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  #61  
Old 29-06-2015, 09:47 AM
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Very interesting thread.

Just wanted to add, that for drizzle algorithm to work dithering may not be needed with less than perfect guiding, as there will be some variations between the sub frames anyway.

Also, although it does not relate to sensitivity, but it might be useful to be aware of the fact that binning will also increase well depth given that camera allows to adjust gain. It can come in handy when using a CCD with small pixels and thus shallow well depths.
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  #62  
Old 29-06-2015, 10:20 AM
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Originally Posted by Slawomir View Post
Very interesting thread.

Just wanted to add, that for drizzle algorithm to work dithering may not be needed with less than perfect guiding, as there will be some variations between the sub frames anyway.

Also, although it does not relate to sensitivity, but it might be useful to be aware of the fact that binning will also increase well depth given that camera allows to adjust gain. It can come in handy when using a CCD with small pixels and thus shallow well depths.
agreed, but the practicality of binning (as Greg pointed out) is compromised by chip design. Although adding together the outputs from 4 full pixels increases the well depth, that extra charge often cannot be squeezed through the transfer registers. Chip designers seem to compromise by making sure that the registers can carry about half of the 2x2 binned full charge and they can also halve the gain to protect the dynamic range in the ADC. The read noise is also generally compromised a bit (it is usually more than that from a 1x1 read). What you often end up with is about 4x the charge at low levels, more read noise, half the gain and about 2x the effective well depth.
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  #63  
Old 29-06-2015, 10:26 AM
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Couple of things worth throwing into the soup

I think there are two different routes - one for science purposes and the other for artistic and visual interpretation purposes.
Binning to gain an improvement in the SNR of the RGB channels especially for faint nebulae in the RGB colour channels and then non binning in the L channel for fine detail will give you "the cake and eat it too" !

The reason we appear to get something for nothing is the eye is very sensitive to luminous and contrast changes (more so than colour detail) - so the L channel will "trick" the eyes into seeing the enhanced colour detail in the RGB channels.

Something I have often wondered about is the actual size of the telescopes image circle, not just the useful spot size.
So whilst the focal length of the scope and the f ratio are known (and for higher end scopes so is the image circle) the image circle and its effect on light gathered at the CCD is often ignored.
I think its assumed that all light from the scopes aperture at the focus point is being harnessed - but is that the case for all scopes ?
ie is there light that is simply being focussed that is not on the CCD at all.

If that is the case ? (and I am asking the question rather than stating a fact) - is this not a factor that will affect efficiency ?
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  #64  
Old 29-06-2015, 10:37 AM
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Originally Posted by rally View Post
Couple of things worth throwing into the soup

I think there are two different routes - one for science purposes and the other for artistic and visual interpretation purposes.
Binning to gain an improvement in the SNR of the RGB channels especially for faint nebulae in the RGB colour channels and then non binning in the L channel for fine detail will give you "the cake and eat it too" !

The reason we appear to get something for nothing is the eye is very sensitive to luminous and contrast changes (more so than colour detail) - so the L channel will "trick" the eyes into seeing the enhanced colour detail in the RGB channels.

Something I have often wondered about is the actual size of the telescopes image circle, not just the useful spot size.
So whilst the focal length of the scope and the f ratio are known (and for higher end scopes so is the image circle) the image circle and its effect on light gathered at the CCD is often ignored.
I think its assumed that all light from the scopes aperture at the focus point is being harnessed - but is that the case for all scopes ?
ie is there light that is simply being focussed that is not on the CCD at all.

If that is the case ? (and I am asking the question rather than stating a fact) - is this not a factor that will affect efficiency ?
Hi rally. good points.

the original post deals purely with broadband pixel-level sensitivity with the assumption that the designer will take care of how many pixels there are and what sampling is used based on other criteria - I guess that could have been made clearer.

However, you could define system efficiency in other ways and the fraction of total "controlled" photons actually put to good use might be one way of doing so - in which case total chip area would come into it. Have you seen any such definitions of sensitivity? - might be useful to compare. I guess that, with a small object like a galaxy, most of the image is a relatively uninteresting star field, but with a large nebula or glob, the field of view is possibly more important than pixel resolution. And for any sort of survey work, field of view is an essential part of the efficiency equation.

