So, I'm trying to wrap my head around the implications of all this.
I have been imaging mostly with a reduced C8, ~f/6.3, ~1300mm, ASI183MC, native image scale 0.38"/pixel, FOV 0.58x0.39 degrees. This setup is primarily used for lunar imaging, while it excels at, at least in my opinion. It's also pretty good for planetary nebulae, globulars, galaxies, small stuff. But I obviously can't fit large extended nebulae, M31, etc.
I've been advised to switch to a short focal length "fast" refractor. I've done some practice with a cheap 60mm guide scope while I mull over buying an APO.
What you're telling me is that, by binning, I could produce an equivalent SNR "tile" at the same image scale as the refractor in a much shorter time, and could regain the FOV by building a mosaic, which I am very familiar with from lunar imaging.
Say, for example, I had a 60mm refractor at f/4.8, 288mm. With my camera, 1.72"/pixel, 2.625x1.75 degree FOV.
To recover this FOV would require 25 tiles. To achieve the same image scale would require resampling 4.5x4.5.
So, if my math is correct, the ratio of the pixel signals between the two systems (after binning to achieve the same pixel scale), neglecting any potential difference in transmittance between the two, simply goes as the ratio of the two areas, yes?
The C8 has a central obstruction of 35%, so an effective area of 0.65 x pi x (203.2mm / 2)^2, while the refractor has an area of pi x (60mm / 2)^2, so the ratio of the two is 0.65 x (203.2 / 60)^2 = 7.45.
So for the same pixel scale, the C8 is 7.45 times "faster" than the 60mm refractor. And that's true regardless of camera, as long as you are comparing the same camera in both scopes. But it would take ~25 tiles to achieve the same field of view. Also independent of camera (just the ratio of the fields of view rounded up, squared).
Edited by Borodog, 16 January 2022 - 05:47 PM.