Exposure time question

[markse68]

Hi all,

if you had a telescope A with an aperture of x cm^2 and another telescope B with an aperture of 10x cm^2, and both telescopes were fitted with camera sensors that gave the exact same fov, would telescope B require 1/10th the exposure time of scope A on the same target irrespective of f ratio (ignoring obstructions from secondary mirror etc)? I'm trying to work out if it's possible to estimate exposures from looking at images on Astrobin and scaling them to the aperture. My feeling is yes but I'm not entirely sure and of course other things like sensor QE etc will have an effect. If they both had the same sensor format and the same number of pixels just scaled to get the same fov?

Thanks, Mark

[The Lazy Astronomer]

I tried to think about this, but my thought processes were only capable of going as far as "something, something... pixel scale... something... seeing conditions" 😃

This seems like the sort of question @vlaiv will answer comprehesively.

[vlaiv]

15 hours ago, markse68 said:

if you had a telescope A with an aperture of x cm^2 and another telescope B with an aperture of 10x cm^2, and both telescopes were fitted with camera sensors that gave the exact same fov, would telescope B require 1/10th the exposure time of scope A on the same target irrespective of f ratio (ignoring obstructions from secondary mirror etc)?

No.

15 hours ago, markse68 said:

My feeling is yes but I'm not entirely sure and of course other things like sensor QE etc will have an effect. If they both had the same sensor format and the same number of pixels just scaled to get the same fov?

Yes, all of the things will have impact. First place is pixel size.

10x aperture scope will collect 10x light from same FOV. But how will all that light be divided into chunks? Smaller number of chunks means higher value per chunk.

If pixels are the same - then yes - 10x aperture scope will have 10x light fall on each pixel.

Next comes QE. Say first camera has 60% QE while second has 80% QE - that is 1/4 increase in QE - not insignificant.

Given that all parameters are equal - then yes, 10x aperture will have x10 higher signal than x aperture scope for same exposure time. Problem is - parameters are never equal. Even night to night differences can be significant in terms of achieved SNR. Transparency can change as much as 0.5 mags or more between the nights. That is 50% increase / reduction in signal strength from target! - Yep, almost the same as difference between 4h of imaging and 6h of imaging.

Not to mention LP.  One mag of difference in SQM can be roughly x2.5 in imaging time! Yes, what you can achieve in 1h in SQM 20 sky - you'll need 2.5h of exposure in SQM19 sky and 6.25h in SQM18 sky.

[Tommohawk]

I think you also need to consider that if scope has 10x aperture and same focal length then above is true. But if has same F value then it will have 10x the FL, in which case you don't have any extra light per pixel.

Doubtless someone will correct me if I'm wrong!

 

[markse68]

8 hours ago, The Lazy Astronomer said:

I tried to think about this, but my thought processes were only capable of going as far as "something, something... pixel scale... something... seeing conditions" 😃

Haha- it made my head hurt a bit!

 

8 hours ago, The Lazy Astronomer said:

This seems like the sort of question @vlaiv will answer comprehesively.

yes- I thought he might know the answer, although I'm still a little confused- first he said no and then he said yes- I think

 

8 hours ago, vlaiv said:

No.

 

8 hours ago, vlaiv said:

Yes

Thanks Vlaiv I think

8 hours ago, vlaiv said:

If pixels are the same - then yes - 10x aperture scope will have 10x light fall on each pixel.

but I guess the larger pixels needed to match fov would have corresponding larger wells so would need the extra light to reach similar adu reading? So maybe it all cancels out? Your other points all make excellent sense too- calculating exposure sounds pretty impossible really

26 minutes ago, Tommohawk said:

But if has same F value then it will have 10x the FL, in which case you don't have any extra light per pixel

Thanks Tommohawk, I think that would be the case if using the exact same sensor but then the fov would change so I was wondering what if the sensor was scaled to match the fov.

Thanks all for input- I really must read that book I bought ages ago

Mark

[vlaiv]

7 minutes ago, markse68 said:

Thanks Vlaiv I think

I said no because in first sentence you said that we only have matching FOV and 10X vs X aperture size. If only those two conditions are met - then answer is no, there is no guarantee that same SNR will be reached in x10 less time with 10X aperture vs X aperture. In fact - odds are that it will not.

I answered yes to the second part where you started questioning pixel size and quantum efficiency and all the rest - that yes refers to the fact that all conditions must be met for SNR to be the same in x10 less time. Only then you can say 10x aperture will have same SNR in x10 less time.

By the way - I did not specify but its worth mentioning - X and 10X are aperture area not diameter.

11 minutes ago, markse68 said:

but I guess the larger pixels needed to match fov would have corresponding larger wells so would need the extra light to reach similar adu reading? So maybe it all cancels out? Your other points all make excellent sense too- calculating exposure sounds pretty impossible really

No canceling out and you don't need to have larger pixel size.

Here is what happens - telescope aperture gathers light and pixel size divides light. Say you have target - how many photons will be collected by telescope - depends on aperture size. Larger aperture size - more photons collected in same amount of time. Number of photons is proportional to aperture area.

All photons from target get divided by all pixels covering the target. If you have smaller pixels - more of them is needed to cover target, if you have larger pixels - less of them is needed to cover the target. All the light from target is spread over those pixels covering the target - and no photons end up elsewhere (some of them actually do end up elsewhere but we model that by other means - like reflectivity of mirrors / transmission of glass, scatter and such).

Say your target gives off 10000 photons per time period. If it is covered by large pixel - only 100 of them is enough to cover the target - we end up with 100 photons per each pixel (10000/100).

If we use smaller pixels and this time we need 400 pixels to cover the target - now we get only 25 photons per pixel as 10000/400 = 25.

Smaller pixels - mean less signal and poorer SNR.

Well depth really does not come into this. It somewhat dictates single exposure length and not total imaging time and only if you want to avoid saturation, but in reality - you'll base your exposure length on read noise and not FW. Any saturated parts you can always "fill in" with handful of short exposures at the end of the session.

20 minutes ago, markse68 said:

Thanks Tommohawk, I think that would be the case if using the exact same sensor but then the fov would change so I was wondering what if the sensor was scaled to match the fov.

You not only have to scale sensor to match FOV - you also need to "scale pixels" to match resolution. With larger sensor you can always get larger pixels by binning them.

If you want to compare two systems very basically - use "aperture at resolution" or "aperture at resolution times QE" if you want to include QE of sensor (it's significantly different). Just be careful that both aperture and resolution need to be squared (both need to be area rather than length - like diameter or pixel size) if you want to compare it to time (have linear dependence).

Here is quick example. How does 100mm at 2"/px and 60% QE compare to 150mm at 1.6"/px and 80% QE in terms of "speed"?

100^2 * 2^2 * 0.6 = 10000 * 4 * 0.6 = 24000

150^2 * 1.6^2 * 0.8 = 22500 * 2.56 * 0.8 = 46080

150mm setup is x1.92 (46080/24000) faster than 100mm setup - or 150mm in 1h will have same signal as 100mm in 1.92h if they image on the same night under same conditions.

 

[markse68]

Thanks Vlaiv for going to so much trouble with your explanation. I think i’m starting to understand it better as a result- not really there yet- some experimenting will help I think

Mark