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Converting surface brightness in stellar magnitudes to photon flux?

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#51 Shiraz

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Posted 25 May 2016 - 06:49 AM

Got to agree Frank - measurement is clearly the best way, especially under heavy pollution.

 

However, if you don't have any measurements, for whatever reason, an extrapolation from the best available sky data will still be useful for "what if" type questions when designing a system. 

 

Ray


Edited by Shiraz, 25 May 2016 - 06:51 AM.


#52 ecorm

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Posted 23 March 2019 - 04:28 PM

I was wondering the same thing as the OP, and stumbled upon this thread when I searched. From the book jhayes_tucson suggested, they give the formula for the photon rate at a focal plane from a star of magnitude m (which I post here under the doctrine of Fair Use for the purpose of commentary and research):

 

(1)    S = NTπ/4(1-ε2)D2 Δλ10-0.4m

 

where:

  • S is the photon flux in photons/second
  • N is the irradiance of a magnitude-zero reference star
  • T, ε, D, and Δλ are parameters related to the telescope, atmospheric transmittance, and bandpass
  • m is the magnitude

Instead of guessing/researching a bunch of parameters to use the above equation, one could use their own telescope/camera to establish a simple relationship between magnitude and electron flux.

The photon flux can be expressed as the electron flux, I, divided by the quantum efficiency η, so:

(2)    I / η = NTπ/4(1-ε2)D2 Δλ10-0.4m
 

where I is in electrons/second. Rearranging to isolate I:

 

(3)    I = ηNTπ/4(1-ε2)D2 Δλ10-0.4m

 

If we lump the quantum efficiency, reference star irradiance, transmittance, telescope, and bandpass parameters into a single constant K, we get a simple relationship between magnitude and electron flux:

(4)    I = K 10-0.4m

 

If one uses their own telescope/camera to measure the electron flux of an object with known magnitude, mtest, they can calculate the constant K for their specific optical/imaging train for a certain degree of atmospheric extinction. For example, I could point my telescope near the zenith away from extended objects, and measure the sky fog with a Sky Quality Meter. The electron flux of the sky could be calculated by using the mean value of a test exposure, adjusted for bias and dark current.

So, for the test measurement, we have:

 

(5)    Itest = K 10-0.4mtest

 

rearranging gives:

 

(6)    KItest / 10-0.4mtest

Once we've calculated K, we can thereafter use equation (4) to estimate the electron flux for a different object at the same atmospheric extinction with known magnitude m.

To adjust for atmospheric extinction, you can reduce m by 0.2 magnitudes per airmass, or use whatever other extinction model you see fit to adjust m.

Atmospheric transparency varies from night-to-night, so I assume the electron flux estimate from equation (4) would only be a rough estimate if K was calculated on a different night. I suppose one could calculate K for nights of different transparency, and use the one that most closely matches the night when one wants to estimate the electron flux for an object of known magnitude.

 

The formula given by the book is for the entire focal plane, but I'm pretty sure the relationship that I derived at equation (4) applies at the per-pixel level.

 

Disclaimer: I'm still fairly new at AP, so it's quite possible I made a mistake in my reasoning above. I'd be happy to hear critiques or corrections.


Edited by ecorm, 23 March 2019 - 04:33 PM.

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#53 freestar8n

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Posted 23 March 2019 - 06:07 PM

I think you are essentially deriving what I show in post #47 - except that I include pixel size and f/ratio as parameters.

 

And I point out that you can find a constant for each filter if you want to break it down across the spectrum.

 

I don't think I ever did a calibration but it would be interesting to compare numbers people get.

 

If you explicitly include pixel size and f/ratio then the constants would be more directly comparable among different systems.

 

Frank



#54 ecorm

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Posted 23 March 2019 - 06:38 PM

I think you are essentially deriving what I show in post #47 - except that I include pixel size and f/ratio as parameters.

 

And I point out that you can find a constant for each filter if you want to break it down across the spectrum.

 

I don't think I ever did a calibration but it would be interesting to compare numbers people get.

