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# Formula to estimate exposure from aperture & F ratio

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Mercury-Atlas

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Posted 17 July 2019 - 04:22 PM

Maybe you can help me:

I understand that the amount of photons that will hit a sensor will depend on the aperture (more aperture meaning more photons, shorter exposures) and on the f-ratio (longer f ratio meaning fewer photons, longer exposures).

What are the formulas to quantify this flux of photons?

What would allow for shorter exposures, a 5.5" F6 scope, or a 6.1" F5.5 scope?

Edited by Adun, 17 July 2019 - 04:23 PM.

### #2 kathyastro

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Posted 17 July 2019 - 05:03 PM

Linear aperture does not affect the amount of light per unit area (i.e. per pixel) of the sensor.  The only variable that affects the illumination of the sensor is "photographic aperture", i.e. focal ratio.  Indeed this is why photographers specify aperture in focal ratio rather than in millimetres.

The f/5.5 scope will have the shorter exposure times.  The required exposure time will be 84% of what the f/6 scope requires for the same image brightness.

The ratio is (f1 / f2) ^ 2.  In this case, (5.5 / 6.0) ^2 = 0.84

Edited by kathyastro, 17 July 2019 - 05:04 PM.

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### #3 ericthemantis

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Posted 17 July 2019 - 05:11 PM

You gotta know how many photons are coming from your target to calculate exact exposures.

What you can do pretty easily is make relative comparisons. And you don't need to know the aperture to do it, just the F-ratio. Full stops are in increments of 1/sqrt(2), and either double or halve the amount of light passing through. Examples of "full-stops" are f/1, f/1.4, f/2.0, f/2.8, f/4, f/5.6, f/8.0, f/11.0 etc. An f/4.0 will only need half the time of a f/5.6 scope for the same image brightness. But that same f/4.0 scope would need twice as long an exposure as an f/2.8 for the same brightness.

Bigger aperture will give you greater resolution (a "bigger picture"), but exposure lengths is strictly determined by f-ratio in this case.

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### #4 gezak22

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Posted 17 July 2019 - 05:26 PM

Exposure also depends on what you are trying to capture. Galaxy cores? Shorter exposures are totally fine. Faint background nebulas? Expose for as long as your mount and sky conditions allow.

### #5 Stephen Kennedy

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Posted 17 July 2019 - 05:54 PM

It is actually a little more complicated than that.

For point sources such as stars, only aperture matters so for a star:

(5.5"/6.1")^2 = 0.81

The exposure time for the 6.1" telescope is 81% of the exposure time of the 5.5" telescope.

For extended objects such as galaxies both the aperture and focal ratio need to be considered.

(5.5"/6.1")^2 x (5.5/6)^2 = 0.68

For an extended object, to gather an equal amount of photons, the exposure time for the 6.1" F/5.5 telescope will be 68% of that of the 5.5" F/6 telescope.

### #6 tkottary

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Posted 17 July 2019 - 06:01 PM

https://www.cloudyni...due-calculator/

### #7 555aaa

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Posted 17 July 2019 - 06:01 PM

For a star, it's all aperture. For a 20th mag star, it's about 440 photons/sec/square meter, covering broadly the spectral response of most cameras. The f/ratio drops out of the equations, because the total number of photons per second that goes into the instrument is only defined by the aperture.

Every 5 magnitudes is a factor of 100x in brightness = optical flux.

If the object is "the sky", the wider the field of view, the more photons make it to some point somewhere on the detector, and what happens as you are going to a lower f/number is typically that you are lowering the focal length and covering a larger solid angle of sky per unit angle on the detector, which = lowering the image magnification. If you hold image magnification constant (sky angle per mm of sensor), then it goes back to being aperture only.

Sorry people were typing while I was thinking.

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### #8 Jon Rista

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Posted 17 July 2019 - 07:14 PM

It is actually a little more complicated than that.

For point sources such as stars, only aperture matters so for a star:

(5.5"/6.1")^2 = 0.81

The exposure time for the 6.1" telescope is 81% of the exposure time of the 5.5" telescope.

