I Object in or out of a circular field of view? (celestial coordinate system)

AI Thread Summary
In celestial coordinate systems, determining if an object with right ascension (RA) and declination (dec) is within a circular field of view of radius R requires a specific calculation. For small angular separations, the angular distance should be computed using the formula sqrt(δdec^2 + cos^2(dec)*δRA^2), rather than a simple Cartesian distance. This is important because the change in RA varies with declination, especially near the poles. The discussion also references a resource for further understanding of the formula and its derivation. Accurate calculations are essential for precise celestial navigation and observation.
vladivostok
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Check that object with position RA and dec is within circle of radius R (in arcminute) ?
In celestial coordinate system (right ascension/declination), how to check if an object with position RA and dec is within a given circular field of view of radius R (in arcminutes) and centred at (0,0)?
R is small in this case so I assumed that I could compute the distance d of the object from the center as in cartesian coordinates: d = (RA^2+dec^2)^0.5 and check that d is less than R. Is that correct (at least for small angular separation) ?

Thanks !
 
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vladivostok said:
Summary:: Check that object with position RA and dec is within circle of radius R (in arcminute) ?

In celestial coordinate system (right ascension/declination), how to check if an object with position RA and dec is within a given circular field of view of radius R (in arcminutes) and centred at (0,0)?
R is small in this case so I assumed that I could compute the distance d of the object from the center as in cartesian coordinates: d = (RA^2+dec^2)^0.5 and check that d is less than R. Is that correct (at least for small angular separation) ?

Thanks !
Not really. A change δdec is the same size everywhere, but a change δRA gets smaller as you get nearer the poles. So for small changes, the angular separation between two objects is give by sqrt(δdec^2 + cos^2(dec)*δRA^2). Of course this won't make much difference if you are at (0,0), which is on the equator.
 
Thanks a lot and sorry for the late reply. Do you have any reference for the formula that you give? I'd like to see how it is derived.
Thanks again.
 
vladivostok said:
Thanks a lot and sorry for the late reply. Do you have any reference for the formula that you give? I'd like to see how it is derived.
Thanks again.
Here's a good link. The formula I gave is down near the bottom of the page. It also has more general formulae for when the differences of the coordinates are not small.
http://spiff.rit.edu/classes/phys373/lectures/radec/radec.html
 
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Great, thanks !
 
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