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Converting surface brightness in stellar magnitudes to photon flux?

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#26 hytham

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Posted 13 May 2016 - 02:53 PM

This thread is what I love so much about this hobby ...

 

It's not just about pretty pictures. Thank you for the education, everyone!

 

EDIT: Thanked smart people for teaching me stuff.


Edited by hytham, 13 May 2016 - 02:56 PM.

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#27 freestar8n

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Posted 13 May 2016 - 06:00 PM

Ok, so I used the formula John sent me. I stuffed it into Excel and played around with the numbers a bit. I am wondering if someone could check my results. 
 
For a 21.5mag star, although I assume that would work for any mathematical point in space including a point on an extended object that has that magnitude, with a bandpass of 300nm (basically the full visible spectrum, Luminance filter), with 80% transmittance from the atmosphere and my scope combined (honestly not sure what the actual transmittance of the atmosphere may be at the zenith on a clear night, so I am just guessing here), with a 150mm aperture, I get a photon flux of 1.065 γ/s. For a magnitude 24 star, I get a flux of 0.1065 γ/s. For a narrow bandpass of 3nm (wavelength of 0.555 for now, as I have not yet figured out the irradiance of a mag 0 star at other wavelengths), that flux drops to 0.001065 γ/s. If I drop the aperture down to 80mm, the flux drop down to 0.000303 γ/s, and if I increase the aperture to 203mm the flux jumps to 0.002 γ/s. 
 
Do those results sound reasonable?


I'm not sure exactly what you are aiming for. If you don't care about star color and the exact transmission and response of your system, you can estimate the photon count based on Vega using a table like this:

http://www.astronomy...usefuldata.html

The top table shows the photon flux in photons per cm^2 per sec per Angstrom - for Vega. It's different in each band but in each band you can multiply by the width of the filter you are using. In all this stuff you have to keep track of photons per lamda - and be careful with any mention of power, because a given power at one wavelength will have a different photon flux than the same power at another wavelength. For Vega it is a hot star so the blackbody peaks in the B filter - but at the same time a given power in B will have fewer photons than a given power in R - because the B photons have more energy.

And for a cooler star with the same V mag - it will have a different total photon flux because there is more power in R and those photons have less energy.

If you just want to measure directly how your system responds - you can do as I described above and just measure a star of known magnitude and the total electron count. Then given the exposure time you can predict what it will be for anything that is about the same color.

But if you do want to keep track of color you would also need to keep track of transmission in the system and spectral response of the detector. That is more involved - but once you break it down to doing actual numerical integrals - in some sense everything is easier because you are just multiplying factors and integrating. But you need to make sure everything is in photon counts per unit lambda - if you integrate over lambda.

For my study on guidestar color:

http://www.cloudynig...ess-and-number/

I took the magnitude of each star in one filter and deduced its blackbody temperature from the color index, or difference in mags between two filters such as B-V or g'-r'. Once you have the temperature you have the blackbody curve - in photons per lambda - and you can multiply that by the various losses and system response - and get the actual photon count in different filters.

But if you want a simple estimate there are simple ways to do it. And if you want something more exact you would need to handle color and system response properly.

Frank

#28 Jon Rista

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Posted 13 May 2016 - 06:55 PM

I am not really concerned with stars. I am concerned with extended objects, particularly nebula, particularly in narrow bands (Ha, OIII, SII and possibly NII.) I am also more concerned about the photon flux of the objects agnostic of the telescope, really. The goal is planning. I'd like to be able to plan out how much integration time I would need to get a desired SNR on the fainter parts of a nebula in a given narrow band, before I actually start imaging it. So I can figure out how many subs per night over how many nights I'll need to image it to get that SNR on the fainter regions of the nebula.

 

So, really, I'd prefer to know the photon flux of the object through the atmosphere, and just stop there. I could adjust roughly for the specific telescope if I need to. I don't need high precision accuracy here...I mostly need ballpark. I'm not doing photometry, I'm choosing objects to image, and trying to plan around my poor weather and poor seeing to figure out which objects are worth spending time on, and which are not. Or, which may require more than one year to mosaic the entire field I'm interested in. I guess it's a more humble and simplistic goal than others may have.

 

I understand the concept of converting surface brightness to photons per second has applications beyond my simple endeavors. I don't want to shut down discussion on the subject, so feel free to use the thread to continue discussing it and the possible applications.


Edited by Jon Rista, 13 May 2016 - 06:58 PM.


