80 replies to this topic

[prefetch]

i've been reading about etendue recently, and i couldn't find a calculator anywhere, so i made one.

 

i think it's accurate, but maybe not, which is why i thought i'd share it with folks smarter than me. 

 

if you find any bugs or problem, please go ahead and correct them - or post them in this thread.  the file is open to edit for anyone with the link:

 

https://docs.google....bQT-5BqBX2pUC4/

 

some interesting data from it:

 

assuming a 16803 camera, here's how these 3 systems compare:

 

NP127fli @ f/5.4: 71,938

C11 @ f/10: 18,318

C14 @ f/11: 15,346

 

again, if you find something wrong with how i've calculated etendue, please let me know and i'll fix it.  thanks.

[Michael Covington]

I find different definitions of étendue in different places and am still trying to decide how useful it is.  Let me think about that a while...  In any case, thanks for the calculator!

Edited by Michael Covington, 07 May 2019 - 10:05 PM.

[prefetch]

I find different definitions of étendue in different places and am still trying to decide how useful it is.

me too.  i'm using this formula: Etendue = diameter^2 * image scale^2 * QE

 

as for it being useful - i'm not sure yet.  still trying to figure that out as well, but i thought creating the calculator and posting would be a good first step to get some discussion going on the topic.

 

edit: changed area^2 to diameter^2 to reflect what the spreadsheet formula is actually doing, and see discussion below, it doesn't really make a difference.

Edited by prefetch, 08 May 2019 - 01:50 PM.

[Michael Covington]

Your formula has étendue increasing as the square of the image scale.  As I understood it, it depends only on aperture (as area in square cm), field of view (in square degrees), quantum efficiency, and light transmission (the latter two as numbers a bit less than 1).  So the same image, magnified, has the same étendue.  But now I'm not sure what source I followed when making my notes.  As I said, I've found different definitions in different places.

[freestar8n]

Etendue amounts to a spatial area times a solids angle and it can either be the aperture area times the solid angle seen by the detector. Or the sensor area times the solid angle of the pupil. The result is the same. But it’s easier to use the first one since you can the subtract the secondary area.

It normally doesn’t include qe but you can go ahead and include it along with total system transmission.

You can do it for the whole sensor or individual pixels.

In all cases it represents the total flux received on the sensor from a section of an extended object. If you factor in qe and T then it represents total electron flux.

Typical units are meters^2 deg^2

Frank

[prefetch]

thanks frank.

 

seeing how you are a PhD in optical sciences, maybe you could offer your take on this:

 

given the calculator show this:

C11 @ f/10: 18,318

C14 @ f/11: 15,346

 

would it be correct to say "imaging with the C11 is 'faster' than the C14, assuming the same camera"? (noting that the C11 image FOV will be slightly larger than the C14 image because of the focal length difference.)

 

am i understanding this correctly?

[freestar8n]

I haven’t looked at the calculator but if you have a single sensor and two ota’s then the etendue is determined by f ratio and yes the f/10 beats f/11 regardless of aperture.

The pupil makes a wider angle at f/10 as seen by the sensor.

I didn’t check the numbers but yes the c11 would have higher etendue with a given sensor.

“Faster” normally just refers to the f ratio. Etendue combines optical speed (as a solid angle of the pupil) with sensor size. Which is the same as aperture area times solid angle of the field of view.

Frank

[Michael Covington]

Frank, do I have this right? --

Étendue is a solid angle (e.g., in square degrees) times a surface area (e.g., in square cm).

It can be the area of the angular field of view in the sky times the surface area of the aperture of the telescope, or it can be surface area of the sensor times the solid angular extent of the aperture of the telescope as seen from the sensor.  These should be the same.

It can also be multiplied by factors less than 1 to account for quantum efficiency and for light loss (central obstruction, imperfectly transparent lenses, etc.) although for obstruction it is easier to just take it out of the surface area or angular area.

 

It can be computed per pixel or per megapixel rather than per entire sensor.

 

Is all this correct?  Also, what is "point étendue"?

[jhayes_tucson]

Etendue is related to the Lagrange invariant and to the more general optical invariant of the optical system.  It is a quantity that is conserved through an optical system and it can be used to show how radiance is conserved.  Wikipedia defines all this stuff pretty well so if you want to go into the gory details check it out here:  https://en.wikipedia.org/wiki/Etendue.  Computing Etendue is fine, but Etendue doesn't include sensor effects and signal strength is what you really want to know.  To do that, I personally find it more useful to simply compare the ratio of sensor signal strengths for two systems to see how they compare.  Here's how to compute signal ratios:

 

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

 

Where:

S_n = Signal strength

F_n = Focal ratio

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

CO_n = Central obscuration ratio (= D_secondary/D_primary)

R_n = Sensor responsivity (QE)

P_n = 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.

