Jump to content

  •  

CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.

Photo

etendue calculator

  • Please log in to reply
51 replies to this topic

#26 ks__observer

ks__observer

    Viking 1

  • *****
  • Posts: 992
  • Joined: 28 Sep 2016
  • Loc: Long Island, New York

Posted 08 May 2019 - 02:48 PM

Etendue comes from the study of illumination (spotlights, floodlights, skylights, etc.) and in my opinion isn't all that applicable directly to astronomy.  Surface brightness at the sensor depends on f-ratio.

Pixel size comes into the formula because SNR in CCD/CMOS cameras is a function of the number of photons counted. 

Larger pixels will count a larger number of photons than smaller pixels. 

This is with the ever constant trade-off between pixel-resolution and SNR -- and the battle of when we have reached the sampling limit.



#27 spokeshave

spokeshave

    Vanguard

  • *****
  • Posts: 2104
  • Joined: 08 Apr 2015

Posted 08 May 2019 - 02:52 PM

Etendue comes from the study of illumination (spotlights, floodlights, skylights, etc.) and in my opinion isn't all that applicable directly to astronomy.  Surface brightness at the sensor depends on f-ratio.

It is directly applicable to astrophotography and this discussion is intended to help dispel the myth that imaging speed is dependent upon f-ratio alone. I know that's not what you said, but it is what some people will take away from what you said. Brightness at the sensor does not determine imaging speed. Brightness at the pixel does, and that is exactly what "imaging etendue" describes. 

 

Tim


  • Jon Rista likes this

#28 Michael Covington

Michael Covington

    Author

  • *****
  • Vendors
  • Posts: 5523
  • Joined: 13 May 2014
  • Loc: Athens, Georgia, USA

Posted 08 May 2019 - 02:58 PM

Isn't brightness at the pixel the same thing as brightness at the sensor?



#29 Francois

Francois

    Vostok 1

  • -----
  • Posts: 111
  • Joined: 09 Jun 2007
  • Loc: Montreal

Posted 08 May 2019 - 03:18 PM

No, I was not making any esoteric argument. My caveat is purely based on the ratio of signal to background-dominated shot noise in a suboptimal aperture, which is what an oversize pixel is.

 

If you are making field of view filling maps in the signal-dominated regime, system etendue is absolutely the figure of merit. For the background-dominated regime it is still the figure of merit as long as the image is sufficiently sampled.

 

None of this is remotely controversial.



#30 Jon Rista

Jon Rista

    ISS

  • *****
  • Posts: 23745
  • Joined: 10 Jan 2014
  • Loc: Colorado

Posted 08 May 2019 - 03:18 PM

Isn't brightness at the pixel the same thing as brightness at the sensor?

No. F-ratio gives you relative instantaneous intensity at any point on the sensor, but it alone does not tell you how many photons are actually landing on each pixel per unit time. That is why etendue DOES apply in our context, in the astrophotography context, and is actually a very important concept. To understand how quickly a PIXEL will fill up with signal, you must factor in the size of those pixels into the equation. That is what D^2*S^2*QE does. It not only factors the relative angular area of the pixel, but also the quantum efficiency, which is the conversion rate at which photons actually free electrons in the pixel well. 


  • Michael Covington likes this

#31 jhayes_tucson

jhayes_tucson

    Fly Me to the Moon

  • *****
  • Posts: 7157
  • Joined: 26 Aug 2012
  • Loc: Bend, OR

Posted 08 May 2019 - 03:25 PM

Isn't brightness at the pixel the same thing as brightness at the sensor?

Yes it is.

 

It is directly applicable to astrophotography and this discussion is intended to help dispel the myth that imaging speed is dependent upon f-ratio alone. I know that's not what you said, but it is what some people will take away from what you said. Brightness at the sensor does not determine imaging speed. Brightness at the pixel does, and that is exactly what "imaging etendue" describes. 

 

Tim

 

Tim,

At this level, it's helpful to be a bit more precise.  First off, what counts is the signal level and that's given by the average irradiance incident on a pixel (in Watts/m2) multiplied by the area of the pixel (in m2) times the responsivity of the sensor (in Amps/Watt).  Michael is correct that for an extended source, the irradiance in the image plane only depends on the focal ratio.  When you throw out the concept of "imaging speed," you have to also consider how the image is sampled, which sweeps up the size and sensitivity of the pixels.