#29 catalogman

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Posted 13 May 2016 - 08:21 PM

 


 

 

Just be aware that candelas and lumens are part of the photometric system, which is weighted to the response of the human eye and ultimately relates back to the brightness of a candle. I have an easier time dealing with simple radiometry, which uses radiance (W/m^2-str), irradiance (W/m^2), Power (W), and Energy (J).  You can certainly make the conversion between the two systems but I personally find that folding in the responsivity of the eye makes things confusing--particularly when computing a detector response.

 

John

 

 

The OP used candelas. But you're right that a newbie might need a more detailed explanation:

 

- To convert photometric units to radiometric units, use this table:

 

    http://www.gamma-sci...ersion-factors/

 

  and this formula:

 

    (photometric unit) = (radiometric unit) x (683) x (photopic response)

 

- Find the area of the source in steradians:

 

    http://www.calculato...ame=solid angle

 

  then multiply (steradians) X (radiometric units) to find the power P collected.

 

- To convert to photons/sec, another way is to use the relation

 

    1 W = 5.03 * (wavelength in nm) X 1015 photons/sec

 

  or use this on-line calculator:

 

    http://www.calctool....y/power_photons

 

For a heterochromatic source, one method is create a photon counting histogram for a
few colors. A more sophisticated method would be to calculate enough sample photon
counts for many wavelengths to create a distribution so that the area of the histogram
becomes a numerical integral.

                                                                                                     -- catalogman


Edited by catalogman, 14 May 2016 - 05:25 AM.


#30 catalogman

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Posted 13 May 2016 - 08:36 PM

Here is a table of some surface brightnesses in both mag/arcsec2 and
mag/arcmin2 that the OP might like to check for calibration purposes:

 

  http://www.stjarnhim...omp/radfaq.html

 

(Scroll down to #10 and #11.)
                                                       -- catalogman


Edited by catalogman, 13 May 2016 - 08:37 PM.


#31 freestar8n

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Posted 13 May 2016 - 08:47 PM

The OP used candelas.                                                                          -- catalogman


Actually - I think you were the first one to bring up candelas. The OP was asking about photon flux. I think candelas and perceptual response are interesting - but a separate issue.

Since the OP was ultimately asking about SQM and sky brightness - there are a ton of links and references on the SQM site:

http://www.unihedron...ojects/darksky/

This paper: http://unihedron.com...report_v1p4.pdf goes into great detail on calibrating it and the spectral response. Since the SQM response doesn't match any standard system - he basically says it requires its own photometric definition of magnitude - adding to the "226 known" ones. So it's hard to be precise about what it even means by "magnitude" - but it describes ways to map from one system to another if you know the spectrum of what you are measuring.

Frank
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#32 freestar8n

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Posted 13 May 2016 - 09:06 PM

For a 21.5mag star, although I assume that would work for any mathematical point in space including a point on an extended object that has that magnitude, with a bandpass of 300nm (basically the full visible spectrum, Luminance filter), with 80% transmittance from the atmosphere and my scope combined (honestly not sure what the actual transmittance of the atmosphere may be at the zenith on a clear night, so I am just guessing here), with a 150mm aperture, I get a photon flux of 1.065 γ/s. For a magnitude 24 star, I get a flux of 0.1065 γ/s. For a narrow bandpass of 3nm (wavelength of 0.555 for now, as I have not yet figured out the irradiance of a mag 0 star at other wavelengths), that flux drops to 0.001065 γ/s. If I drop the aperture down to 80mm, the flux drop down to 0.000303 γ/s, and if I increase the aperture to 203mm the flux jumps to 0.002 γ/s. 
 
Do those results sound reasonable?


From the table I sent earlier, Vega has 995.5 photons / cm^2 / second / Angstrom in the V-band

So given your numbers for a mag 21.5 star, with Vega having V Mag 0, I get:

Phot/sec = 995.5 * 3000 * 0.8 * pi * 7.5^2 * 10^(-0.4*21.5) = 1.06 photons/sec

which matches your result.

Every 2.5 mags it will go down by a factor of 10 - so mag 24 would have 0.106

If it's a star that will be the rate photons arrive on the detector from the star. If it's sky glow and that is the magnitude per square arc-second - that is the amount that will arrive per square arc-second.

If you have a detector with that telescope and its pixels are one arc-second square - then each pixel will receive 1.06 photons per second from that sky glow. And if you use that same detector under that same sky with any other telescope - the number of photons per pixel will only depend on f/ratio.