 

Finally, if you want to compare a 127 mm scope to the 14" both with a 16803 (as proposed by the OP:)

System 3:  127 mm aperture, F = 5.4, CO = 0, Tranmission = 0.85 (8 surfaces @ 0.98), R = 0.65, P = 9.0 um (NP127 w/16803)

 

S1/S3 = 0.224, which means that the refractor will have a signal almost 4.5x larger than the C14.  If you use F/11 for the C14, you'll get a number that is pretty close to the ratio given by the OP's numbers.  It is important to understand that in this particular case, the reason that the results are so close is because the cameras are the same.  Therefore, all that counts is the irradiance in the focal plane, which is closely related to Etendue (and which doesn't take into account all the other stuff that contributes to the final irradiance at focus.)  Just remember that what you gain in irradiance with the smaller refractor, you lose in image scale (along with sampling in object space.)  To increase both signal strength and image scale, you need a larger aperture.  So, unfortunately, there is no free lunch.  Larger aperture is always a benefit and that's why professional telescopes are so large.

 

John

Edited by jhayes_tucson, 08 May 2019 - 10:38 PM.

[bobzeq25]

me too.  i'm using this formula: Etendue = area^2 * image scale^2 * QE

 

as for it being useful - i'm not sure yet.  still trying to figure that out as well, but i thought creating the calculator and posting would be a good first step to get some discussion going on the topic.

Area squared or diameter squared (which is a measure of area)?

[Jon Rista]

thanks frank.

 

seeing how you are a PhD in optical sciences, maybe you could offer your take on this:

 

given the calculator show this:

C11 @ f/10: 18,318

C14 @ f/11: 15,346

 

would it be correct to say "imaging with the C11 is 'faster' than the C14, assuming the same camera"? (noting that the C11 image FOV will be slightly larger than the C14 image because of the focal length difference.)

 

am i understanding this correctly?

In the case of "same camera", f-ratio applies directly and becomes the dominant factor (not necessarily sole factor...and in this case, CO DOES matter). So the C11 is faster than the C14, and the differences between f-ratios and etendue's are pretty similar. 

 

Where your calculator gets interesting is when you have very dissimilar systems, and you want to compare them. When neither the scope nor camera is the same, it gets a lot harder to easily grasp how two systems may compare. This is where taking the aperture diameter squared times image scale squared times quantum efficiency (which isn't exactly etendue) can be useful, in that it gives you a single number that generally represents system performance for disparate systems for which f-ratio alone may not tell you enough to fully understand how they compare.

 

The intriguing thing you come to learn from these kinds of calculations is that f-ratio may not matter at all in some cases, and that aperture becomes the dominant factor. This is the case where image scale is the same (which implies that cameras are different, or at the very least that binning is employed in one of the cases to negate differences in effective pixel size and thus image scale). A great example here is an 11" f/10 SCT with a KAF-16200 vs. a 10" f/4 Newt with an IMX183. Both systems have the same image scale. The differences in clear apertures is offset (and then some, a bit) by differences in Q.E. So both systems perform the same. This is DESPITE the fact that the SCT has a significantly higher f-ratio. This is also DESPITE the fact that the IMX183 has significantly smaller pixels. We normalized the systems, though, same light grasp and same image scale...so their "etendue" are also normalized.

[freestar8n]

Frank, do I have this right? --

Étendue is a solid angle (e.g., in square degrees) times a surface area (e.g., in square cm).

It can be the area of the angular field of view in the sky times the surface area of the aperture of the telescope, or it can be surface area of the sensor times the solid angular extent of the aperture of the telescope as seen from the sensor.  These should be the same.

It can also be multiplied by factors less than 1 to account for quantum efficiency and for light loss (central obstruction, imperfectly transparent lenses, etc.) although for obstruction it is easier to just take it out of the surface area or angular area.

 

It can be computed per pixel or per megapixel rather than per entire sensor.

 

Is all this correct?  Also, what is "point étendue"?

Hi Michael-

 

Etendue is meant as a measure of optical throughput - and its most basic form it would be aperture area times field of view solid angle.

 

So that by itself is separate from system transmission and vignetting.  And QE applies to a property of the OTA/camera combined system - and wouldn't normally be in the etendue.  But for comparing two different imaging systems it would be useful to do so.

