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Does an inch matter?

Started by atracht, 05/11/2004 05:29PM
Posted 05/11/2004 05:29PM Opening Post
Is there any significant (for either visual or photographic use) difference in light-gathering ability between the 10" Mewlon 250 and the 11" Celestron C-11?

The C-11 with its central obstruction of 34% and XLT coatings (average 400-750nm transmission of 83.5%) should give the equivalent of a perfect 9.45" diameter scope.

11 * sqrt((1-.34^2)*.835) = 9.45

The M250 has a smaller mirror, relatively smaller (29%) central obstruction, no corrector plate, and different mirror coatings. Does anyone know how it compares in light througput? I'm looking for a quantitative answer, if possible.

I'm sure that there are many other considerations, such as optical and product quality, price/performance, and f-ratios, but I'd like to keep this discussion limited to the physics of light throughput.

Thanks,

Allen
Posted 11/14/2005 08:43PM #1
Allen:

The M250 has 96% mirror coatings. So
9.8425^2 - 2.8346^2 = 88.8396
SQRT 88.8396 * 0.96^2 = 8.69" effective aperture.

Paul Krumenacher
Posted 11/14/2005 09:47PM #2
Hi Allen, where did you get that formula?
An 11" aperture has an area of 95.033 sq. in., so factor of 34% area obscuration by the secondary means you have an equivalent of 62.722 sq in. Now take into account an efficiency penalty of 0.835 so you have a final equivalent area of 52.37 sq. in, which is 8.17" of "perfect" aperture. Did I mess up?

Ivan