Arc seconds per pixel and critical sampling - confused

[RobertI]

I’m trying to get my head around a particular aspect of imaging and am going to ask a potentially numpty question  , sorry if it’s been asked many times before.

The formula for arc seconds per pixel from a given scope/camera appears to only involve the pixel size and focal length of the scope. This is often cited as the formula to use to work out the ideal focal length for a given camera to get an arcseconds per pixel to resolve the finest detail possible. This formula also implies that a 70mm refractor can produce the same level of detail as an SCT scope of the same focal length, using the same camera, given long enough to acquire the image (as the 70mm would be much slower). Removing all other factors such as atmosphere, is this this true?

There is another formula for critical sampling which implies that aperture is involved in addition to focal length. Is this because larger apertures produce a smaller airy disks than smaller apertures?

Assuming both formulae are correct, under what circumstances would one use either of the above formulae? Sorry if the question is confusing, I've confused myself!

Thanks 

[coatesg]

3 minutes ago, RobertI said:

I’m trying to get my head around a particular aspect of imaging and am going to ask a potentially numpty question  , sorry if it’s been asked many times before.

The formula for arc seconds per pixel from a given scope/camera appears to only involve the pixel size and focal length of the scope. This is often cited as the formula to use to work out the ideal focal length for a given camera to get an arcseconds per pixel to resolve the finest detail possible. This formula also implies that a 70mm refractor can produce the same level of detail as an SCT scope of the same focal length, using the same camera, given long enough to acquire the image (as the 70mm would be much slower). Removing all other factors such as atmosphere, is this this true?

There is another formula for critical sampling which implies that aperture is involved in addition to focal length. Is this because larger apertures produce a smaller airy disks than smaller apertures?

Assuming both formulae are correct, under what circumstances would one use either of the above formulae? Sorry if the question is confusing, I've confused myself!

Thanks 

The maximum resolution (ie smallest resolvable feature) is proportional to 1/aperture - as you say, the airy disc of a large aperture is smaller than a smaller aperture (ignoring any obstructions for the moment!). 

The critical sampling formula gives the minimum focal length for a given scope/camera combination so that you are not under sampling and losing potential resolution (based on the Nyquist criterion, and assuming you need 3 pixels coverage for the smallest resolvable feature). For the same camera, this will be shorter for a 70mm 'frac than a larger SCT. Going above this will not generate any appreciable gain in resolution - you'll just make the airy disc cover more pixels and the image dimmer.

The pixel resolution formula is looking at the same problem from the other direction - it allows you to work out what focal length/pixel size combination you need to reach a given resolution - this can be as small as you want, but it won't show extra resolution unless the diffraction from the scope (ie the airy disc) is small enough -> ie scope has enough aperture. 

[RobertI]

2 hours ago, coatesg said:

The pixel resolution formula is looking at the same problem from the other direction - it allows you to work out what focal length/pixel size combination you need to reach a given resolution - this can be as small as you want, but it won't show extra resolution unless the diffraction from the scope (ie the airy disc) is small enough -> ie scope has enough aperture.

Many thanks Graeme, at least it wasn't a totally stupid question! So am I right in thinking the pixel resolution formula is a rough guide and the critical sampling formula is an accurate guide? Still a bit confued because on the Astropix website, Jerry Lodriguss explains how to obtain optimum sampling , but there is no mention of aperture. I'm sure I am missing the point somewhere!

[vlaiv]

First formula is giving you resolution that you are sampling at given focal length and pixel size.

This means that sampling resolution is only dependent on these two values. It has nothing to do with maximum level of information you can record with telescope.

Think of it this way: you have two stars that are separated by 20 arc seconds, and you use a telescope and a camera to record those stars. How many pixels will those stars be apart in the image?

One telescope / camera combination can make those two stars be 10 pixels apart in image - this means that telescope and camera give resolution of 2"/pixel.

Other telescope / camera combination can make those two stars be 40 pixels apart in image - this means that telescope and camera give resolution in second case of 0.5"/pixel.

It can also tell you for example how large object will be in your image (sort of "zoom" factor). If you image Jupiter at 1"/pixel - it will be 45 pixels across in your image.

If you image it at 0.25"/pixel it will be 180px across in your image.

So first formula gives you way to calculate "zoom" factor, and it depends on FL and pixel size - aperture does not come into equation for this one. Similar to regular magnification - you divide scope FL with eyepiece FL - zoom is not affected by aperture of telescope, but rather by FL only.

 

Now on to the critical sampling. As you know there is limit to how much telescope, even under ideal conditions (perfect seeing, ...), can magnify image. Second formula is telling you what is the largest focal length for given aperture and camera pixel size, or in another words - what is the biggest zoom factor that is worth using, and if you "zoom in" more than that (and you can by using even longer focal length) - it will not show you any more detail.

So first formula is telling you what "zoom" factor will you have for given scope / camera combination - or to put it in proper terms - it will give you sampling resolution in arc seconds per pixel.

