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Chuck Anstey
scholastic sledgehammer

Reged: 01/17/05
Posts: 960
Re: Question about calculating magnification [Re: wilash]
#1566705 - 04/25/07 05:34 PM

Quote:

I would be very interested to learn the formulas that are used to calculate FOV and arcseconds/pixel if you have some sources.

Go here

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russ_watters
Pooh-Bah

Reged: 11/24/04
Posts: 1340
Loc: Trappe, PA
Re: Question about calculating magnification [Re: wilash]
#1567074 - 04/25/07 08:01 PM

Quote:

Do not confuse linear magnification with angular magnification.

I'm not seeing a difference between the two concepts. When using small angles, a change in the length of the small side is directly proportional to the change in angle.

Ie:
Quote:

Linear magnification is the difference between object and image size.

In astronomy object size is measured in angular diameter. How would you measure object size any other way
?
Quote:

Angular magnification is used at infinity (or undetermined object distances) and that is what is used in observing. Angular magnification describes the difference in viewing angle based on a norm - approximately 52 degrees if I remember correctly.

I'll get into that in my next post...

Quote:

As far as displaying the image, linear magnification does change with display size as it is changing image size. Angular magnification does not because the area that is viewed is not changing.

Yes, the area that is being viewed does change! Take a book and set it across the room from you. The reason you can't read it is that it's angular size decreases.

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Equipment: Orion Atlas 11, ED80, DSI-C, DSI III Pro, Dell Inspiron Laptop.
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russ_watters
Pooh-Bah

Reged: 11/24/04
Posts: 1340
Loc: Trappe, PA
Re: Question about calculating magnification [Re: russ_watters]

I described in words my method for calculating magnification, but let me go through the math and explain exactly what it means....

Attached is my best image of Saturn. In the image, Saturn is 173 pixels across. I said above that a standard monitor is 75 dpi - measuring my 21" monitor at 15.75 and 1280 gives 81 dpi. I mentioned, though, that today people often go with higher, so that is somewhat arbitrary (and a problem when trying to scale an image for the web). Also arbitrary is my estimation that the optimal viewing distance is 24" (yes, I actually just measured my distance to my laptop screen). You probably sit closer to a laptop than a desktop, also. Using trig (the trick here is that since one side is much larger than the other, we can just assume a right triangle), we find that a triangle that is 1/81" on one side and 24" on another side has an angle of .02947 degrees. Multiply by Saturn's diameter and Saturn, on your screen, subtends an angle of 5.10 degrees.

At the time, due to its distance, Saturn was about 20 arcsec in diameter. That's .005556 degrees.

So. If you hold up your monitor to the sky, 24" from your face and happen to have extremely good vision, Saturn on the monitor will look 5.10/.005556= 917 times larger on your monitor than in the sky.

When I first posted the pic and forgot that my laptop has unusually high resolution, when people told me it was "big", I realized that meant it was so big it was starting to get grainy. So I resized it when I posted it to my website.

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Equipment: Orion Atlas 11, ED80, DSI-C, DSI III Pro, Dell Inspiron Laptop.
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russ_watters
Pooh-Bah

Reged: 11/24/04
Posts: 1340
Loc: Trappe, PA
Re: Question about calculating magnification [Re: russ_watters]

Here it is shrunk by 1/3, giving about 610x magnification (colors are a little different too, but it is the same image otherwise): Looks a lot better...

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russ_watters
Pooh-Bah

Reged: 11/24/04
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Re: Question about calculating magnification [Re: imhotep]

Wrt the actual question:

Quote:

Gavin & Russ: I think I need to explain myself a little better. What you've both said is totally correct. I realize that trying to determing magnification based upon a digital sensor is somewhat unrealistic. My goal here is to determine the limits of my scope with regard to using barlows with a CCD. Let me explain.

For simple visual work, I usually go by the rule of thumb - a maximum of 50x per inch of aperture. On my 4.25" scope this comes out to 212.5x. So for visual observing, my 7mm ortho yields 157x, but with a 2x barlow it is at 314x which obviously isn't practical for my small aperture.