 

But none of this tells you anything about resolution or how one will compare to another for specific objects.  It is just about pulling photons from the sky over an area - and for that you need both aperture and optical speed.

 

Here is one example:

 

https://www.lsst.org...ists/keynumbers

 

This is the LSST - and it has the word "survey" in it - so high etendue is important.  They show the etendue as a simple area*solidangle - but the final number includes not only the impact of the secondary obstruction - but also the vignetting over the field.  That makes sense if you want to measure throughput and you have drop off away from the center.  It ends up being an "effective etendue" rather than the definition from geometric optics.

 

I don't know "point etendue."

 

For comparing different imaging systems, including different cameras, it would make sense to include secondary obstruction, transmission loss, vignetting and QE to create a single number that captures all these factors that impact how many electrons you get, in total, when you aim at some wide patch of nebulosity.  It isn't all that matters - but it's a good measure of realized throughput.  I think everything except QE is part of the effective etendue.  But adding QE makes sense as a metric for comparing systems.

Frank

[spokeshave]

I find different definitions of étendue in different places and am still trying to decide how useful it is.  Let me think about that a while...  In any case, thanks for the calculator!

Etendue is useful for clarifying all of the misunderstandings regarding imaging speed, aperture and f-ratio. As most know, threads on that topic rage from time-to-time with a lot of misinformation and no real resolution. 

 

In one of those threads not long ago, I introduced the idea of what I call "imaging etendue" to try to settle the debate on which systems were faster than others. I don't claim to have invented the idea, though I haven't seen it discussed before. I also took some liberties with calculating etendue, with the caveat that the simplified calculation would only be meaningful when comparing one system to another and would be meaningless as a stand-alone number. I also corrupted the pure meaning of etendue by including a term that corrects for QE of the sensor. 

 

Etendue is (loosely) defined as the product of the area of the entrance pupil and the solid angle seen by the sensor. For a telescope, the area of the entrance pupil is the area of the aperture = pi*r^2. That's mathematically the same as (pi/4)*d^2. Since telescope apertures are always given as a diameter not a radius, and the pi/4 term is just a common constant, I simplified the entrance pupil area calculation to just d^2. This is fine as long as the resultant imaging etendue calculation is only used to compare one system to another. 

 

The other etendue term is the solid angle seen by the sensor. For the purposes of comparing the relative speed of imaging systems, it is more important to evaluate the solid angle seen by a pixel. The angle (not solid angle) seen by a pixel is just the image scale in arcseconds-per-pixel, and the solid angle is just the image scale squared. So, the simplified "imaging etendue" is simply:

 

E = d^2*s^2

 

Where "d" is the aperture diameter and "s" is the image scale. Again, units are not important since this is only a comparative metric. This can be further refined by including a QE term to account for quantum efficiency in the overall system imaging speed. Of course other adjustments can also be made that account for central obstruction, telescope throughput, etc. But for comparing imaging systems, I find that the simple E = QE*d^2*s^2 works very well as a reliable first approximation.

 

The end result is a metric that represents the light throughput for each pixel. Imaging etendue is obviously agnostic of focal length and f-ratio (though both are buried in the calculation) and represents a direct linear comparison of imaging speed for different imaging systems. If one system has an imaging etendue of 1000 and another has an imaging etendue of 5000 (again - units are irrelevant in a comparison), the second system will truly image 5 times faster than the first. 

 

I beat this drum at every opportunity I get because there is so much myth and misinformation regarding imaging speed. Some people will incorrectly argue that f-ratio alone determines imaging speed. Others will incorrectly argue that aperture alone determines imaging speed. Still others argue (correctly) that imaging speed depends on both aperture and f-ratio but also must take into account pixel size. Imaging etendue is nothing more than a mathematical representation of the last point. 

 

Tim

Edited by spokeshave, 08 May 2019 - 08:04 AM.

[prefetch]

Area squared or diameter squared (which is a measure of area)?

yes, diameter, not area.  

[Michael Covington]

I note that you are using d^2 to measure in units of pi/4 of a square degree...  as you said, units don't matter for comparison, but this reminds me of the early 1980s, when I was in graduate school.  Back then, pizzas were sold by diameter in inches, and we quickly decided to measure pizza in round inches (e.g., a 10-inch pizza is 100 round inches) to compare them.

[spokeshave]

Area squared or diameter squared (which is a measure of area)?