Frank

Edited by freestar8n, 13 May 2016 - 09:11 PM.

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#33 catalogman

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Posted 13 May 2016 - 10:17 PM

<snip>

I'd like to be able to plan out how much integration time I would need to get a desired SNR on the fainter parts of a nebula in a given narrow band, before I actually start imaging it. So I can figure out how many subs per night over how many nights I'll need to image it to get that SNR on the fainter regions of the nebula.

<snip>

 

For another example of calculating exposure time from known radiant intensity, scroll down to Section 13.3:

 

http://www.stjarnhim...omp/radfaq.html

 

But see the warning about reciprocity failure. So if you're using film, your approach may not even be

valid for the faint regions.

 

                                                                                                                                       -- catalogman


Edited by catalogman, 13 May 2016 - 10:19 PM.


#34 freestar8n

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Posted 13 May 2016 - 11:10 PM

If you want to know the number of electrons per pixel you can find it just knowing the pixel size and the f/ratio.

e/s/pixel = 3.3E6 * (bandwidth in nm) * (pix size in um / FRatio)^2 * 10^(-0.4 * mag/sqarcsec)

Again this assumes V band. You would then multiply by transmission loss - etc.

That would be the sky background signal - as a rate in e/s. So you would multiply by exposure time - and take the sqrt of the result to find the sky background noise.

And it would add in quadrature to read noise, dark current noise - etc.

Frank

Edited by freestar8n, 13 May 2016 - 11:15 PM.

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#35 Jon Rista

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Posted 17 May 2016 - 02:41 PM

Frank, thanks. That last formula might do the job I am looking for.

I am curious, when you say V-band. Is that the 40-75Ghz frequency range? Or are you using that term to refer to "visual" band? If it does not apply to the visible spectrum, I assume the 3.3E6 coefficient might need to change?

#36 freestar8n

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Posted 17 May 2016 - 04:38 PM

Yes - Visual band - V filter in Johnson-Cousins system, centered at about 550 nm. You can't really talk about "magnitude" without an implied filter response. The SQM measures magnitude with its own filter response - so it is hard to map directly to other filters. The most common one to use is the V magnitude.

From the useful astronomical data link I provided above, the photon flux factor for V filter is 995.5, while for B it is 1392.6 and for R it is 702.0. So the 3.3E6 factor would increase for B and decrease for R - in direct proportion.

If you want to know the full photon count from a given star with a Lum filter you would need to know its color and add up the counts in each filter - and at the same time figure in the spectral response of the optical system and detector. You can see examples of that in the thread I point to above where I calculate flux from guidestars.

But you should be able to get a decent estimate using the counts in B V R filters. A star with Vega's color will have more counts in B - over a given bandwidth - than in V or R - assuming the system response is about the same.

Frank
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#37 Shawnxc

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Posted 17 May 2016 - 07:52 PM

If you want to know the number of electrons per pixel you can find it just knowing the pixel size and the f/ratio.

e/s/pixel = 3.3E6 * (bandwidth in nm) * (pix size in um / FRatio)^2 * 10^(-0.4 * mag/sqarcsec)

Again this assumes V band. You would then multiply by transmission loss - etc.

That would be the sky background signal - as a rate in e/s. So you would multiply by exposure time - and take the sqrt of the result to find the sky background noise.

And it would add in quadrature to read noise, dark current noise - etc.

Frank

I had been looking into this topic also, this equation is the clearest and simplest so far. Thanks.

 

Can you help me to clarify a few things? the output e/s/pixel, is it possible it should be incoming flux in photons/pixel? And then multiply by QE, transmission losses to get e/s/pixel? Doing it this way, numbers matches with my calculations better.

 

how to estimate transmission loss? 

 

If not too much trouble, can you show a few steps leading to this equation? Especially the pixel size/f. Just want to understand a bit better


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#38 freestar8n

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Posted 18 May 2016 - 04:20 PM

I had been looking into this topic also, this equation is the clearest and simplest so far. Thanks.
 
Can you help me to clarify a few things? the output e/s/pixel, is it possible it should be incoming flux in photons/pixel? And then multiply by QE, transmission losses to get e/s/pixel? Doing it this way, numbers matches with my calculations better.
 
how to estimate transmission loss? 
 
If not too much trouble, can you show a few steps leading to this equation? Especially the pixel size/f. Just want to understand a bit better


Hi-

The numbers at http://www.astronomy...usefuldata.html say that Vega at mag 0 will produce 995.5 photons per cm^2 per second per Angstrom of bandwidth.