With that said, all of my images taken with the SPC900NC coupled to the same 2x barlow have been fine. I would like to figure out how to determine the same magnificaiton limits for a given optics/ccd combination.

Yeah, I know exactly what you are talking about. When you look up a telescope's maximum magnification and compare it to what looks good on a ccd (using my calculation), most people conclude that you can exceed the maximum theoretical resolution when using a ccd. Note: I'm not there yet with my new C11, but in my defense, the higher you go, the more the atmosphere matters.

Here's my best Saturn with my ETX-105, a little smaller than your scope, but a decent basis for comparison. At 90 pixels across, using my above established thumbrule of .025 degrees per pixel (or even easier, 4.4x per pixel), that's 400x magnification.

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Equipment: Orion Atlas 11, ED80, DSI-C, DSI III Pro, Dell Inspiron Laptop.
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wilash
Fairy Godmother

Reged: 09/30/03
Posts: 5746
Re: Question about calculating magnification [Re: russ_watters]
#1567224 - 04/25/07 09:15 PM

Quote:

Quote:

Do not confuse linear magnification with angular magnification.

I'm not seeing a difference between the two concepts. When using small angles, a change in the length of the small side is directly proportional to the change in angle.

Linear magnification is the difference in size between the object and image. At 1x the image is exactly the same size as the object - the image of a one inch insect is one inch long.

Angular magnification is in relation to a particular angle of view. At 1x, the world looks like what we see. The image size of the one inch insect will change with object distance even though the angular magnification remains 1x.

Quote:

Ie:
Quote:

Linear magnification is the difference between object and image size.

In astronomy object size is measured in angular diameter. How would you measure object size any other way
?

You can measure an object by its size. The sun is 1,392,000km in diameter.

Quote:

Quote:

Angular magnification is used at infinity (or undetermined object distances) and that is what is used in observing. Angular magnification describes the difference in viewing angle based on a norm - approximately 52 degrees if I remember correctly.

I'll get into that in my next post...

Quote:

As far as displaying the image, linear magnification does change with display size as it is changing image size. Angular magnification does not because the area that is viewed is not changing.

Yes, the area that is being viewed does change! Take a book and set it across the room from you. The reason you can't read it is that it's angular size decreases.

You are confusing linear and angular magnification. The linear magnification of the book is changing in your eye, but the angular magnification of the picture in the book remains the same.

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wilash
Fairy Godmother

Reged: 09/30/03
Posts: 5746
Re: Question about calculating magnification [Re: russ_watters]
#1567276 - 04/25/07 09:37 PM

Quote:

I described in words my method for calculating magnification, but let me go through the math and explain exactly what it means....

Attached is my best image of Saturn. In the image, Saturn is 173 pixels across. I said above that a standard monitor is 75 dpi - measuring my 21" monitor at 15.75 and 1280 gives 81 dpi. I mentioned, though, that today people often go with higher, so that is somewhat arbitrary (and a problem when trying to scale an image for the web). Also arbitrary is my estimation that the optimal viewing distance is 24" (yes, I actually just measured my distance to my laptop screen). You probably sit closer to a laptop than a desktop, also. Using trig (the trick here is that since one side is much larger than the other, we can just assume a right triangle), we find that a triangle that is 1/81" on one side and 24" on another side has an angle of .02947 degrees. Multiply by Saturn's diameter and Saturn, on your screen, subtends an angle of 5.10 degrees.

At the time, due to its distance, Saturn was about 20 arcsec in diameter. That's .005556 degrees.

So. If you hold up your monitor to the sky, 24" from your face and happen to have extremely good vision, Saturn on the monitor will look 5.10/.005556= 917 times larger on your monitor than in the sky.

When I first posted the pic and forgot that my laptop has unusually high resolution, when people told me it was "big", I realized that meant it was so big it was starting to get grainy. So I resized it when I posted it to my website.

Actually, you have assumed viewing distance does not change, which of course it can and does, and so your conclusion and idea does not work. You also can connect the virtual image you see in a telescope with a displayed image. Given the same angular size from the viewer, the display image and the scope image will be identical and have the same angular magnification; this has no baring on the size of the display image. Since viewing distance is important in angular resolution, simply using display size only does not work - how far away the image is makes a difference. (Actually, when you come to viewing display images you can no longer talk about angular "magnification," but just angular size.)