 

John

Attached Thumbnails

  • Screen Shot 2019-05-08 at 1.22.26 PM copy.jpg

  • bsavoie, Jon Rista and roofkid like this

#32 jhayes_tucson

jhayes_tucson

    Fly Me to the Moon

  • *****
  • Posts: 7157
  • Joined: 26 Aug 2012
  • Loc: Bend, OR

Posted 08 May 2019 - 03:28 PM

...

 

None of this is remotely controversial.

 

lol.gif lol.gif lol.gif

 

Francois,

You must be new around here!   Almost everything is controversial--no matter how long or how well it has been understood.

John

 

 

PS  Joking aside, it's nice to have more professional folks who can help explain some of this stuff...so welcome.


Edited by jhayes_tucson, 08 May 2019 - 03:31 PM.

  • Michael Covington and roofkid like this

#33 Francois

Francois

    Vostok 1

  • -----
  • Posts: 111
  • Joined: 09 Jun 2007
  • Loc: Montreal

Posted 08 May 2019 - 03:38 PM

lol.gif lol.gif lol.gif

 

Francois,

You must be new around here!   Almost everything is controversial--no matter how long or how well it has been understood.

John

 

 

PS  Joking aside, it's nice to have more professional folks who can help explain some of this stuff...so welcome.

New to renewed participation. I joined way back when I was an undergrad, and read through lots of eyepiece reports when I was in charge of acquisitions for an EPO program.


  • Michael Covington likes this

#34 roofkid

roofkid

    Viking 1

  • -----
  • Posts: 641
  • Joined: 13 Jan 2016

Posted 08 May 2019 - 04:05 PM

Hi prefetch,

 

I'm very surprised that you did not come across my work while searching. You can find one information thread here: https://www.cloudyni...cts-more-light/ including a calculator.

 

I'm getting different results than you from the original post and the similar results to John when he replied.

 

Cheers,

Sven


  • bobzeq25 likes this

#35 prefetch

prefetch

    Viking 1

  • -----
  • topic starter
  • Posts: 522
  • Joined: 24 Apr 2013

Posted 08 May 2019 - 11:00 PM

Hi prefetch,

 

I'm very surprised that you did not come across my work while searching. You can find one information thread here: https://www.cloudyni...cts-more-light/ including a calculator.

 

I'm getting different results than you from the original post and the similar results to John when he replied.

 

 

i didn't find that page during my searches.  i found this page: https://www.cloudyni...-of-telescopes/ but the calculator gives me a permission denied error, which is when i decided to just make my own quick and dirty version.

 

your calculator is very nice!  

 

i think the reason that mine does not match yours is i am not using FOV in my formula, but instead i am using the arcsec/pixel in my formula.  i suppose that i may be in error here.  i've only recently (the last day or so) learned about the concept of etendue and i'm just familiarizing myself with it by jumping headlong into this.  ;-)



#36 prefetch

prefetch

    Viking 1

  • -----
  • topic starter
  • Posts: 522
  • Joined: 24 Apr 2013

Posted 08 May 2019 - 11:39 PM

Here's how to compute signal ratios:

 

S1/S2 = (F2/F1)2 (T1/T2) (1-CO12)/(1-CO22) (R1/R2) (P1/P2)2

 

Where:

Sn = Signal strength

Fn = Focal ratio

Tn = Optical transmission (due to surface coatings taking into account the number of surfaces)

COn = Central obscuration ratio (= Dsecondary/Dprimary)

Rn = Sensor responsivity (QE)

Pn = Pixel dimension

 

Here's an example:

System 1:  14" aperture, F = 10.8, CO = 0.32, Transmission = 0.85 (8 surfaces @ 0.98), R = 0.65, P = 9.0 um  (C14 w/16803)

System 2:  11" aperture, F = 10.0, CO = 0.31, Transmission = 0.85 (8 surfaces @ 0.98), R = 0.75, P = 3.8 um  (C11 w/ASI-1600MM-C)

 

Plugging in the numbers:

S1/S2 = 4.1, which says that the 14" system will produce 4.1 times as much signal as the 11" system.

