big man said:
Thanks, that is useful to know.
I found the limiting magnitude of the 16-inch to be about 12.4 or so from comparing my images to star catalogues and finding the faintest star on the image. But I'm taking about 25 images per second.
So I'm wanting to know how I can find the limiting magnitude of the 12-inch from this.
I don't think you can from just that information. There are several things to consider in telescope resolution (R) and limiting magnitude (M). Usually, most of what is in the "common" definitions for both R and M are referring to
visual limits, not photography.
For
Visual, the standards are that:
R=(4.56/D), where D is the scope's aperture in inches.
and:
M=10+(5(log(D)*0.4342945))
But, actual visual M depends on seeing conditions, type of telescope, power of eyepiece, angle from zenith, etc., etc. There are calculators for this at:
http://www.go.ednet.ns.ca/~larry/astro/maglimit.html and:
http://www.geocities.com/catskills_astronomy_club/calculator.htm
Which appear to be the same and even try (not accurately) to factor in your age.
The "standard" formulae first shown above show that:
For a 16" scope: M=16.02 and R=0.28 arcseconds.
For a 12" scope: M=15.40 and R=0.38 arcseconds.
But, you're doing digital photography (CCD) with 25 images per second (and stacking??). So, the only way to figure your M limit in the photos is as you do in your post quoted above and compare faintest stars to a catalog. Then jiggle several of the inputs in one of the web calculators to match what your 16" can see as a limit and then
just change the aperture only to 12" leaving all the other parameters alone to see what you could expect for the 12" scope in the same conditions.
Of course you could change exposure time and stack fewer or more images or move to a darker sky location and all of that would change. With the CCD and exposure/stacking/location variables, the answers would near infinite.