 

If you explicitly include pixel size and f/ratio then the constants would be more directly comparable among different systems.

 

Frank

 

Oops! I admit I was skimming this thread once I had reached page 2, so I somehow missed that post.

 

It appears your model assumes the transmittance and obstruction parameters are the same (or close enough) with other systems having different focal ratios.



#55 freestar8n

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Posted 23 March 2019 - 06:48 PM

Oops! I admit I was skimming this thread once I had reached page 2, so I somehow missed that post.

 

It appears your model assumes the transmittance and obstruction parameters are the same (or close enough) with other systems having different focal ratios.

You can go ahead and include obstruction, system transmission, and QE to make it more complete.  The last two depend on wavelength - and transmission can be hard to know very well.

 

But I can see if I can get values from some of my exposures.

 

For the specific example of trying to map an SQM value to the flux with an Ha filter - it is fundamentally hard because the SQM doesn't distinguish red glow from blue.  But once you calibrate it for the filter, it should apply across different systems as long as the sky spectrum doesn't change much.

 

Frank


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#56 ecorm

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Posted 23 March 2019 - 07:22 PM

Is AperturePhotometry the correct way one estimates the flux of stars in PixInsight? According to the documentation, the table it spits out contains the magnitude and flux of the stars detected in the image. One could use that table to calculate their system's "constant" using several stars averaged out.



#57 freestar8n

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Posted 23 March 2019 - 07:55 PM

Hi-

 

I don't know exactly how it works in PI, but what you describe is fairly standard.  You get the integrated flux from the star as measured in electrons above the background across the central aperture.

 

For several stars you would get an idea of how magnitude maps to flux with the system.

 

For more accurate results you would do it with two different filters and factor in the role of color and atmospheric extinction - to correct for reddening that would make blue stars dimmer than expected due to the airmass.

 

Photometry gets fairly involved if you want to get accurate results.  But just using visual magnitude and a single filter should be in the ballpark.

 

Frank


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#58 ecorm

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Posted 23 March 2019 - 08:43 PM

For more accurate results you would do it with two different filters and factor in the role of color and atmospheric extinction - to correct for reddening that would make blue stars dimmer than expected due to the airmass.

 

Photometry gets fairly involved if you want to get accurate results.  But just using visual magnitude and a single filter should be in the ballpark.

 

For the AperturePhotometry, I would choose a starfield near the zenith to simplify the atmospheric extinction aspect. I think I would perform the test for only my luminance filter, since luminance is where I care the most about achieving high SNR. I guess it's about time I list my gear in my signature so that others on this forum can better help me.

 

A ballpark figure is all I need. My goal is similar to Jon's, where I'd like to be able to estimate in advance the total integration time I need to reach a certain SNR goal.

 

I don't know yet what constitutes a "good" SNR for the purpose of pretty pictures, but I hope I'll get a feeling for that as I gain experience. I suppose I can play around with stacking increasing numbers of light frames until I no longer see aesthetic improvements. I understand that SNR only makes sense in terms of a specific part of the image: the background sky, a galaxy arm, a wisp of nebulosity, etc.

When it comes to narrowband imaging, I'm a bit lost when it comes to figuring out the relationship between photon flux and the published magnitudes of emission nebulae. I'll read over this thread more carefully, and will do some more studying.

 

I truly appreciate you helping me understand stuff here and in other threads.


Edited by ecorm, 23 March 2019 - 08:45 PM.


#59 freestar8n

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Posted 24 March 2019 - 01:29 AM

For the AperturePhotometry, I would choose a starfield near the zenith to simplify the atmospheric extinction aspect. I think I would perform the test for only my luminance filter, since luminance is where I care the most about achieving high SNR. I guess it's about time I list my gear in my signature so that others on this forum can better help me.

 

A ballpark figure is all I need. My goal is similar to Jon's, where I'd like to be able to estimate in advance the total integration time I need to reach a certain SNR goal.

