exposure times in relation to scopes?

[Whippy]

I use the word 'exposure' in the loosest sense, my afocal fun and games basically is an excersize in getting as much light in as quickly as I can! Anyways, my C8N is obviously bigger and faster than my MN56 but by how much? Is there an equation that you can use to work this out?

Tony..

[rawhead]

Is the table on this thread any help?

http://stargazerslounge.com/index.php/topic,20020.msg206713.html#msg206713

[cygnusx1]

the exposure time are primarily (disregarding obstruction, etc ) related to the aperture and hence the primary mirror area by an inverse square relation.

i.e a 8" will gather light (8*8)/(4*4) = 2 times faster than a 4".

In other words for each doubling of aperture exposure time is reduced by a factor of 4.

(Edit: SteveL- turned off smilies in this post so the equations dont get messed up)

[Andrew*]

i.e a 8" will gather light (8*8)/(4*4) = 2 times faster than a 4".

erm, little slip on the maths there Kevin:

8*8 (64) / 4*4 (16) = 4 times faster.

Of course, it also has a lot to do with focal ratio....?

[Whippy]

That's what I was thinking Andrew, I was thinking something along the lines of Kevin's equation plus (?) something to do with either focal length or ratio.

Tony..

[Kaptain Klevtsov]

Using the focal ratio idea, an f/8 'scope takes 4X as long as an f/4 'scope to gather the light. (8/4)^2 so its the (f/A)/(f/B) squared.

Kaptain Klevtsov

[Whippy]

Cheers for that KK.

So...(bear with me, my maths is fairly shaky)

Lets take for example my C8N, 8" aperture with a f5 ratio and my MN56, 5" aperture with a f6 ratio.

Going by aperture: (8x8)64/(5x5)25=2.56

Focal ratio: 6/5 squared = 1.44

Ok, so going by aperture the C8N is 2.56 times faster and 1.44 by ratio. How to combine the two....Hm...

Tony..

BTW, if anyone think I'm chasing my own tail for no good reason, feel free to chip in anytime .

[cygnusx1]

Using the focal ratio idea, an f/8 'scope takes 4X as long as an f/4 'scope to gather the light. (8/4)^2 so its the (f/A)/(f/B) squared.

Kaptain Klevtsov

Incorrect.

e.g a f/8 scope f/l 800 , aperture 100 + f/4 scope f/l 400, aperture 100

they both have same light gathering capabiliities i.e its only aperture dependent!

[Andrew*]

No, Kevin - you HAVE to take into account the focal length at which you are imaging, because this determines the area of sky from which you are collecting photons. Of course it also depends on the aperture.

How do you connect aperture and focal length? Focal ratio!

Exposure times relate for the most part to the focal ratio more than the aperture. Why do people get such excellent results on faint DSOs from 66mm scopes if exposure times are to do with aperture??

Andrew

[MikeP]

Just looked in my New CCD Astronomy book ...

Two telescopes with the same focal ratio will require the same exposure times, even if there is a difference in aperture. The scope with larger aperture will provide more magnification.

As a consequence, you'll see a larger image but a smaller FOV.

Mike

[Whippy]

Aperture doesn't make a scrap of difference, only in that it will affect the image size? Really?

So what you're saying is that I could get an image from my C8N, and I would get the same level of detail taking the same pic with a say, ST80 (using the same exposure time) and the only difference would be the size of the target? That can't be right surely!!

Tony..

[MikeP]

There is an example in the book that should make it clearer (sorry about the mixed units). It compares scope A - 8" SCT f/5 with scope B - SCT 16" f/5.

It assumes about a 30% obstruction, so the light gathering areas are about 50 - 18 = 32 sq in and 200 - 72 = 128 sq in.

Focal lengths are about 1000mm and 2000mm.

Area of sky covered by the chip (ST-237 camera assumed) is 12 x 16 arc minutes and 6 x 8 arc minutes

Area of sky covered by each pixel is 1.5 arc seconds and 0.75 arc seconds.

So, comparing B with A - light gathering is quadrupled but the area of sky covered by the chip is a quartered. The net result is that 4x the light is spread out over 4x the area, so there is no net change in the amount of light hitting the CCD chip.

It goes on to consider using B with a hypothetical camera with pixels twice the size and concludes it would allow shorter exposure time for the same result or more detail captured for the same exposure.

Not sure I totally understand myself, but it does make sense. Hope it helps.

Mike

[SteveL]

That kind of makes sense when its explained like that.

[dph1nm]

This one always causes no end if fun - I think the confusion often arises because changing f-number in a terrestrial camera does change the exposure (so f16 in bright sun, f4 in twilight etc). But what one tends to forget is that the lens aperture changes.

For telescopes, all that really matters is the number of photons collected from a square arcsecond on the sky - this is determined by aperture. Bigger is better! Focal length just alters how these are spread out on your detector. But a galaxy is the same number of arcseconds across whether at at f8 or at f4 - so if the aperture is the same you get the same number of photons making up the image of that galaxy and it has the same signal-to-noise.

NigelM

[MartinB]

Nigel, I'm not quite sure what you are suggesting. The focal ratio is the critical figure for determing how much exposure you need for a given signal to noise ratio. A 0.5 focal reducer will quarter the exposure time you need and will give you a field of view 4 times as large in terms of area due to reduced magnification. It's very similar to thinking about how the image dims when you up the magnification visually despite the fact that the aperture is the same.