For extended objects such as galaxies both the aperture and focal ratio need to be considered.

(5.5"/6.1")^2 x (5.5/6)^2 = 0.68

For an extended object, to gather an equal amount of photons, the exposure time for the 6.1" F/5.5 telescope will be 68% of that of the 5.5" F/6 telescope.

For a true point source, only aperture would matter. Stars are not necessarily true point sources all the time, and exactly how they cover pixels can make how quickly they saturate depend...it may depend just on aperture, maybe both aperture and f-ratio.

For extended objects, aperture is actually accounted for in f-ratio, and since the signal of extended objects is distributed over the field only f-ratio really matters.

Now, these apply for a given camera, used across different scopes.

When you get into comparing different scopes AND different cameras, then things get more complicated. A lot of factors come into play, so overall if you want exact results it can get quite complex. But for more general comparisons, the following formula will give you a simple "performance" factor that can then be compared across systems:

P = A^2 * S^2 * Q

Where A is aperture, S is image scale and Q is quantum efficiency. It won't be absolutely perfect...there are other things like transmission and filter bandpass to consider, as well as the more true nature of quantum efficiency (which is an integration over the band), but the formula above will usually work well enough for most comparisons.

### #9 Stephen Kennedy

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Posted 17 July 2019 - 09:08 PM

Focal ratio only accounts for aperture when two different telescopes have the same focal length which they do not in the example given. When comparing exposure times for two telescopes you have to know both the aperture and focal length to be able to calculate the focal ratio of each.

For example, if two telescopes both have a focal length of 1000 mm but one has an aperture of 200 mm and the other 100 mm.  Then they will be F/5 and F/10 respectively with F/5 being faster.  In order to have both telescopes at the same focal ratio the 200 mm aperture telescope would have to have a focal length of 2,000 mm.

Focal ratio can only be calculated if both the focal length and the aperture are known.

When comparing two telescopes of different apertures and different focal ratios, both have to be accounted for in exposure time for extended objects.

### #10 DmitriNet

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Posted 17 July 2019 - 09:19 PM

Both aperture and f-ratio may be influencing exposure time

Even for a star, there are roughly two cases:

1.  PSF size (width at half-flux) > pixel size

2.  PSF size (width at half-flux) < pixel size

In general, exposure = 0.8* full-well-capacity/number_of_photons_absorbed_by_pixel

Also, 50% of photons entering aperture are concentrated in in the  PSF size (width at half-flux) which in case of ideal for the scope seeing is Airy disk diameter

So, in 1.  number_of_photons_absorbed_by_pixel ~ D*D (area of aperture) *pixel_X_size*pixel_X_size/(Airy_disk_diameter*Airy_disk_diameter)

Since  Airy_disk_diameter ~ f-ratio,

exposure_time ~ (f-ratio/(pixel_size*aperture))^2

Only in case 2 f-ratio may not matter, but aperture always matters.

### #11 ks__observer

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Posted 18 July 2019 - 12:35 AM

Extended object:

JR performance factor.

Star point source intensity:

(Aperture/f-ratio)^2

Exposure time:

https://www.cloudyni...d-maybe-qhy163/

https://www.cloudyni...strodons/page-3

### #12 Jon Rista

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Posted 18 July 2019 - 12:37 AM

Focal ratio only accounts for aperture when two different telescopes have the same focal length which they do not in the example given. When comparing exposure times for two telescopes you have to know both the aperture and focal length to be able to calculate the focal ratio of each.

For example, if two telescopes both have a focal length of 1000 mm but one has an aperture of 200 mm and the other 100 mm.  Then they will be F/5 and F/10 respectively with F/5 being faster.  In order to have both telescopes at the same focal ratio the 200 mm aperture telescope would have to have a focal length of 2,000 mm.

Focal ratio can only be calculated if both the focal length and the aperture are known.

When comparing two telescopes of different apertures and different focal ratios, both have to be accounted for in exposure time for extended objects.

