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# convert SQM reading to photon flux

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### #1 morten

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Posted 20 January 2019 - 08:00 AM

Hi there

Many of us dream of a setup in a remote observatory, or even buying a property in a really dark area. The investment is substantial, and the question is how much better you images will be in terms of SNR. Of course there are issues with clouds seeing etc, but here I'm only interested in SNR.

One way is to use an on-line remote scope, but there are differences in equipment, that complicate this (a bit). In this post I tried a "dry" approach bycalculating how much better my setup would perform at some of the best sites possible. I need you guys to tell me if I screwed up in some way in my calculations. I'm fully aware that this approach is not accurate (or even remotely scientific), but I hope it will end be right ballpark. So opinions on the following are very welcome (I'm aware that the approach is quite banal, and someone probably have posted or published something similar previously - in which case this is just at testimony to my lack of Google skills). Here it goes:

Assumption 1) the SQM(L) near zenith reading is roughly proportional to the skyMag

Assumption 2) My SQM(L) readings can be compared with other peoples SQM(L), if the measurements are done near zenith

Skymag difference between A and B = 2.5log(A/B)

(log is logarithm to the base of 10)

My sky has SQM between 21.1 and 21.5, averagely around 21.2

My single pixel sky background photon-flux  with the L filter averages 1.25 e-/sec (:=SBF(21.2))

Hence under another sky, with SQM = A, the photon flux should be

SBF(A)= exp10((21.2-A)/2.5+log(SBF(21.2)))

Somewhat to my surprise the results showed an more powerful effect of the sky background, that I expected. This graph shows SNR for a very weak object (one of the arcs in the gravitational lens of Abell 2218)

The 21.2 reading is from my Danish countryhouse, around 65km from Copenhagen city centre (app 1 mio inhabitants and capital). In my former (inner) suburban home 7 km north of the city centre  the SQM is around 19.2 (the house was around 1.3 km from the sea, with some woods nearby - this is probably why the SQM was not worse). I have imaged from there, and gotten decent images. The SBF from those images fits with the SQM of 19.1-19.3. However, according to the graph, it should have been more difficult to image than remember. The question is what is wrong here, or is it just the fact that I never attempted to image very weak objects before I started out in the country, so my memory is skewed towards the ease of brighter objects.

Morten

Edited by morten, 20 January 2019 - 08:53 AM.

### #2 sharkmelley

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Posted 20 January 2019 - 08:14 AM

You seem to have done something wrong but I don't know what is wrong because you haven't explained your calculation of SNR.

A change in SQM of 1.0 is a factor of 2.5x in sky brightness (i.e. photon flux) and will lead to a factor of sqrt(2.5) in background noise.  So the SNR for a very faint object will also change by a factor of sqrt(2.5)=1.58x because the signal of the faint object remains constant.

But in your graph (at the 300subs point) the change from SQM 21.8 to 20.8 leads to SNR changing from 6.3 to 0.7 which is a factor of around 9x.  This is far too large.

Mark

Edited by sharkmelley, 20 January 2019 - 08:15 AM.

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### #3 morten

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Posted 20 January 2019 - 08:29 AM

Thank you Mark

You are right, there is an error in the graph (a rather simple one in the cell programming)

You are also right about the lack of specification of the SNR calculation, sorry about that, and thank you for pointing this out-

The correct graph is here.

SNR is calculated as target flux (TF = 0.115 e-/sec) divided by the noise of a stack of  300sec subs. The noise of each sub is sqrt(SBF(A) + TF + read noise). The readnoise is 3.9 e
Camera QSI660 wsg, Filter Astrodon Gen2E tru balance L filter, scope 300 mm f/4 newt with 100mm central obstruction. Images taken at alt >60 degrees, no correction for extinction.

Sorry for leaving this out, I was so obsessed about what was wrong, and looked at it for a while. I now realize that I had entered one of the formula wrongly in the spreadsheet. Now, I do think it is right.

Edited by morten, 20 January 2019 - 08:53 AM.

### #4 sharkmelley

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Posted 20 January 2019 - 08:36 AM

When SQM changes from 20.8 to 21.8 the difference in SNR shouldn't increase by more than a factor of 1.58x

So there's still a problem with your graph.