So for a telescope with Aperture D and focal length f the number of photons arriving per second from a star of magnitude M is

photons/sec = 995.5 * (bandwidth in Angstrom) * pi * (D(cm)/2)^2 * 10^(-0.4*M)

If instead of talking about a star, we have nebulosity that has mag M per square arc-second, and we have pixels of width w, then first we can find how many square arc-seconds each pixel captures.

The width of the pixel in arc-seconds is

w(arc-sec) = w(microns)/f(mm) * 206.265

so the area in square arc-sec of that pixel is just that number squared

Multiplying the two above and converting bandwidth to nm and diameter to mm we get

photons/pixel/sec = 9955 * (bandwidth in nm) * pi * (D(mm)/2)^2 * 10^(-0.4*M(mag/sqarcsec)) * [w(microns)/f(mm)*206.265]^2

When you multiply that out you get the simple version I show above. This is for the V magnitude and you would just change that front factor according to the link for B and R mag.

This corresponds to all photons entering the aperture that will be imaged onto the pixel. If you want electron count instead - you would multiply by transmission losses and QE of the sensor.

The link I show above to the thread on guidestar color has transmission curves for a typical sct - along with response curves for some sensors. That should help give an idea of losses across the spectrum. And you also have to factor in star color.

But as a ballpark figure for background sky glow on a sensor - this should work well. For a given sensor it is set by f/ratio rather than aperture.

Frank
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#39 Shawnxc

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Posted 19 May 2016 - 02:51 AM

Thanks, Frank. Excellent math with a very useful equation. It makes exposure SNR calculation a lot easier 



#40 freestar8n

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Posted 19 May 2016 - 05:02 AM

Thanks Shawnxc-

Here is a plot showing the main factors at work. Transmission of B, V, R filters, transmission of XLT SCT, QE of Sony 694 ccd, and photon flux (per nm) of two star types: Sun at 5500K, and Vega at 9600K.

You have to visualize how a given star would have a certain V magnitude - and how much flux it would end up contributing across the visible spectrum. The peak in Vega flux is in B, but the purple curve begins to plummet in the B.

Frank

JCSCT.png
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#41 josh smith

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Posted 20 May 2016 - 09:16 AM

Favorite thread I've read in some time! Thanks for all the useful info. :)

#42 Shawnxc

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Posted 20 May 2016 - 09:52 AM

Favorite thread I've read in some time! Thanks for all the useful info. :)

 

agree



#43 Jon Rista

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Posted 20 May 2016 - 05:08 PM

 

I had been looking into this topic also, this equation is the clearest and simplest so far. Thanks.
 
Can you help me to clarify a few things? the output e/s/pixel, is it possible it should be incoming flux in photons/pixel? And then multiply by QE, transmission losses to get e/s/pixel? Doing it this way, numbers matches with my calculations better.
 
how to estimate transmission loss? 
 
If not too much trouble, can you show a few steps leading to this equation? Especially the pixel size/f. Just want to understand a bit better


Hi-

The numbers at http://www.astronomy...usefuldata.html say that Vega at mag 0 will produce 995.5 photons per cm^2 per second per Angstrom of bandwidth.

So for a telescope with Aperture D and focal length f the number of photons arriving per second from a star of magnitude M is

photons/sec = 995.5 * (bandwidth in Angstrom) * pi * (D(cm)/2)^2 * 10^(-0.4*M)

If instead of talking about a star, we have nebulosity that has mag M per square arc-second, and we have pixels of width w, then first we can find how many square arc-seconds each pixel captures.

The width of the pixel in arc-seconds is

w(arc-sec) = w(microns)/f(mm) * 206.265

so the area in square arc-sec of that pixel is just that number squared

Multiplying the two above and converting bandwidth to nm and diameter to mm we get

photons/pixel/sec = 9955 * (bandwidth in nm) * pi * (D(mm)/2)^2 * 10^(-0.4*M(mag/sqarcsec)) * [w(microns)/f(mm)*206.265]^2

When you multiply that out you get the simple version I show above. This is for the V magnitude and you would just change that front factor according to the link for B and R mag.

This corresponds to all photons entering the aperture that will be imaged onto the pixel. If you want electron count instead - you would multiply by transmission losses and QE of the sensor.

The link I show above to the thread on guidestar color has transmission curves for a typical sct - along with response curves for some sensors. That should help give an idea of losses across the spectrum. And you also have to factor in star color.

