Jared
Post Laureate
Reged: 10/11/05
Posts: 4554
Loc: Piedmont, California, U.S.
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Quote:
Jarad, when you were looking at Saturn, which scope had more contrast? Which showed the details clearer? Diffraction is not just a matter of seeing the smallest detail. It affects contrast as well. Look at Qkslvr's images. You can say the detail is about the same in both. But are they the same? Seeing is not the issue. And it has nothing to do with whether the camera is used on this planet or for something further away. You could say diffraction affects the efficiency of being able to reproduce detail. Even making the 50mm system work at 150 line/mm is not as effective as the 16" working at 32 line/mm. The 16" is always going to be ahead. Certainly, placing the sensor that is on the 16" on the 50mm will improve things greatly for the refractor, regardless of FOV - what is the "ideal" FOV anyway (it would depend what you want to photograph). You may say contrast is not important, but you can only compensate for that so far.
I agree the difference betweeen a 50mm and 16" scope is large, but I have never heard of exceptions in the amateur astrophotography hypothosis of "sampling." If it works for one, it should work for the other. (And I do have a 50mm scope.) Perhaps it is a rule of thumb, but that is not the feeling I am getting from the folks posting here - it sound as if it were science. I know it is not science.
There is another problem. Science says it is not the pixel pitch that is the determining factor to match a sensor to optics. It is the sensor/format size. So when using an APS-C size sensor, nothing changes the requirements of the optics from a 6MP chip to a 12MP chip. Regardless of seeing, small chips require the optics to work harder.
So, can you slap any camera on any scope and get an image? Yes. Will it be the "best" image you can get? That depends. Diffraction is real. It is not simply limiting resolution. It is affecting contrast across a range of spacial frequencies.
As it happens, in the opinions of multiple observers last night the contrast was better in my 110mm than in the 20". I believe that is primarily due to the fact that my 110mm is an APO, so the color correction is better than in the 20" achromat. The lack of chromatic aberration may not impact resolution significantly, but it certainly does improve contrast, so the image looked less washed out in my scope (despite being dimmer). In addition, poor seeing often affects larger scopes more than smaller scopes depending on the average size of the thermal cells of air.
I also understand the concepts of modulation transfer function, contrast, and acutance. As far as Qkslvr's images go... As I mentioned in more than one of my posts, none of what I have discussed applies to planetary photography where you run at much higher focal ratios. In those cases, certainly diffraction and MTF will have a significant affect on image quality. All I'm suggesting--and I believe practical experience bears this out--is that for prime focus astrophotography resolution is typically limited by seeing conditions, not diffraction. I never suggested, and I do not believe, that the images would be identical between a large scope and a small scope. Certainly there would be differences in contrast, noise, and image intensity. The larger scope wins on all these counts. I only suggested that when choosing a CCD camera for a telescope, in order to maximize spatial resolution in particular you need not take into account focal ratio--only focal length is relevant.
You mentioned in your post that seeing is not the issue and that overall image quality had nothing to do with whether the photographic system was used for terrestrial photography or for astrophotography. If that is the case, then why does the Hubble space telescope outperform much larger ground based telescopes? Isn't there an advantage to getting above the Earth's turbulent atmosphere? Isn't that Hubble's primary edge? Isn't that why professional telescopes now employ adaptive optics--to alleviate the effects of poor seeing?
The way many astronomers match a CCD to their telescope is actually quite practical: first you pick the best compromise of spatial resolution and sensitivity for the objects you wish to image (pixel size), then you pick the largest field of view you can afford! CCD's go up in cost pretty dramatically with the number of pixels/field of view.
-------------------- - Jared Willson
- TV 60iS refractor
- 80mm SV/LOMO refractor
- A-P 130 EDFS refractor
- Planewave 12.5" CDK
- A-P 900 mount
- A-P Mach1 GTO mount
- Takahashi Teegul SP travel mount
- SBIG STL-11000 C1 camera with AO-L
- Some heavily light polluted skies
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ScopenChuck
super member
Reged: 11/04/04
Posts: 163
Loc: Billings, Montana
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Geez, What ever happen to point and shoot.
-------------------- 16" f/6 truss tube ( still working on it)
10" f/4.5 Meade Starfinder Eq
6" f/8 Criterion Dynascope (1970 & 1974)
5" f/9 Meade LXD75
60mm Meade ETX
Box of spare parts
Chuck
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Jared
Post Laureate
Reged: 10/11/05
Posts: 4554
Loc: Piedmont, California, U.S.
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You're right! I've gotten excited and drifted off topic.
