Converting from longitude latitude to Cartesian coordinates

Question

I have some earth-centered coordinate points given as latitude and longitude  WGS-84    How can i convert them to Cartesian coordinates  x y z  with the origin at the center of the earth

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You can do it this way on Java   public List lt Double gt  convertGpsToECEF double lat  double longi  float alt         double a 6378 1      double b 6356 8      double N      double e  1- Math pow b  2  Math pow a  2        N  a  Math sqrt 1 0- e Math pow Math sin Math toRadians lat    2          double cosLatRad Math cos Math toRadians lat        double cosLongiRad Math cos Math toRadians longi        double sinLatRad Math sin Math toRadians lat        double sinLongiRad Math sin Math toRadians longi        double x   N 0 001 alt  cosLatRad cosLongiRad      double y   N 0 001 alt  cosLatRad sinLongiRad      double z    Math pow b  2  Math pow a  2   N 0 001 alt  sinLatRad       List lt Double gt  ecef  new ArrayList lt  gt         ecef add x       ecef add y       ecef add z        return ecef

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In python3 x it can be done using       Converting lat long to cartesian import numpy as np  def get cartesian lat None lon None       lat  lon   np deg2rad lat   np deg2rad lon      R   6371   radius of the earth     x   R   np cos lat    np cos lon      y   R   np cos lat    np sin lon      z   R  np sin lat      return x y z

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Theory for convert GPS WGS84  to Cartesian coordinates https   en wikipedia org wiki Geographic coordinate conversion From geodetic to ECEF coordinates  The following is what I am using    Longitude in GPS WGS84  and Cartesian coordinates are the same    Latitude need be converted by WGS 84 ellipsoid parameters semi-major axis is 6378137 m  and    Reciprocal of flattening is 298 257223563    I attached a VB code I wrote   Imports System Math   Input GPSLatitude is WGS84 Latitude h is altitude above the WGS 84 ellipsoid  Public Function GetSphericalLatitude ByVal GPSLatitude As Double  ByVal h As Double  As Double          Dim A As Double   6378137  semi-major axis          Dim f As Double   1   298 257223563   1 f Reciprocal of flattening         Dim e2 As Double   f    2 - f          Dim Rc As Double   A    Sqrt 1 - e2    Sin GPSLatitude   PI   180    2            Dim p As Double    Rc   h    Cos GPSLatitude   PI   180          Dim z As Double    Rc    1 - e2    h    Sin GPSLatitude   PI   180          Dim r As Double   Sqrt p   2   z   2          Dim SphericalLatitude As Double    Asin z   r    180   PI         Return SphericalLatitude End Function   Please notice that the h is altitude above the WGS 84 ellipsoid   Usually GPS will give us H of above MSL height   The MSL height has to be converted to height h above the WGS 84 ellipsoid by using the geopotential model EGM96  Lemoine et al  1998   This is done by interpolating a grid of the geoid height file with a spatial resolution of 15 arc-minutes   Or if you have some level professional GPS has Altitude H  msl heigh above mean sea level  and UNDULATION the relationship between the geoid and the ellipsoid  m  of the chosen datum output from internal table  you can get h   H msl    undulation    To XYZ by Cartesian coordinates   x   R   cos lat    cos lon   y   R   cos lat    sin lon   z   R  sin lat

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Here s the answer I found   Just to make the definition complete  in the Cartesian coordinate system    the x-axis goes through long lat  0 0   so longitude 0 meets the equator     the y-axis goes through  0 90   and the z-axis goes through the poles     The conversion is   x   R   cos lat    cos lon   y   R   cos lat    sin lon   z   R  sin lat    Where R is the approximate radius of earth  e g  6371 km    If your trigonometric functions expect radians  which they probably do   you will need to convert your longitude and latitude to radians first  You obviously need a decimal representation  not degrees minutes seconds  see e g  here about conversion    The formula for back conversion      lat   asin z   R     lon   atan2 y  x    asin is of course arc sine  read about atan2 in wikipedia  Don   t forget to convert back from radians to degrees    This page gives c  code for this  note that it is very different from the formulas   and also some explanation and nice diagram of why this is correct

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The proj 4 software provides a command line program that can do the conversion  e g   LAT 40 LON -110 echo  LON  LAT   cs2cs  proj latlong  datum WGS84  to  proj geocent  datum WGS84   It also provides a C API   In particular  the function pj geodetic to geocentric will do the conversion without having to set up a projection object first

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Why implement something which has already been implemented and test-proven   C   for one  has the NetTopologySuite which is the  NET port of the JTS Topology Suite   Specifically  you have a severe flaw in your calculation  The earth is not a perfect sphere  and the approximation of the earth s radius might not cut it for precise measurements   If in some cases it s acceptable to use homebrew functions  GIS is a good example of a field in which it is much preferred to use a reliable  test-proven library

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Coordinate   coordinates   new Coordinate 3   coordinates 0    new Coordinate 102  26   coordinates 1    new Coordinate 103  25 12   coordinates 2    new Coordinate 104  16 11   CoordinateSequence coordinateSequence   new CoordinateArraySequence coordinates    Geometry geo   new LineString coordinateSequence  geometryFactory    CoordinateReferenceSystem wgs84   DefaultGeographicCRS WGS84  CoordinateReferenceSystem cartesinaCrs   DefaultGeocentricCRS CARTESIAN   MathTransform mathTransform   CRS findMathTransform wgs84  cartesinaCrs  true    Geometry geo1   JTS transform geo  mathTransform

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If you care about getting coordinates based on an ellipsoid rather than a sphere  take a look at Geographic coordinate conversion - it gives the formulae  GEodetic Datum has the WGS84 constants you need for the conversion  The formulae there also take into account the altitude relative to the reference ellipsoid surface  useful if you are getting altitude data from a GPS device

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I have recently done something similar to this using the  quot Haversine Formula quot  on WGS-84 data  which is a derivative of the  quot Law of Haversines quot  with very satisfying results  Yes  WGS-84 assumes the Earth is an ellipsoid  but I believe you only get about a 0 5  average error using an approach like the  quot Haversine Formula quot   which may be an acceptable amount of error in your case   You will always have some amount of error unless you re talking about a distance of a few feet and even then there is theoretically curvature of the Earth     If you require a more rigidly WGS-84 compatible approach checkout the  quot Vincenty Formula  quot  I understand where starblue is coming from  but good software engineering is often about trade-offs  so it all depends on the accuracy you require for what you are doing    For example  the result calculated from  quot Manhattan Distance Formula quot  versus the result from the  quot Distance Formula quot  can be better for certain situations as it is computationally less expensive   Think  quot which point is closest  quot  scenarios where you don t need a precise distance measurement  Regarding  the  quot Haversine Formula quot  it is easy to implement and is nice because it is using  quot Spherical Trigonometry quot  instead of a  quot Law of Cosines quot  based approach which is based on two-dimensional trigonometry  therefore you get a nice balance of accuracy over complexity  A gentleman by the name of Chris Veness has a great website that explains some of the concepts you are interested in and demonstrates various programmatic implementations  this should answer your x y conversion question as well

[mapping] Converting from longitude\latitude to Cartesian coordinates

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