How to make random positions in Equatorial coordinate system?
If I divide Dec between -90 and 90 evenly, the space distribution is not random.
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asked Jun 25, 2016 at 16:25
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1 Answer
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Well, your problem is that you have a sphere. To compensate for the polar declination skew, you just calculate $$\sin^{-1}(\mathtt{rand})$$ Where $\mathtt{rand}$ is a number in an evenly divided -1 to 1 range. Alternatively, you can use another trigonometric function based on what you have available, but you get the idea.
RA can obviously be divided evenly.
answered Jun 25, 2016 at 17:39
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$\begingroup$ @hohmannfan You mean I need to make random Decs, then change them to sin-1(dec)? or sin-1(dec) is a weighing coefficient which correlates with area? $\endgroup$ Commented Jun 29, 2016 at 9:54
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$\begingroup$ @questionhang You need to make random numbers evenly in the -1 to 1 area, and pipe them through the sin-1 formula. The output can be used together with random RA's to get random coordinates on a sphere, without any skew. $\endgroup$ Commented Jun 29, 2016 at 10:52
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$\begingroup$ @@hohmannfan I do not understand my question now. What is a random distribution on a 2-D sphere? $\endgroup$ Commented Jul 1, 2016 at 16:23