Your previous post said you needed to know "aperture and f-ratio", which is what I was responding to. If you know f-ratio, you do not need to know aperture, because f-ratio is focal length/aperture. F-ratio is all you need to know for extended objects, assuming the same camera on each scope.

Edited by Jon Rista, 18 July 2019 - 12:39 AM.

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Mercury-Atlas

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Posted 18 July 2019 - 09:40 AM

In the link tkottary shared, John Hayes explains with details his formula:

... 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:
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.

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

So assuming the same camera, we'd get:

Sa/Sb = (Fb/Fa)2 (Ta/Tb) (1-COa2)/(1-COb2)

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)

This agrees with those of you saying that only focal ratio (not aperture) matters for exposure time of extended objects.

But this is against advise I was told 2 years ago in the EAA forum. I got to believe aperture did matter, but the lack of consensus in your answers makes me wonder.

Perhaps the assertions not "the only variable that matters for exposure is focal ratio" is assuming that we are comparing at a fixed focal length? Because if there is such assumption, then I need to keep looking for answers, for my question wants to compare scopes of different focal lengths, focal ratios, and apertures (but assuming the same camera, so, different image scale too)

Edited by Adun, 18 July 2019 - 09:40 AM.

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### #14 bobzeq25

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Posted 18 July 2019 - 11:34 AM

Maybe you can help me:

I understand that the amount of photons that will hit a sensor will depend on the aperture (more aperture meaning more photons, shorter exposures) and on the f-ratio (longer f ratio meaning fewer photons, longer exposures).

What are the formulas to quantify this flux of photons?

What would allow for shorter exposures, a 5.5" F6 scope, or a 6.1" F5.5 scope?

The difference is trivial.

What's not trivial is:

1) Either scope is awfully big (heavy, longish focal length) for a starter AP scope.  You'd like something less than 600mm and 10 pounds.

2) The dependence of proper subexposure on your light pollution level.

### #15 jhayes_tucson

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Posted 18 July 2019 - 01:02 PM

Maybe you can help me:

I understand that the amount of photons that will hit a sensor will depend on the aperture (more aperture meaning more photons, shorter exposures) and on the f-ratio (longer f ratio meaning fewer photons, longer exposures).

What are the formulas to quantify this flux of photons?

What would allow for shorter exposures, a 5.5" F6 scope, or a 6.1" F5.5 scope?

=========================

Putting aside a few stupendously incorrect answers, there are a number of good comments here.  I'd like to expand a bit on how this stuff works.  Right up front you need to understand that there is a difference in the radiometry of extended sources (pretty much everything but stars) and extended sources.  In normal, every day photography, we only deal with extended sources.  In the night sky, we deal with both stars, which are point sources, and extended sources--such as nebula, galaxies, and solar system objects.  The equations from my Etendue post comparing two systems are valid for extended objects where the irradiance in the focal plane varies only as the inverse of the focal ratio squared.  That means that for any given object, an 8 m telescope operating at say F/4 will produce an image that has exactly the same irradiance in the focal plane as an 80 mm, F/4 telescope.

That is an amazing result!  The 8 m telescope is 100 times larger with 10,000 times more collecting area so how can that be?  The answer lies in how the light is spread out in the image plane.  Remember that the 8m scope has a focal length that is 100 times larger so the image will also be 100 times larger in the focal plane.  That means that even though the large scope gathers 10,000 times more photons, it spreads them out over 10,000 times more area in the focal plane.  That will illuminate the focal plane with the same irradiance (in W/m2) for each of the two scopes.  The big (and I mean really big) difference will be the image scale.  That big scope produces a much bigger image than the image produced by the 80 mm scope even though they are at the exact same irradiance.

When it comes to stars, things are different because diffraction, rather than first order properties, determines the size of the point spread function.  First, forget about the atmosphere and think of a scope in a vacuum.  The maximum irradiance of a star image from a perfect telescope varies with the (diameter/focal ratio)2 = (D/F)2.  That means that for any given focal ratio, a larger telescope will pick up fainter stars than a smaller telescope.  Double the size and you quadruple peak stellar irradiance.

