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Old 23-06-2009, 08:21 AM
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toryglen-boy (Duncan)
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Exposure difference with different scopes.

I use an ED80 to image, although i am looking at something additional for two reasons

1. the ED80 is widefield and i also need something with a smaller frame
2. I want something with a larger aperture to capture more photons!

and it got me thinking, i can work out the frame size with a CCD calc utility, but can someone give me a comparison, like for like, with exposures on two differing scopes?

Let me give you an example, would a 20 sec exposure of say, the Lagoon Nebula on an 8" reflctor, be about as bright as 60 second exposure on the ED80? i know the resolution, amount of info captured, and frame size would be different, but you can see what i am asking here, just for some general info on the differences with exposures lengths as a comparison.

its a bit covoluted i know.

thanks for any answers.

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Old 23-06-2009, 09:00 AM
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Good question Duncan, I'm looking forward to the answers.
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Old 23-06-2009, 05:05 PM
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It all works on f number. Two scopes with the same f number need the same exposure for extended objects. As an example take an 8" f5 (40" focal length) and a 4" f5 (20" focal length). The 8" gathers 4 times as much light as the 4", but because the image scale is doubled this light is spread over 4 times the area, so the extra light is exactly cancelled by the extra area over which it is spread.

The general result is this: If you take an exposure of length t1 with a scope of focal ratio f1 and then you go to a scope of focal ratio f2, you will need an exposure t2 given by

t2=t1*(f2/f1)^2

So just work out the focal ratios of your two scopes and plug into this result

Last edited by Geoff45; 23-06-2009 at 05:50 PM.
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Old 23-06-2009, 06:25 PM
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Quote:
Originally Posted by ghsmith45 View Post
It all works on f number. Two scopes with the same f number need the same exposure for extended objects. As an example take an 8" f5 (40" focal length) and a 4" f5 (20" focal length). The 8" gathers 4 times as much light as the 4", but because the image scale is doubled this light is spread over 4 times the area, so the extra light is exactly cancelled by the extra area over which it is spread.

The general result is this: If you take an exposure of length t1 with a scope of focal ratio f1 and then you go to a scope of focal ratio f2, you will need an exposure t2 given by

t2=t1*(f2/f1)^2

So just work out the focal ratios of your two scopes and plug into this result

many thanks for that, but its far too much like hard work !!

can you post a practical example?
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Old 23-06-2009, 07:41 PM
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Originally Posted by toryglen-boy View Post
many thanks for that, but its far too much like hard work !!

can you post a practical example?
Example 1:
8" scope of 64" focal length, so has a focal ratio of 8
3" scope of 15" focal length, so has a focal ratio of 5

Say you take a 2 min exposure with the 3".
Then you need a 2*(8/5)^2=2*64/25=5.12 min with the 8". Bear in mind though that you'll have an image about 4 times bigger in scale.

Example 2:
8" of 40" focal length (so it has a focal ratio of 5)
4" of focal length 28" (so a focal ratio of 7)
5 min exposure with 4" equates to 5*25/49=2.5min (approx) with 8"


Note added: Like many things you can't just take such calculations as gospel. In the first example the exposure in the 8" may be fine for the neb, but since stars are point sources, you may end up overexposing the stars and so it may be better to stack two 2.5 min exposures.

Last edited by Geoff45; 23-06-2009 at 11:35 PM.
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Old 24-06-2009, 07:45 PM
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Duncan check this post I did a while back.

http://www.iceinspace.com.au/forum/s...ad.php?t=45876

Same Camera and mount as yours. The f5 image would almost be the same image scale as your ed80, but slightly faster (due to same fl but more apeture).

These are beginner images, but they should show the image scale differences.

Brett
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Old 25-06-2009, 05:19 PM
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phew .. my head is spinning from some of the intellectual replies to this .. must be the end of the day.

my understanding is pretty simplistic but works for me ...

Same F-stop = same exposure length, regardless of aperture, focal length or colour of the OTA

So if the 50" you're about to buy is F/6 and your 80mm is F/6, then exposure times would be the same to achieve the same image brightness.

If the larger scope you're looking at has a slower F-stop (F/10, F/12, rather than F/4, F/6 etc) then that larger scope will require longer exposures to get the same brightness of detail.

eg. If you go from an F/6 scope to an F/8 scope you'll go from 60 second to a longer exposure time (80 secs??) to achieve the same brightness of image.



Roger.
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Old 25-06-2009, 06:07 PM
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Dunan

Download this app, its simple, free and insanely usefull, I use it all the time. Its a CCD calculator and shows the results from every concievable combination of scope/cam. Its by Ron Wodaski.

http://www.newastro.com/book_new/camera_app.php
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Old 26-06-2009, 08:32 AM
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Quote:
Originally Posted by Bassnut View Post
Dunan

Download this app, its simple, free and insanely usefull, I use it all the time. Its a CCD calculator and shows the results from every concievable combination of scope/cam. Its by Ron Wodaski.

http://www.newastro.com/book_new/camera_app.php

Hi Fred, yeah i have been using that for ages, i was really after more info on the brightness of the image captured, and length of exposure, rather than size of frame etc.

many thanks though !!

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Old 27-06-2009, 04:20 PM
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Quote:
Originally Posted by rogerg View Post
phew .. my head is spinning from some of the intellectual replies to this .. must be the end of the day.

my understanding is pretty simplistic but works for me ...