 

john,

 

this is interesting to me - and i want to understand that i'm doing this right.  to follow your example:

 

S1

Fn = 10.8
Tn = 85.1%
COn = 32%
Rn = 65% (QE)
Pn = 9 um

 

S2

Fn = 10
Tn = 85.1%
COn = 31%
Rn = 75% (QE)
Pn = 3.8 um

 

and so:

(F2/F1)2 = .857

(T1/T2) = 1

(1-CO12)/(1-CO22) = .993

(R1/R2) = 87%

(P1/P2)2 = 5.6

 

am i doing that right?  and if you multiply all that together, you get 4.14 (which i'm assuming is that same as your 4.1)

 

and i guess the takeaway is that this concept is really about the ratio between two system as opposed to some absolute signal strength?



#37 bobzeq25

bobzeq25

    Hubble

  • *****
  • Posts: 16935
  • Joined: 27 Oct 2014

Posted 09 May 2019 - 12:04 AM

i didn't find that page during my searches.  i found this page: https://www.cloudyni...-of-telescopes/ but the calculator gives me a permission denied error, which is when i decided to just make my own quick and dirty version.

 

your calculator is very nice!  

 

i think the reason that mine does not match yours is i am not using FOV in my formula, but instead i am using the arcsec/pixel in my formula.  i suppose that i may be in error here.  i've only recently (the last day or so) learned about the concept of etendue and i'm just familiarizing myself with it by jumping headlong into this.  ;-)

arc sec/pixel works fine.  While you could calculate things for the whole sensor, the formulas used here have been about what's going on on a per pixel basis.


Edited by bobzeq25, 09 May 2019 - 12:20 AM.

  • jhayes_tucson likes this

#38 jhayes_tucson

jhayes_tucson

    Fly Me to the Moon

  • *****
  • Posts: 7157
  • Joined: 26 Aug 2012
  • Loc: Bend, OR

Posted 09 May 2019 - 12:18 AM

john,

 

this is interesting to me - and i want to understand that i'm doing this right.  to follow your example:

 

S1

Fn = 10.8
Tn = 85.1%
COn = 32%
Rn = 65% (QE)
Pn = 9 um

 

S2

Fn = 10
Tn = 85.1%
COn = 31%
Rn = 75% (QE)
Pn = 3.8 um

 

and so:

(F2/F1)2 = .857

(T1/T2) = 1

(1-CO12)/(1-CO22) = .993

(R1/R2) = 87%

(P1/P2)2 = 5.6

 

am i doing that right?  and if you multiply all that together, you get 4.14 (which i'm assuming is that same as your 4.1)

 

and i guess the takeaway is that this concept is really about the ratio between two system as opposed to some absolute signal strength?

 

 

Yes, your numbers look correct and I rounded the result (4.138899351) to one decimal point, which is 4.1.  You could certainly calculate the actual signal in Amps if you have the absolute value of the responsivity (rather than normalized value given on most data sheets,) the radiance of the source (in Watts/m2/str,) and a few normalizing constants; but, that's not a particularly useful number for most of what we normally want to do in imaging.  It is almost always more useful to compare two systems as I did in my example.  The question is what happens if we use a different telescope or choose a different camera--compared to what we might be currently doing.  

 

Here's a real example.  In the next year I plan to move from my C14 Edge to a PW 20" system and this calculation easily shows that with my existing camera, I'll get a signal that's 2.48x stronger.  That implies that I can either achieve a better SNR in my subs with the same 20 minute exposures that I commonly use now (by about 1.6x) or that I can cut my sub-exposure time to 8 minutes to get subs with the same SNR that I currently get now.  (Before anyone jumps all over me, RN plays a role here so 8 minutes might not be exactly accurate--but, it should be close.)  Either way, I'll spend considerably less time achieving the same level of SNR in my stacked images, which means having the ability to image a lot more objects each year.  Oh, and these two systems conveniently have a similar EFL so the image scale is similar (the 20" actually has a slightly wider field.)  In this case, I don't care about absolute signal values.  I really only care about how much better the new system should be compared to the old system and that's what this calculation provides.

 

John


  • prefetch and roofkid like this

#39 Michael Covington

Michael Covington

    Author

  • *****
  • Vendors
  • Posts: 5523
  • Joined: 13 May 2014
  • Loc: Athens, Georgia, USA

Posted 09 May 2019 - 08:44 AM

arc sec/pixel works fine.  While you could calculate things for the whole sensor, the formulas used here have been about what's going on on a per pixel basis.