Last edited by Shiraz; 29-06-2015 at 10:52 AM.
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  #65  
Old 29-06-2015, 11:03 AM
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Good point Rally.

Pixels with microlenses are sensitive to the angle of incidence of the light rays. So most camera sensors require light to strike the pixel within a certain angle.

So faster scopes with light rays at a steeper angle may tend to vignette more.

Centre sharpness, corner performance, vignetting,varying spot sizes over the field of view would come into it. Some scopes have greater vignetting than others. Some have a bit of a bright spot in the centre and vignette out at the corners more.

Some sensors will cope with steeper rays than others. Some filters are less effective with steeper rays.

I recall with Planewave that the designer has a few parameters they can play with but they are conflicting. F ratio, vignetting, corner correction may be conflicting targets.

Gapless microlenses, no cover slip glass affect QE.

A lot of scopes barely cover a 16803 chip. That's where Tak FSQ, BRC250 and AP scopes are so good. AP in particular have huge illuminated corrected circles. Something like 100mm. Massive and would handle a Medium Format camera.

You are not going to be able to use a full frame or above sensor on anything with less than a 3.5 inch focuser so that is a more practical aspect of the comparison sheet results. Which of those combinations can actually handle that sort of sensor?

Greg.
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Old 29-06-2015, 01:16 PM
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Ray,

I only mention this because your formula includes F No. as one of the primary terms.

But since incoming light based purely on aperture and focal length may not have as much relevance as the percentage incoming illumination of the CCD chip, I think its worth considering.

Say for example we have two OTAs with the same focal length and aperture and therefore F No, but one has an image circle of say 100mm and the other has an image circle of 50mm
Both in theory support a large chipped camera - but one is spreading that light at the focus plane over an area that is 4 times larger than the other the pi.r^2 does this - double the radius - quadruple the area
eg 50mm image circle yields roughly 2000mm2
100mm yields roughly 8000mm

So our efficiency calculation really needs to consider just how the incoming light is being distributed at the focal plane.
In my example that factor is 400% difference - so its not insignificant and if real world examples are used some scopes have image circles well under 50mm - often under 30mm so if for example it was say 25mm then the variation is a factor of 16 times (1600%) difference

So are we are worrying ourselves about small percentage variations yet this is potentially an order of magnitude difference !

I am not sure there is a uniform standard for defining the image circle of a telescope but its usually some notional diameter at which the amount of vignetting falls off below a certain level and or the spot size and chromatic aberration grows to a dimension that makes it unuseable.
But I think manufacturers have left it to their own discretion as to how this is interpreted.
Which makes it very difficult to quantify using a generic formula.

All of this ignores the quality of light that different telescopes are capable of focussing in the first place - strehl ratio, longitudinal plot characteristics in terms of the useable spectral width of a given scope.
Its one thing to illuminate a CCD with incoming photons its another entirely to ensure that only photos that are "supposed" to be received by a given well (pixel) are in fact received at that point and not spread out all over the place.

So the "noise" of the system is entirely relevant to the efficiency since at the end of the day if the signal we are trying to record is lost in the noise of poor optical performance or poor CCD performance (eg low dynamic range, high noise) etc we are losing efficiency in swathes.

Food for thought - its a complex problem methinks.

Rally
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  #67  
Old 29-06-2015, 01:47 PM
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Quote:
Originally Posted by rally View Post
Ray,

I only mention this because your formula includes F No. as one of the primary terms.

But since incoming light based purely on aperture and focal length may not have as much relevance as the percentage incoming illumination of the CCD chip, I think its worth considering.

Say for example we have two OTAs with the same focal length and aperture and therefore F No, but one has an image circle of say 100mm and the other has an image circle of 50mm
Both in theory support a large chipped camera - but one is spreading that light at the focus plane over an area that is 4 times larger than the other the pi.r^2 does this - double the radius - quadruple the area
eg 50mm image circle yields roughly 2000mm2
100mm yields roughly 8000mm

So our efficiency calculation really needs to consider just how the incoming light is being distributed at the focal plane.
In my example that factor is 400% difference - so its not insignificant and if real world examples are used some scopes have image circles well under 50mm - often under 30mm so if for example it was say 25mm then the variation is a factor of 16 times (1600%) difference

So are we are worrying ourselves about small percentage variations yet this is potentially an order of magnitude difference !

