Getting my head round it ...

[Demonperformer]

Just want to clarify a couple of things, which I think are right, but still seem a little weird to me:

1. It is the f/ratio of the scope that is important for capture times, so a 3" f/5 will only take the same amount of time to capture an object as a 6" f/5, irrespective of the increased aperture (although it would affect image scale)?

2. It is the surface brightness of an object, rather than its magnitude, that determines how long capture time you need. So it would take longer to get an image of NGC598 (mag 5.8, surface brightness 14.4) than to get an image of NGC689 (mag 10.9, SB 12.3)?

As I say, from what I have been reading this is what I understand, but it seems a little weird. However, if anyone can confirm I have understood correctly, then I can proceed accordingly.

Thanks for your help.

[steppenwolf]

1. You are right, F5 is F5 whether you are using a huge telescope or a small camera lens - the different aperture comes into play with the calculation of the F ratio which is the Focal Length divided by the Aperture. If the aperture increases but the F Ratio remains the same then it must be because the Focal Length has increased and an increase in focal length means a smaller field of view and vice versa.

[Demonperformer]

Thanks for that, Steve.

Have I got my second point right?

Thanks

[sgazer]

1. The way I now look at imaging scopes is to ignore the aperture! Basically I think about how 'fast' a focal ratio I want the scope (an f5 scope will require half the exposure time of a f7 scope irrespective of aperture) and then think about the focal length which will determine the FOV for a given camera, also taking into account the difficulty in tracking accurately at longer focal lengths and the increase in weight and size of the scope for longer focal lengths at a given focal ratio. The two of these will then automatically give you a certain aperture

eg. an f5 scope, 750mm focal length will produce a 150mm aperture scope. If I thought I wanted half the FOV for a given camera (to zoom in a bit more) but the same fast focal ratio, then I would need a 1.5m focal length scope with a 300mm aperture - quite a beast!

Of course the flip side of this is determining what camera to use. An f5 scope/lens will always be f5, but the focal length comes into play here. It and the camera sensor size and pixel size will determine the FOV and resolution.

I'll stop now!

2. Good question, not sure. Could the magnitude and surface brightness refer to the brightest part and the average brightness overall respectively?

[Demonperformer]

Thanks for that, it helps. Basically, I have the scopes that I have, but am always interested to relate the pictures I see on SGL with what I could (potentially!) achieve. So as I understand what you have said, a picture taken with an 8" f6.3 sct would require me to use the same exposure on my 4" f6.6 refractor to get similar (albeit smaller scale) results.

I must admit to selecting fairly extreme examples on point 2, but it strikes me as strange that it could potentially take six times the exposure (12.3 > 14.4) to get a picture of an object that is supposedly 100x brighter (5.8 > 10.9).

Hopefully, one of the other SGL imaging gurus can come up with an answer.

[sgazer]

So as I understand what you have said, a picture taken with an 8" f6.3 sct would require me to use the same exposure on my 4" f6.6 refractor to get similar (albeit smaller scale) results.

Yep, that's it. So a tiny little f5 lens on your camera will produce as bright an image as a stonking great 16" f5 dob!... only it will appear much smaller with the lens and the resolving power will be less. However, even a small 60mm refractor can resolve down to the limit of typical UK skies (apart from moments of exceptional seeing).

[FraserClarke]

2. It is the surface brightness of an object, rather than its magnitude, that determines how long capture time you need. So it would take longer to get an image of NGC598 (mag 5.8, surface brightness 14.4) than to get an image of NGC689 (mag 10.9, SB 12.3)?

Magnitude is the total brightness of the object (i.e. all the light). Surface brightness the brightness in a given area (usually a square arcsecond). So NGC598 in your example is bright, but also very large, hence low surface brightness. NGC689 on the other hand is relatively compact.

Low surface brightness == harder to image generally, because you're getting less light per pixel.

[sgazer]

it's like these rear brake lights and indicators on cars these days. The output brightness is specified over a certain area, so as long as it's within that then it's legal. However, the LEDs give a very high brightness over a very small area so whilst still meeting the requirement, they still dazzle you if you are stopped up close behind them at night!

[Demonperformer]

Thanks, TD. Makes sense. Now I can start planning some 'long-duration' imaging.

[brantuk]

Does this mean that If I'm photographing NGC689 whilst driving, my stopping distance is reduced and I'm gonna crash? And will the number of pixels determine how serious the injuries are? lol

(sorry - couldn't resist a little fun - the thread has been extremely useful to me)