Linear magnification does not depend on viewing distance. This is why display size changes linear magnification as it is simply the difference in dimensions between the object and image. So an 8x10 print has a higher linear magnification than a 4x5 print regardless of the viewing distance.

Edited by wilash (04/25/07 09:48 PM)

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gavinm
Pooh-Bah

Reged: 08/26/05
Posts: 1449
Loc: Auckland New Zealand
Re: Question about calculating magnification [Re: wilash]
#1567545 - 04/26/07 12:09 AM

Imhotep, the other thing to remember (actually answering your question ) is that limiting magnification really has nothing to do with resolution, but light gathering ability. As you magnify, using smaller focal length eyepieces, you "spread out" the light so the image becomes dimmer. Thus magnification is limited because the image is too dim to see properly.

A CCD, being an integrating (ie storing) device is not limited in this way (unlike the eye). Magnification is limited by resolution, seeing and diffraction, not 'dimness'.

--------------------
Gavin

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russ_watters
Pooh-Bah

Reged: 11/24/04
Posts: 1340
Loc: Trappe, PA
Re: Question about calculating magnification [Re: wilash]
#1567605 - 04/26/07 01:26 AM

Quote:

Quote:

Quote:

Do not confuse linear magnification with angular magnification.

I'm not seeing a difference between the two concepts. When using small angles, a change in the length of the small side is directly proportional to the change in angle.

Linear magnification is the difference in size between the object and image. At 1x the image is exactly the same size as the object - the image of a one inch insect is one inch long.

Well, that's kinda meaningless since the concept of magnification is always distance dependent. By your logic, you could show an inch-long photo of an inch-long insect and another photo of a dot and call them the same magnification. They aren't.

Alternately.....
Quote:

You can measure an object by its size. The sun is 1,392,000km in diameter.

Well actually, no, you cannot measure an object in space by it's "size" unless you have a really really long tape measure. The only "size" you can actually measure with a telescope is angular size. Calculating an object's actual dimensions requires a knowledge of the distance and....the magnification of the telescope, which is a physical property of the telescope.

And even if you could....then you could draw a 1km image of the sun and say it's magnification is 1/1,392,000? That's not just illogical, it's rediculous. That simply isn't what the word "magnification" means. Heck, what you are saying suggests that the magnification depends on the object and we all know that it doesn't. A telescope with a 1000mm objective and a 10mm eyepiece produces a magnification of 100x, period.
Quote:

You are confusing linear and angular magnification. The linear magnification of the book is changing in your eye, but the angular magnification of the picture in the book remains the same.

 I see that the concept exists - it just isn't useful/doesn't apply to telescopes.

--------------------
Equipment: Orion Atlas 11, ED80, DSI-C, DSI III Pro, Dell Inspiron Laptop.
www.russsscope.net

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russ_watters
Pooh-Bah

Reged: 11/24/04
Posts: 1340
Loc: Trappe, PA
Re: Question about calculating magnification [Re: wilash]
#1567609 - 04/26/07 01:32 AM

Quote:

Actually, you have assumed viewing distance does not change, which of course it can and does

Reread it: I took into account both the object distance and the image distance.
Quote:

Given the same angular size from the viewer, the display image and the scope image will be identical and have the same angular magnification

Well, yes, since what you just said is redundant (given the same angular size you will get the same angular size), it is true....but it doesn't mean anything.
Quote:

Since viewing distance is important in angular resolution, simply using display size only does not work - how far away the image is makes a difference.

....which is why I used it in the calculation.
Quote:

Linear magnification does not depend on viewing distance. This is why display size changes linear magnification as it is simply the difference in dimensions between the object and image. So an 8x10 print has a higher linear magnification than a 4x5 print regardless of the viewing distance.

I see what you are saying, but how is that concept at all useful?

Have you ever seen an 8x10 of the moon labeled 1/17,000,000th magnification?

Again, do you have any references to the use this concept in astronomy?