So what happens when we throw in the atmosphere?  When light passes through the atmosphere it adds a time varying deviation to the inbound plane wavefront from a star.  That means the instantaneous star image will no longer be a perfect Airy pattern and that the form of the pattern will vary with time.  When we take a long time exposure of a star, we record the time averaged pattern, which will always be larger than what we would get in a vacuum.  That means that the light energy from a star will be smeared over a larger area on the sensor and we'll record a lower peak signal compared to what we might expect.  A star image will still not behave like an extended source but it won't vary perfectly as (D/F)2 either.  It will be close...just not perfect and how imperfect will depend on bad the seeing might be.  You can see this effect for yourself by imaging an area of the sky dense in stars and carefully comparing the faintest stars detectable on nights of good seeing verses nights with poor seeing.  So, a larger diameter telescope will always produce star images with higher peak irradiance in the focal plane than a smaller scope with the same focal ratio even when you consider atmospheric seeing.

Remember that so far I've only mentioned the effect on the optical irradiance in the focal plane.  To get at the signal you measure, you then have to fold in the sensor characteristics.  The size of the pixels, the responsivity (Q.E.) and the exposure length will determine the ultimate signal level that you'll record.  Remember that larger pixels always record a higher signal but if you make the pixels too big, you risk under-sampling the image and throwing away detail.  However, this is where sampling plays its most important role in this whole thing.  Realize that if you take two telescopes of different focal ratios and use equal sampling in object space, the signal strength will be exactly the same!  That's why I'm content with my slow F/10.8 system for imaging.  It produces exactly the same signal as an F/5 system that uses a camera with the exact same angular sampling relative to the sky.  I've attached a slide to show how this works.  Just remember that the two scopes have to have different cameras--you can't just remove the camera from the f/10.8 and plop it onto the F/5 scope (that's equal sampling in image space.)  You only get the same signal when the sampling relative to the sky is constant (assuming the QE of the two cameras is equal.)

Finally, to directly answer your question, the larger faster F/5.5, 6.1" scope will produce around 19% more irradiance in the focal plane for extended objects than the F/6, 5.5" scope.  It will also image slightly fainter stars for the same camera and exposure time.  That's not a huge difference.  In my view, differences like this don't become significant until they reach a level of around 30% - 50% since other factors can may overwhelm their relative importance.  Portability, weight, optical quality, available accessories and stuff like that may be more important than that 19%.  On the other hand, differences of 200% - 300% really get my attention and that's where I'm much more willing to put my efforts.  Everything being equal, I will personally always opt for the larger aperture.

John

#### Attached Thumbnails

Edited by jhayes_tucson, 19 July 2019 - 10:47 AM.

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### #16 Jon Rista

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Posted 18 July 2019 - 02:44 PM

The equations from my Etendue post comparing two systems are valid for extended objects where the irradiance in the focal plane varies only as the inverse of the focal ratio squared.  That means that for any given object, an 8 m telescope operating at say F/4 will produce an image that has exactly the same irradiance in the focal plane as an 80 mm telescope.

John, before you lose the chance to edit, you may want to clarify the above to make sure everyone understands that it isn't just any 80mm telescope, but an 80mm f/4 telescope. (Beginners forum and all...and the fact that most 80mm refractors are around f/6, and most don't even reduce to f/4...maybe f/4.5, f/4.8...)

Otherwise, excellent post, as always.

Edited by Jon Rista, 18 July 2019 - 02:47 PM.

### #17 jhayes_tucson

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Posted 18 July 2019 - 04:09 PM

John, before you lose the chance to edit, you may want to clarify the above to make sure everyone understands that it isn't just any 80mm telescope, but an 80mm f/4 telescope. (Beginners forum and all...and the fact that most 80mm refractors are around f/6, and most don't even reduce to f/4...maybe f/4.5, f/4.8...)

Otherwise, excellent post, as always.

Thanks for the catch Jon!  Yes, that's what I meant and I fixed it.