Ah. A key point.  We have to decide whether to calculate the étendue of the whole sensor or of a single pixel.  



#40 jhayes_tucson

jhayes_tucson

    Fly Me to the Moon

  • *****
  • Posts: 7157
  • Joined: 26 Aug 2012
  • Loc: Bend, OR

Posted 09 May 2019 - 10:46 AM

Ah. A key point.  We have to decide whether to calculate the étendue of the whole sensor or of a single pixel.  

 

Etendue is a quantity that is conserved through an optical system and it doesn’t matter how you define the chief ray.  Whether the chief ray goes to the edge of a pixel or the edge of the field, etendue is still conserved.  What folks here really have to decide is whether we are talking about Etendue as it is actually defined or if it makes sense to accept some other definition for some other purpose that may or may not make any sense.  As I said a couple of times, it makes more sense to accept that etendue is conserved, put that aside, and to compare signal levels.  The relative signal is what really matters.

 

John


  • Michael Covington, bobzeq25 and roofkid like this

#41 roofkid

roofkid

    Viking 1

  • -----
  • Posts: 641
  • Joined: 13 Jan 2016

Posted 09 May 2019 - 11:45 AM

Thanks for the additional information John, I learned something from your posts :)



#42 freestar8n

freestar8n

    Vendor - MetaGuide

  • *****
  • Vendors
  • Posts: 8842
  • Joined: 12 Oct 2007

Posted 09 May 2019 - 05:15 PM

Hmm - this is going all over the place.  I thought my LSST reference would help out.

 

Etendue is a measure of optical throughput - and is an *extrinsic* quantity.  It absolutely depends on the chief ray in terms of defining the field of view either as a solid angle or as a detector with some area.

 

The units of etendue are Length^2 * SolidAngle.  For large telescopes it could be m^2*deg^2

 

You can calculate that for the entire sensor or for an individual pixel.  But it represents the total flux that would be collected over some area in the image.  It is not a point measure of anything.  If you want a point measure of irradiance, i.e. an intrinsic quantity, all you need is the solid angle of the pupil.

 

But in both cases there are other factors that would reduce etendue - the main ones being vignetting and system transmission loss.  Both of those can be approximated for the overall field size - or for pixels in different parts of the field.  If you factor those things in, you get the "effective etendue".

 

There are no properties of the sensor in this at all - except the area of the entire sensor or a pixel.  No single number will tell you everything about an image or its snr.  But the point of etendue is to capture the flux that would be received by the entire optical system - including the size of the image you are talking about - whether it's the sensor or a pixel.

 

So - in an etendue calculator all you need is either the sensor area and the solid angle of the pupil - or the aperture area and the solid angle of the field of view the sensor sees.

 

For effective etendue you would add in transmission loss and overall vignetting.

 

If you want to do it for both the sensor and a pixel in the sensor - you would need to specify both sizes.  But neither one by itself will tell you about snr.  For that you would need read noise, sky transparency, sky background, object brightness - etc.  If you want to fold in these other things - please don't call it etendue.

 

Frank



#43 ks__observer

ks__observer

    Viking 1

  • *****
  • Posts: 992
  • Joined: 28 Sep 2016
  • Loc: Long Island, New York

Posted 09 May 2019 - 05:30 PM

If you want to do it for both the sensor and a pixel in the sensor - you would need to specify both sizes.  But neither one by itself will tell you about snr.  For that you would need read noise, sky transparency, sky background, object brightness - etc.  If you want to fold in these other things - please don't call it etendue.

Read noise would drop out if you assume each camera option being compared will be used at about the same "swamp factor."



#44 Organic Astrochemist

Organic Astrochemist

    Viking 1

  • *****
  • Posts: 891
  • Joined: 10 Jan 2015

Posted 09 May 2019 - 05:37 PM

No, I was not making any esoteric argument. My caveat is purely based on the ratio of signal to background-dominated shot noise in a suboptimal aperture, which is what an oversize pixel is.

I wasn’t trying to be esoteric or controversial. Just trying to understand.

 

If you are making field of view filling maps in the signal-dominated regime, system etendue is absolutely the figure of merit.

