Finding straight line distance between an ellipsoid and a point

In summary: The centre of the ellipsoid is not a centre of gravity.In summary, there are various methods to calculate the straight line distance from a point to the surface of an ellipsoid representing the Earth, such as using the Bowring algorithm or closed form solutions. However, it is important to note that this distance may not necessarily be the same as the gravitational height above the surface due to the non-uniform density of the ellipsoid.
  • #1
Shadowsteps
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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.
 
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  • #2
Shadowsteps said:
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.


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
 
  • #3
DonAntonio said:
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
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.
 
  • #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.
 
  • #5
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.
 

FAQ: Finding straight line distance between an ellipsoid and a point

1. What is the formula for finding the straight line distance between an ellipsoid and a point?

The formula for finding the straight line distance between an ellipsoid and a point is:
d = √((x-x0)2 + (y-y0)2 + (z-z0)2)
where (x,y,z) is the coordinates of the point and (x0,y0,z0) is the center of the ellipsoid.

2. How is the center of the ellipsoid determined in this calculation?

The center of the ellipsoid is typically determined by the equation of the ellipsoid, which is of the form:
((x-x0)/a)2 + ((y-y0)/b)2 + ((z-z0)/c)2 = 1
where (x0,y0,z0) is the center and a, b, and c are the semi-axes of the ellipsoid.

3. Can this formula be used for any shape of ellipsoid?

Yes, the formula for finding the straight line distance between an ellipsoid and a point can be used for any shape of ellipsoid, as long as the center and semi-axes are known.

4. How does this calculation differ from finding the distance between a point and a sphere?

This calculation differs from finding the distance between a point and a sphere in that the ellipsoid is an oblate or prolate spheroid, meaning it is not perfectly round like a sphere. Therefore, the semi-axes of the ellipsoid are not equal, whereas the radius of a sphere is the same in all directions.

5. What are some practical applications of finding the straight line distance between an ellipsoid and a point?

This calculation has many practical applications, such as in geodesy for measuring distances on the Earth's surface, in navigation for calculating the distance between a ship or aircraft and a point on the Earth's surface, and in satellite positioning systems for determining the distance between a satellite and a point on the Earth's surface.

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