John

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Mercury-Atlas

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Posted 18 July 2019 - 09:37 PM

The equations from my Etendue post comparing two systems are valid for extended objects where the irradiance in the focal plane varies only as the inverse of the focal ratio squared.

Thank you John. I appreciate you taking your time to write here.

What are those objects you speak of? Or more importantly: which objects have a different irradiance, for which your equations aren't valid?

Before today, I believed that aperture (not only F ratio) did matter for making exposures shorter, this was from an answer I got in the EAA forums about two years ago.

Your formulas and some people's answers to my question, state that aperture matters not in reducing exposure times (given a fixed camera, so differing image scales).

However other answers point to the opposite (just like the one I learned from two years ago). Take Tim's answer in the other thread:

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.
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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 .
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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. .
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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.

At first glance, such diverging opinions confused me, with John's formula featuring no aperture (only focal ratio) while Tim's formula features Aperture and image scale (but no Focal Ratio).

But then again, if I expand image scale on Tim's formula I get: E ~= Diameter2 * (Pixel_Size / FL)2

And since Focal Ratio = FL / Diameter, Tim's formula can be elaborated as:

E ~= Pixel_Size2 / FR2

Which then agrees with what others are saying: exposure time is inversely proportional to the square of F Ratio. It seems both  'sides' are right (although then proportionality to pixel size seems odd).

For my own personal comparisons, I found Tim's formula more useful, if only because it makes it easier to compare 3 or 4 scopes / setups, than only 2 scopes. I did enhance it as he mentioned, to account for central obstruction and light transmission throughput so as to reducing "aperture" to "effective aperture". In an Excel comparing my scopes with different reducer combinations I got some very interesting results.

Edited by Adun, 18 July 2019 - 09:45 PM.

### #19 freestar8n

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Posted 19 July 2019 - 04:07 AM

This topic comes up regularly in various forms - and there is a way to answer it with no math at all - and instead to appreciate certain fundamental things about how imaging works.

My answer to these and related questions is always - consider the pupils - and understand how they work.

These topics also steer towards "etendue" - but it isn't needed when considering simply how bright an image is in the image plane.  Etendue refers to the throughput of a complete optical system - with a sensor of some size specified.  It could be the complete sensor or just a pixel - but it involves how flux is collected by the system over a specified area.

But the actual brightness of the image is independent of the area over which it is measured.  Irradiance is an intrinsic quantity - whereas etendue is extrinsic.

A good way to imagine the local brightness in the image plane is to imagine a tiny bug walking around and looking at how light is falling on the image plane.  It falls from the sky above it - in a big circular patch - or annulus in the case of an sct.  The bug has no idea what the aperture or focal length is - it just sees light coming in from above over some solid angle of the sky.

And the important fundamental thing to realize is - that glowing area above it is glowing at exactly the brightness of the source object - whether it is the moon, a faint nebula, or the sun.  There is a uniform glow coming down - and its brightness is the same as the source (at each point in the image).  (Actually the brightness is reduced by any losses of transmission in the optics - but ignore that for now).

So - the light locally falling in the image plane all comes from this glowing disc or annulus above - and each point in the image will see the same annulus glowing uniformly - but with a different brightness corresponding to each local part of the object.

As a result - what determines how bright the image is at a given point?  All that matters is the the size of the solid angle of that glowing patch above the image.  And that is set by the cone angle of the arriving light - which is set by the f/ratio and the secondary obstruction.  The size and distance of these things doesn't matter - only the solid angle does.

People naturally think the pupil is concentrating light into it or something - but it isn't.  The irradiance is made big with fast optics by increasing that solid angle of light shining on each part of the image.  The solid angle can be large with a small aperture if the focal length is short; and it can be small if the aperture is large but the focal length is long.  But fast or slow, the pupil is glowing with the brightness of the source - and it can't be boosted.

Once you know how the brightness in the image plane behaves you can talk about how much light is collected by a pixel or a sensor - and it just scales with the area.  Then you can talk about etendue.