 

I don't know yet what constitutes a "good" SNR for the purpose of pretty pictures, but I hope I'll get a feeling for that as I gain experience. I suppose I can play around with stacking increasing numbers of light frames until I no longer see aesthetic improvements. I understand that SNR only makes sense in terms of a specific part of the image: the background sky, a galaxy arm, a wisp of nebulosity, etc.

When it comes to narrowband imaging, I'm a bit lost when it comes to figuring out the relationship between photon flux and the published magnitudes of emission nebulae. I'll read over this thread more carefully, and will do some more studying.

 

I truly appreciate you helping me understand stuff here and in other threads.

Sounds great!  I will aim to see if I can get numbers from my system.

 

The main thing complicating magnitude calculations is that ccd's record photons and don't care if they are blue or red.  But magnitudes of stars are always described in terms of some assumed passband.  That makes it hard to convert an SQM visual magnitude to whatever photon count you might get in a given filter.

 

One thing to remember is that any time you have two different fluxes in e/s, you can take the ratio and express it as a magnitude difference.  Even if you don't know what the magnitudes mean on some absolute scale - it can be handy to know the difference in terms of a delta magnitude.

 

If your sky background is 1 e/s at one site - and 100 e/s at another with the same setup - you know it is 5 magnitudes brighter - without needing to know a lot of other details.

 

Frank



#60 ecorm

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Posted 24 March 2019 - 02:20 AM

The main thing complicating magnitude calculations is that ccd's record photons and don't care if they are blue or red.  But magnitudes of stars are always described in terms of some assumed passband.  That makes it hard to convert an SQM visual magnitude to whatever photon count you might get in a given filter.

Yeah, I've just been reading-up on the UBVRI photometric system, as well as B-V color indices. I've seen someone (perhaps you) in another thread mention that SQM has its own passband standard. I now understand that star magnitudes in catalogues are specified for the V band, and have a certain B-V color index. The way I understand it, to convert from V magnitude to some luminance filter passband, one would need to figure out a conversion factor for their specific luminance filter, based on the blackbody spectral curve for that particular star. This conversion factor would be different for stars having a different spectral curves!

At least with the SQM, the conversion factor to a luminance filter should be consistent if one assumes the sky color remains the same.

This is getting a lot more complicated than I thought. I might as well just estimate the electron flux directly for my object of interest using a test exposure. During my imaging sessions, I should try to reserve some time to grab a few test exposures for my "next" target. This way, I have test data that I can use to plan for the next imaging run. If the telescope ends up being different on the next imaging run, I could use the formulas in this thread to convert the measured electron flux accordingly (the filters would remain the same).

My brain hurts from trying to give myself a crash course in basic photometry, so I hope what I wrote above is somewhat coherent.  lol.gif 
 

If your sky background is 1 e/s at one site - and 100 e/s at another with the same setup - you know it is 5 magnitudes brighter - without needing to know a lot of other details.

 

My imaging rig would effectively become an oversized/overpriced SQM after it's been "calibrated" for the sky in my backyard!


Edited by ecorm, 24 March 2019 - 02:26 AM.


#61 Jon Rista

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Posted 24 March 2019 - 03:57 AM

This is getting a lot more complicated than I thought. I might as well just estimate the electron flux directly for my object of interest using a test exposure. During my imaging sessions, I should try to reserve some time to grab a few test exposures for my "next" target. This way, I have test data that I can use to plan for the next imaging run. If the telescope ends up being different on the next imaging run, I could use the formulas in this thread to convert the measured electron flux accordingly (the filters would remain the same).

My brain hurts from trying to give myself a crash course in basic photometry, so I hope what I wrote above is somewhat coherent.  lol.gif

 

Welcome to the club. :p You are now where I ended up a couple years ago when I asked this question. It is pretty complicated...more complicated than I cared to deal with once I really got into it. 

 

I have such variable skies that photometry never really became a thing for me. Too many variables. 

 

In the end, it doesn't take much experimentation with a few test exposures to figure out what you need to do to create pretty pictures, if that is your goal...and you don't really need to deal with all the math at all (unless you want to.)