[trevboyd]

This one always causes no end if fun - I think the confusion often arises because changing f-number in a terrestrial camera does change the exposure (so f16 in bright sun, f4 in twilight etc). But what one tends to forget is that the lens aperture changes.

For telescopes, all that really matters is the number of photons collected from a square arcsecond on the sky - this is determined by aperture. Bigger is better! Focal length just alters how these are spread out on your detector. But a galaxy is the same number of arcseconds across whether at at f8 or at f4 - so if the aperture is the same you get the same number of photons making up the image of that galaxy and it has the same signal-to-noise.

NigelM

I don't think I agree with this last statement. Its true that for the same aperture, the same number of photons will be used to make the image, but as you said, the focal length determines how spread out they are (i.e. a shorter focal length results in less magnification and so the galaxy is projected onto less of the detector.) This results in a brighter image of the galaxy (more photons per pixel) and means that the signal-to-noise will be better surely?

I agree with Andrew (AstroPhethean) - the focal ratio is surely the best indicator here. Take a simple example of two scopes:

A: 200mm aperture, F=1000mm (f/5)

B: 100mm aperture, F=1000mm (f/10)

With the same focal length, images taken through these scopes will be the same "size" (cover the same area of sky) - this keeps things simple to explain.

Clearly, scope A will require less exposure than B due to its larger aperture, which is reflected in its faster focal ratio.

Scopes of different focal lengths introduce the issue I discussed above, different sized images and thus the photons being "spread out" over different areas on the chip. The more area they are spread over, the dimmer the image will be.

Trev

[trevboyd]

Nigel, I'm not quite sure what you are suggesting. The focal ratio is the critical figure for determing how much exposure you need for a given signal to noise ratio. A 0.5 focal reducer will quarter the exposure time you need and will give you a field of view 4 times as large in terms of area due to reduced magnification. It's very similar to thinking about how the image dims when you up the magnification visually despite the fact that the aperture is the same.

Martin, you took the words out of my mouth (and explained them better and more succinctly!)

Trev

[dph1nm]

Oh dear, this opened a can of worms.

Right - I assumed in making my statement above that CCD readout noise was negligible. If so, then the signal-to-noise does not depend on the number of pixels. So a long f-ratio telescope may have e.g. 16 pixels covering 1 sq arsec, whereas a short f-ratio may only have 4 covering the same area. If the apertures are the same then, yes, the short f-ratio telescope will have a 4 times the number of photons per pixel if we use the same CCD. BUT - I can just bin up 2x2 (in software) the pixels from the long f-ratio shot, and then I get identical number of photons in identically sized pixels (and identical s/n). The only difference will be that the short f-ratio shot will cover a larger area of sky. And what matters for a picture of an object is the s/n per arcsecond on the sky. An image does not get dimmer in terms of s/n per arcsecond when you up the magnification.

To put it another way: if I image M81 with the same expoure time on the same CCD on two telescopes of identical aperture but different f-ratio, then the resulting images will be identical in terms of s/n (and if I bin them to the same pixel size in microns they will look identical).

Now, if read noise is signifcant then larger pixels (so small f-ratio) can win. And if the pixels become larger than your object of interest then smaller pixels start to win. However, at the end of the day professional astronomers use ever larger aperture telescopes for a good reason - and that is to image fainter objects in the same exposure time.

NigelM

[MartinB]

I agree that binning 2x2 will give you the same sensitivity and plate scale as halving the focal length. However the long focal length binned image will have 1/4 the field of view compared with the shorter focal length image using the same chip.

To keep things very simple and sticking to issues relating to the scope:-

F ratio determines exposure length - an F5 scope will require a quarter of the exposure time as any F10

Focal length determines the magnification/image scale

[cygnusx1]

I agree that binning 2x2 will give you the same sensitivity and plate scale as halving the focal length. However the long focal length binned image will have 1/4 the field of view compared with the shorter focal length image using the same chip.

To keep things very simple and sticking to issues relating to the scope:-

F ratio determines exposure length - an F5 scope will require a quarter of the exposure time as any F10

Focal length determines the magnification/image scale

So are you saying 2 F/5's f/l 1000 and 2000 both have the same exposure lengths/light collecting capacity?

Goes against intuition as the latter ones has 400mm aperture and the former only 200!!

[themos]

Ah, but it's not absolute number of photons that we care about but number of photons per second PER UNIT AREA. That's what makes a bright image.

[MartinB]

It's true Kevin and it does go against intuition. Themos explains it perfectly.

[cygnusx1]

Roger, now its clear

# photons/ unit area/sec :thumbup:

[themos]

To flesh it out a bit, the difference in focal lengths of f/5 scopes will result in different size images on the chip. A really small scope would make a nice bright image (of Mars, say) but 1x1 pixel across, the largest would make the same nice bright image but 100x100 pixels across. Here, "nice bright image" means using a lot of the pixel's dynamic range.

[dph1nm]

What matters is photons/unit area on the sky/sec

"F ratio determines exposure length - an F5 scope will require a quarter of the exposure time as any F10 " - sorry, no - assuming the scopes have the same aperture then you get the same image s/n from the same exposure time. All that happens is the F5 covers 4 times the area of sky (so might be desirable for other reasons).

NigelM