Mark

### #5 morten

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Posted 20 January 2019 - 09:30 AM

Thank you again Mark

I have assumed that SQM(A) - SQM(B) = -2.5log(SBF(B)/SBF(A))

If SQM changes by one from 20.8 to 21.8, then 1= 2.5log(SBF(20.8)/SBF(21.8)), or the flux ratio SBF(20.8)/SBF(21.8) should be 10^(0,4) = 2,51 (0,4 = 1/2.5). This is the number in the graph. Why should it be 1.58?

Edited by morten, 20 January 2019 - 09:32 AM.

### #6 sharkmelley

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Posted 20 January 2019 - 09:44 AM

Now I understand.  You are plotting flux ratios but your graph title is "SNR as a function of SkyMag (SQM)"  so I thought (maybe wrongly) that it was intended to be a plot of signal-to-noise.

If you are plotting flux ratio then just change the graph title to make this clearer.

Mark

Edited by sharkmelley, 20 January 2019 - 10:12 AM.

### #7 ks__observer

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Posted 20 January 2019 - 10:11 AM

Also remember the SNR is different at each point on the target.

The brighter target points will have more snr than the dimmer areas.

The max benefit jumping each SQM is sqrt(2.5)=1.58, which is the benefit for dimmest target area.

Edited by ks__observer, 20 January 2019 - 10:12 AM.

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Posted 20 January 2019 - 10:12 AM

A change in SQM of 1.0 is a factor of 2.5x in sky brightness (i.e. photon flux) and will lead to a factor of sqrt(2.5) in background noise.  So the SNR for a very faint object will also change by a factor of sqrt(2.5)=1.58x because the signal of the faint object remains constant.

Edit:  I guess my original comment below was wrong.  My plans to get to dark sites are still on.

I was under the (mistaken) impression that for each 1.0 change in SQM, imaging time needed to increase 2.5x to obtain an equivalent SNR.  For example, in my mag 20.0 skies, I thought I needed 6.25 times as much imaging time to reach the same SNR as mag 22 skies.

I'm glad to hear that's not the case.  In order to reach the same SNR in mag 20 as I would at mag 22, apparently I only need 2.5x as much imaging time.  Knowing this greatly reduces my urgency to get to darker skies for imaging.

Edited by chadrian84, 20 January 2019 - 10:38 AM.

### #9 ks__observer

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Posted 20 January 2019 - 10:21 AM

You do need 2.5time/SQM:

1.58xSNR=sqrt(2.5*N)=1.58*sqrt(N)

### #10 sharkmelley

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Posted 20 January 2019 - 10:56 AM

Edit:  I guess my original comment below was wrong.  My plans to get to dark sites are still on.

I was under the (mistaken) impression that for each 1.0 change in SQM, imaging time needed to increase 2.5x to obtain an equivalent SNR.  For example, in my mag 20.0 skies, I thought I needed 6.25 times as much imaging time to reach the same SNR as mag 22 skies.

On the contrary, your original comment was correct.  It is true that for each 1.0 change in SQM, imaging time needs to increase 2.5x to obtain an equivalent SNR.

It is also true that for your mag 20.0 skies, you need 6.25 times as much imaging time to reach the same SNR as mag 22 skies.

The reason is as follows:

Each 1.0 change in SQM changes SNR by a factor of 1.58 ( the square root of 2.5)

Increasing the imaging time by 2.5x also changes SNR by a factor of 1.58.

So the increased imaging time cancels out the SNR change caused by SQM.

Mark

Edited by sharkmelley, 20 January 2019 - 11:32 AM.

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### #11 morten

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Posted 20 January 2019 - 12:45 PM

Thank you Mark for helping me getting this right. I really appreciate the help

It turns out that I forgot to take the square root of the background that increases with a factor or 2.51 (10^0,34) for each increase by one in the SQM. Hence, the shot noise from the background will increase with the square root of that, i.e. 1.58 as you said.

It now makes sense with my prior experience, and the advantage of a truely dark site is still impressive, but not as extreme as my erroneous calculation suggested.

The corrected graph is here

And the factor that the integration time need to be increased as a function of the change in SQM reading is shown here

Thanks to Mark for correcting multiple errors

Edited by morten, 21 January 2019 - 02:07 AM.

### #12 Stelios

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Posted 20 January 2019 - 01:49 PM

It would be interesting if you could extend this to lower SQM's, down to 18 or at the very least 19, as there are several who photograph from light-polluted regions with low SQM's. (I could probably do the calculation myself if I shake the cobwebs off, but you've already set up for it).

Incidentally you probably mean the inverse of the factor that integration needs to be increased. At 22 SQM you need 2.5 *less* integration time than at 20 SQM. Unless I'm totally confused...

