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How much 'faster' is my scope than my SCT?


blinky
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I just got hold of my new F4 newt last night and am wondering what the forumla is for working out how much faster this is than my SCT with F6.3 reducer on it?

Say I took a 10 min exposure with my Newt, what exposure length would I have had to use with my F6.3 SCT to capture the same amount of photons?

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to compare any two focal ratios, square them both, then divide one by the other.

In this instance, 6.3x6.3/4x4 = 2.48, so the newt is 2.48x faster than the SCT, or, the SCT needs 2.48x the exposure length of the newt to get the same exposure.

It doesn't matter which way you divide them, since if you do 4x4/6.3x6.3 you get 0.4, which means the SCT is 0.4x as fast, or, the newt needs 0.4x the esposure length of the SCT.

Does that all make sense??

Andrew

Edited by Andrew*
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That all goes assuming the aperture is the same.

If not, that needs to be factored in too.

I was wondering how my new 11 inch F10 Edge compares to my F8 6 inch frac.

The Edge is almost 4 times the area but not that much slower.

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Also you can't be sure you will really get F6.3 in the SCT. It may be nearer F7, depending on how you arrange what goes where in the train.

Rob, on non point sources why do you need to factor in the aperture? I though it just went with F ratio?

Olly

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Rob, on non point sources why do you need to factor in the aperture? I though it just went with F ratio?

Olly

Thats what I thought as well, I dont see where aperture comes into it perhaps some more explanation would be good

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Hi

There is a good explanation of this in Ron Wodaski's The New CCD Astrophototgraphy, but as it's not available these days i'll summarise and quote...

"Comparing two f5 SCTs one 8" one 16" using ST237 camera

For the 8", you light gathering area is approx 32 sq inches

Focal length is 1000mm

Area of sky covered by chip is 12*16 arc mins

Area of sky covered by each pixel is 1.5 *1.5 arcsecs

For 16", light gathering area is 128sq inches approx

Focal length is 2000mm,

Area of sky covered = 6 *8 arc mins

Area of sky covered by each pixel is 0.75* 0.75 arcsecs

Comparing focal length has doubled, aperture doubled

Light gathering ability is 4 time greater in 16" (128 vs 32)

Area of sky covered is 1/4 in of the 8"

"Hence four times the light is spread over four times the area, so there is no net change in the amount of light hitting the CCD - light per pixel remains the same, but amount of sky per pixel reduces"

Therefore this is why all things being equal aperture does not effect how fast a scope is, the focal ratio is the important bit.

Does that make sense.

Ian

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Thanks for the answers guys - just what I wanted to know - a single 10min sub is equal to (roughly) a 25min sub with my SCT.

So it follows....... That if I grab 6X10 min subs i.e. 1hr, that would be equivalent to 2.5hrs with the SCT - Happy days:hello2:

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The more imaging I do, the more I want F ratio, F ratio, F ratio... That is what gets you deep.

Olly

:) I can see Olly will be visting the Orion Optics website a few more times in the near future... :p

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The maths makes sense, but when I look in, for example, a 12 inch F8 scope compared with a 6 inch F8 scope, even though the image is ispead over 4 times the area, it is brighter.

Having had large aperture scopes, and having just had this reinforced by my recent addition, a bigger scope definately bangs in the photons better than a smaller one.

I have no idea why this is, nor do I question the science, but it is what I find to be true.

Cheers

Rob

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Rob, visually it's the other way around. Looking through a larger scope will be brighter at the same magnification, regardless of focal length. Also, an 8" f/5 will be just as bright at say 100x as an 8" f/10 at 100x.

Andrew

Edited by Andrew*
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