#62 Mert

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Posted 24 March 2019 - 05:33 AM

Maybe I'm totally off, but there is an option in meteoblue.com if you select

meteorological maps just before special options, you get another screen and

on the right side and select Air quality, there is an option of Aerosol

Optical Depth which more or less gives you the transparency of the

atmosfere.

 

Here is the descriptin of what they define:

Aerosol optical depth
The optical depth is a measure of how well electromagnetic waves can pass through a medium. Therefore, the aerosol optical depth is the measure for the reduction of light transmission caused by atmospheric aerosols. It describes the total light extinction in the vertical atmospheric column, which depends on the light's wavelength and the amount of atmospheric aerosols. The greater the value of optical depth, the greater the aerosol concentration. Sources of aerosol can be diverse: wild fire, desert dust or anthropogenic air pollution. The aerosol optical depth is dimensionless.

Maybe it helps a little?

 

Regards,

Mert



#63 ManojK

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Posted 10 November 2019 - 11:42 AM

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

Resurrecting this very old thread. Recently I have been trying to figure out the same thing (how many electrons from target, compared to Read Noise and Sky Glow) and stumbled upon this thread. 

 

Had mostly arrived at what Frank has above from first principles, but noticed two things that I am doing which seems to differ from this formula and wanted to bounce it off all of you.

 

First, photon flux from a target (assume all V band. An approximation, but should suffice).

This is the formula I am starting with:

mt-mstd = -2.5 log10(Ft/Fstd)

 

mstd here is Vega. Apparent mag is 0, and Fstd = 995.5

 

Rearranging 

-0.4*mt = log10(ft/995.5)

ft = 995.5*10^(-0.4*mt)

 

Above it is listed as

ft = 10^(-0.4*mt)

Am I missing something?

 

Second, assuming optics are seeing limited (and not diffraction limited), the energy collected by the scope should be spread out over what seeing gives. Implying what the pixel collects is the following:

[w(microns)/f(mm)*206.265]^2/s^2

Where s seeing

 

Approximating photon flux from Vega to 1000, this gives me

1000*10 * (bandwidth in nm) * pi * (D(mm)/2)^2 * 1000*10*10^(-0.4*M(mag/sqarcsec)) * [w(microns)/f(mm)*206.265]^2/s^2

 

Pulling out all the constants, we have

1000*10*1000*10*pi*206.265*206.265/4 = 3.34*10^12

Even removing one fstd from the above, I get 3.34*10^8, while frank has 10^6

Again wondering what I am missing.

 

Continuing on, the final simplified formula, I am getting, is

3.34*10^12 * (bandwidth in nm) * (pix size in um / FRatio)^2 * 10^(-0.4 * mag/sqarcsec) / s^2

 

Would love your thoughts. 

 

thx!

Manoj



#64 ManojK

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Posted 10 November 2019 - 02:16 PM

Never mind. I found the folly in my equations.

 

Ft = 995.5*10^(-0.4*mt)

 

But this is in cm2. Converting it to mm2 and assuming Vega is 1000 photons/angstrom/sec/cm2

 

Ft = 1000*10^(-0.4*mt)/100 = 10*10^(-0.4*mt)

 

Again this is in Angstroms. Converting to nm we get

 

10* 10*10^(-0.4*mt) = 100*10^(-0.4*mt) photons/nm/sec/mm2

 

Photons captured by Telescope = Ft * (bandwidth in nm=b) * aperturearea

= Ft * b * pi*(D/2)^2

Substituting, we get

= 100*10^(-0.4*mt) * b * pi * (D^2)/4

 

Per pixel = area of pixel * Photons captured by Telescope

= (pixel*206.265/FL)^2 *  100*10^(-0.4*mt) * b * pi * (D^2)/4

 

Now D^2/FL^2 = 1/FR^2. Pulling all the constants together, we get

 

(100*3.14*206.265^2/4) * b * 10^(-0.4*mt) * (pixel/FR)^2 = 3.341*10^6 * b * 10^(-0.4*mt) * (pixel/FR)^2

 

Which is the same as what Frank had to begin with. 

 

The only question I have then is, does it make sense to adjust this for seeing?

