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Mean Latitude/Longitude

Date: 07/10/2003 at 14:21:30
From: Peter Richard
Subject: Spherical Trigonometry

If we have three points on the earth measured in latitude/longitude, 
such as 'A'= 33S54;151E12 / 'B'= 37S49;144E58 / 'C'= 51N30;0W10 
what is the formula to calculate a mean latitude/longitude for this 
group of 3?

When I say 'mean latitude/longitude' I am referring to a 'midpoint in 
space' between the three. So perhaps if I used the term 'equidistant' 
rather than 'mean' it might make it clearer.


Date: 07/15/2003 at 14:47:57
From: Doctor Rick
Subject: Re: Spherical Trigonometry

Hi, Peter.

I might approach this as follows: 

Convert the lat-long positions into x,y,z coordinates, find the mean 
position in 3-dimensional space, and project this point back onto the 
sphere, and convert back to latitude and longitude. (I am assuming - 
and hoping - that a spherical approximation to the earth's surface 
is sufficient.)

This method will give the correct result in the "flat-earth 
approximation" in which the points are close enough together that the 
curvature of the earth can be ignored. It seems like a reasonable 
operational definition of an average position on the earth, but I'm 
being rather intuitive at this point.

To convert each location to (x,y,z) coordinates, use the formulas

  x_n = cos(lat_n)*cos(lon_n)
  y_n = cos(lat_n)*sin(lon_n)
  z_n = sin(lat_n)

where location n has latitude lat_n and longitude lon_n. I have set 
the radius of the earth equal to 1 since we aren't interested in 
actual distances; the radius would cancel out eventually anyway.

Average the coordinates independently:

  x = (x_1 + x_2 + x_3)/3
  y = (y_1 + y_2 + y_3)/3
  z = (z_1 + z_2 + z_3)/3

Now convert to latitude and longitude using the inverse transformation

  r = sqrt(x^2 + y^2 + z^2)
  lat = arcsin(z/r)
  lon = arctan(y/x)

I haven't tested this, so let me know if you have any trouble with 
it. If you're programming, you should use the atan2() function found 
in many programming languages, spreadsheets etc. Otherwise you'll 
have to do some extra stuff if x = 0 or if the longitude is more than 
90 degrees from the Prime Meridian.

The "mean" in any ordinary sense can be quite different from the 
point that is equidistant from three points. To demonstrate, draw a 
circle and pick three points on the circle, all on the right side. 
The point that is equidistant from the three points is the center of 
the circle. I doubt that this is what you mean by "mean"! (However, 
it could be what you want: If you expect a structure to be circular 
and you have found three points on the edge of the structure, you 
could use such a method to find the center of the structure.)

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Higher-Dimensional Geometry
High School Trigonometry

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