See my discussion above about why d^2 is fine to use for the aperture area. The actual area is really irrelevant. What matters is the comparison of etendue for two (or more) different imaging combinations. For each, if you used the actual area calculation (pi*r^2) instead of the simplified version (d^2), the final result will be different for each by a factor of pi/4. Since that is a constant, the relative imaging speed of each is unchanged. It just makes the calculation that much easier.

 

Tim

[spokeshave]

I note that you are using d^2 to measure in units of pi/4 of a square degree...  

Actually, it would be pi/4 of a square inch, or millimeter, or centimeter, or furlong - whatever your aperture is measured in.

 

Tim

[bobzeq25]

See my discussion above about why d^2 is fine to use for the aperture area. The actual area is really irrelevant. What matters is the comparison of etendue for two (or more) different imaging combinations. For each, if you used the actual area calculation (pi*r^2) instead of the simplified version (d^2), the final result will be different for each by a factor of pi/4. Since that is a constant, the relative imaging speed of each is unchanged. It just makes the calculation that much easier.

 

Tim

Agreed.  I was just questioning a (now edited) post where he said "area squared".

[jhayes_tucson]

Etendue is useful for clarifying all of the misunderstandings regarding imaging speed, aperture and f-ratio. As most know, threads on that topic rage from time-to-time with a lot of misinformation and no real resolution. 

 

In one of those threads not long ago, I introduced the idea of what I call "imaging etendue" to try to settle the debate on which systems were faster than others. I don't claim to have invented the idea, though I haven't seen it discussed before. I also took some liberties with calculating etendue, with the caveat that the simplified calculation would only be meaningful when comparing one system to another and would be meaningless as a stand-alone number. I also corrupted the pure meaning of etendue by including a term that corrects for QE of the sensor. 

 

Etendue is (loosely) defined as the product of the area of the entrance pupil and the solid angle seen by the sensor. For a telescope, the area of the entrance pupil is the area of the aperture = pi*r^2. That's mathematically the same as (pi/4)*d^2. Since telescope apertures are always given as a diameter not a radius, and the pi/4 term is just a common constant, I simplified the entrance pupil area calculation to just d^2. This is fine as long as the resultant imaging etendue calculation is only used to compare one system to another. 

 

The other etendue term is the solid angle seen by the sensor. For the purposes of comparing the relative speed of imaging systems, it is more important to evaluate the solid angle seen by a pixel. The angle (not solid angle) seen by a pixel is just the image scale in arcseconds-per-pixel, and the solid angle is just the image scale squared. So, the simplified "imaging etendue" is simply:

 

E = d^2*s^2

 

Where "d" is the aperture diameter and "s" is the image scale. Again, units are not important since this is only a comparative metric. This can be further refined by including a QE term to account for quantum efficiency in the overall system imaging speed. Of course other adjustments can also be made that account for central obstruction, telescope throughput, etc. But for comparing imaging systems, I find that the simple E = QE*d^2*s^2 works very well as a reliable first approximation.

 

The end result is a metric that represents the light throughput for each pixel. Imaging etendue is obviously agnostic of focal length and f-ratio (though both are buried in the calculation) and represents a direct linear comparison of imaging speed for different imaging systems. If one system has an imaging etendue of 1000 and another has an imaging etendue of 5000 (again - units are irrelevant in a comparison), the second system will truly image 5 times faster than the first. 

 

I beat this drum at every opportunity I get because there is so much myth and misinformation regarding imaging speed. Some people will incorrectly argue that f-ratio alone determines imaging speed. Others will incorrectly argue that aperture alone determines imaging speed. Still others argue (correctly) that imaging speed depends on both aperture and f-ratio but also must take into account pixel size. Imaging etendue is nothing more than a mathematical representation of the last point. 

 

Tim

 

Tim,

I hadn't given it much thought before now, but if you look at the expression that I provided in my previous post for signal strength and you simply rearrange the terms, it reduces to the same expression that you've provided (if you ignore the throughput terms.)

 

John

[spokeshave]

Tim,

I hadn't given it much thought before now, but if you look at the expression that I provided in my previous post for signal strength and you simply rearrange the terms, it reduces to the same expression that you've provided (if you ignore the throughput terms.)

 

John

Thanks John. Yours is clearly the more rigorous expression, but it all boils down to the same concept. In fact, yours is the correct expression of etendue whereas mine essentially co-opts the concept of etendue to develop a metric that is suitable only for comparing the relative imaging speeds of systems. But mine is simpler. 

 

Tim

[Francois]

...

