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Galactic distance equation

  1. Jan 13, 2009 #1
    I'm currently working on my final year project, and one of the little bits to do is to see if certain data points fall within a circle with my own defined radius and central co-ordinates. I've been given the equation to use:

    [tex]d = \frac{\sqrt{(l-l_{0})^2 + (b-b_{0})^2}}{r}[/tex]

    where [tex]l_{0}[/tex] and [tex] b_{0}[/tex] are my central galactic co-ordinates, r is my radius (in degrees) and d is the distance the data point is from the central co-ordinates (also in degrees) - if d is less than 1, then the point is within my radius.

    Although I don't need to, I'd just like to know where the equation comes from! I can see how the right hand side of the equals is a rearrangement of the equation of a circle, but I don't see how the distance is put in. Any help is appreciated!
  2. jcsd
  3. Jan 13, 2009 #2


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  4. Jan 13, 2009 #3
    I realised that it must be Pythagoras, as that's where the equation for a circle comes from, it was more the fact that - when rearranged from the equation - the hypotenuese would be [tex]r^2d^2[/tex]. Since it is normally just [tex]r^2[/tex], I wasn't sure if you were able to just put the distance ([tex]d^2[/tex]) in.
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