Consider two systems A & B. A has 20% higher etendue, but the target exceeds the field of view so a mosaic of two images is required unlike B for which the target fills the field of view. Which is better?

 

For the background-dominated regime it is still the figure of merit as long as the image is sufficiently sampled.

Consider two systems A & B. A has 20% higher etendue, but lacks the image scale to resolve a dust lane in a faint galaxy. System B shows the dust lane but requires longer acquisition time. Which is better?

None of this is remotely controversial.

John’s example seems intuitively obvious that, if one has the mount, that the PW 20” with larger aperture and faster f-ratio is superior to the C14 with smaller aperture and slower f-ratio. The calculator quantifies this.



#45 freestar8n

freestar8n

    Vendor - MetaGuide

  • *****
  • Vendors
  • Posts: 8842
  • Joined: 12 Oct 2007

Posted 09 May 2019 - 06:23 PM

John’s example seems intuitively obvious that, if one has the mount, that the PW 20” with larger aperture and faster f-ratio is superior to the C14 with smaller aperture and slower f-ratio. The calculator quantifies this.

If you want to compare etendue of two systems with the same sensor - and if you ignore vignetting, system transmission and secondary obstruction - all you need to do is compare f/ratios.  The faster system will have higher etendue.

 

If you also have two different sensor sizes - and perhaps much more vignetting - or a larger secondary - you would need to factor those things in also - for a comparison of effective etendue.

 

In all this stuff etendue only tells you about the total throughput.  It doesn't tell you about snr or image quality expected with some sensor and its associated noise.  Or resolution.

 

And if you keep the units well defined - it is a useful absolute quantity that is good for calculating actual number of photons received - if you include the brightness of the object and total exposure time.  Etendue isn't just useful for relative comparisons.

Frank


  • Michael Covington likes this

#46 jhayes_tucson

jhayes_tucson

    Fly Me to the Moon

  • *****
  • Posts: 7157
  • Joined: 26 Aug 2012
  • Loc: Bend, OR

Posted 09 May 2019 - 07:05 PM

If you want to compare etendue of two systems with the same sensor - and if you ignore vignetting, system transmission and secondary obstruction - all you need to do is compare f/ratios.  The faster system will have higher etendue.

 

If you also have two different sensor sizes - and perhaps much more vignetting - or a larger secondary - you would need to factor those things in also - for a comparison of effective etendue.

 

In all this stuff etendue only tells you about the total throughput.  It doesn't tell you about snr or image quality expected with some sensor and its associated noise.  Or resolution.

 

And if you keep the units well defined - it is a useful absolute quantity that is good for calculating actual number of photons received - if you include the brightness of the object and total exposure time.  Etendue isn't just useful for relative comparisons.

Frank

 

Right.  Etendue is a radiometric quantity but it doesn't say anything about signal level.

 

John



#47 james7ca

james7ca

    Fly Me to the Moon

  • *****
  • Posts: 7415
  • Joined: 21 May 2011
  • Loc: San Diego, CA

Posted 10 May 2019 - 02:17 AM

Here is a post that shows how the aperture-only and f-ratio-only proponents are actually saying the same thing (well, usually, but in truth it just depends upon which variables you decide to represent in your equations). This post also does a breakdown on Roger Clark's take on etendue.

 

  https://www.cloudyni...dpost&p=8574023



#48 Organic Astrochemist

Organic Astrochemist

    Viking 1

  • *****
  • Posts: 891
  • Joined: 10 Jan 2015

Posted 10 May 2019 - 06:34 PM

If you want to compare etendue of two systems with the same sensor - and if you ignore vignetting, system transmission and secondary obstruction - all you need to do is compare f/ratios. The faster system will have higher etendue.

If you also have two different sensor sizes - and perhaps much more vignetting - or a larger secondary - you would need to factor those things in also - for a comparison of effective etendue.

In all this stuff etendue only tells you about the total throughput. It doesn't tell you about snr or image quality expected with some sensor and its associated noise. Or resolution.

And if you keep the units well defined - it is a useful absolute quantity that is good for calculating actual number of photons received - if you include the brightness of the object and total exposure time. Etendue isn't just useful for relative comparisons.
Frank

Thank you. All I was trying to say (inartfully) was that etendue cannot be the sole measure of an optical system.

