How do you calculate camera magnification?

[Moonshed]

I understand that by dividing the focal length of the ‘scope by the focal length of the eyepiece we arrive at the amount  of magnification that combination gives.

Today I was asked what the magnification was of an image of Mars I took recently and I have no idea! I used a planetary camera and do not have a clue as to how to work it out.

Could somebody please tell me what the magic formula is?

Thanks in advance.

[AKB]

You don't.

For a camera, there are really two numbers of note:

• FOV – Field of view (in degrees / minutes / seconds of arc)
• Resolution (angular resolution per pixel)

FOV in radians = (chip dimension) / (focal length)

Resolution = FOV / (number of pixels in that camera dimension, ie. width or height)

That's all there is.

Tony

Edit: omitted to say that 1 radian ~ 57.3 degrees

Edited November 16, 2020 by AKB
Correct units

[vlaiv]

There is no magnification associated with camera.

Take the same image of mars, and look at it from 20cm - it will have one size, step back 20 meters - it will become much smaller.

Magnification (as in with eyepiece) of a telescope magnifies angles - keep eye close to eyepiece and start moving it away from eyepiece - planet size won't change.

Telescope + sensor is projection device - angles are converted into linear distance on sensor. Formula is rather simple - (distance on sensor from optical axis) / (focal length of telescope) = tan(angle)

Since most of the time we don't work with actual size of the sensor but rather with pixels on sensor (which also have size), here is formula for that:

(source: http://www.wilmslowastro.com/software/formulae.htm#ARCSEC_PIXEL)

This is basically the same as above - with the use of small angle approximation and angles being in arc seconds and not radians - hence constant. It gives you resolution of the image.

Your mars is 80px across? Your sampling rate is 0.25"/px? - well, total angular size of mars is 80px * 0.25"/px = 20"

This is how "magnification" works for camera sensors

[Gfamily]

As has been said, it's not really a valid statement to say "an image has a magnification" 

However, if you know the diagonal size of the sensor, you can say the field of view is roughly comparable to that size of eyepiece. 

So, for example, the ASI 120mm camera has a sensor size that's 4.8mm x 3.6mm, so has a diagonal of about 6mm. If you were imaging Mars with that camera, then it would have a similar field to that you would get with a 6mm eyepiece. 

Yes, it's not an exact match, but as a rule of thumb, it'll do  

[vlaiv]

1 minute ago, Gfamily said:

As has been said, it's not really a valid statement to say "an image has a magnification" 

However, if you know the diagonal size of the sensor, you can say the field of view is roughly comparable to that size of eyepiece. 

So, for example, the ASI 120mm camera has a sensor size that's 4.8mm x 3.6mm, so has a diagonal of about 6mm. If you were imaging Mars with that camera, then it would have a similar field to that you would get with a 6mm eyepiece. 

Yes, it's not an exact match, but as a rule of thumb, it'll do  

This is true - if diagonal of sensor is the same as diameter of field stop - one gets in principle same field of view (bar the fact that rectangle and circle are of different shape).

We can go on to argue that eyepiece with 6mm field stop will have such and such magnification - and there is definite relationship between field stop / magnification and apparent field of view, but this still won't provide us with magnification as image on the screen will depend on the way it has been presented - as well as size of the screen.

There is a way to answer someone's question: "How much magnification is this" - when they are watching image of the planet on the screen, but it is a bit more involved.

We need to know sampling rate at which planet was captured (arc seconds per pixel) and we also need to know DPI of display device and distance of observer to display device. From that - we can calculate apparent size of the planet in the sky and apparent size of the planet on the screen as seen by viewer.

Divide the two and it will give you magnification. Then you can safely answer: "If you were observing with a telescope on that particular night when image was taken, and were using X magnification - you would see the same planet size as you are seeing now".

[Moonshed]

Thank you Tony and vlaiv,

I just knew it would be dead easy stuff you could do in your head.

When I asked the question I didn’t realise the complexities involved with pixels, radians, arc seconds and mystical Druid symbols. That’ll teach me!

At least I know the answer to give now: “Magnification? Yes of course, it’s somewhere between humungous and stupendous, depending on the temperature of the chip’s pixels.” That should do it.

But seriously thank you so much for giving me the information, it’s much appreciated! It will take me a while to digest it, I may be gone some time...

 

Edited November 16, 2020 by Moonshed

[vlaiv]

4 minutes ago, Moonshed said:

Thank you Tony and vlaiv,

I just knew it would be dead easy stuff you could do in your head.

When I asked the question I didn’t realise the complexities involved with pixels, radians, arc seconds and mystical Druid symbols. That’ll teach me!

At least I know the answer to give now: “Magnification? Yes of course, it’s somewhere between humungous and stupendous, depending on the temperature of the chip’s pixels.” That should do it.

But seriously thank you so much for giving me the information, it’s much appreciated! It will take me a while to digest it, I may be gone some time...

 

Now, now, don't get disheartened.

I'm looking at your semi circular Mars from the other thread on my computer screen right now. I'm guessing it was captured couple of days ago when the Mars was 18 arc seconds in diameter. It measures 102px across.

This gives me resolution of ~ 0.1765"/px.

My computer screen has 96 dpi. We can also calculate it like this - it has diagonal of 23 inches and it is 1920 x 1080. There are sqrt(1920^2 + 1080^2) = 2203px along the diagonal that is 23" long, so 2203 / 23 = 95.78 DPI.

