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Maximum Magnification When Imaging

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#1 paulgo12

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Posted 03 July 2020 - 01:18 AM

Hi! Just wondering, when imaging, is the maximum magnification the same with visual? That is, x2 of aperture in mm?

Or can I go higher? I will be using an ASI224MC to image Jupiter. Scope is Celestron Omni 150/750.

Thanks!

#2 Jenz114

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Posted 03 July 2020 - 01:58 AM

Hello! Hopefully I can help a bit. When imaging we usually try to match the focal length to aperture ratio of the telescope to the camera's pixel size to avoid over and under sampling. To do this there is a general rule of thumb to multiply your camera's pixel size by 5, assuming excellent seeing conditions, to get a rough estimate of what your target focal length to aperture ratio should be. In your case using the ASI224MC this would be 3.75 multiplied by 5 for a guideline of f/18.75. Your telescope's native ratio is f/5, so using a 4x barlow would get your ratio to f/20, which is very close to the "ideal" f/18.75. Seeing conditions greatly influence the final outcome of an image, and you can always try using a smaller focal length to aperture ratio, but it will be at the expense of some detail.


Edited by Jenz114, 03 July 2020 - 04:20 AM.

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#3 drd715

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Posted 03 July 2020 - 02:02 AM

Not exactly a relationship between visual and imaging.   

 

Magnification in visual is presented by the eyepiece presentation to the eye. Visually it can surpass in power of magnification the maximum that the lens diameter can resolve details. Basically limits in resolution such as Dawes and other definitions based upon objective diameter will determine the most detail you can get for a given scope; more diameter more resolution. At some point the maximum resolving limit of the objective size will in practice be limited to the maximum "seeing" limits of that night's sky conditions - in other words not just the telescope size is the limiting factor in resolution. Some nights can be much better than others and dark sky sites are a big help.  

 

The camera imaging sensor is not like the human eye. "Power" is not really a concept of imaging. Image framing size is somewhat similar to the concept of visual power as it provides a defined size of sky that will illuminate the sensor. Sized in arc minutes/ seconds  of view.  This is directly defined by the focal length of the scope and the imaging chip sensor physical size. Combined they produce a captured image of defined arc seconds. As for resolution in imaging this gets more complicated but it is still basically the same as visual resolution. It just involves the ratio of Dawes limit over the number of pixels this covers on the sensor. That discussion gets complicated and somewhat controversial. The  math is fairly well defined. In the end though it comes down to matching the pixel size to the lens resolution limits. 

 

Basically it is more about resolution than ultimate magnification power. So trying to compare visual power to imaging scale is like comparing apples to oranges. They are both fruit, just different. 



#4 sg6

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Posted 03 July 2020 - 03:56 AM

No magnification, you plonk the prime image directly on the sensor.

Seems that the image size matters more or less. Again a rule that is fairly broad and often ignored or has exceptions - one has to question the title "Rule".

 

As a fair guid take the focal lengthj of the scope, assume a 1 arc sec target and depermine the size of one pixel the match the result:

Pixel = Tan(1 arcsec)*Focal length.

It will be small.

 

That is generally as small as it is suggested you go (exception to follow).

Would say determine that pixel size and use a pixel 1-2x that value, but people use 4x and more.

 

The exception is planets, people go down to half that pixel size per arcsec. Yes planets are brighter but the image is bigger and so dimmer and in one way they are all just an image. Should what the image is of make a difference?

 

Work the numbers through, find out what 1 Arcsec is and double it to for 2 Arcsec and get something in that general region is likely the easy solution.

 

But there is no magnification as such.


Edited by sg6, 03 July 2020 - 03:57 AM.


#5 MalVeauX

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Posted 03 July 2020 - 10:01 AM

Hi! Just wondering, when imaging, is the maximum magnification the same with visual? That is, x2 of aperture in mm?

Or can I go higher? I will be using an ASI224MC to image Jupiter. Scope is Celestron Omni 150/750.

Thanks!

Hi,

 

As others have stated, they're not related. What you're looking to learn about is sampling. This is based on aperture (defines resolution potential), wavelength of light (angular resolution), focal length and pixel pitch (size of pixel). The assumptions are that seeing is perfect for the image scale and resolution potential of the system (usually its not, but with lucky imaging, we can assume as much for this purpose), and that we want the limit to be the airy disc. It's typical to want a few overlapping distances of diameters of airy discs to be able to differentiate a point light source or signal. There are many different theories and opinions on this, but this is where the "5 times the pixel size" rule comes from essentially, or you can calculate it for each wavelength and go for the value(s) that ideally sample or critically sample the wavelengths you're interested in.

 

I have attached an excel calculator with values already input for your setup (150mm aperture, 3.75um pixels). You can change values if you want to see how it works. The formula is included. Simply input values in the units labeled next to the boxes called input. Then hit enter or click somewhere and it will update the calculations. It's crude but works.