I am not sure there is a uniform standard for defining the image circle of a telescope but its usually some notional diameter at which the amount of vignetting falls off below a certain level and or the spot size and chromatic aberration grows to a dimension that makes it unuseable.
But I think manufacturers have left it to their own discretion as to how this is interpreted.
Which makes it very difficult to quantify using a generic formula.

All of this ignores the quality of light that different telescopes are capable of focussing in the first place - strehl ratio, longitudinal plot characteristics in terms of the useable spectral width of a given scope.
Its one thing to illuminate a CCD with incoming photons its another entirely to ensure that only photos that are "supposed" to be received by a given well (pixel) are in fact received at that point and not spread out all over the place.

So the "noise" of the system is entirely relevant to the efficiency since at the end of the day if the signal we are trying to record is lost in the noise of poor optical performance or poor CCD performance (eg low dynamic range, high noise) etc we are losing efficiency in swathes.

Food for thought - its a complex problem methinks.

Rally
Hi Rally

I think that what you are saying is encapsulated in the formula. by way of explanation, imagine a 2mfl f8 system and a 1mfl f4 system. The f8 system has 9micron pixels and the f4 system has 4.5 micron pixels. Both have the same aperture and angular sampling.
according to the formula, both systems have the same pixel level sensitivity (which is right). However, for the same field of view, the f8 system will need a corrected focal plane of 4x the area of that of the f4 system, because the f8 system pixels are larger - which I think is the point you are getting at - but the larger focal plane does not affect the operation of the formula or change the system sensitivity.

Focal length has as much effect on sensitivity as aperture, since it determines the fraction of the incoming photons that actually gets into a pixel of a given size (the aperture determines how many there are in total). That is why ("myth" papers notwithstanding) FNo rather than aperture is vitally important for fixed pixel size. If you doubt this, try changing your scope FNo with a Barlow and see what happens (it will not change the aperture or the pixel size, but it will drastically reduce the sensitivity).

regards Ray

Last edited by Shiraz; 29-06-2015 at 02:42 PM.
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Old 29-06-2015, 02:25 PM
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Hi Ray,

Well I think I get it !

So a larger Image circle is simply the scopes greater ability to see more of the sky at the same focal length ?

The image scale will of course remain the same !

So a given pixel receives much the same light as another one on a different system (with a different image circle), just that it can illuminate a larger imaging chip/plate/film.
So its capturing more light overall and making it useable, but not at the expense of a given pixel
So one scope must still be more efficient in terms of light gathering power even though

So it begs the question (at least for me) - Is F Ratio actually the FL divided by scope Aperture ? - Or is it focal length divided by some sort of "Effective" scope Aperture"
Or have I confused your F No. with F ratio ? - Maybe thats what I have done !
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  #69  
Old 29-06-2015, 02:51 PM
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Originally Posted by rally View Post
Hi Ray,

Well I think I get it !

So a larger Image circle is simply the scopes greater ability to see more of the sky at the same focal length ?

The image scale will of course remain the same !

So a given pixel receives much the same light as another one on a different system (with a different image circle), just that it can illuminate a larger imaging chip/plate/film.
So its capturing more light overall and making it useable, but not at the expense of a given pixel
So one scope must still be more efficient in terms of light gathering power even though

So it begs the question (at least for me) - Is F Ratio actually the FL divided by scope Aperture ? - Or is it focal length divided by some sort of "Effective" scope Aperture"
Or have I confused your F No. with F ratio ? - Maybe thats what I have done !
I think that we agree Rally - what's the fun in that...

A system with a wider angular field of view will make more use of the available light by accepting it from a wider field of view, but if you look at the signal to noise level within any part of the image, the total field of view has no effect at all. ie, there are two measures of sensitivity, one that is based on how much signal to noise you get in the background sky in an image and the other that measures the total number of photons that you make use of - they are different measures.

