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  #21  
Old 28-03-2013, 07:36 PM
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Bassnut (Fred)
Narrowfield rules!

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Originally Posted by rmuhlack View Post
I assume that as Mr "Narrowfield rules" that your statement of "anything under 2m FL is rubbish" is just hyperbole. Surely FOV comes into "image impact" as well. I mean, if we were to use an Orion Optics ODK12, that has a FL of 2040mm. A KAF16803 and a KAF1603 both have similar well depth and QE of 100,000 and 55-60% and the same pixel size of 9micron, yet the 1603 on that scope will give a FOV of 23.5'x15.6' where as the KAF16803 will give more than 10x the area at a FOV of 62'x62'

Sure, as long as you crop the hell out of the 16803 pic to maintain the "ascetics", that works. But why not spend the money on even higher QE instead of boring multi megapixel WF FOV?.
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  #22  
Old 28-03-2013, 10:30 PM
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Shiraz (Ray)
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Hi Richard

FWIW, my take on your original question:

The signal in a pixel from an extended object (eg nebula or galaxy) is:

S = B*A*pixangle*opticsefficiency*time* QE

Where B is the object surface brightness, A is the aperture area and pixangle is the solid angle subtended by a pixel.

If you keep pixangle the same for the two systems (by using different cameras) and assuming all else is equal, increasing the aperture area A will proportionally increase the pixel signal – nothing else changes much.

The resolution of the two systems will be largely determined by the atmospheric seeing. Ignoring tracking blur and charge diffusion, the total PSF for each system is the convolution of the atmospheric PSF and the optics PSF. Assuming Gaussian approximations apply to the PSFs and that seeing is specified in angular FWHM terms,

FWHMtotal = SQRT(FWHMseeing^2+FWHMoptics^2)

Now, FWHMoptics = 1.03*lambda/aperturediameter, (eg 0.58 arcsec for the 200mm)

so for the 200mm system in 2 arcsec seeing, FWHMtotal = 2.08 arcsec
and for the 300mm system, FWHMtotal = 2.04 arcsec

ie, there is not much difference in FWHM of the combined PSFs – scope resolution is not a significant factor and both systems are basically seeing limited in 2 arcsec seeing. Since the pixels are 1x1 arc sec in each case, the sampling is also similar at about 2 pixels/FWHM (the ideal seems to be somewhere around 2.5-3, so the images will be slightly undersampled).

entirely different answer if you kept the same camera. And of course this does not include any consideration of SNR and dynamic range.

regards Ray

Last edited by Shiraz; 29-03-2013 at 06:51 AM. Reason: make clear its surface brightness
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  #23  
Old 29-03-2013, 01:15 AM
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AstroJunk (Jonathan)
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Re Aperture: 2 arc second seeing will result in a many more arc second 'spludge' on the ccd due to the scintillation. The resolving power of the larger scope should still come into play in producing a smaller star image all things being equal.

As a general rule, I would be planning for a perfect night, not a rubbish one - we get plenty of those. When the conditions are just right you need the best equipment you can stretch to so that you can make it stick, and I'm sorry, that means selling your car...
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  #24  
Old 29-03-2013, 06:38 AM
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rmuhlack (Richard)
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Originally Posted by AstroJunk View Post
Re Aperture: 2 arc second seeing will result in a many more arc second 'spludge' on the ccd due to the scintillation. The resolving power of the larger scope should still come into play in producing a smaller star image all things being equal.

As a general rule, I would be planning for a perfect night, not a rubbish one - we get plenty of those. When the conditions are just right you need the best equipment you can stretch to so that you can make it stick, and I'm sorry, that means selling your car...
Haha that's what my wife is afraid of !
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  #25  
Old 29-03-2013, 06:41 AM
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rmuhlack (Richard)
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Quote:
Originally Posted by Shiraz View Post
Hi Richard

FWIW, my take on your original question:

The signal in a pixel from an extended object (eg nebula or galaxy) is:

S = B*A*pixangle*opticsefficiency*time* QE

Where B is the object brightness, A is the aperture area and pixangle is the solid angle subtended by a pixel.

If you keep pixangle the same for the two systems (by using different cameras) and assuming all else is equal, increasing the aperture area A will proportionally increase the pixel signal nothing else changes much.

The resolution of the two systems will be largely determined by the atmospheric seeing. Ignoring tracking blur and charge diffusion, the total PSF for each system is the convolution of the atmospheric PSF and the optics PSF. Assuming Gaussian approximations apply to the PSFs and that seeing is specified in angular FWHM terms,

FWHMtotal = SQRT(FWHMseeing^2+FWHMoptics^2)

Now, FWHMoptics = 1.03*lambda/aperturediameter, (eg 0.58 arcsec for the 200mm)

so for the 200mm system in 2 arcsec seeing, FWHMtotal = 2.08 arcsec
and for the 300mm system, FWHMtotal = 2.04 arcsec

ie, there is not much difference in FWHM of the combined PSFs scope resolution is not a significant factor and both systems are basically seeing limited in 2 arcsec seeing. Since the pixels are 1x1 arc sec in each case, the sampling is also similar at about 2 pixels/FWHM (the ideal seems to be somewhere around 2.5-3, so the images will be slightly undersampled).

entirely different answer if you kept the same camera. And of course this does not include any consideration of SNR and dynamic range.

regards Ray
Thanks Ray, that is extremely helpful. I will be plugging all those equations into my model.
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  #26  
Old 29-03-2013, 07:01 AM
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Quote:
Originally Posted by rmuhlack View Post
Thanks Ray, that is extremely helpful. I will be plugging all those equations into my model.
modelling is a great tool - all the best with it.