 

In which case we would divide this by seeing, because photons captured by telescope is spread over the seeing resolution. And a pixel captures a ratio of that. 

 

i.e photons from target on a pixel = (3.341*10^6 * b * 10^(-0.4*mt) * (pixel/FR)^2)/s^2


Edited by ManojK, 10 November 2019 - 03:40 PM.


#65 freestar8n

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Posted 10 November 2019 - 04:04 PM

Never mind. I found the folly in my equations.

 

Ft = 995.5*10^(-0.4*mt)

 

But this is in cm2. Converting it to mm2 and assuming Vega is 1000 photons/angstrom/sec/cm2

 

Ft = 1000*10^(-0.4*mt)/100 = 10*10^(-0.4*mt)

 

Again this is in Angstroms. Converting to nm we get

 

10* 10*10^(-0.4*mt) = 100*10^(-0.4*mt) photons/nm/sec/mm2

 

Photons captured by Telescope = Ft * (bandwidth in nm=b) * aperturearea

= Ft * b * pi*(D/2)^2

Substituting, we get

= 100*10^(-0.4*mt) * b * pi * (D^2)/4

 

Per pixel = area of pixel * Photons captured by Telescope

= (pixel*206.265/FL)^2 *  100*10^(-0.4*mt) * b * pi * (D^2)/4

 

Now D^2/FL^2 = 1/FR^2. Pulling all the constants together, we get

 

(100*3.14*206.265^2/4) * b * 10^(-0.4*mt) * (pixel/FR)^2 = 3.341*10^6 * b * 10^(-0.4*mt) * (pixel/FR)^2

 

Which is the same as what Frank had to begin with. 

 

The only question I have then is, does it make sense to adjust this for seeing?

 

In which case we would divide this by seeing, because photons captured by telescope is spread over the seeing resolution. And a pixel captures a ratio of that. 

 

i.e photons from target on a pixel = (3.341*10^6 * b * 10^(-0.4*mt) * (pixel/FR)^2)/s^2

Glad you got it working.  I think this thread is just about nebulosity, which acts as an extended object, so seeing wouldn't play a role.  Every patch of light is blurred by seeing but it just gets spread out locally and there is no change in flux.

 

But for stars the adu count in a pixel would depend critically on seeing - not just the fwhm of the star but the nature of the star profile and how steeply it drops to zero - which is captured by the Moffat beta term.

 

Frank



#66 ManojK

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Posted 10 November 2019 - 09:25 PM

Thanks Frank, that makes sense. So ditch the seeing. 

 

Now for the next step. Lets take pinwheel galaxy as an example. 

 

Apparent Magnitude m = 7.86

Apparent size s = 28.8'*26.9'

 

This gives me a surface brightness of m * 2.5*log10(28.8*26.9*60*60) = 23.97 

 

A system I am considering:

 

QHY600M camera (IMX455) + FSQ106 EDXIV

 

pixel size = 3.76 microns

Focal ratio of 5

QE = 85%

 

Filter bandwidth of 5nm (Narrowband)

 

Signal photons = 3.341*10^6 * 5 * 10^(-0.4*23.97) * (3.76/5)^2 = 0.00243 photos/sec/pixel

 

Signal EPS = 0.85 * 0.00243 = 0.00212 eps

 

 

readnoise rn = 3.6 EPS

Sky glow from http://tools.sharpcap.co.uk/ is 0.21 eps (Mag 19 skies. Suburban)

Subs = 600s

 

Lets determine how many subs (n) we need. 

 

Shot noise from target st = sqrt(600*n*0.00212)

Shot noise from skyglow ss= sqrt(600*n*0.21)

 

SNR = (0.00212 * 600*n)/ sqrt(st^2 + ss^2 + n*rn^2)

= 1.272*n/sqrt(1.272*n + 126*n + 12.96)

 

Lets say we want an SNR of 2. Squaring everything

 

4 = 1.62*n^2/(n * 140.232)

 

n = 347

 

So around 350 subs of 10 minutes each for a total integration time of 58 hrs!!

