Equation used: TFOV = 15.04*T*Cos(delta)

Where T is drift time and Cos(delta) is cosine of Regulus declination at 11.9 degrees.

According to the math, if one knows TFOV and AFOV, one should be able to work most of the numbers associated with a scope, right? For example, if TFOV = AFOV/Mag, then Mag = AFOV/TFOV. Once you know Mag and eyepiece focal length, you're off and running. From there, you can calculate your effective focal length.

My data is surprising, so can someone check the logic based on the actual TFOV figures I trust to be good? Something bugs me about the data, though. According to the 12mm w/1.6 power barlow, I should be observing Jupiter at nearly 300x. Maybe, but it does not feel right. Maybe something is wrong with the math or logic.

Now, I calculated the Barlow power, which should be 2x inserted into the diagonal and 3x before the diagonal. I took the ratio of drift times both with and without the diagonal and found the drift ratio to be 1.5x (with the barlow after the diagonal) and 2.2x (before the diagonal.) I think this is valid math.

I also found my scope to be operating effectively at f/15.5 to f/16 (~2300mm) with the diagonal. This seems a bit long, though not too surprising, since the scope is advertised at f/12 (~1800mm.) But, f/16 is the derived figure based on TFOV measurements and the best AFOV eyepiece data I could find.

Any discussion to lock this down would be greatly appreciated. Thanks in advance.

Edit: I understand TFOV = AFOV/Mag is an approximation and the eyepiece focal length can be off by some fraction of a mm. Also, the AFOV listed may not be accurate, either. But, I am just looking for a good ball park figure for visual work, to be comfortable saying the scope really was at ~300x.

I do not have a CCD to work the pixel method. Field stop information can vary, too, and is often hard to come by, IME. So, I prefer the star drift method, as it is reasonably accurate despite distortions.

Another method that sounds promising is knowing the lunar diameter and comparing it's size to that on the focal plane.