This means that single pixel is 25.4mm / 95.78 = ~0.2652mm

I'm viewing the screen from roughly 600mm of distance.

According to this calculator: http://www.1728.org/angsize.htm

angular size of single pixel on my screen from this distance is  91.169 arc seconds. Image of Mars has 102px across - this means that it totals at 91.169 x 102 = ~ 9300"

And finally, I'm seeing object that is in reality 18" across to be 9300" across, so my magnification at the moment is 9300 / 18 = x516.666

Mars looks to me as if I was observing it with a telescope on the night it was recorded and I used x516.666 magnification whilst doing it.

I feel it is too much magnification - I'll step away from my computer screen now to at least double that distance for comfortable ~ x260 magnification

[Moonshed]

1 hour ago, vlaiv said:

Now, now, don't get disheartened.

I'm looking at your semi circular Mars from the other thread on my computer screen right now. I'm guessing it was captured couple of days ago when the Mars was 18 arc seconds in diameter. It measures 102px across.

This gives me resolution of ~ 0.1765"/px.

My computer screen has 96 dpi. We can also calculate it like this - it has diagonal of 23 inches and it is 1920 x 1080. There are sqrt(1920^2 + 1080^2) = 2203px along the diagonal that is 23" long, so 2203 / 23 = 95.78 DPI.

This means that single pixel is 25.4mm / 95.78 = ~0.2652mm

I'm viewing the screen from roughly 600mm of distance.

According to this calculator: http://www.1728.org/angsize.htm

angular size of single pixel on my screen from this distance is  91.169 arc seconds. Image of Mars has 102px across - this means that it totals at 91.169 x 102 = ~ 9300"

And finally, I'm seeing object that is in reality 18" across to be 9300" across, so my magnification at the moment is 9300 / 18 = x516.666

Mars looks to me as if I was observing it with a telescope on the night it was recorded and I used x516.666 magnification whilst doing it.

I feel it is too much magnification - I'll step away from my computer screen now to at least double that distance for comfortable ~ x260 magnification

Don’t get disheartened he says, then proceeds to explain why the 11 dimensions of the space time continuum combined with string theory is the reason why a solution to the seeming disparities between Quantum Mechanics and Relativity may lie within the merging of chaos theory and Mandelbrot sets of infinite complexity. 😂

Personally I think the magnification of my Mars image is somewhere between 10 and 1,000  depending on how far you sit from your computer screen and whether you use metric or imperial measurements.

Vlaiv, I bet you’re a Grand Master chess player 🏆🥇🏅🎖🏅

[vlaiv]

8 hours ago, Moonshed said:

Vlaiv, I bet you’re a Grand Master chess player

I would possibly call myself just a fairly good player, but I have not played in a long time ...

[Moonshed]

9 hours ago, vlaiv said:

I would possibly call myself just a fairly good player, but I have not played in a long time ...

I bet you watched “Queen’s Gambit” though.

[vlaiv]

34 minutes ago, Moonshed said:

I bet you watched “Queen’s Gambit” though.

No, but I have seen it being advertised and planned on watching it sometimes.

[Moonshed]

8 minutes ago, vlaiv said:

No, but I have seen it being advertised and planned on watching it sometimes.

Make sure you do, it’s very good and you will love it!

[JamesF]

I really think it's best to avoid the entire topic of magnification when considering imaging.  I've struggled to convince myself that magnification is useful in any way as an absolute value whether imaging or visual.  As a means of comparing different optical configurations it might be handy, but if someone says they're viewing Mars at 100x magnification does the information actually have any useful meaning beyond that?

James

[tooth_dr]

On 16/11/2020 at 22:03, vlaiv said:

Your mars is 80px across? Your sampling rate is 0.25"/px? - well, total angular size of mars is 80px * 0.25"/px = 20"

This is how "magnification" works for camera sensors

Vlad, I've just been reading on this thread, and trying to work this out. Would you be able clarify if I'm right here>

If I now use a 2x barlow:

Mars is now 160px across.

Sampling rate is now 0.125"/px.

 

So angular size of Mar is 160px x 0.125"/px = 20"

It remains unchanged?

[JamesF]

15 minutes ago, tooth_dr said:

Vlad, I've just been reading on this thread, and trying to work this out. Would you be able clarify if I'm right here>

If I now use a 2x barlow:

Mars is now 160px across.

Sampling rate is now 0.125"/px.

 

So angular size of Mar is 160px x 0.125"/px = 20"

It remains unchanged?

Seems perfectly reasonable to me.  The only way the angular diameter of Mars can change is if it gets closer or further away.

James

[vlaiv]

37 minutes ago, tooth_dr said:

Vlad, I've just been reading on this thread, and trying to work this out. Would you be able clarify if I'm right here>

If I now use a 2x barlow:

Mars is now 160px across.

Sampling rate is now 0.125"/px.

 

So angular size of Mar is 160px x 0.125"/px = 20"

It remains unchanged?

Yep, like @JamesF noted above - using 2x barlow is like using x2 longer focal length - sampling rate increases and number of pixels increases.

[tooth_dr]

29 minutes ago, JamesF said:

Seems perfectly reasonable to me.  The only way the angular diameter of Mars can change is if it gets closer or further away.

James

Thanks James, that makes perfect sense when you say it like that.

[tooth_dr]

7 minutes ago, vlaiv said:

Yep, like @JamesF noted above - using 2x barlow is like using x2 longer focal length - sampling rate increases and number of pixels increases.

Thanks Vlad, makes total sense now.