Figuring magnification with a DSLR or webcam

[Kirscovitch]

Is there some sort of formula for figuring the magnification with a DSLR or webcam? I know for figuring the mag with an EP is FL/ EP length. So I imagine that there is a similar one for a camera. 

When showing some of my DSO shots to friends they always want to know what magnification it is and I always have to tell them I don't know.

What would the mag be with my C6N (750mm) and a Canon 1100D?

[wxsatuser]

Try this.....

http://www.astro.shoregalaxy.com/dslr_calc.htm

[ronin]

There isn't a magnification.

The image has a size in mm.

This is related directly to the focal length of the scope and the anglular size of the object.

Double the focal length and you double the image size.

If you had something like M42 or M45 both are about 1degree and a focal length scope of 600mm, you get:

Tan(1) = I/F,  where I - image size and F= focal length

so Tan(1)*F = I, the other way: I = Tan(1)*F

I = ~10.47mm

Most are a fair bit smaller then 1 degree across so you get a smaller image.

The moon would come out about 3.34mm across.

[idigitize]

I don't think its a number you can really say,

You could say the projected size on the chip is x (probably about 30x) but you don't show them the chip to look at.

The magnification would vary depending on the media you used to show them on. Ie.. A monitor, how big are the pixels, how big is your monitor, are you viewing it 100% etc. If you showed them a print, how big is the print, 7x5, a4, or an a1 poster print. I think it becomes a very complicated subject very quickly.

I found telling them how old the object is, changed the subject quickly

[Kirscovitch]

There isn't a magnification.

The image has a size in mm.

This is related directly to the focal length of the scope and the anglular size of the object.

Double the focal length and you double the image size.

If you had something like M42 or M45 both are about 1degree and a focal length scope of 600mm, you get:

Tan(1) = I/F,  where I - image size and F= focal length

so Tan(1)*F = I, the other way: I = Tan(1)*F

I = ~10.47mm

Most are a fair bit smaller then 1 degree across so you get a smaller image.

The moon would come out about 3.34mm across.

I appreciate the help but that just confused me even more lol!  

Try this.....

http://www.astro.shoregalaxy.com/dslr_calc.htm

And maybe I just didn't put in the specs correctly but the results were NA.

[Kirscovitch]

I don't think its a number you can really say,

You could say the projected size on the chip is x (probably about 30x) but you don't show them the chip to look at.

The magnification would vary depending on the media you used to show them on. Ie.. A monitor, how big are the pixels, how big is your monitor, are you viewing it 100% etc. If you showed them a print, how big is the print, 7x5, a4, or an a1 poster print. I think it becomes a very complicated subject very quickly.

I found telling them how old the object is, changed the subject quickly

Good point Ben. In the past whenever I showed anyone something through the EP they wanted to know what mag it was at. Mostly the people I would show have some experience with hunting and are familiar with hunting scopes where the magnification is set or somewhat variable (ie 3-9x)

[sharkmelley]

In microscopy you can print the image of a flower or fly on a piece of paper and the concept of magnification makes sense.  For instance if the fly is 2mm long and its image on paper is 200mm then that is 100x magnification.  In telescopy if you print an image of Jupiter 200mm diameter on a piece of paper then you have shrunk it quite considerably from its true size!!

However, the disc of Jupiter on paper (or on the screen) held at arm's length, say, appears much larger then the bright dot you see in the sky.  So maybe you can compare the size of Jupiter in the image at arm's length with the size of Jupiter you see through a powerful telescope at a certain magnification - it's all about the angle the image subtends at the eye.  But the equation is not easy - it depends on imaging scope focal legth, pixel size of camera, the size of each pixel on the display device and the distance the display device is held from the eye.

But if you really want it, I'm sure I could come up with the formula!

Mark

[brantuk]

With imaging it's not so much a case of figuring the magnification, but rather a case of working out the image scale for your camera chip - so you need to know the width and depth of the camera sensor and the focal length of the scope so you can image at the correct field of view for it. The formulae for this are:

FOV width = 2*ATAN(sensor width) / 2* focal length))

FOV depth = 2*ATAN(sensor depth) / 2*focal length))

Where the sensor width and focal length are in mm and the result is in degrees.

I'm just starting this imaging stuff myself and those formulae are lifted straight from MEPC which I've been reading avidly recently. Alternatively and for the easy way out, you can just pop your measurements into a point and click FOV calculator like CCDCalc or 12DString and it will give you the result for any camera/scope combination.

http://www.newastro.com/book_new/camera_app.php

http://www.12dstring.me.uk/fov.htm

Both free to download - but CCDCalc runs independent of the internet. Hth