Need Formula for C8 Mag. x's

Just waiting for my C8 to arrive. Question is: how do I arrive at what Magnification I am at with different EP's I have? I do not know the Focal length of the 8" My TV 102 is 880mm. I divide my EP. size lets say my Rad 4mm. into 880 and I am at 220 X. my 55mm. Plossl gives me 16 x.
can someone give me the formula I will need with my SCT? thanks so much for your time.. John M

The focal length of your new C8 is 2032 mm. The math is the same as your TV 102.

Walter

The problem with mirror-focusing scopes is that their effective focal length changes with the location of focus.
For c8, the f/10 (f.l.=2030mm) is likely to be for the focus located at the top of 1.25" diagonal. If it is so, switching to an average 2" diagonal would result in extending the focus location by ~1.5". This would be "produced" by moving the primary ~1mm closer to the secondary, with resulting secondary magnification (and effective focal length) increased by 4-5%. So you could count with more likely f.l. of 2100+mm.

The effective f.l. would also vary with different eyepieces, since their effective field stop location also vary. But it shouldn't exceed a couple of percent.

If you want to find out what is your scope's f.l. you could measure the exit pupil, devide the aperture with it, and multiply the result with the eyepiece's f.l. But for this you need to be sure what is the eyepiece's f.l. and you can't be unless you measure it up yourself. Another method is to take the eyepiece off and, say, have the Sun - **with the sun-filter on** - projected onto semi-transparent paper screen with engraved mm-scale located at the chosen location behind the back opening and measure up the diameter of its focused image.

With Sun's angular disc going between 31.6' in the summer to 32.6' in the winter, it will focus into ~19mm diameter disc, and your f.l. will be given by 57.3S/0.54, where "S" is the diameter of Sun's image on the projection screen. Assuming you can mesure Sun's image to within 1/4mm accuracy, the measured f.l. will have accuracy somewhat less than +- 1/100.