The equation is based on an assumption that you will want to maximise the signal to noise ratio in any part of an image that your system will support. It is based on the standard definition of FNo (fl/diam).

regards Ray

Last edited by Shiraz; 29-06-2015 at 03:22 PM.
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Old 29-06-2015, 03:52 PM
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Okay, few things to cover now that the daily grind is out of the way!

Quote:
Originally Posted by rmuhlack
What I am trying to get at is what is the effect (with respect to both Ray's original formula and to the resulting resolution of the final image) of performing:
1) a 2x drizzle integration on dithered sub-frames that were captured with 2x2 binning, compared with
2) integration (without drizzle) of dithered sub-frames that were captured with 1x1 binning
The reason I said that you cannot have your cake and eat it too comes down to physics. Going back to "Astronomy 101", your aperture is what determines the number of photons entering the optical system. Let's say for instance that you have a FLI 16803 (because it is sheer awesome) and a telescope that allows you to image the entire M42 complex in one shot.

Let's say that when imaging M42 there are 100 million photons/second entering the telescope. This means that on average we are getting 6 photons/second/pixel. If we do a 2x2 bin this increases to about 24 photons/second/pixel, this is why imaging times can be shorter. What you do need to remember though is that in both images, the same amount of photons were present (if both images have the same overall imaging time).

The next step is if we have a 2x2 bin and that is drizzled. Without going into too much detail, drizzling works by basically creating a pseudo unbinning (just changing the image scale) and then when combining the images runs a mathematical calculation on a pixel to pixel basis of the statistical weighted average of what each pixels value should be. Sorry if that sounds like a messy explanation!
The point though is that there ends up potentially being a considerable amount of lost data through this process. Because every image has a lot of information that isn't of use due to being over sampled, it is just lost forever.

Of course, without doing some testing when I get the time and weather permits, I would say that imaging at 2x2 bin and then drizzling would be a waste of time. I doubt you would get fully back to unbinned resolution and from what I can tell, you would lose SNR on the inbinned images anyway.

Quote:
Originally Posted by rally
Well I think I get it !

So a larger Image circle is simply the scopes greater ability to see more of the sky at the same focal length ?

The image scale will of course remain the same !

So a given pixel receives much the same light as another one on a different system (with a different image circle), just that it can illuminate a larger imaging chip/plate/film.
So its capturing more light overall and making it useable, but not at the expense of a given pixel
So one scope must still be more efficient in terms of light gathering power even though

So it begs the question (at least for me) - Is F Ratio actually the FL divided by scope Aperture ? - Or is it focal length divided by some sort of "Effective" scope Aperture"
Or have I confused your F No. with F ratio ? - Maybe thats what I have done !
I understand where you are coming from with that but luckily telescopes don't work that way. What a telescope does is collect light from a large portion of sky and then focus it towards a point. As was previously mentioned, the "better" telescopes have a larger corrected field BUT there isn't a finite quality of light that is able to enter the optical train.

To give an example. Let's say you put a KAF-8300 on two 8" F/10 telescopes, one with a corrected field of 40mm and one with 100mm. Neither will suffer too much of vignetting although realistically the 100mm one should fair better simply because it has a larger corrected field. As both telescopes have the same focal length, image scale is exactly the same but with an ironic twist, the 40mm corrected telescope should be ever so slightly faster

The reason for this is that to have larger corrected field you must by definition have a larger secondary mirror. Granted, it won't make much of a difference. Now to get back to what you asked, the reason it doesn't make a big difference between a 40mm & 100mm corrected field is that the photons aren't being scattered across the entire corrected field. The 100mm one can have a larger chip and get a larger FOV but it doesn't have any effect on what is in the centre of view, because, by definition, all of the photons are concentrated into the centre of view

Hope that helps.
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  #71  
Old 29-06-2015, 04:40 PM
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Hi Ray,

Late to the party (I've been away for a few weeks) but your formula is effectively the same as one I've been using in a spreadsheet for comparing scope/camera combinations

I don't see that Drizzle is relevant to the calculation as it is something that happens after data capture and it doesn't violate the TANSTAAFL principle. For those interested in the theory: http://www.stsci.edu/hst/HST_overvie...34.html#385457

Cheers,
Rick.
thanks Rick - exactly what was needed.
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