#### FOV Equations Question stargatecolorado

Patrick,

I have a spreadsheet that allows me to calculate the field of view for my different equipment configurations.  I have been comparing it to the results I get from CduC.  At long focal lengths, my values are almost identical to CduC, but when I move towards shorter focal lengths, my values begin to vary quite a bit.  That tells me your equations are different (and probably more complex) from mine. And, I'm assuming mine are incorrect because I have also tested my equations against an online FOV calculator.  The online version differs slightly from yours, but to a very small degree.

The Excel equations I use are as follows: FOV (H or V in Minutes) = DEGREES(2*ARCTAN((<HS or VS in mm>/2)/<FL in mm>))*60, where
* HS is the horizontal dimension of the sensor chip
* VS is the vertical dimension of the sensor chip
* FL is the focal length of the scope or camera lens

I have a book on three-dimensional trigonometry, but since the sensor is flat, it seemed logical to use two-dimensional trigonometry.  That may be where I'm going wrong.

Any help you could provide would be greatly appreciated.  If you would rather discuss this off-line, let me know and I'll send you an email. Patrick Chevalley

CdC use a similar formula, but first compute the FOV for a single pixel then multiply by the number of pixel.
This is very similar for long focal but give significant difference for short focal.
But for short focal both are wrong. The CdC method will be right for a curved sensor and your method for an uncorrected lens with extreme coma.

The only way to get a realistic FOV in this case (and the best in all case) is to take an image of a star field and plate-solve.
If you solve with CCDciel/ASTAP you can directly send the solved image frame to CdC. This define the "rectangle" number 10 with the exact FOV and rotation.

Patrick Michael Poxon

Hi, somewhere you would need to take into account the convergence of RA values away from the celestial equator? Just a thought.
Mike

On Sat, Jun 25, 2022 at 5:29 AM stargatecolorado <eponline01@...> wrote:
Patrick,

I have a spreadsheet that allows me to calculate the field of view for my different equipment configurations.  I have been comparing it to the results I get from CduC.  At long focal lengths, my values are almost identical to CduC, but when I move towards shorter focal lengths, my values begin to vary quite a bit.  That tells me your equations are different (and probably more complex) from mine. And, I'm assuming mine are incorrect because I have also tested my equations against an online FOV calculator.  The online version differs slightly from yours, but to a very small degree.

The Excel equations I use are as follows: FOV (H or V in Minutes) = DEGREES(2*ARCTAN((<HS or VS in mm>/2)/<FL in mm>))*60, where
* HS is the horizontal dimension of the sensor chip
* VS is the vertical dimension of the sensor chip
* FL is the focal length of the scope or camera lens

I have a book on three-dimensional trigonometry, but since the sensor is flat, it seemed logical to use two-dimensional trigonometry.  That may be where I'm going wrong.

Any help you could provide would be greatly appreciated.  If you would rather discuss this off-line, let me know and I'll send you an email. stargatecolorado

Patrick,

Thanks for the clarification! stargatecolorado

Mike,

The FOV rectangle is simply a rectangle projected on to the celestial sphere; it is the same size regardless of where on the sphere it appears.  That's why the rectangle calculations have no reference to coordinates.  The center of the rectangle will be placed at the coordinates of the selected object, but it's position will not change  the size of the rectangle.

I hope that helps. Michael Poxon

ah yes, I wondered why there was no reference to coordinates - and presumably the actual size of the FOV can be given in declination degrees anyway (which dont change with declination of course)

On Mon, Jun 27, 2022 at 7:27 AM stargatecolorado <eponline01@...> wrote:
Mike,

The FOV rectangle is simply a rectangle projected on to the celestial sphere; it is the same size regardless of where on the sphere it appears.  That's why the rectangle calculations have no reference to coordinates.  The center of the rectangle will be placed at the coordinates of the selected object, but it's position will not change  the size of the rectangle.

I hope that helps.