-------------------- - Jared Willson
- TV 60iS refractor
- 80mm SV/LOMO refractor
- A-P 130 EDFS refractor
- Planewave 12.5" CDK
- A-P 900 mount
- A-P Mach1 GTO mount
- Takahashi Teegul SP travel mount
- SBIG STL-11000 C1 camera with AO-L
- Some heavily light polluted skies
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imhotep
Vendor - Optical Supports
Reged: 02/14/07
Posts: 1718
Loc: Floriduh
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Quote:
Since Takahashi makes a 60mm f/8.3 scope, let's use that.
Let's say I want to target an sampling rate (image scale) of 1"/pixel, which would give best sampling for a given OBJECT during a typical "best" seeing night for many people. In other words, 'N' number of pixels will cover that given object, which will strike the best balance between resolution vs. angular coverage. This can be accomplished with a theoretical 2.5 micron pixel camera. To get the same number of 'N' pixels covering that same object for optimium sampling with a 16" f/5 Newtonian, you'd need a 9.5 micron pixel camera.
Which fills those 'N' number of pixels faster, assuming the same QE chip? The 16" Newt, by virtue of its much greater aperture area...more photons are being delivered to the same area of the chip.
Man Jay, I was with you up until this point Stick with me, I think I'm almost there to understanding. Doesn't FOV factor into determining how many arcseconds you are covering per pixel? Humor me, let's use my current scope and sensor:
Objective = 4.25 inches or 107.95mm Max. theoretical resolution = 1.07" SPC900NC pixel size = 7.5 microns SPC900NC sensor size = 4.8mm x 3.6mm
What else do I need to plug into a formula that will calculate arcsec/pixel? If I have that formula I'll be able to use it with future scope/camera combos.
Thanks for your help!
-------------------- Curt
www.opticalsupports.com
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budongbait
newbie
Reged: 05/13/07
Posts: 2
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Quote:
Man Jay, I was with you up until this point Stick with me, I think I'm almost there to understanding. Doesn't FOV factor into determining how many arcseconds you are covering per pixel? Humor me, let's use my current scope and sensor:
Objective = 4.25 inches or 107.95mm Max. theoretical resolution = 1.07" SPC900NC pixel size = 7.5 microns SPC900NC sensor size = 4.8mm x 3.6mm
What else do I need to plug into a formula that will calculate arcsec/pixel? If I have that formula I'll be able to use it with future scope/camera combos.
Thanks for your help!
Hi imhotep,
Thanks for starting this thread. It was very helpful for me. I myself am a n00b (actually I make n00bs look good...) and have tried to answer this same question of late, perhaps what I've come up with will help you out, it seems to have helped me out with a similar question...
Basically my question was this. If I hook a CCD (659 x 494 / 5.6u per pixel) up to my telescope (Celestron 8i SE = 2030 mm / 203 mm = f/10) how much sky would I image and what eyepiece (magnification) would that be equal to if the CCD is a rectangle inscribed inside said eyepiece's view.
So the from the discussion in the thread it was obvious to me that the only common factor here is arcminutes/arc seconds. So I googled (and googled and googled) until I found the following link.)
Telescope Eyepieces
And it says to find the field of view through an eyepiece you should take the field of view for the eyepiece and divide by the magnification you are using.
For my setup I'm using a 6 mm "expanse" with a 66° degree FOV on my 2032mm focal length 8i SE, with a 2x barlow, resulting in a FOV of... 66*(2 * 2032)/6 = 0.0974 degrees or 5.8 arcmin. (Right everyone?)
Ok cool. Now there are lots of CCD calculators on the web, but only a few places where the original equations are given. I found this one useful.
CCD University
And it says: Sampling in arcseconds = (206.265 / (focal length in mm) )* (pixel size in microns)
So for my fire-i camera (659 x 494 @ 5.6 um/pixel) and my 2032 mm focal length telescope: (206.265/2032) * 5.6 = 0.903 arcsec per pixel, so the total amount of sky my telescope + CCD will pick up will be roughly 0.903 * 659 = 9.92 arcmin (595 arcsec/60) by 0.903 * 494 = 7.4 arcmin (206 arcsec/60)
Now for a rectangle inscribed in a circle: The CCD's diagonal dimension is 12.368 arc min by Pythagorean theorem
So my CCD to eye piece FOV ratio is 12.368 : 5.8 and if I translate that over to the concept of magnification it would be x : 677 so x = (677*5.8/12.368) = 317 So the effective magnification of my CCD/telescope combination is 317x.
Ok so once again I make n00bs look good, and I haven't bought the fire-i yet so I can't check to see if this is actually what happens. But it sounds good on paper. And honestly I'm posting this so the pro's here can correct my mistakes.
Now if I run with the numbers you gave above and earlier... you have a 4.5" telescope with a 1100mm focal length - as your 7mm eyepiece gives a 157x magnification.
