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Finding straight line distance between an ellipsoid and a point

  1. Oct 31, 2012 #1
    I have an ellipsoid representing the Earth (WGS84) and the current location of a spacecraft (somewhere above the surface). I am trying to find a method that allows me to calculate the straight line distance from the point to the surface of the ellipsoid.


    Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Oct 31, 2012 #2

    There are several ways to do it:

    1) Calculate the formula of the straightline passing through the spacecraft (point S) and the ellipsoid's center, and then find out

    the intersection point P between that line and the ellipsoid. Finally, just calculate the distance between S and P

    2) Take a generic point on the ellipsoid ( according to its formula ) and find its distance to point S. Use now differential calculus to find

    the minimum of the distance function you got.

    DonAntonio
     
  4. Oct 31, 2012 #3

    D H

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    Method 1 is incorrect. Method 2 doesn't help much.

    Method 1 is the starting point of the iterative Bowring algorithm, see Borkowski (1989). The Bowring algorithm and an improved version are easy to code up, but they're iterative. There are also a couple of closed form solutions. These aren't quite so easy to encode but they closed form. See Rey-Jer You (2000), W. E. Featherstone and S. J. Claessens (2008).

    K. M. Borkowski (1989), "Accurate algorithms to transform geocentric to geodetic coordinates", Bulletin Géodésique (Journal of Geodesy), 63:1

    Rey-Jer You (2000), "Transformation of Cartesian to Geodetic Coordinates Without Iterations", Journal of Surveying Engineering, 126:1

    W. E. Featherstone and S. J. Claessens (2008), "Closed-form transformation between geodetic and ellipsoidal coordinates", Studia Geophysica et Geodaetica, 52:1

    All of the cited papers can be found online, but I don't know if they are legit links (don't violate copyright), so you'll have to find them yourself if you are interested.
     
  5. Oct 31, 2012 #4
    Well both the ellipsoid and the spacecraft position must be specified in the same coordinate system. Then you can perform the calculation in that coordinate system.
     
  6. Oct 31, 2012 #5

    haruspex

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    Using a Lagrange multiplier, λ, you can easily do the calculus to get [itex]x = \frac{x'}{1+\lambda/a^2}[/itex] etc. where (x', y', z') is the position of the craft and the equation of the ellipsoid is [itex]\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1[/itex]. Substituting back for x, y, z in the ellipsoid equation produces a cubic in λ.
    Note that this is to get the nearest point on the ellipsoid. If the ellipsoid has uniform density this will not be the same as the gravitational height above the surface. In general, the force of gravity will not be towards the nearest point. OTOH, neither will it be towards the centre of the ellipsoid.
     
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