There is an error in my post - B in the first equation is "surface brightness" not "brightness" as originally described - have edited the post. Apologies.

regards ray

Last edited by Shiraz; 29-03-2013 at 09:33 AM.
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  #27  
Old 29-03-2013, 10:16 AM
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multiweb (Marc)
ze frogginator

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Originally Posted by rmuhlack View Post
So - in what respect will the images from these two setups be different?
I imaged a lot on the 500-600mm FL ball park with different size scopes. With aperture comes better resolution on small details and obviously less time involved in data acquisition. Some of the issues with bigger aperture are collimation, thermal equilibrium, more prone to seeing conditions, weight, etc...

But in the end aperture will always give you more resolution that you just won't get with smaller systems even at the same image scale.
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  #28  
Old 29-03-2013, 10:29 AM
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Marc,
Larger apertures always gather more light and can show fainter objects.
Large apertures have a higher resolution than smaller apertures.

However, in an imaging set-up the seeing, plate scale and pixel size will definately impact on resolution. (The OP example was 2" seeing which is MUCH larger than the Airy disks, it needs around f25-f30 to image at the Airy disk plate scale, hence this choice for solar, lunar and planetary)
We also have to distinguish between stellar images and extended surface objects. Longer focal lengths favour star images, faster f ratios favour extended objects (In the OP example the focal lengths varied, but not the f ratios)
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  #29  
Old 01-04-2013, 10:35 AM
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When I was considering the purchase of the RH200 all these variables influenced the final image train.

For a perfect optical system the Airy Disc size is only dependant on F number. From Clarkvision here



http://www.clarkvision.com/articles/...ml#Diffraction

Table 6
=================================== =============
red= Green= Blue=
0.6 0.53 0.47
micron micron micron
=================================== =============
f/ratio diffraction spot diameter in microns
=================================== =============
2 2.9 2.6 2.3
2.8 4.1 3.6 3.2
4 5.9 5.2 4.6
5.6 8.2 7.2 6.4
8 11.7 10.3 9.2
11 16.1 14.2 12.6
16 23.4 20.7 18.3
19 27.8 24.6 21.8
22 32.2 28.5 25.2
32 46.8 41.4 36.7
45 65.9 58.2 51.6
64 93.7 82.8 73.4
=================================== =============


It is a bit more complicated than this as the Airy Disc is not uniform in brightness across its diameter. Otherwise planetary imagers would have no hope of seeing detail when the Airy disc is 60 micron in diameter at f/40.

The diffraction pattern of a circular aperture is a Bessel function which has a pronounced peak. It is this peak that is just above the noise that planetary imagers are recording to give them contrast. The same thing is happening with extended objects as it is a one to one mapping of the point spread function of the optic to each sensor pixel.

Selecting a camera with small pixels to improve resolution is a bad choice as the stars will saturate very quickly due to small well depth.

The reason bright stars are bigger than dim ones is that the very weak higher orders of the Bessel Function are saturating the sensor.

It is basically a matter of balance. With about 2" seeing my system with 9 micron pixels at 600mm FL gives me 3.09" per pixel. With dithering this can improved by a factor of the inverse of root two back to about the seeing. So it has to be a very bad night for my system to be limited by seeing. The actual resolution is at least twice your pixel resolution. In practice it is far worse.

There is no such thing as a perfect optic. I would go for a high Q sensor with bigger pixels than the so called matching to optic. There is no point having resolution that cannot be recorded for dim objects above the noise and fail to meet resolution because of saturation of brighter stars due to well depth limitations.



Bert
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  #30  
Old 01-04-2013, 10:55 AM
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Merlin66 (Ken)
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Bert,
""The reason bright stars are bigger than dim ones is that the very weak higher orders of the Bessel Function are saturating the sensor."

Not strictly true....
All star images will have the same FWHM...
What is happening is that the brighter star image becomes saturated and the dimmer images only record the brighter core "peak".
The PSF of the telescope can effectively destroy the neat Airy disk pattern.....
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  #31  
Old 01-04-2013, 11:01 AM
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Quote:
Originally Posted by Merlin66 View Post
Bert,
""The reason bright stars are bigger than dim ones is that the very weak higher orders of the Bessel Function are saturating the sensor."

Not strictly true....
All star images will have the same FWHM...
What is happening is that the brighter star image becomes saturated and the dimmer images only record the brighter core "peak".
The PSF of the telescope can effectively destroy the neat Airy disk pattern.....
Quite correct. I was considering perfect optics. Not the usual result that real optics give, a distorted image that can be approximated by a Gaussian function at best.

Central obstructions throw energy into the higher orders of the Bessel Function. Wavefront abberations just add to the Point Spread Function.

bert
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