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Does aperture "rule" in AP?

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#76 HenkSB

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Posted 18 April 2024 - 12:18 PM

Getting back on topic, the mixture of responses about pixel scale and focal ratio is confusing enough to try to make it precise along with tracking errors, seeing, the Rayleigh criterion and the resolution of the human eye.  It goes in 2 steps:

 

1) Calculate the standard deviation of the angle RMS due to the Rayleigh criterion, seeing, tracking and pixel scale.  For convenience we simply do an RMS of these quantities since they are basically independent.

 

2) Convert the object with the resolution from (1) into a picture of a certain diameter (like a laptop screen) placed at a suitable viewing distance, viewed by the human eye.  Find out how this resolution on the picture, as viewed by the user, compares to the resolution of the human eye.

 

Step 1:  The Rayleigh resolution is rho = 1.22*lambda/D where lambda is the wavelength (say, green=5000 Angstrom and D is the diameter), the tracking error RMS is tau, the atmospheric (seeing) RMS is alpha, and the pixel scale is sigma=mu/f where mu is the pixel size and f is the focal length.  The resulting angular resolution is theta = sqrt(rho^2 + tau^2 + alpha^2 + sigma^2).

 

Step 2:  Scale the radius omega of the object to the viewing angle phi = r/d (small angle approximation) by a factor phi/omega and scale the resolution theta accordingly: the viewing resolution vRes = theta*phi/omega.  Compare this to the resolution eta of the human eye.  If vRes is larger than eta you will notice blurring.

 

As you can see, the effect of pixel scale only involves the focal length.  The aperture only enters through the Rayleigh criterion.   Binning is not considered explicitly but you can simply use a larger pixel size.  That's all as far as resolution goes.

 

The focal ratio R=f/D has nothing to do with the resolution directly; however, the number of photons received on the object is proportional to the 1/R^2, which is very important and determines how grainy the picture looks.

 

I think that's all there is to it.  Here are some results, one for a 12" F/4 Newt, one for a 100 mm F/7.5 frac.  We scale the number of photons to 1 for the Newt and see how much the frac gets.  "Pixels in point image" means how many pixels are occupied by a point source.

I missed a few things: I forgot to mention that r=picture radius, d=viewing distance.  I also forgot to print the object radius, which is 10 arc minutes.  I changed the RMS of pixel pitch to 0.5*mu/f instead of mu/f, which seems to be better, statistically.

 

I re-ran the numbers also printing the RMS fraction of the various components.  Note that the RMS fractions don't add up to 100% as they contribute in an RMS sense. I also added a C14.

 

For the frac and Newt the Rayleigh criterion contributes the most, for the Newt less than the frac.  For the C14 the tracking and seeing are the biggest contributors.  The pixel scale impact is the least with the C14, of course, by far.  As could be expected, for the Newt the image scale is the largest fraction, while for the frac it is the Rayleigh criterion.  The Newt picture viewing resolution is 2.3 times worse than that of the human eye, for the frac it is 4 times, for the C14 it is 1.8 times.

 

(Newt)

Aperture=300 mm, focal length=1200 mm, pixel pitch=3.76 micron,
seeing=0.5 arcsec, tracking RMS=0.5 arcsec, wavelength=5000 Angstrom
Object radius = 10 arcmin
RMS fraction of Rayleigh = 47%, tracking = 56%, seeing = 56%, image scale = 36%
Picture radius = 20 cm viewed at a distance of 50 cm
Viewing resolution = 121.471 arcsec
Human eye resolution = 60 arcsec
Pixels in point image = 2.54413
Photons received = 1

 

(frac)

Aperture=100 mm, focal length=750 mm, pixel pitch=3.76 micron,
seeing=0.5 arcsec, tracking RMS=0.5 arcsec, wavelength=5000 Angstrom
Object radius = 10 arcmin
RMS fraction of Rayleigh = 82%, tracking = 32%, seeing = 32%, image scale = 33%
Picture radius = 20 cm viewed at a distance of 50 cm
Viewing resolution = 210.818 arcsec
Human eye resolution = 60 arcsec
Pixels in point image = 2.79148
Photons received = 0.284444

 

(C14)

Aperture=356 mm, focal length=3910 mm, pixel pitch=3.76 micron,
seeing=0.5 arcsec, tracking RMS=0.5 arcsec, wavelength=5000 Angstrom
Object radius = 10 arcmin
RMS fraction of Rayleigh = 44%, tracking = 62%, seeing = 62%, image scale = 12%
Picture radius = 20 cm viewed at a distance of 50 cm
Viewing resolution = 109.556 arcsec
Human eye resolution = 60 arcsec
Pixels in point image = 7.97083
Photons received = 0.132638

 

The numbers make sense to me so it's not too difficult to predict how a telescope performs under various conditions.



#77 imtl

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Posted 18 April 2024 - 12:40 PM

Seeing 0.5 arcsec?? On what planet?
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#78 Oort Cloud

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Posted 18 April 2024 - 06:37 PM

Seeing 0.5 arcsec?? On what planet?


Nepal


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