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Latitude and Longitude, GPS Conversion


Date: 12/07/2001 at 15:12:47
From: M.S.
Subject: GPS conversion

Hello,

What is the equation to convert latitude/longitude/altitude (LLA)
data into earth-centered/earth-fixed (ECEF) data? 

Thanks.


Date: 12/08/2001 at 11:07:30
From: Doctor Fenton
Subject: Re: GPS conversion

Dear M.S.,

Thanks for writing to Dr. Math.  

I'm assuming that "altitude" in your question refers to "altitude 
above the reference ellipsoid." If you are obtaining your altitude 
from a GPS receiver, I believe that this would be the correct 
interpretation. Many altitude specifications are "altitude above mean 
sea level," however, and this can make a difference of up to nearly 
100 meters. You would have to get information on the height of the 
mean sea level surface (called the "geoid") relative to the reference 
ellipsoid in the area of interest. The defining document for WGS84 
(World Geodetic System 1984 - the Department of Defense geodetic 
coordinate system used by GPS) gives a geoid map showing contours of 
geoid height and a table of geoid heights on a latitude-longitude 
grid.

The basic formulas for converting from latitude, longitude, altitude
(above reference ellipsoid) to Cartesian ECEF are given in the 
Astronomical Almanac in Appendix K.  They depend upon the following
quantities:

     a   the equatorial earth radius
     f   the "flattening" parameter ( = (a-b)/a ,the ratio of the
         difference between the equatorial and polar radii to a;
         this is a measure of how "elliptical" a polar cross-section
         is).

The eccentricity e of the figure of the earth is found from
     
    e^2 = 2f - f^2 ,  or  e = sqrt(2f-f^2) .

For WGS84, 

         a   = 6378137 meters
       (1/f) = 298.257224

(the reciprocal of f is usually given instead of f itself, because the
reciprocal is so close to an integer)

Given latitude (geodetic latitude, not geocentric latitude!), compute

                                  1
        C =  ---------------------------------------------------
             sqrt( cos^2(latitude) + (1-f)^2 * sin^2(latitude) )

and
        S = (1-f)^2 * C .

Then a point with (geodetic) latitude "lat," longitude "lon," and 
altitude h above the reference ellipsoid has ECEF coordinates

       x = (aC+h)cos(lat)cos(lon)
       y = (aC+h)cos(lat)sin(lon)
       z = (aS+h)sin(lat) 

The Almanac also gives an iterative procedure for the inverse
conversion from ECEF to Lat/lon/alt, although there is a method due
to Bowring that may be better, and the Almanac also gives a reference 
to a closed form (non-iterative) method due to Borkowski.

If you need any further information, please write us again.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/   


Date: 11/12/2002 at 19:41:14
From: George Thomas
Subject: GPS conversion

Dear Dr. Math,

You stated the equation for C to be as follows:

                              1
   C =  ---------------------------------------------------
        sqrt( cos^2(latitude) + (1-f)^2 * sin^2(latitude) )

I just want to verify that longitude is not necessary in the above 
equation.

Thanks,
GPS challenged


Date: 11/12/2002 at 21:38:00
From: Doctor Fenton
Subject: Re: GPS conversion

Hi George,

Thanks for writing to Dr. Math. 

The formula for C is indeed independent of longitude, because the
Earth is modeled as an (oblate) spheroid, which is a surface of
revolution. Every plane containing the polar axis cuts the spheroid
in exactly the same elliptical shape, so the expressions that 
determine the radius at a given latitude depend only upon the 
latitude, not the longitude.

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

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