 

Tim

With the very important caveat that this is true in when noise is dominated by the source's shot noise. Once you get into the skyglow-limited regime for features smaller than the DIQ (like, say, stars), pixel size does matter. For a dark site with ~1" seeing and seeing-dominated DIQ, that's going to happen for stars of 20th magnitude. At that point you will absolutely want to sample the image properly if you're interested in stars or small features. For large scale features that does not apply.

 

As for etendue, it's not like anything is new there. The astronomers from the 1930s were already well aware.

[Jon Rista]

With the very important caveat that this is true in when noise is dominated by the source's shot noise. Once you get into the skyglow-limited regime for features smaller than the DIQ (like, say, stars), pixel size does matter. For a dark site with ~1" seeing and seeing-dominated DIQ, that's going to happen for stars of 20th magnitude. At that point you will absolutely want to sample the image properly if you're interested in stars or small features. For large scale features that does not apply.

 

As for etendue, it's not like anything is new there. The astronomers from the 1930s were already well aware.

While etendue is not new overall, I think the point Tim was trying to make, that many of us try to make when these topics come up as these are not necessarily as well-known in amateur and particularly beginner circles...is that neither f-ratio alone nor aperture alone determines system performance.  That, overall, across the broad range of possible system combinations, performance is better and more accurately approximated (in the sky limited regime) by etendue...or Tim's formula, which I like because it is simple and convenient, and relatively easy to calculate for pretty much any system.

 

This matters, because as Tim said it is frequently stated...here on CN and in other forums (both online and in the real world)...that performance is either solely determined by f-ratio, that f-ratio is the only thing that matters, or that performance is solely determined by aperture, or that aperture is the only thing that matters. Neither are in fact true, and when such statements are made it misleads and confuses the subject. That then leads to things like broad-scale demonizing of any larger telescope because "big is bad" and "big is hard" and that only smaller telescopes are truly easier (when, in fact, a larger telescope could, paired with the right image scale, TROUNCE a smaller scope.) 

Edited by Jon Rista, 08 May 2019 - 01:02 PM.

[Francois]

While etendue is not new overall, I think the point Tim was trying to make, that many of us try to make when these topics come up as these are not necessarily as well-known in amateur and particularly beginner circles...is that neither f-ratio alone nor aperture alone determines system performance.  That, overall, across the broad range of possible system combinations, performance is better and more accurately approximated (in the sky limited regime) by etendue...or Tim's formula, which I like because it is simple and convenient, and relatively easy to calculate for pretty much any system.

 

This matters, because as Tim said it is frequently stated...here on CN and in other forums (both online and in the real world)...that performance is either solely determined by f-ratio, that f-ratio is the only thing that matters, or that performance is solely determined by aperture, or that aperture is the only thing that matters. Neither are in fact true, and when such statements are made it misleads and confuses the subject. That then leads to things like broad-scale demonizing of any larger telescope because "big is bad" and "big is hard" and that only smaller telescopes are truly easier (when, in fact, a larger telescope could, paired with the right image scale, TROUNCE a smaller scope.) 

Yes, as I said it's a caveat. Not a rebuttal. In the bright source regime for wide-scale mapping speed it's a good characterization (which was very much on the mind of astronomers going to image the southern hemisphere back before there was infrastructure and had to carry all the stuff from the coast to the site on mule-back; no, I'm not kidding, that's how it was done).

[Organic Astrochemist]

While etendue is not new overall, I think the point Tim was trying to make, that many of us try to make when these topics come up as these are not necessarily as well-known in amateur and particularly beginner circles...is that neither f-ratio alone nor aperture alone determines system performance.  That, overall, across the broad range of possible system combinations, performance is better and more accurately approximated (in the sky limited regime) by etendue...or Tim's formula, which I like because it is simple and convenient, and relatively easy to calculate for pretty much any system.

 

This matters, because as Tim said it is frequently stated...here on CN and in other forums (both online and in the real world)...that performance is either solely determined by f-ratio, that f-ratio is the only thing that matters, or that performance is solely determined by aperture, or that aperture is the only thing that matters. Neither are in fact true, and when such statements are made it misleads and confuses the subject. That then leads to things like broad-scale demonizing of any larger telescope because "big is bad" and "big is hard" and that only smaller telescopes are truly easier (when, in fact, a larger telescope could, paired with the right image scale, TROUNCE a smaller scope.) 

I think Francois's point is that even etendue doesn't fully capture performance. Using larger pixels or binning will always increase your etendue, but it won't always improve your image.

 

I don't think I fully understand all the issues, but it seems that diameter and image scale are inversely related and that suggest to me that there might be some optimum values for a given target size.

[Michael Covington]

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.