The field of view for your CCD is
(206.265/1100) * 7.5 = 1.4" / pixel (" = arc seconds) 640*1.4" = 14.9' (' = arc minutes) and 480 * 1.4" = 11.2'
So your field of view is 14.9' x 11.2' with a diagonal of 18.6'
Now I don't know the field of view for your 7mm orthoscopic eyepiece but say it is 60° that means your FOV for your eyepiece is 22.4' so using the same (probably flawed) logic as above I calculate 18.6' : 22.5' as x : 157 so x = 189x I would hazard to say that your CCD is roughly a 189x.
Now I realize that the original question was geared towards knowing if using a barlow would be a waste because you are at a magnification that is above the diffraction limit of the telescope. As as most people point out the atmospheric conditions at the time limit us well below the diffraction limit. And while I am good at Fourier transforms I'm also lazy and a n00b so I'm just going with the effective magnification and that tells me I'm ok ish for my 8 inch telescope.
However as I'm trying to image Saturn at the moment I know that I need to push higher than the diffraction/atmosphere limit because I want Saturn to occupy more than ~30 pixels on my telescope. If you look at the CCD University link above my 5.6um pixels are just oversampling for my telescope, but I'm going to push to an even higher magnification and way over-sample the image so as to reduce the quantization noise of my CCD. And as I can pull in a good amount of light (ahhh only a n00b would think 203 mm pulls in a good amount of light) I feel some what safe in doing this.
All the calculations also help with figuring out if I need a reducer to see Andromeda, which given its dimensions it turns out that I'll need to buy a bigger CCD (actually I'll need to take multiple photos and stitch them together.)
From wikipedia: M42 Orion Nebula is 65 x 60 arc minutes M31 Andromeda 190 x 90 arc minutes M16 Egal nebual 7x2 (?) arc minutes..
Ok I hope this helps. Budong Bait... (the n00b)
PS: Actually I secretly hope that I'm 100% wrong and some kind soul comes in and schools me by correcting my math. All I ask is that they be gentle. (Spelling and grammar included)
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imhotep
Vendor - Optical Supports
Reged: 02/14/07
Posts: 1718
Loc: Floriduh
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Budong Bait,
Thanks for the post. Your math-work there is easy to follow. To me, your final answers seem correct. I'm also very interested to see what the pros think. This thread is getting old though, so let's wait and see if anyone else posts. If not, I'd like to paste your work on a new thread and hopefully generate a new discussion.
Again, thanks for your tremendous efforts here! Curt
-------------------- Curt
www.opticalsupports.com
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Chuck Anstey
scholastic sledgehammer
Reged: 01/17/05
Posts: 960
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The trouble with the above calculations are they don't have any real meaning because it depends on the EP used. I have a 9mm TMB (AFOV 30d), 9mm UO Orthoscopic (AFOV 42d), 9mm TMB/BO Planetary (AFOV 60d), and 9mm Nagler (AFOV 82d). They all have the same magnification but different TFOV. No matter which one you pick for your FOV calcualtion, you will get a different answer for the magnification.
I have a CCD camera of 3M pixels and a DSLR of 6M pixels, the individual pixel size is identical. Does the DSLR, with twice the area FOV, have more or less magnification over the CCD camera?
Now change the problem. I have a CCD camera of 3M pixels and a DSLR of 6M pixels, the FOV is identical because the DSLR camera's pixels are smaller. Does the DSLR, with smaller pixels, have more or less magnification over the CCD camera?
The only numbers that are consistent and repeatable are arc-seconds per pixel and TFOV. Those two values tell you what you need to know about the camera / OTA combination. From here you can argue about the correct amount of oversampling, diffraction limited, seeing limited, and everything else all you want.
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budongbait
newbie
Reged: 05/13/07
Posts: 2
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Quote:
The trouble with the above calculations are they don't have any real meaning because it depends on the EP used. I have a 9mm TMB (AFOV 30d), 9mm UO Orthoscopic (AFOV 42d), 9mm TMB/BO Planetary (AFOV 60d), and 9mm Nagler (AFOV 82d). They all have the same magnification but different TFOV. No matter which one you pick for your FOV calcualtion, you will get a different answer for the magnification.
I have a CCD camera of 3M pixels and a DSLR of 6M pixels, the individual pixel size is identical. Does the DSLR, with twice the area FOV, have more or less magnification over the CCD camera?
Now change the problem. I have a CCD camera of 3M pixels and a DSLR of 6M pixels, the FOV is identical because the DSLR camera's pixels are smaller. Does the DSLR, with smaller pixels, have more or less magnification over the CCD camera?
