FOV calculations

[PKay (Peter)]

The original question was (edited):

Does any one know how to calculate the FOV from the specs?


For example, with my Orion EON:
Aperture 85 mm, focal length 560 mm, f-ratio 6.6
The FOV seen on the screen is about 3.4 X 2.6 degrees.

With the Skywatcher Esprit 150mm
Aperture 150 mm, focal length 1050 mm, f-ratio 7

What would be the FOV you end up with (aside from the answer...smaller)??

I suppose that the term 'field radius' might come into it as well??

I just searched astrobin for Esprit 150mm.
The average field radius was 0.5 - 0.6 degrees.

[PKay (Peter)]

Joes reply:

FOV (degrees) = (114.6 x sensor size) / (2 x focal length)

-------------------------------------------------------------------

Worked example:

ZWOASI1600MC has a sensor size of 17.6 x 13.3mm.
My focal length (refractor) = 560mm.

I converted mm to meters (just guessing the right thing to do).

Long dimension of FOV:

(114.6 x 0.0176) / (2 x 0.560) = 1.8 degrees

--------------------------------------------------------------------------

The answer above (1.8 degrees).
Measured dimension taken from my images is 3.4 degrees.

Not sure about this. Might be the field flattener??

[OzEclipse (Joe Cali)]

Peter,

I fixed this in my other post. I first wrote a trig formula that works for almost anything, then changed my mind and typed a simplified formula that only works for focal lengths longer than about 100mm that doesn't use trig.

Seeing any trig function makes some people switch off and you need to determine if your computer program/ app or calculator is working in degrees or radians. Excel uses radians.

Unfortunately I didn't edit the formula properly and left the formula looking a bit unwieldy but nonetheless correct. With the simplification to eliminate the trig, it should still be within 0.03 degrees of the true field at your focal length.

Here is an even simpler version of the same formula : -
FOV (degrees) = (57.3 x sensor size/focal length)

The full trig version is
FOV = 2*arctan(sensor dimension÷(2*focal length))

If you use EXCEL to do this, the result will be in radians not degrees. To convert, multiply the result by 57.3. If you use a scientific calculator, it can be set to degrees.

I can't explain your own scope result but the formula works. If your scope uses a reducer flattener then yes the focal length is shorter and the field would be wider. Even so a factor of 2 seems a bit extreme.

For the Esprit 150mm at f7
FL = 1000mm Field of view with your camera = 1.0 x 0.8 degrees


For the Esprit 150mm at f5.25
FL = 787mm Field of view with your camera = 1.3 x 1.0 degrees

regards

Joe

Originally Posted by PKay

Joes reply:

FOV (degrees) = (114.6 x sensor size) / (2 x focal length)

-------------------------------------------------------------------

Worked example:

ZWOASI1600MC has a sensor size of 17.6 x 13.3mm.
My focal length (refractor) = 560mm.

I converted mm to meters (just guessing the right thing to do).

Long dimension of FOV:

(114.6 x 0.0176) / (2 x 0.560) = 1.8 degrees

--------------------------------------------------------------------------

The answer above (1.8 degrees).
Measured dimension taken from my images is 3.4 degrees.

Not sure about this. Might be the field flattener??

[Merlin66 (Ken)]

Why not just enter your data into CCDCalc, it gives all the answers...
http://www.newastro.com/book_new/camera_app.html

This shows a FOV of 82 x 108 arc min (1.36 x 1.8 deg)
(For the 150/ f7 the FOV would be 44 x 58 arc min)

EDIT: to achieve your 3.4 deg field, you must have a reducer fitted which brings the EON focal ratio down to f3.5...

[PKay (Peter)]

Thanks heaps for the info.

My personal calculations for my FOV were way off. Don't know what I was thinking.

Ron Wodaski's CCD calculator is amazing!
It agreed very well with images I have taken of the same targets.

It turns out my Orion EON refractor 85mm , F/L 560mm (with ZWO ASI1600) has a FOV 81.9 x 108.3 arc minutes.

This agrees with Joe's formula.

Very clever Joe!

It also means I need another scope

[Atmos (Colin)]

When I am doing my calculations I use:
Pm= Pixel size in microns
FL= Your focal length
Px= Number of pixels in X axis
Py= Number of pixels in Y axis

((((Pm*206.265))/FL)*Px)/60

So with my QHY163 and refractor it would be:
(3.8*206.265)/677= 1.1577
(1.1577*4656)/60= 89.84 arcminutes
(89.84/4656)*3522= 67.96 arcminutes

[Merlin66 (Ken)]

Colin,
Yeah but....
It's much easier just to plug the numbers into CCDCalc for each telescope/ lens, camera combo ;-)

[Atmos (Colin)]

Not when you’re on the run without a computer
That and it takes me less than 10s

[rustigsmed (Russell)]

i like using this one
https://astronomy.tools/calculators/field_of_view/
nice and easy

[PKay (Peter)]

It just keeps getting better.

Astronomy.tools link from Russ is unreal!

thanks brainy ones :-)

[JA]

1. Punch in the focal length
2. Select the Sensor type & Size (even custom size)
3. Hit Compute ........

https://www.scantips.com/lights/fieldofview.html

Best
JA

[Atmos (Colin)]

And if you want one on your iPhone/iPad, I’ve been using CCDFieldCalc for a few years. It was released by ATIK so it only has their cameras but it’s pretty cool.