But as a ballpark figure for background sky glow on a sensor - this should work well. For a given sensor it is set by f/ratio rather than aperture.

Frank

 

 

Thank you, Frank. Extremely useful! :) Exactly what I needed. You are a wealth of information on these kinds of things. I really appreciate that, and the time you take to share your knowledge.


Edited by Jon Rista, 20 May 2016 - 05:12 PM.


#44 freestar8n

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Posted 20 May 2016 - 07:49 PM

Thanks for the comments everyone. I have done this stuff in many ways over the years but this thread motivated me to put things in these simple terms - and it is useful for me also.

I got a little concerned that the estimated flux was too high - but I just did some checks and things seem ok.

Here are some more useful expressions - starting with the original:

s (sky flux in e/s) = T(Transmission loss factor) x 3.3E6 x bandwidth(nm) x [pixelsize(um) / FRatio]^2 x 10^(-0.4M (mag/sqarcsec))

From the graphs above, a rough value for T would be 0.5 due to optical loss and ccd QE.

For a given setup everything would be fixed except M - so you can find s based on M or vice versa.

Here is M as a function of s:

M = 2.5 x log10[(3.3E6 x bandwidth x p^2) / (T x F^2)] - 2.5 x log10(s)

Again the first part is a constant for a system and you just need to subtract 2.5*log10(s) to find the sky background mag per square arc-sec.

One final thing to calculate is the sub-exposure length at which the contribution from read noise is equal to the contribution from sky and dark noise. That is the time when r^2 = (s+d)*t - so t = r^2/(s+d)

If you know r and d for your camera you can estimate the sky background signal and know this crossover time beyond which sky and dark noise begin to win over read noise. Nothing magically changes at that point and read noise continues to be a factor - but you need to go at least that far if you want to have read noise not be the dominant noise term.

I recently changed my EdgeHD from f/7 to f/10, with 8300ccd and 5.4 um pixels. I operate at -10C so the dark current is 0.02e/s and read noise is fairly high at about 10.5e.

My location has fairly clear skies in a suburban setting but there are bright streetlights around so it is hard to know the limiting mag. But it is probably about mag 5 limiting mag skies.

Using a sloan rp filter with bandwidth 150nm and a transmission factor of 0.5 my system yields

M = 19.6 - 2.5xlog10(s)

I measured the sky background in images of 20s and 600s as about 1.9 e/s, so I get

M = 18.9 mag/sqarcsec

And this corresponds to about a mag 4.7 sky, which is about right.

For that sky and that setup, the crossover exposure time is 10.58^2/(1.9+0.2) = 53 seconds

So - dark current is negligible here and it is sky background limited - and I need to go about a minute for sky background to begin winning over read noise.

I also did some 3nm Ha images - again at f/10 - and the story there is quite different. In that case my sky background is only 0.03 e/s - which is just larger than the dark current at 0.02 e/s. The crossover exposure time is then:

t = 10.58^2/(0.03+0.02) = 37 minutes

This means I need very long exposures to avoid being completely dominated by read noise. And if I reduced the temperature more the noise would go down a bit - but it would also be even more read noise dominated - and I would need to go even longer to 47 minutes.

This last point causes a lot of confusion because it makes it seem like things are worse because you need to expose longer. But you don't *need* to expose longer because more cooling actually reduces the noise and your images improve. What it does mean is that you now need to go even longer for sky and current noise to dominate read noise.

For bright Ha objects I don't need very long exposures because the whole field is glowing with strong object signal. But for faint objects with dark background, f/10 narrowband really does benefit from long exposures.

But even with RGB imaging at f/10, typical sky glow may be fine with shortish exposures. It depends on the setup and the sky - but a lot of this can be easily estimated.

Frank
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#45 Shawnxc

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Posted 20 May 2016 - 08:08 PM

I checked the output against a few sets of on camera measurements also. Seems to be pretty accurate. This formular is one of the most important formulars to understand astrophotography and interaction between the factors, in my humble opinion. And beautifully simple. Nice work Frank!


Edited by Shawnxc, 20 May 2016 - 08:19 PM.


#46 Alph

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Posted 21 May 2016 - 11:03 AM

Anyone interested in the subject should read this reference.
 
http://www.eso.org/o...book/node6.html
 
It has been there for almost 17 years.  There is no need to reinvent the wheel here. That site contains a lot of useful details, like tabularized filter profiles which come handy if you want to integrate over the filter bandwidth (for higher accuracy) instead of just using the filter FWHM. You will also find there the equations for point sources and extended sources side by side. This should be relevant to the OP who keeps mentioning object SNR; study it carefully.

