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3D Geometry: Calculate the position of the north star

  1. Jan 31, 2012 #1

    ElijahRockers

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    Gold Member

    1. The problem statement, all variables and given/known data

    Ok I'm able to track one star in the sky, over a period of one hour. We use three measurements to find the angle of the arc traced by the star. The three measurements also constitute two vectors. We can take the cross product of those vectors and it will give us a vector that is perpendicular to both of those, and hence points in the direction of the north star. Divide by its magnitude for a unit vector. Then, we can use the formula [itex]z=\rho*cos\varphi_{c}[/itex] Where rho is arbitrary (in this case, 1) and phi is the colatitude of the north star.

    2. Relevant equations

    Measurement 1: Longitude=0, Colatitude=45
    Measurement 2: Longitude=-8.840 Colatitude=48.729
    Measurement 3: Longitude=-15.561 Colatitude=51.95

    [itex]x=sin\varphi_{c}cos\theta[/itex]
    [itex]y=sin\varphi_{c}sin\theta[/itex]
    [itex]z=cos\varphi_{c}[/itex]

    3. The attempt at a solution

    Using the right hand rule, I deduce that M2M1 would be vector A and M2M3 should be vector B. AxB should point fron the origin (earth) in the direction of the north star.

    AxB/||AxB|| = <-.24957, -.43297, .8662>

    These are the euclidean coordinates for the north star, but the answer is needed in sphereical coords, [itex](\theta,\varphi)[/itex]

    I get the correct latitude, 60°, but the longitude (which is supposed to be 240°) does not come out right. Instead I get 300°.

    Any help would be much appreciated...

    OK. When I use the formula [itex]x=sin\varphi_{c}cos\theta[/itex] I get 120 degrees for theta, and I notice that 360-120 is 240. What am I missing here? When I use [itex]y=sin\varphi_{c}sin\theta[/itex] I get like -60°. What's going on?
     
    Last edited: Jan 31, 2012
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