My rubbish Saturn Image from the 8th Oct

[Nigella Bryant]

Posting my first image of Saturn, rubbish I know, to low really from here. C11 plus 2.5x powermate and zwo asi 462mc camera.

Edited October 24, 2023 by Nigella Bryant

[dweller25]

Some detail, but I think poor seeing scuppered your efforts there, hopefully you will get some steadier skies soon….. and Jupiter will be much higher shortly 🙂👍

Edited October 24, 2023 by dweller25

[Nigella Bryant]

1 hour ago, dweller25 said:

Some detail, but I think poor seeing scuppered your efforts there, hopefully you will get some steadier skies soon….. and Jupiter will be much higher shortly 🙂👍

It was rubbish seeing and low still for me between hedge and tree's 😭

[Neil H]

Well it's 90% better than what i got 

[vlaiv]

4 hours ago, Nigella Bryant said:

C11 plus 2.5x powermate and zwo asi 462mc camera.

Do read this short wiki article : https://en.wikipedia.org/wiki/Spatial_cutoff_frequency , and try to spot the problem in above sentence

 

[Nigella Bryant]

2 minutes ago, vlaiv said:

Do read this short wiki article : https://en.wikipedia.org/wiki/Spatial_cutoff_frequency , and try to spot the problem in above sentence

 

It is a question of seeing and atmospheric disturbance. I know my scope can do better as Jupiter isn't that bad. Attached Jupiter using the same set up. 

[vlaiv]

2 minutes ago, Nigella Bryant said:

It is a question of seeing and atmospheric disturbance.

Of course, but that is very much related.

Spatial cutoff frequency is well defined for perfect optical system (so perfect telescope and absence of seeing aberrations) and sampling above it just has negative impact on final image.

It is simply waste as no detail above spatial cut off frequency can be recorded - but it lowers SNR per pixel and thus forces longer exposure. Longer exposure often goes above atmospheric coherence time and along seeing aberrations we also get motion blur (different seeing wavefronts end up superimposed on single exposure).

For 2.9um pixel and if imaging in visible light - meaning 400-700nm, following applies:

F/ratio = 1 / lambda * sampling_frequency

Sampling frequency is 1 / 2 * pixel size (two samples per wavelength corresponding to cutoff frequency), so we end up with:

F/ratio = 2 * pixel_size / lambda => 2 * 2.9um / 0.4um => F/14.5 (we use 400nm as lower bound for most detail).

That is highest you really need to go in order to capture all the available detail that perfect telescope can provide in ideal conditions using this camera in visible spectrum.

You are sampling at ~F/25 (C11 is F/10 and x2.5 telecentric amplifier gives F/25), so your exposures need to be (25/14.5)^2 = ~3 times longer to achieve the same SNR / signal level per sub.  This can easily push you over coherence time for given seeing and you enter region where most of the subs are not only distorted by atmosphere but also blurred by moving atmosphere (motion blur).

[Nigella Bryant]

3 minutes ago, vlaiv said:

Of course, but that is very much related.

Spatial cutoff frequency is well defined for perfect optical system (so perfect telescope and absence of seeing aberrations) and sampling above it just has negative impact on final image.

It is simply waste as no detail above spatial cut off frequency can be recorded - but it lowers SNR per pixel and thus forces longer exposure. Longer exposure often goes above atmospheric coherence time and along seeing aberrations we also get motion blur (different seeing wavefronts end up superimposed on single exposure).

For 2.9um pixel and if imaging in visible light - meaning 400-700nm, following applies:

F/ratio = 1 / lambda * sampling_frequency

Sampling frequency is 1 / 2 * pixel size (two samples per wavelength corresponding to cutoff frequency), so we end up with:

F/ratio = 2 * pixel_size / lambda => 2 * 2.9um / 0.4um => F/14.5 (we use 400nm as lower bound for most detail).

That is highest you really need to go in order to capture all the available detail that perfect telescope can provide in ideal conditions using this camera in visible spectrum.

You are sampling at ~F/25 (C11 is F/10 and x2.5 telecentric amplifier gives F/25), so your exposures need to be (25/14.5)^2 = ~3 times longer to achieve the same SNR / signal level per sub.  This can easily push you over coherence time for given seeing and you enter region where most of the subs are not only distorted by atmosphere but also blurred by moving atmosphere (motion blur).

So are you saying image at f20 instead? 

[vlaiv]

Just now, Nigella Bryant said:

So are you saying image at f20 instead? 

Where does it say that?

I think that best results can be obtained provided you do following:

- use barlow and "dial in" F/ratio of your optical system to match pixel size you are using. Formula can easily be derived from above spatial cut off frequency and goes like F/ratio = 2*pixel_size / minimum_wavelength - where pixel size and minimum wavelength are in same units (meters, nanometers,  micrometers ...) and minimum wavelength is smallest wavelength in range of wavelengths you are recording. If using OSC sensor - use stacking software that supports bayer drizzle (AS!3).

- use as short exposure length as possible. This will be governed by QE of sensor and its read noise. Use highest QE and lowest read noise sensor you can get and don't look at histogram to set your exposure length, just use as short as possible. Only use longer exposure length if seeing allows, but most times it will be around 5-6ms per exposure.