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Equipment: Orion Atlas 11, ED80, DSI-C, DSI III Pro, Dell Inspiron Laptop.
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russ_watters
Pooh-Bah

Reged: 11/24/04
Posts: 1340
Loc: Trappe, PA
Re: Question about calculating magnification [Re: russ_watters]
#1567618 - 04/26/07 01:39 AM

....er, wait.... are you a biologist? I could see how this concept might make sense for a microscope, where the objects you image are always smaller than the image and the viewing distances are short. It doesn't make any sense at all for a telescope - and it isn't used.

Here's wik on the subject: http://en.wikipedia.org/wiki/Magnification

Here's a link describing linear magnification. Notice in the example, the object distance is shorter than the image distance. So angular magnification is only useful when enlarging small objects, not magnifying distant ones.
http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lensdet.html

Here's a link saying what I'm saying: the concept of linear magnification is meaningless for telescopes: http://www.ux1.eiu.edu/~cfadd/1160/Ch26OIn/Tele.html

Quote:

A telescope is used for viewing far distant objects. Therefore, we are interested in increasing the anlge subtended by an object or we are interested in the angular magnification with a telescope. The linear magnification is somewhat meaningless in this context.

[and on the previous page it describes using linear magnification in a microscope]

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Equipment: Orion Atlas 11, ED80, DSI-C, DSI III Pro, Dell Inspiron Laptop.
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russ_watters
Pooh-Bah

Reged: 11/24/04
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Re: Question about calculating magnification [Re: gavinm]
#1567630 - 04/26/07 01:51 AM

Quote:

Imhotep, the other thing to remember (actually answering your question ) is that limiting magnification really has nothing to do with resolution, but light gathering ability. As you magnify, using smaller focal length eyepieces, you "spread out" the light so the image becomes dimmer. Thus magnification is limited because the image is too dim to see properly.

A CCD, being an integrating (ie storing) device is not limited in this way (unlike the eye). Magnification is limited by resolution, seeing and diffraction, not 'dimness'.

Actually, no, magnification is limited by your aperature due to diffraction. You can always brighten a dim image by lengthening your exposures. The nearer planets and moon are so bright that even a small telescope would be capable of thousands of times magnification (visual or ccd) if brightness was the limiting factor.

--------------------
Equipment: Orion Atlas 11, ED80, DSI-C, DSI III Pro, Dell Inspiron Laptop.
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gavinm
Pooh-Bah

Reged: 08/26/05
Posts: 1449
Loc: Auckland New Zealand
Re: Question about calculating magnification [Re: russ_watters]
#1567656 - 04/26/07 02:48 AM

Only on high-power subjects, like the moon etc. But on your average DSO, brightness and contrast (as well as all you mentioned) limit magnification.

And looking back, I didnt answer the question either. The question was what is the effective magnification when using CCD's. The answer is there is no answer (without considering everything such as how far you are standing away from your monitor etc)

--------------------
Gavin

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imhotep
Vendor - Optical Supports

Reged: 02/14/07
Posts: 1718
Loc: Floriduh
Re: Question about calculating magnification [Re: wilash]
#1567934 - 04/26/07 09:14 AM

Will, I guess if I understood everything perfectly then I wouldn't be asking questions.

--------------------
Curt
www.opticalsupports.com

Edited by imhotep (04/26/07 09:36 AM)

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imhotep
Vendor - Optical Supports

Reged: 02/14/07
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Re: Question about calculating magnification [Re: wilash]
#1567954 - 04/26/07 09:29 AM

Quote:

OK. Your magnification limit in an imaging system is very complex - how good are you with Fourier Transforms? Basically, you need to think about sensor size and diffraction of the optical system. If you push the magnification and end up with a slow focal ratio, you are also losing to diffraction. Depending on format, you could be limited to a f/11 system, you could also get away with something higher. Naturally, it depends what you do with the data and how "good" you think the result is.

I'm actually glad that you brought this up. In the past I have used formulas to determine the limiting f-ratio with my digital censors. For example, with the CMOS censor in my Rebel XT it is around f/11. I'm not sure what it is for my SPC900NC.

I agree that it is important to know the limits of your censor with respect to f-ratio in order to make sure you are not losing sharpness to diffraction. Regardless, this has nothing to do with my question about magnification.