So - no math and no explicit aperture, or focal length or a concept of "collecting" light and then "focusing" it.  It's just a pupil glowing with the exact brightness of the source - and illuminating each point in the image.  Bigger pupil solid angle means more brightness and shorter exposure.

As for how stars behave - they are completely different and impossible to predict in general.  They depend on the seeing and aberrations of the system - and how exactly the image of a point source is formed.  Extended objects are much simpler to predict.

Frank

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### #20 Jon Rista

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Posted 19 July 2019 - 09:53 AM

So - no math and no explicit aperture, or focal length or a concept of "collecting" light and then "focusing" it.  It's just a pupil glowing with the exact brightness of the source - and illuminating each point in the image.  Bigger pupil solid angle means more brightness and shorter exposure.

If you had exclude the part about "shorter exposure" I would have agreed. But since you have included shorter exposure, that is demonstrably untrue. Pixel size DOES play a role in the required exposure length. If you configure two systems that have the same image scale, then f-ratio does not actually tell you anything. Aperture, in fact, tells you everything.

Consider the following:

System A: KAF-16200, Celestron EdgeHD 1100 f/10

System B: IMX183, TS 10" f/4 Newtonian

These two systems have similar image scale:

System A: 0.46

System B: 0.49

They have similar apertures:

System A: 11" w/ 30% CO

System B: 10" w/ 25% CO

They have dissimilar quantum efficiencies in reality:

System A: 60%

System B: 84%

But lets assume they have the same Q.E. and eliminate this factor.

In this case, the f-ratios of f/10 and f/4 tell us nothing. Or, in fact, they lie to us...since one would generally assume, if f-ratio was truly everything, that the f/4 system would be able to use much shorter exposures. This is plain and simply false. The two systems can, assuming identical Q.E., use the same exposure! The f/10 system is 6.25x slower than the f/4 system: (10/4)^2. However the f/10 system is paired with pixels that are 6.25x larger than the f/4 system: 6^2/2.4^2! These counter-balanced traits completely nullify the f-ratio. F-ratio doesn't mean squat here, and it certainly doesn't mean that the f/4 system can use shorter exposures...neither on a sub-exposure basis nor an integrated exposure basis.

In point of fact, the reality of these two systems is the Newtonian system with the IMX183 is actually faster, thanks to the higher quantum efficiency. So despite the fact that the bigger pixels of the KAF-16200 nullify the loss of light at f/10, the IMX183 is 1.4x faster. Further, it so happens that the IMX183 ends up covering a larger field of view as well.

This is why people are talking about Etendue. F-ratio is, as we've found in the past every time this topic comes up, insufficient on its own to fully describe the performance of a given system...system, including both scope and camera where cameras may differ from system to system. The only time f-ratio alone is sufficient is if you are assuming the same camera is always used on every telescope...which is a limited use case.

Edited by Jon Rista, 19 July 2019 - 09:53 AM.

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### #21 freestar8n

freestar8n

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Posted 19 July 2019 - 05:31 PM

Jon I'm trying to get a fundamental point across about how all this stuff works.

My statement about exposure applies perfectly well if you leave the sensor fixed and only compare different ota's of different aperture and f/ratio - which is the OP's question.

If you want to talk about SNR in detail everything unravels and there are a million variables involved.  I have also dealt with that case in a way that goes beyond simply "snr" and instead looks at how deep you can go to achieve a given snr in a given time.  That will involve sensor properties, light pollution, time lost between exposures, and the role of pattern noise.  It is provided here:

https://freestar8n.s...ps.io/StackSNR/

The OP is asking for a formula for how exposure time relates to aperture, f/ratio, and flux.  For a given camera and with no consideration for resolution, it only depends on the irradiance in the image plane - and that is set by the solid angle subtended by the pupil - because the pupil is always glowing with the brightness of the source - except for transmission loss and other small terms.  No need for math or formulae if you can appreciate this fundamental principle of how pupils work.

If you vary the sensor and other properties - I think the most meaningful way to compare two systems is how deep they can go to achieve a given snr in a given time.  There are no simple equations to represent all that's involved in that case - hence the tool I provided.

Frank