Same F-stop = same exposure length, regardless of aperture, focal length or colour of the OTA

So if the 50" you're about to buy is F/6 and your 80mm is F/6, then exposure times would be the same to achieve the same image brightness.

If the larger scope you're looking at has a slower F-stop (F/10, F/12, rather than F/4, F/6 etc) then that larger scope will require longer exposures to get the same brightness of detail.

eg. If you go from an F/6 scope to an F/8 scope you'll go from 60 second to a longer exposure time (80 secs??) to achieve the same brightness of image.



Roger.
This is essentially true but it is the image size that will change as you change the size of the scope.
An example to describe this is if you use a 200mm diameter f8 scope it will have a focal length of 1600mm. If you take a 1 min exposure you will get a certain brightness in your image. It image will be somewhere between 0.5 and 1 deg across depending on your camera.
If you swap to a 50mm camera lens at f4 (2 stops wider than f8) it will only take 15 sec exposure to achieve the same level of brightness in your image as the 200mm scope image.
The difference is the size of the field. It will be a 60deg wide image.
The point to this is example is that it isn't very good to compare an 80mm diameter fast scope with a short focal length with a bigger (ie >200mm) scope with a longer focal length but a slower f ratio.
The 80mm scope will give you a brighter image more rapidly but only because the image is much smaller.
It depends on what you want to image. Little objects like most pns and most galaxies can't be imaged with a decent size with a little 80mm refractor.
If you want to image widefield nebs than the little scope is good.
By the way. None of this applies to imaging stars as they are essentailly point sources and f ration makes little difference for them. Only light capture matters hence a big scope will image dimmer stars than a little scope.
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Old 27-06-2009, 04:48 PM
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Quote:
Originally Posted by Terry B View Post
Only light capture matters hence a big scope will image dimmer stars than a little scope.
Hi There

as stated, yes i know about the field changing size, and TBH i didnt want to be rude at all the replies, but i kinda had a rough idea about all that to, but what i asked hasnt really been tackled by any reply, so i will try to explain it more simply

say i am taking an image of the Eta Carina. Ragardless of image size, or any of that, i have 2 telescopes an 80mm refractor at F7.5 and a 200mm reflector at around F5. now then, purely based on telescope aperture and the telescopes light gathering power what would be the eqivelant exposure time for an image of equal brightness? thats equal brightness, not detail captured, or area captured, or size of image.

hope that explains it.

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Old 27-06-2009, 05:04 PM
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Quote:
Originally Posted by Terry B View Post
This is essentially true but it is the image size that will change as you change the size of the scope.
An example to describe this is if you use a 200mm diameter f8 scope it will have a focal length of 1600mm. If you take a 1 min exposure you will get a certain brightness in your image. It image will be somewhere between 0.5 and 1 deg across depending on your camera.
If you swap to a 50mm camera lens at f4 (2 stops wider than f8) it will only take 15 sec exposure to achieve the same level of brightness in your image as the 200mm scope image.
The difference is the size of the field. It will be a 60deg wide image.
The point to this is example is that it isn't very good to compare an 80mm diameter fast scope with a short focal length with a bigger (ie >200mm) scope with a longer focal length but a slower f ratio.
The 80mm scope will give you a brighter image more rapidly but only because the image is much smaller.
It depends on what you want to image. Little objects like most pns and most galaxies can't be imaged with a decent size with a little 80mm refractor....
Of course Terry, but Duncan specifically said he had an understanding of the FOV changes, but was asking about exposure differences. Good to clarify the whole situatio I suppose though
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Old 30-06-2009, 07:02 PM
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Quote:
Originally Posted by toryglen-boy View Post
Hi There

as stated, yes i know about the field changing size, and TBH i didnt want to be rude at all the replies, but i kinda had a rough idea about all that to, but what i asked hasnt really been tackled by any reply, so i will try to explain it more simply

say i am taking an image of the Eta Carina. Ragardless of image size, or any of that, i have 2 telescopes an 80mm refractor at F7.5 and a 200mm reflector at around F5. now then, purely based on telescope aperture and the telescopes light gathering power what would be the eqivelant exposure time for an image of equal brightness? thats equal brightness, not detail captured, or area captured, or size of image.

hope that explains it.

The 80mm scope will require 2.25 times the exposure to give the same brightness for the nebula, which you get by calculating (7.5/5)^2.


The stars, being point sources follow a different rule. They will be brighter in the 200mm pic (by a factor of a little less than 3) if the 80mm exposure is 2.25 times longer than the 200mm exposure.

So in fact there has to be a compromise. Using two different aperture scopes you can arrange for the stars to be the same brightness with different exposures, or you can have the neb the same brightness in different exposures, but you can't have the stars and the neb the same brightness in different exposures.

Last edited by Geoff45; 30-06-2009 at 07:13 PM.
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Old 01-07-2009, 02:46 PM
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Quote:
Originally Posted by ghsmith45 View Post
The 80mm scope will require 2.25 times the exposure to give the same brightness for the nebula, which you get by calculating (7.5/5)^2.


The stars, being point sources follow a different rule. They will be brighter in the 200mm pic (by a factor of a little less than 3) if the 80mm exposure is 2.25 times longer than the 200mm exposure.

So in fact there has to be a compromise. Using two different aperture scopes you can arrange for the stars to be the same brightness with different exposures, or you can have the neb the same brightness in different exposures, but you can't have the stars and the neb the same brightness in different exposures.

Thanks Geoff, thats just what i was looking for, so i could take an exposure that would be shorter, and have less noise in a bigger scope?

now you can see my reason for asking...

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