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

In summary, 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), you need to compute the distance d of the object from the center using the formula d = (RA^2+dec^2)^0.5 and check that d is less than R. However, this may not be accurate for small angular separations, as a change in RA gets smaller as you get nearer to the poles. A more accurate formula is sqrt(δdec^2 + cos^2(dec)*δRA^2), which can be found in the provided reference.
  • #1
TL;DR 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 !
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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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  • #5
Great, thanks !
 

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