 

The following PDF succinctly summarizes basic facts from  the ESO formula book

 

https://www.lsw.uni-...s/docu_html.pdf


Edited by Alph, 21 May 2016 - 04:33 PM.


#47 freestar8n

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Posted 24 May 2016 - 05:16 PM

If it's ok with the OP I wanted to give an updated suggestion for using SQM mag values to estimate sky flux per pixel.  What I show above lets you estimate the flux given the sky mag, pixel size, and f/ratio - plus some assumptions about the sky color and OTA transmission.  But for a given system you can calibrate it to the sqm measurement for each filter - and it should apply well to other systems under the same sky.

 

Ultimately all that matters on a per-pixel basis if the ratio of pixel width to f-number, or p/F.  This isn't a surprise because (p/F)^2 is related to the optical throughput of each pixel, and also tied to the Lagrange invariant or optical invariant for each pixel.  Only the ratio matters - and if you change pixel size or f/ratio you don't need to know each value - just the ratio.

 

The sky flux in e/s can then be written as:

 

s = C (p/F)^2 10^(-0.4M)

 

where p is the pixel size in microns, F is the f/ratio, and M is the SQM sky magnitude.

 

C is a constant that will depend on the filter used, the sky color, the system transmission, and detector response - but for typical amateur systems in typical settings - the values of C may not differ much among users.  That's what I'd be interested to find out.

 

If you take an SQM measurement and then do exposures with RGBL filters, each filter will have a corresponding C value based on the measured sky flux in exposures.  So you would need to image a dark part of the sky and subtract off the bias and dark current to find the flux in e/s.  Then you can find C with:

 

C = s(F/p)^2 10^(0.4M)

 

There would be a CR, CG, CB, CL value for each filter.  But once you know those values for your site - they should apply for other systems involving different sensors and OTA's - and they may apply for different locations - especially with the green filter since it is a better match to SQM response.

 

I can't make measurements right now - and I am currently using Sloan filters so they wouldn't match other systems.  But I'd be interested to see what values of C people get for different systems and different light pollution.

 

Just measure s in e/s (so it can be compared to dark current) and measure p in microns so the numbers are consistent.

 

Frank


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#48 Jon Rista

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Posted 24 May 2016 - 07:07 PM

Frank, sounds very interesting. I have some AstroDon LRGB and 3nm Ha, SII and OIII filters. If I ever get another clear sky, I can get some data, and I would be happy to share it with you if you are interested. 



#49 Shiraz

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Posted 24 May 2016 - 11:06 PM

It's a minefield Jon. 

 

I used the sky data available at https://www.gemini.e...-sky-background which gives fluxes for a true dark sky. Also found some data from an Australian site that agreed pretty well, so these figures should apply reasonably well for a dark site anywhere.

 

Integrating over the visible band yielded approximately Vmag/arcsec^2 of 20.7 == 200 photons/arcsec^2/m^2/sec as a starting point for any magnitude calculation. This figure is the photon flux over a 300nm visible bandwidth, not just the Vband. For narrowband, the flux is roughly 0.5 p/nm/arcsec^2/m^2/sec at the standard NB wavelengths (can read the exact figures in the data (all relate to Vmag 20.7) - but there are so many assumptions here that 0.5 is probably a good rule of thumb).

 

I think that its also a pretty good approximation to what you get with an SQM (or the skybrightness maps) - where SQM 20.7 would give you about 200 photons/arcsec^2/m^2/sec broadband.

 

I have a spreadsheet that does SNR calculations for any system - will put it up on a local site and let you know, if you would be interested. Regards Ray


Edited by Shiraz, 25 May 2016 - 06:53 AM.


#50 freestar8n

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Posted 25 May 2016 - 05:10 AM

I used the sky data available at https://www.gemini.e...-sky-background which gives fluxes for a true dark sky.


For the purpose of calibrating an SQM reading to a given system at a given location - I don't think you can do better than measuring with your sqm at your site - and calibrating it with the sky background you read with your equipment. If you then change the optics or sensor - you would just apply the (p/F)^2 factor.

And if the color of the sky glow is similar at two different sites - the same factors should work pretty well. That's why I'm interested in seeing different C values people get.

In my case - and for many others - I am somewhat near a streetlight, so the only meaningful way to know the "true" values here are with my own SQM, and with my own OTA/ccd.

Frank


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