Second formula is telling you for a given scope / camera - what is the largest focal length that you can use that will yield meaningful detail (everything "zoomed in" after that focal length is pointless). For this formula - aperture plays part because it defines maximum detail obtainable for telescope. In case you are wondering how to change focal length of a telescope - by using suitable barlow lens - that second formula is guide for choosing barlow lens for given camera and scope for optimum planetary imaging - hence the name critical - meaning larger than that is pointless. This is similar to max useful magnification rule in visual - so you can use max x50 per inch of aperture, by selecting appropriate eyepiece. Same thing here but for given camera you vary magnification by choosing different focal length.

HTH

[coatesg]

Definitely not a daft question! 

You do need to consider the potential resolution from the available aperture, and he kinda hints at it, but the figures he uses are fairly reasonable for the examples he gives.

The Airy disc isn't the best measure in most cases - the Dawes limit is perhaps better a better calc to use (=4.56/Aperture in inches). For planetary, if there is doubt, it's perhaps better to slightly oversample than undersample - for me my seeing is rarely anywhere near good enough to take full advantage though!! For the 130mm scope in the link, the Dawes limit is about 0.9", so 0.33"/px is great.

The first formula in the original post is useful, but you need to use it in conjunction with your aims - it may be that you choose to image at large pixel scales (2-4"/px) for widefield work as it gives the required field of view and makes guiding a lot more forgiving!!

 

 

[RobertI]

Thankyou so much @coatesg and @vlaiv, I think I have got it now, your detailed and clear explanations much appreciated. I have dived down the rabbit hole and now out the other side!

[ollypenrice]

The question is an excellent one. It can be answered (incorrectly and/or correctly) from either theory or practice. Because I take pictures I'm going to go with practice. By going with practice you don't have to worry your head about why the theory isn't working - and my experience is that it doesn't. And I think it doesn't because there are far too many loose variables at play. For example, theory states clearly that the large SCT should give smaller stars than the small refractor. But it just doesn't - in my experience. It then gets very complicated because the SCT can, very obviously, out resolve the small refractor for planetary imaging - by a country mile.

In the end I'm just going to look at the pictures.

Olly

[kens]

Its worth noting that because the Dawes limit depends on aperture and your pixel scale depends on focal length than you can express the Dawes limit in terms of the spatial resolution on your sensor and it is totally dependent on F ratio (FL/aperture).

So, for example, the Dawes limit of a F/5 scope is 2.8um and an F/8 scope is 4.5um

So with an F/5 scope you can theoretically resolve objects that are 2.8um apart but if your pixels are, say, 5.6um, then that would not be achievable.

So the upshot is that increasing the aperture wont improve resolution on the sensor unless it also gives a lower F ratio. 

 

[RobertI]

1 hour ago, ollypenrice said:

The question is an excellent one. It can be answered (incorrectly and/or correctly) from either theory or practice. Because I take pictures I'm going to go with practice. By going with practice you don't have to worry your head about why the theory isn't working - and my experience is that it doesn't. And I think it doesn't because there are far too many loose variables at play. For example, theory states clearly that the large SCT should give smaller stars than the small refractor. But it just doesn't - in my experience. It then gets very complicated because the SCT can, very obviously, out resolve the small refractor for planetary imaging - by a country mile.

In the end I'm just going to look at the pictures.

Olly

Very interesting observations Olly, thanks for sharing you real world experience, which I thinks demonstrates that the formulae can only take you so far. I too shall continue to enjoy the images, as my brain is full.

[RobertI]

1 hour ago, kens said:

Its worth noting that because the Dawes limit depends on aperture and your pixel scale depends on focal length than you can express the Dawes limit in terms of the spatial resolution on your sensor and it is totally dependent on F ratio (FL/aperture).

So, for example, the Dawes limit of a F/5 scope is 2.8um and an F/8 scope is 4.5um

So with an F/5 scope you can theoretically resolve objects that are 2.8um apart but if your pixels are, say, 5.6um, then that would not be achievable.

So the upshot is that increasing the aperture wont improve resolution on the sensor unless it also gives a lower F ratio. 

 

Another interesting perspective, thanks for explaining Ken.

[Merlin66]

Ken,

I have an issue with the f ratio solution....

The angular size of the Airy disk varies with aperture, the Linear size of the Airy disk varies with focal length.

A 200 mm f5 will have a longer focal length than a 100 mm f5  and hence the linear Airy disk (or the seeing disk in reality) will be larger.

For instance using the above and a 1 arc sec (seeing) the 200mm produces a 4.8 micron (FWHM) image whereas the 100mm image is 2.4 micron.

To optimise the imaging sampling, completely different pixel sizes are required. 

[kens]

For a fixed F ratio and pixel size the Airy disk angular size will be smaller for the larger aperture.

• theta = 0.252 lambda / d

for theta in arcsec,  lambda (wavelength)  in nm and d (aperture diameter) in mm

But  the focal length will also be longer so a given size of disk will cover more pixels as you point out

• Linear scale  = FL/206.3

for linear scale in um/arcsec and FL in mm

So the linear size (linear scale x theta) is a function of FL / d which is the F-ratio

For the Airy disk the linear size in um is given by 0.635 * F-ratio for a wavelength of 520nm

So as far as the optical resolution goes the F-ratio is what matters. The seeing is of course another consideration

[Merlin66]

OK.

49 minutes ago, kens said:

The seeing is of course another consideration

Yes,  when taken into account the FL/D formula needs to be modified.