (Do) of the telescope is 304.8mm:

Now the effective f-ratio (Fn) is given by Ft/Do

Fn = 2276/304.8 = 7.5

Using the 26mm eyepiece which has the same AF, I get fn = 7.7

Comments? Questions? Feedback?

--
Michael Cook

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