--------------------
Curt
www.opticalsupports.com

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imhotep
Vendor - Optical Supports

Reged: 02/14/07
Posts: 1718
Loc: Floriduh
Re: Question about calculating magnification [Re: imhotep]
#1568006 - 04/26/07 09:57 AM

Gavin & Russ, thank you both for your perseverance with this thread. I plugged my numbers into the calculator that chuck linked to and here's what I got:

Telescope focal length: 1080mm (close, it is actually 1097.5)
Arc-seconds per pixel for this combo: 1.43
Field of view in arc-minutes: 11.4' x 15.3'

The above was based upon the dimensions of the SPC900NC sensor:

4.8mm x 3.6mm
640x480
Pixel size = 7.5 microns

--------------------
Curt
www.opticalsupports.com

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wilash
Fairy Godmother

Reged: 09/30/03
Posts: 5746
Re: Question about calculating magnification [Re: imhotep]
#1568426 - 04/26/07 01:33 PM

Russ, I am not saying linear magnification is useful in astonomy. As I have said, it is not useful. I am saying there is a confusion in the posts here between angular and linear magnification. The qualities of one is being applied to the other and it won't work.

I am not a biologist. My field is imaging systems.

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wilash
Fairy Godmother

Reged: 09/30/03
Posts: 5746
Re: Question about calculating magnification [Re: imhotep]
#1568449 - 04/26/07 01:47 PM

Quote:

Quote:

OK. Your magnification limit in an imaging system is very complex - how good are you with Fourier Transforms? Basically, you need to think about sensor size and diffraction of the optical system. If you push the magnification and end up with a slow focal ratio, you are also losing to diffraction. Depending on format, you could be limited to a f/11 system, you could also get away with something higher. Naturally, it depends what you do with the data and how "good" you think the result is.

I'm actually glad that you brought this up. In the past I have used formulas to determine the limiting f-ratio with my digital censors. For example, with the CMOS censor in my Rebel XT it is around f/11. I'm not sure what it is for my SPC900NC.

I agree that it is important to know the limits of your censor with respect to f-ratio in order to make sure you are not losing sharpness to diffraction. Regardless, this has nothing to do with my question about magnification.

And to bring in one more factor, processing can affect this as well. I wish there was a simple answer to knowing where diffraction is going to be a hinderance, but there are so many things that go into making that final image that there is no pat answer.

One advantage to astrophotography is that the subject is usually not something within common experience and so you can push things to more exptremes and get a pleasing image - we are so used to the things around us on this planet that if we did the same things to them as we do to our astro pics, we would probably end up with a strange looking image.

Be careful of the "pixel sampling" size in amateur astophotography, it is a fuzzy concept that I can see no valid reason for. It sounds cool, but on closer inspection you will find it completely ignores diffraction. I would just get your camera and start using it. Experience will show what is what.

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Chuck Anstey
scholastic sledgehammer

Reged: 01/17/05
Posts: 960
Re: Question about calculating magnification [Re: wilash]
#1568517 - 04/26/07 02:18 PM

Quote:

Be careful of the "pixel sampling" size in amateur astophotography, it is a fuzzy concept that I can see no valid reason for. It sounds cool, but on closer inspection you will find it completely ignores diffraction.

Have to disagree with you on this one. Now sampling size tends to be a more advanced concept but it does take into account diffraction. It also takes into account you are taking an image with a very discrete sensor array and not something that is more continuous like film (but still essentially discrete at the grain level). For beginning imagers if they are sampling any where near their diffraction limit on DSOs then they are probably beyond the limits of some other part of their imaging system. For planetary and lunar I have seen talk and results showing that some over-sampling brings in more detail with rapidly diminishing returns.

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Qkslvr
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Reged: 06/23/06
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Loc: NE Ohio, US
Re: Question about calculating magnification [Re: imhotep]
#1568772 - 04/26/07 04:17 PM

Quote:

For example, with the CMOS censor in my Rebel XT it is around f/11.

You're not limited to f/11, Using a Tele-Extender and a 26mm eyepiece I've had mine up to f/35 or so.

--------------------
Mike
Onyx 80ED/N8/CG-5/40D

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