[knobby]

I think Vlaiv sometimes forgets we aren't all maths experts but I believe he is saying aim for around F15, derived from F/ratio = 2 * pixel_size / lambda => 2 * 2.9um / 0.4um => F/14.5

This works out the same as using the easier (maybe less accurate) recommendation of 5 x pixel size (or 2 x in poor seeing / 7 x in excellent seeing)  as F number or in your case 5 x 2.9 = F15 🙂

 

 

[Nigella Bryant]

4 hours ago, knobby said:

I think Vlaiv sometimes forgets we aren't all maths experts but I believe he is saying aim for around F15, derived from F/ratio = 2 * pixel_size / lambda => 2 * 2.9um / 0.4um => F/14.5

This works out the same as using the easier (maybe less accurate) recommendation of 5 x pixel size (or 2 x in poor seeing / 7 x in excellent seeing)  as F number or in your case 5 x 2.9 = F15 🙂

 

 

Thanks, no, I'm far from being a math's expert despite having two degrees, still didn't understand it, lol.

[Soupir94]

23 hours ago, Nigella Bryant said:

Posting my first image of Saturn, rubbish I know, to low really from here. C11 plus 2.5x powermate and zwo asi 462mc camera.

I would be delighted with that as a first image. If I ever get a chance!

[vlaiv]

5 hours ago, knobby said:

I think Vlaiv sometimes forgets we aren't all maths experts but I believe he is saying aim for around F15, derived from F/ratio = 2 * pixel_size / lambda => 2 * 2.9um / 0.4um => F/14.5

This works out the same as using the easier (maybe less accurate) recommendation of 5 x pixel size (or 2 x in poor seeing / 7 x in excellent seeing)  as F number or in your case 5 x 2.9 = F15 🙂

 

 

Yes, that recommendation 5 x pixel size is directly derived from formula I wrote and article on wiki I linked to.

55 minutes ago, Nigella Bryant said:

Thanks, no, I'm far from being a math's expert despite having two degrees, still didn't understand it, lol.

Given your solar setup and work you do with it, I did suspect that you have academic background. That is why I just linked the article in the first post.

Math for calculation of spatial cutoff frequency is straight forward. Math behind wave nature of light producing that cut off frequency is not as straight forward, but it's not very difficult for someone with masters degree or higher in sciences. It boils down to interference effects which turn out to have the same form as Fourier transform of aperture (there is whole field of optics named Fourier optics because of this). This is similar to how Fourier transform represents filters (convolution in spatial / temporal domain is the same as multiplication in frequency domain and vice verse - convolution theorem). In any case - there is clear cutoff point due to circular aperture in frequency domain and there is limit to how much optics can resolve because of this (this is why we need to use radio telescopes in different spots around the globe to be able to get good resolution since radio waves have much much longer wavelengths).

Sampling finer than this limit produces no additional detail - same data will be recorded, but using higher F/ratio simply spreads the light (aperture gathers only so much light per unit time since photon flux does not change) over more pixels and signal per pixel gets lower - so does SNR (for same exposure time).

[Nigella Bryant]

38 minutes ago, vlaiv said:

Yes, that recommendation 5 x pixel size is directly derived from formula I wrote and article on wiki I linked to.

Given your solar setup and work you do with it, I did suspect that you have academic background. That is why I just linked the article in the first post.

Math for calculation of spatial cutoff frequency is straight forward. Math behind wave nature of light producing that cut off frequency is not as straight forward, but it's not very difficult for someone with masters degree or higher in sciences. It boils down to interference effects which turn out to have the same form as Fourier transform of aperture (there is whole field of optics named Fourier optics because of this). This is similar to how Fourier transform represents filters (convolution in spatial / temporal domain is the same as multiplication in frequency domain and vice verse - convolution theorem). In any case - there is clear cutoff point due to circular aperture in frequency domain and there is limit to how much optics can resolve because of this (this is why we need to use radio telescopes in different spots around the globe to be able to get good resolution since radio waves have much much longer wavelengths).

Sampling finer than this limit produces no additional detail - same data will be recorded, but using higher F/ratio simply spreads the light (aperture gathers only so much light per unit time since photon flux does not change) over more pixels and signal per pixel gets lower - so does SNR (for same exposure time).

Thanks vlaiv, appreciate your input. 

[MartianHill]

22 hours ago, Nigella Bryant said:

It is a question of seeing and atmospheric disturbance. I know my scope can do better as Jupiter isn't that bad. Attached Jupiter using the same set up. 

That is superb Nigella 

[Nigella Bryant]

1 hour ago, MartianHill said:

That is superb Nigella 

Thanks MartianHill, much appreciated. 

[Pete Presland]

Not rubbish at all.

Saturn is a tough target even at the correct sampling rate, but it is certainly miles better than my first Saturn image .

Vlaiv is certainly right about the sampling rate, but i think most of us end slightly over-sampled when planetary imaging

I use a X1.8 Barlow with my Asi224mc which should give me F16, but with the focuser, ADC, is probably nearer F20. And with my Asi462mm i am definitely way over-sampled due the smaller pixel size.

Edited October 31, 2023 by Pete Presland