The only numbers that are consistent and repeatable are arc-seconds per pixel and TFOV. Those two values tell you what you need to know about the camera / OTA combination. From here you can argue about the correct amount of oversampling, diffraction limited, seeing limited, and everything else all you want
Hi Chuck,
Thanks for pointing out the problems and inherent inconsistencies with the calculations I ran through. Fundamentally I needed to go through those calculations to provide myself with a frame of reference to understand how a CCD works when at prime focus. I've been using telescopes with eyepieces from a young age but never really understood the math behind them.
After doing these calculations I find I'm better able to understand the answers and topics that were posted in the thread, and they make allot of sense now.
I have a good grasp of sampling (quantization) theory and have a decent grasp on diffraction limitations. I'm now trying to put these concepts together in a way that makes sense for me as my frame of reference once again is using my telescope with an eye piece (and using a web cam through my eyepiece which I believe is call image projection).
I would really appreciate it if you have any simplified (CCD for n00bs) descriptions for the major concepts in CCD prime focus use - and in general figuring out what I would expect to see as compared to staring through my eyepiece (my mantra).
Thanks again Chuck (and to everyone who has helped in the previous posts.) -Gus
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Chuck Anstey
scholastic sledgehammer
Reged: 01/17/05
Posts: 960
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Gus, I think you have everything you need to figure things out, you just need to change your way of thinking. Imaging means thinking "Do I have enough FOV to cover the object?" and "Am I sufficiently sampled for the object?" I no longer think about trying to relate EP view with imaging, even though I did that at the beginning.
Another point is to not put the cart before the horse. If you are doing long exposure imaging, it takes a lot of practice and skill to legitimately use 1 arc-second or better sampling. I know it may seem like you are under-sampled but you really need to have it together and great seeing to show the difference between 1.25" and 0.8" arc-seconds per pixel. Even though it seems like the first thing to worry about, image scale for high-resolution is actually one of the last things to worry about. There are so many other issues that destroy attempts at long exposure high resolution that they need to addressed first.
The real reference point you need is to understand what a high quality 1" per pixel image looks like. Just look at Florent's M51 in the CCD forum. I believe his image scale is 1.25" per pixel and my best M51 in fantastic seeing conditions at 0.79" per pixel could not compete in real resolution. Now he does have a better scope so his pixels are of higher quality but the rest of the equipment is identical so higher sampling wasn't the solution to higher resolution.
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imhotep
Vendor - Optical Supports
Reged: 02/14/07
Posts: 1718
Loc: Floriduh
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Chuck, thanks for your comments. Henceforth I will think of my imaging capabilities in terms of FOV and arcsec/pixel as you suggest. But I'm still stuck on this question - what formula will alllow me to calculate arcsec/pixel of my current setup?
Here's what I have so far.
Objective = 4.25 inches or 107.95mm
Max. theoretical resolution = 1.07"
SPC900NC pixel size = 7.5 microns
SPC900NC sensor size = 4.8mm x 3.6mm
-------------------- Curt
www.opticalsupports.com
Edited by imhotep (05/15/07 02:24 PM)
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Chuck Anstey
scholastic sledgehammer
Reged: 01/17/05
Posts: 960
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The freebie CCDCalc is what I use. Plus it has pictures of many of the common DSOs so you can see what it really means.
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imhotep
Vendor - Optical Supports
Reged: 02/14/07
Posts: 1718
Loc: Floriduh
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::sigh:: Ok I guess I'll keep looking. I just feel like for me personally, knowing that formula would be a whole lot simpler than having to rely upon an online tool. Plus I'll learn more that way too.
-------------------- Curt
www.opticalsupports.com
Edited by imhotep (05/15/07 02:47 PM)
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imhotep
Vendor - Optical Supports
Reged: 02/14/07
Posts: 1718
Loc: Floriduh
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budongbait, I just reread your posting and read up on the formulas in the CCD University link.
Sampling in arcseconds = (206.265 / (focal length in mm) )* (pixel size in microns )
So for my setup...
Sampling = (206.265 /(1097))* (7.5) = 1.41 arcsec/pixel
Bingo. Now if someone can correct if my wrong, I could wisely afford to oversample my images a little bit further couldn't I?
-------------------- Curt
www.opticalsupports.com
Edited by imhotep (05/15/07 02:45 PM)
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Chuck Anstey
scholastic sledgehammer
Reged: 01/17/05
Posts: 960
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imhotep, The CCDCalc program can be downloaded to your machine and run locally. Once you know the answer you don't need to keep looking it up.
As far as your sampling size, I think you are pushing that scope pretty hard now so it doesn't seem likely you will achieve any real increase in resolution by a smaller sampling size. You might for solar system objects so you will just have to try it with and without a barlow.
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jmoura
newbie
Reged: 01/18/08
Posts: 2
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Dear Gavin Based on this approach, with some minor modifications, I made a little spreadsheet to calc amplification. If you want it, send me an email and I`ll send it to you.
Good skies
José Moura
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