### #13 sharkmelley

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Posted 20 January 2019 - 02:13 PM

And the factor that the integration time need to be increased is here

No.

For each 1.0 change in SQM, integration time must change by a factor of 2.5.

As I explained earlier, the reason is as follows:

Each 1.0 change in SQM changes SNR by a factor of 1.58 ( the square root of 2.5)

Increasing the imaging time by 2.5x also changes SNR by a factor of 1.58.

So the increased imaging time cancels out the SNR change caused by SQM.

Mark

Edited by sharkmelley, 20 January 2019 - 02:14 PM.

### #14 morten

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Posted 20 January 2019 - 03:40 PM

You’re right, again
Thank you

I enter correct the previous post with the correct graph, not to clutter this thread more with.

Stelios the new graph just show the relative changes and can thus be extended to other SQM values.

The idea was to present a simple way to transfer ones system to another sky and see what happened.

Somehow I did this too carelessly and ended up with a number of errors, Thanks to Mark these (I hope) now got corrected.

Edited by morten, 20 January 2019 - 04:56 PM.

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Posted 20 January 2019 - 05:46 PM

Are these the correct numbers for how 0.25 increments in SQM affect integration time?  I'm not confident I calculated it right.

### #16 morten

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Posted 21 January 2019 - 02:20 AM

No, as Mark said for each decrease by one in SQM you have to increase the exposure by a factor of 2.51 to obtain the same SNR. The correct data are below (and I have changed the curve accordingly in the previous post)

In my experience there is a small penalty as you do not really get the theoretical increase in SNR once you integrate a large number of subs, probably because of normalization issues, or despite meticulous checking inclusion of some subpar subs.

As an example, here is an example of an image of Abell2218 where I directly have measured the SNR

Here is a comparison between the theoretical SNR curve (red) and the measured data blue. Once I get more than a couple of hundred subs the SNR increase starts to loose pace. I've seen this in many different projects not only Abell2218

Finally the table

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Posted 22 January 2019 - 07:15 AM

Ok, got it.  Below is the formula along with an updated chart of.25 SQM intervals.

### #18 555aaa

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Posted 22 January 2019 - 10:42 AM

Don't forget that seeing can have as large of an effect on SNR as sky brightness once you are imaging below the seeing limit as most are.

### #19 sharkmelley

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Posted 22 January 2019 - 02:02 PM

Don't forget that seeing can have as large of an effect on SNR as sky brightness once you are imaging below the seeing limit as most are.

Can you explain why?

Mark

### #20 555aaa

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Posted 22 January 2019 - 03:58 PM

Can you explain why?

Mark

See this paper. The better the seeing, the more each unit area of sensor contains only star+sky or only sky, and not a smear of both. The paper below was written for stars, for which there is a well-defined SNR measurement, but I believe that the same concept applies to extended objects. What is the SNR of some nebula? If you take all the light of that nebula and lump it into one fuzzy blob, you will have blob + sky background, and then with the sky you have sky background, so you have a single number SNR. But you also have a very bland image of a grey blob. It contains information, but only one value. The image becomes interesting and appealing when it contains the information of where are the edges or contours of the nebula versus where is the nebula not present. That means high spatial frequencies with low noise, and so such an image has a lot higher information content, and it also requires better seeing. If you look at the paper or Wei-Hao's paper that he cited you can see that seeing is critical to get high SNR on stars and I believe that the same really applies for extended objects. This is why I always cringe when people say "my stars are round, I don't need to improve" because they don't know how much deeper they can image with better guiding (or focusing or better control of tube currents etc.).

Wei-Hao's formula is in this thread:

Edited by 555aaa, 22 January 2019 - 05:11 PM.

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### #21 choward94002

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Posted 22 January 2019 - 05:11 PM

Can you explain why?

Mark

555aaa has a much more technical explanation ... mine is more empirical.  When seeing is bad the object will be "jumping" around (that's the twinkle) which will smear the fine details over time ... I consider SNR to not only be the calculated values but also the level of detail that I can see in the picture ... and that level of detail (which the paper describes as high spatial frequency) will vary based on the seeing.

What that means for me is that generally when my seeing is 1.75-1.5 I'm only going to be going after big things like HorseHead or Running Man (which don't have much fine detail), when it's 1.5-1.25 I'm going to be going after the larger galaxies like M33 or planetaries like M1 and when it's 1.25-1.0 (which happens maybe three or four times a year for me) it's time to hope that Sombrero or Cigar galaxies are around ...