 

Does that sound right?


Edited by ManojK, 10 November 2019 - 09:26 PM.


#67 Jon Rista

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Posted 11 November 2019 - 12:35 AM

My original goal was mostly around extended object signal, not stars, so I don't think seeing really matters.

 

That said...58 hours for an SNR of just 2 around a red/orange zone border seems rather extreme... I use 3nm filters myself, in a red/white zone. Now I am at f/4, but my SNR is many multiples of 2 with about 10-15 hours of signal per channel (usually 30-40 hours total per image, but that is also usually 3 channels.) I usually stack 60-90x600s subs per channel.



#68 freestar8n

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Posted 11 November 2019 - 03:17 AM

I'm not sure what happened here but the first step taking log10 of the area in arc-sec doesn't look right.  You can see the pinwheel by eye from a dark site so it must be much brighter than that in mag per arc-sec squared.

 

I didn't go through the math but it should take much less time to achieve snr of 2.

 

Frank

 

Addendum - I think the mag value may be ok but something else is wrong.  I'll try to look at it soon.

 

Addendum2:  Oh - I see you are using a 5nm filter with a galaxy.  That will capture almost no signal - so yes that time is probably about right.  You would only use 5nm to capture Ha signal - and it is much brighter on a per nm scale within the passband.


Edited by freestar8n, 11 November 2019 - 04:26 AM.


#69 ManojK

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Posted 11 November 2019 - 02:41 PM

My original goal was mostly around extended object signal, not stars, so I don't think seeing really matters.

 

That said...58 hours for an SNR of just 2 around a red/orange zone border seems rather extreme... I use 3nm filters myself, in a red/white zone. Now I am at f/4, but my SNR is many multiples of 2 with about 10-15 hours of signal per channel (usually 30-40 hours total per image, but that is also usually 3 channels.) I usually stack 60-90x600s subs per channel.

Agreed Jon. Which is why I picked Pinwheel, which most definitely does not need 58 hours. 

 

 

I'm not sure what happened here but the first step taking log10 of the area in arc-sec doesn't look right.  You can see the pinwheel by eye from a dark site so it must be much brighter than that in mag per arc-sec squared.

 

I didn't go through the math but it should take much less time to achieve snr of 2.

 

Frank

 

Addendum - I think the mag value may be ok but something else is wrong.  I'll try to look at it soon.

 

Addendum2:  Oh - I see you are using a 5nm filter with a galaxy.  That will capture almost no signal - so yes that time is probably about right.  You would only use 5nm to capture Ha signal - and it is much brighter on a per nm scale within the passband.

Thanks Frank. That makes sense. if I swap the bandwidth for that of a RGB filter (100nm) and change the SNR to about 10, I get about 10 hours. which seems more in line.

 

Jon, that is probably around the SNR you see for the faint regions on pinwheel? Maybe 10 to 15?

 

One problem I have is that given my light polluted skies, for continuum objects, I actually don't use RGB filters.

 

Instead I use a combination of Ha20nm (Red), Stromgren_Y(Green) and Stromgren_V(Blue). And then use photometric calibration in pixinsight to get the color weights correct. This works great, but given only 20nm bandwidth a target like pinwheel needs insanely high total integration time.

 

Best captured under a darker sky, so I can use more bandwidth on the filters, I guess. 

 

In any case, these are the sort of heuristics I am wanting to glean from this exercise and it seems to work for that. 

 

Thanks again all!

 

Fantastic discussion again and reminds me why I love this hobby. 


Edited by ManojK, 11 November 2019 - 02:41 PM.


#70 Jon Rista

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Posted 12 November 2019 - 02:28 AM

Ah, yeah...Frank is correct, if you are using narrow band filters on broadband objects that don't emit a lot in those narrow bands, then you'll need some rather immense integrations. I don't bother much with broadband objects from my back yard, however if/when I do, I don't generally need 60 hours to get something visible. The images, even with 10-20 hours, do tend to be noisy thanks to all the additional shot noise, but 60 hours is not really a requirement (probably wouldn't hurt, though!)




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