resolution in arc seconds per pixel

[Pete Presland]

This a follow on from a discussioin i had in the solar imaging section http://stargazerslounge.com/topic/227689-televue-x5-powermate/#entry2456456  as to the use of a x5 powermate for imaging.

i have done quite a bit of reading since and the " 0.25" arcseconds per pixel " is the phrase that seems to come up regularly.

for my particular set ups i use the following

Solar imaging - Coronado PST/DMK21au618, pixel size 5.6um

with a x2.5 Powermate (5.6 x 206)/1000 = 0.86

with a x4 imagemate (5.6 x 206)/1600 = 0.72

Planetary imaging - Celestron C9.25/ASI120mm, pixel size 3.75um

with a x2.5 Powermate (3.75 x 206)/5875 = 0.13

the questions i have are,

a, is my maths correct? apologies if not.

b, For the PST with the x2.5 Powermate my images are good, with the x4 image mate not as good and no real extra detail showing. is the small aperture the reason for not being able to get near the 0.25" arcseconds per pixel?

c, For the C9.25 0.13 seems lower than the optimum size, but i am reasonably happy with the overall detail on the images. is this the opposite of the above, with the larger apperture allowing me to go lower than the 0.25" arcseconds per pixel?

i have a fair idea of the limits of my equipment, but would like to try and understand the reasons why better. hopefully to allow me to make more rational decissions in the future.

thanks for reading.

[JB80]

This is something that interests me also, up until now I have always been in the camp of just trying a bunch of things and see what looks the best.

Now though I'm chasing that final touch, I guess it would be getting the mix right and then truly having a HD image?

I'm going to follow this with interest and maybe do some reading too because I would like to understand it better as well.

[JamesF]

I was never impressed with the performance of the 4x ImageMate to be honest.  I could never get a image that I felt was acceptable and I'm fairly sure I'm not alone in that respect.

James

[JamesF]

Solar imaging - Coronado PST/DMK21au618, pixel size 5.6um

with a x2.5 Powermate (5.6 x 206)/1000 = 0.86

with a x4 imagemate (5.6 x 206)/1600 = 0.72

Planetary imaging - Celestron C9.25/ASI120mm, pixel size 3.75um

with a x2.5 Powermate (3.75 x 206)/5875 = 0.13

the questions i have are,

a, is my maths correct? apologies if not.

I always start with "arcseconds per mm = 206265 / focal length (in mm)".  The 206265 constant is the number of arcseconds in a radian, as the basic trigonometry on which this is based is most easily expressed using radians.

So, for the native PST you have 206265 / 400 = 516 arcseconds per mm (to the nearest arcsecond).  Add a 2.5x barlow/powermate and that reduces by the same factor, so 206 arcseconds per mm (and similarly 129 arcseconds per mm for a 4x multiplier).

Given these figures, the number of arcseconds per pixel is then found by multiplying by the pixel size (in mm).  So with a 5.6um pixel you get 516 * 0.0056 = 2.89 arcseconds per pixel for the native PST, 206 * 0.0056 = 1.15 when a 2.5x barlow is added and 0.72 arcseconds per pixel with a 4x barlow.

The C9.25 is a little more tricky because its focal length changes depending on where the primary mirror is.  I suspect, though I have no definitive information, that th focal length is only 2032mm when the image plane (and therefore the camera sensor) is about 100mm behind the back of the OTA.

However using exactly the same maths and assuming you really do have that focal length, the number of arcseconds per mm without a barlow should be 206265 / 2032 = 102.  With the 2.5x barlow that comes down to 41 arcseconds per mm.  Multiplying by the pixel size again gives 0.38 arcseconds per pixel without a barlow and 0.15 arcseconds per pixel with the 2.5x.

With one minor error therefore I'd say your maths looks sound.

One of the problems I found when I was starting planetary imaging was that many of these figures were talked about, yet I couldn't find any justification for them, even in some books that give them.  I wasn't prepared to say they were wrong, but I couldn't see why they might be right.  I think in some cases that's because many imagers aren't necessarily that comfortable with the maths involved so its easier to remember the important numbers and not worry too much about the rest.  In other cases I think they're just "rules of thumb" that people have found to generally work out.  There are also, to be fair, some things I've come across that aren't really based on any solid foundation and can sometimes serve to mislead more than inform.  The old chestnut about a webcam being the same as a 6mm eyepiece being one.

For that reason I actually sat down one evening and worked through the maths for matching the best possible resolution of a telescope with the resolution of a camera.  My reasoning was that there's no point extending the focal length beyond the point where they're matched because you won't actually capture more detail.  You'll just get the same detail, but covering more pixels.  And that you can do in processing.  Because I understand that I'm now fairly comfortable with it and those are the figures I give to other people.  It's not a perfect answer because the maximum resolution of a telescope depends on the wavelength of light you're interested in and there's quite a significant difference between the wavelengths of red and blue light, but it's a good place to start.  The figures make no allowance for the quality of the atmosphere however, and on most nights that will be the limiting factor.  I try to work to "what will give the best possible results given the opportunity" rather than "what's probably good enough".

I don't honestly know where the 0.25 arcseconds per pixel figure comes from and I don't recall seeing an explanation of it anywhere.  My suspicion is that it's somehow related to some assumed atmospheric limit on seeing, but I really don't know.

James

[JamesF]

Something else that occurred to me whilst I was shutting the chickens in for the night...

Because the effective focal length of a Mak or SCT varies as described above (actually it depends on the distance between the primary and secondary) the only way I've found to be sure about the focal length in use was to take an image of something I knew the size of and work backwards.  For example measure the number of pixels across an image of Jupiter use the pixel size to calculate the physical size of the image on the sensor and look up the angular size of Jupiter at the time the image was taken to get an arcseconds per mm figure from which I could get the focal length.  FireCapture can do a rough calculation of this sort for you and save it in the .txt file associated with the capture.

When I had a flip mirror, filter wheel, barlow and ASI120MM hanging off the back of my C9.25 I decided on the basis of the calculated focal length that actually I had a far longer focal length than I really needed, mostly I think because the length of all the kit meant the effective focal length of the OTA was a fair chunk more than the quoted 2032mm.  Temporarily I resolved that by dropping the flip mirror out though I'd really like to have that back and have therefore been looking for a much lower magnification barlow of decent quality.

James

[Pete Presland]

thank you to Mike for this link i had to print the page off and read it a few times    http://www.tanstaafl.co.uk/2012/03/focal-ratio-capture-length-planetary-imaging/

well having given this some thought over the last week or so and apologising for any mathmatical errors i have come up with this.

ignoring the OTA used and only considering the wavelength imaged in.

For the DMK21au618/Coronado PST @ HA with a pixel size of 5.6um and following the 5.6 x 10 -6/656 x 10 -7 = 8.53 then x2 (to give 2 pixels per smallest detail) then gives F17.06 x 1.2 (to allow for pixel diagonal) gives me F20.47 so the F25 i have been imaging at is just a little over the optimum F ratio.

With the ASI120mm/C9.25 @ R, G, B following the above maths for each wavelength it works out as F13.84, F16.05, F22.48, which does seem to make matters more tricky. so at F10 that i have been doing most of my imaging at is just a little below the optimum F ratio, for the R channel but well under for the B channel. but with the x2.5 powermate a little over the optimum F ratio, for the B channel but well over for the R and G channels.

summing up it certainly seems correct that the x2.5 powermate is all i am going to need for HA & R, G, B imaging. with both imaging OTA's AT f10, possibly even this is a little overpowered.