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Find the perifocal

  1. Sep 23, 2013 #1
    In Kepler:
    Given: x0, y0, z0, vx0, vy0 and vz0

    I can find Longitude of the Ascending Node(N), inclination(i), Argument of Periapsis(w) and True Anomaly(TA).

    I can find the perifocal(the plane of the conic(ellipse, circle, hyperbola or parabola)) coordinates from the above.

    rp = x * (Cos(N) * Cos(w) - Sin(N) * Sin(w) * Cos(i)) + y * (Sin(N) * Cos(w) + Cos(N) * Sin(w) * Cos(i)) + z * (Sin(w) * Math.Sin(i))
    rq = x * (-Cos(N) * Sin(w) - Sin(N) * Cos(w) * Cos(i)) + y * (-Sin(N) * Sin(w) + Cos(N) * Cos(w) * Cos(i)) + z * (Cos(w) * Sin(i))
    rw = x * (Sin(N) * Sin(i)) + y * (-Cos(N) * Sin(i)) + z * (Cos(i))

    I get a different answer for the Perifocal if a use the Equations:
    rp = r*cos(TA)
    rq = r*sin(TA)
    rw = 0

    How can I reconcile?
     
    Last edited: Sep 23, 2013
  2. jcsd
  3. Sep 23, 2013 #2
    I solved it myself.

    rq = - r*sin(TA)
     
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