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
Quote:
Originally Posted by PKay
Joes reply:
FOV (degrees) = (114.6 x sensor size) / (2 x focal length)
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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
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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??
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