Edited by choward94002, 22 January 2019 - 05:13 PM.

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### #22 morten

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Posted 23 January 2019 - 10:24 AM

See this paper. The better the seeing, the more each unit area of sensor contains only star+sky or only sky, and not a smear of both. The paper below was written for stars, for which there is a well-defined SNR measurement, but I believe that the same concept applies to extended objects. What is the SNR of some nebula? If you take all the light of that nebula and lump it into one fuzzy blob, you will have blob + sky background, and then with the sky you have sky background, so you have a single number SNR. But you also have a very bland image of a grey blob. It contains information, but only one value. The image becomes interesting and appealing when it contains the information of where are the edges or contours of the nebula versus where is the nebula not present. That means high spatial frequencies with low noise, and so such an image has a lot higher information content, and it also requires better seeing. If you look at the paper or Wei-Hao's paper that he cited you can see that seeing is critical to get high SNR on stars and I believe that the same really applies for extended objects. This is why I always cringe when people say "my stars are round, I don't need to improve" because they don't know how much deeper they can image with better guiding (or focusing or better control of tube currents etc.).

Wei-Hao's formula is in this thread:

Thank you for the reference. Isn't this the old discussion about single pixel SNR versus area SNR (Patrick Moore vs. Craig Stark f.ex).

It is clear that if the stars images are disturbed by turbulence their light is distributed to more pixels with lower levels in each pixel. Hence the star is more likely to drown in the background noise. For a given exposure you will therefore see more stars when the seeing is good than when the seeing is bad. We probably all use the number of stars as an sort of through-the-night ongoing quality parameter for our images, and know when it drops it is either cloud or bad seeing.

The single pixel SNR that my post was about is mathematically, and conceptually simpler, but it ignores area SNR, and as you said, it therefore ignores something important

BTW Viking 1 you are really blessed with good seeing. I'm happy wien mine is 1.6, and extremely rarely it is lower than 1.5.

### #23 ks__observer

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Posted 12 February 2019 - 06:59 AM

It is true that for each 1.0 change in SQM, imaging time needs to increase 2.5x to obtain an equivalent SNR.

It is also true that for your mag 20.0 skies, you need 6.25 times as much imaging time to reach the same SNR as mag 22 skies.

I believe this off by a sqrt factor.

You just need 1.58x more target signal to wash out 2.5x more sky noise.

SNR = T/sqrt(T + S)

Say for dimmest part of target T=1, and say S=400:

SNR = 1/sqrt(1+400) = 1/sqrt(401)

Now say you drop one SQM:

SNR = 1/sqrt(1 + 400×2.5) = approx 1/(sqrt(400)*sqrt(2.5))= 1/(sqrt(400)×1.58)

So to match the previous sky you need 1.58x more target signal.

Chadrian was correct in his original post ....

Edited by ks__observer, 12 February 2019 - 07:19 AM.

### #24 sharkmelley

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Posted 12 February 2019 - 02:19 PM

I believe this off by a sqrt factor.

You just need 1.58x more target signal to wash out 2.5x more sky noise.

SNR = T/sqrt(T + S)

Say for dimmest part of target T=1, and say S=400:

SNR = 1/sqrt(1+400) = 1/sqrt(401)

Now say you drop one SQM:

SNR = 1/sqrt(1 + 400×2.5) = approx 1/(sqrt(400)*sqrt(2.5))= 1/(sqrt(400)×1.58)

So to match the previous sky you need 1.58x more target signal.

Chadrian was correct in his original post ....

No, that's incorrect. What your equations show is that a change of 1 magnitude in light pollution alters the SNR by a factor of 1.58 - we all agree with that.

But you then draw the wrong conclusion.  If you increase your exposure time by 1.58x to give more target signal then the background noise also increases so you don't actually achieve a 1.58x increase in SNR.

In fact you need to increase imaging time by 2.5x to change SNR by a factor of 1.58.

The increased imaging time then cancels out the SNR change caused by increase in light pollution.

Mark

Edited by sharkmelley, 12 February 2019 - 02:20 PM.

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### #25 ks__observer

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Posted 12 February 2019 - 03:24 PM

You are correct.

T = target flux

t = time

S= sky flux

Tt1/sqrt(Tt1 + St1) = Tt2/sqrt(Tt2 + SQM-Delta*St2)

squaring both sides

t1/(T+S) = t2/(T+delta*S)

Assume T is zero

delta*t1 =t2

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