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James R
Jul14-08, 12:52 AM
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Quote Quote by B. Elliott View Post
Just wondering, but how did you get those numbers for the cross sectional area and why couldn't you just calculate for the available area?

6" diameter: Area = 28.26 in2
8" diameter: Area = 50.24 in2
My mistake. I used for the area of a circle [itex]A=\pi d^2[/itex] (i.e. squared the diameter instead of the radius). That means I was out by a factor of 4 on both the areas. But that doesn't affect the ratio of the areas, which is still 1.8, as I said.

Quote Quote by russ_watters View Post
Yeah, that post had a few things wrong with it. Something else in there is that there is a central obstruction from 10-15% and usually the fraction is bigger for a smaller aperture.

That said, the focal ratios on the two scopes are different - the 8" scope has a higher focal ratio, so things will actually appear dimmer than in the smaller scope. But the magnification is much higher.
Magnification isn't determined by the focal ratio. The magnification you get is the focal length of the scope divided by the focal length of the eyepiece. Essentially, magnification is not the biggest issue for a telescope, anyway. Often the best views are at lower magnifications (depending on what you're looking at, of course). But changing eyepieces can give you whatever magnification you want, up to a practical limit determined by the scope aperture.

I'm not sure why you think things will be dimmer in a larger scope. Given the same magnification, things will always look brighter in the larger scope.