Based on what you said (and the calculator) a 70 mm f/4 finder has higher etendue than a C14!

james7ca, as ever, is helpful
“What I think I've just shown is that it depends upon how you "slice and dice" the variables in your exposure calculations. You can measure exposure with just aperture (diameter) and plate scale (or pixels scale) or you can simply use f-ratio and pixel size, but I'd expect that most people would find it easier to use f-ratio and pixel size since that means that you don't have to calculate the plate scale. In any case, you always need to know the aperture, the focal length, and the pixel size.”

#49 freestar8n

freestar8n

    Vendor - MetaGuide

  • *****
  • Vendors
  • Posts: 8842
  • Joined: 12 Oct 2007

Posted 10 May 2019 - 06:44 PM

Etendue is a measure of system throughput. That is all. But it is important.

If you have a wide nebula in the sky and you aim different imaging systems at it - the one with highest etendue will collect the most photons across the whole sensor. It could be smallest aperture or it could also be the slowest. It combines f ratio with sensor size.

 

High etendue could mean a fast system with a large sensor - with no specific reference to aperture or field of view.

 

Or it could mean a large aperture system with a wide field of view - with no specific reference to sensor size or f/ratio.

 

In both cases it combines a spatial area (sensor size or aperture) with a solid angle (either field of view or the solid angle represented by f/ratio).

 

If you use the same units for area and solid angle in both cases - you get the same value for etendue.

Frank


Edited by freestar8n, 10 May 2019 - 08:55 PM.


#50 Coconuts

Coconuts

    Mariner 2

  • *****
  • Posts: 225
  • Joined: 23 Sep 2012

Posted 14 September 2019 - 07:50 AM

While the concept of etendue comes at work in relation to microscopy illumination systems, I only recalled that it could be helpful in comparing astro imaging configurations when I read a Celestron white paper on their RASA astrographs, and again when CN poster rockstarbill posted the "Performance = aperture^2 * scale^2 * QE" formula, here: https://www.cloudyni...x455/?p=9639554.

 

In that post, and among the systems rsb chose to compare, the FSQ106 with the 0.72x reducer won, with an "Imaging Etendue" (QE factored in) of 37,064.  Via private message, rsb shared a link to this thread, and the cool Google Sheets calculator by the OP.  Thanks to you both!  With such a nice tool, I plugged in one my Sigma Art lenses, the 105 mm f/1.4, to see how it would score (with my Canon 6D, for which I guestimated a QE of 50%).  Wow, even stopped down a bit to f/2.0 to reduce vignetting, the imaging etendue was surprisingly high: 168,734!  I then entered values for some of my other Sigma Art lenses, all at f/2.0.  That got my attention: their scores were all an identical 168,734, irrespective of their very different focal lengths/diameters!  That shouldn't actually have been much of a surprise, though, and certainly won't be to the more technical posters here. 

 

On reflection, though, the "Performance = aperture^2 * scale^2 * QE formula" formula might suggest to those without optical training that the first term, aperture^2, is more valuable than it really is.  This lead me to reformulate the basic etendue formula in terms that are perhaps more germane to astro-imagers. 

This one is:

 

Imaging Etendue equals 424.36 * QE * pixel size^2  / focal ratio^2

 

For reflectors, you also have to multiply by 1 - secondary/primary ratio^2. 

This and the QE factor aren't in the formal etendue calculation, but they apply for our purposes, hence the term "Imaging Etendue".

Pixel size is in um.  The normalized version (QE = 100, pixel size 1 um, focal ratio 1.0) has an imaging etendue of 42,436.

 

From this perspective, imaging throughput on extended objects is all about high QE, larger pixels, and faster optics.  Obviously, you can easily get into trouble here with an undersampled image, and similarly, the "right" field of view is completely target dependent.  But I at least find this reformulation to be helpful.

 

Edit:  Going back through the thread in detail, I see where jhayes tucson had already expressed it this way, albeit in longer form: https://www.cloudyni...ator/?p=9349503

 

All the best,

 

Kevin


Edited by Coconuts, 14 September 2019 - 09:40 AM.



CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.


Recent Topics






Cloudy Nights LLC
Cloudy Nights Sponsor: Astronomics