 

Since you are imaging in color, let's look at RGB (red, green and blue). That's red (650nm +/-), green (510nm +/-) and blue (475nm +/-). Since you are capturing all of these wavelengths in one capture with a color sensor, you want to sample ideally for the shortest wavelength which has the most angular resolution. The longer wavelengths have less angular resolution and will undersample a little, but that's ok, the point is to make sure you're at the critical sampling point or above it, and not below that point, so that you're getting the most recordable resolution for your system. Using those wavelengths and the imaging train parameters of your system: 150mm aperture, 3.75um pixel pitch, and those wavelengths we calculate that:

 

150mm aperture

Pixel pitch: 3.75um

650nm (red) will critically sample at F14.18 (F14)

510nm (green) will critically sample at F18.08) (F18)

475nm (blue) will critically sample at F19.41 (F19)

 

So to critically sample all wavelengths in your system of RGB, you need to target F19 essentially or the nearest value that you can realistically achieve, such as F20 as an easier target. This will critically sample all three wavelengths.

 

Since your scope is F5, that means you want a 4x barlow or extender/amplifier to take it to F20 with your 224MC.

 

Notice, if you were to use the "rule of thumb" of 5x the pixel size (3.75um x 5) it would give you F18.75 and it would be pretty close (F19 rounded up). So it does work essentially. But I wanted to help you with knowing why and how it gets there more specifically.

 

Very best,

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#6 Dartguy

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Posted 03 July 2020 - 10:30 AM

Hi,

 

As others have stated, they're not related. What you're looking to learn about is sampling. This is based on aperture (defines resolution potential), wavelength of light (angular resolution), focal length and pixel pitch (size of pixel). The assumptions are that seeing is perfect for the image scale and resolution potential of the system (usually its not, but with lucky imaging, we can assume as much for this purpose), and that we want the limit to be the airy disc. It's typical to want a few overlapping distances of diameters of airy discs to be able to differentiate a point light source or signal. There are many different theories and opinions on this, but this is where the "5 times the pixel size" rule comes from essentially, or you can calculate it for each wavelength and go for the value(s) that ideally sample or critically sample the wavelengths you're interested in.

 

I have attached an excel calculator with values already input for your setup (150mm aperture, 3.75um pixels). You can change values if you want to see how it works. The formula is included. Simply input values in the units labeled next to the boxes called input. Then hit enter or click somewhere and it will update the calculations. It's crude but works.

 

Since you are imaging in color, let's look at RGB (red, green and blue). That's red (650nm +/-), green (510nm +/-) and blue (475nm +/-). Since you are capturing all of these wavelengths in one capture with a color sensor, you want to sample ideally for the shortest wavelength which has the most angular resolution. The longer wavelengths have less angular resolution and will undersample a little, but that's ok, the point is to make sure you're at the critical sampling point or above it, and not below that point, so that you're getting the most recordable resolution for your system. Using those wavelengths and the imaging train parameters of your system: 150mm aperture, 3.75um pixel pitch, and those wavelengths we calculate that:

 

150mm aperture

Pixel pitch: 3.75um

650nm (red) will critically sample at F14.18 (F14)

510nm (green) will critically sample at F18.08) (F18)

475nm (blue) will critically sample at F19.41 (F19)

 

So to critically sample all wavelengths in your system of RGB, you need to target F19 essentially or the nearest value that you can realistically achieve, such as F20 as an easier target. This will critically sample all three wavelengths.

 

Since your scope is F5, that means you want a 4x barlow or extender/amplifier to take it to F20 with your 224MC.

 

Notice, if you were to use the "rule of thumb" of 5x the pixel size (3.75um x 5) it would give you F18.75 and it would be pretty close (F19 rounded up). So it does work essentially. But I wanted to help you with knowing why and how it gets there more specifically.

 

Very best,

Thanks for the calculator, MalVeauX!

 

So, for my C8 (2000) and Neximage 5 (2.2), I get F11.4.  Should I just use it without a barlow at F10?

 

Thanks!

 

Jon



#7 MalVeauX

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Posted 03 July 2020 - 10:37 AM

Thanks for the calculator, MalVeauX!

 

So, for my C8 (2000) and Neximage 5 (2.2), I get F11.4.  Should I just use it without a barlow at F10?

 

Thanks!

 

Jon

Hi Jon,

 

C8 (200mm aperture)

2.2um pixels (Neximage 5)

Blue (475nm) critically samples at F11.4 (F11)

Green (510nm) critically samples at F10.6 (F11)

Red (650nm) critically samples at F8.3 (F8)

 

So really the native F10 of your C8 is very close already to critically sampling the shorter wavelengths. You'd want to get it to F12 or so or even F15 to ensure you're completely past the critically sampling point. If you cared to. Otherwise, you are close enough that I wouldn't be too worried about imaging at F10 with that system without any extra glass, it's "close enough."

 

Very best,


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#8 paulgo12

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Posted 03 July 2020 - 08:04 PM

Whoa. Thank you for your answers! And thanks for the calculator, MalVeauX! 

Your answers are very helpful!

I'm a visual observer but I'm trying my hands on EAA and planetary imaging due to light pollution in my new home. 

Thanks everyone!


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