blocks|key|1279443|text|在python3.x中，可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279444|#+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|code-block|syntax|javascript|1279445|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In python3.x it can be done using : 

<pre><code># 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
</code></pre>

blocks|key|1494174|text|proj.4软件提供了可以进行转换的命令行程序，例如|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1494175|LAT=40
LON=-110
echo+$LON+$LAT+%7C+cs2cs+%2Bproj=latlong+%2Bdatum=WGS84+%2Bto+%2Bproj=geocent+%2Bdatum=WGS84|code-block|syntax|javascript|1494176|它还提供了一个C+API。具体地说，函数pj_geodetic_to_geocentric将执行转换，而不必首先设置投影对象。|style|CODE|1494177|entityMap|0|LINK|mutability|MUTABLE|url|http://proj4.org/|1|https://github.com/OSGeo/proj.4/blob/master/src/proj_api.h^0|0|6|0|0|0|K|P|7|5|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|11|8|@$A|12|B|13|K|L]]|9|@$A|14|B|15|1|16]]|C|$]]|$1|M|3|-4|5|6|7|17|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

The <a href="http://proj4.org/" rel="noreferrer">proj.4</a> software provides a command line program that can do the conversion, e.g.

<pre><code>LAT=40
LON=-110
echo $LON $LAT | cs2cs +proj=latlong +datum=WGS84 +to +proj=geocent +datum=WGS84
</code></pre>

It also provides a <a href="https://github.com/OSGeo/proj.4/blob/master/src/proj_api.h" rel="noreferrer">C API</a>. In particular, the function <code>pj_geodetic_to_geocentric</code> will do the conversion without having to set up a projection object first.

blocks|key|1494115|text|如果你想获得基于椭球体而不是球体的坐标，可以看看Geographic_coordinate_conversion+-它给出了公式。GEodetic+Datum具有转换所需的WGS84常量。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1494116|其中的公式还考虑了相对于参考椭球曲面的高度(如果您从GPS设备获取高度数据，则非常有用)。|1494117|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Geographic_coordinate_conversion#From_geodetic_to_ECEF_coordinates|1|https://en.wikipedia.org/wiki/Geodetic_datum#World_Geodetic_System_1984_(WGS_84)^0|O|W|0|1T|E|1|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]|$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

If you care about getting coordinates based on an ellipsoid rather than a sphere, take a look at <a href="https://en.wikipedia.org/wiki/Geographic_coordinate_conversion#From_geodetic_to_ECEF_coordinates" rel="nofollow noreferrer">Geographic_coordinate_conversion</a> - it gives the formulae. <a href="https://en.wikipedia.org/wiki/Geodetic_datum#World_Geodetic_System_1984_(WGS_84)" rel="nofollow noreferrer">GEodetic Datum</a> 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).

blocks|key|1494137|text|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);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1494138|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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);
</code></pre>

blocks|key|1279485|text|我用Python做了一个函数，考虑到地球不是一个完美的球体这一事实。参考资料在评论中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279486|++++#+this+function+converts+latitude,longitude+and+height+above+sea+level+
++++#+to+earthcentered+xyx+coordinates+in+wgs84,+lat+and+lon+in+decimal+degrees+
++++#+e.g.+52.724156(West+and+South+are+negative),+heigth+in+meters
++++#+for+algoritm+see+https://en.wikipedia.org/wiki/Geographic_coordinate_conversion#From_geodetic_to_ECEF_coordinates
++++#+for+values+of+a+and+b+see+https://en.wikipedia.org/wiki/Earth_radius#Radius_of_curvature

from+math+import+*

def+latlonhtoxyzwgs84(lat,lon,h):


++++a=6378137.0+++++++++++++#radius+a+of+earth+in+meters+cfr+WGS84
++++b=6356752.3+++++++++++++#radius+b+of+earth+in+meters+cfr+WGS84
++++e2=1-(b**2/a**2)
++++latr=lat/90*0.5*pi++++++#latitude+in+radians
++++lonr=lon/180*pi+++++++++#longituede+in+radians
++++Nphi=a/sqrt(1-e2*sin(latr)**2)
++++x=(Nphi%2Bh)*cos(latr)*cos(lonr)
++++y=(Nphi%2Bh)*cos(latr)*sin(lonr)
++++z=(b**2/a**2*Nphi%2Bh)*sin(latr)
++++return([x,y,z])|code-block|syntax|javascript|1279487|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I made a function in Python that takes into account the fact that the earth is not a perfect sphere. References are in the comments:
<pre class="lang-py prettyprint-override"><code> # this function converts latitude,longitude and height above sea level 
 # to earthcentered xyx coordinates in wgs84, lat and lon in decimal degrees 
 # e.g. 52.724156(West and South are negative), heigth in meters
 # for algoritm see https://en.wikipedia.org/wiki/Geographic_coordinate_conversion#From_geodetic_to_ECEF_coordinates
 # for values of a and b see https://en.wikipedia.org/wiki/Earth_radius#Radius_of_curvature

from math import *

def latlonhtoxyzwgs84(lat,lon,h):


 a=6378137.0 #radius a of earth in meters cfr WGS84
 b=6356752.3 #radius b of earth in meters cfr WGS84
 e2=1-(b**2/a**2)
 latr=lat/90*0.5*pi #latitude in radians
 lonr=lon/180*pi #longituede in radians
 Nphi=a/sqrt(1-e2*sin(latr)**2)
 x=(Nphi+h)*cos(latr)*cos(lonr)
 y=(Nphi+h)*cos(latr)*sin(lonr)
 z=(b**2/a**2*Nphi+h)*sin(latr)
 return([x,y,z])
</code></pre>

blocks|key|1279403|text|你可以在Java上这样做。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279404|public+List<Double>+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%2B0.001*alt)*cosLatRad*cosLongiRad;
++++double+y+=(N%2B0.001*alt)*cosLatRad*sinLongiRad;
++++double+z+=((Math.pow(b,+2)/Math.pow(a,+2))*N%2B0.001*alt)*sinLatRad;

++++List<Double>+ecef=+new+ArrayList<>();
++++ecef.add(x);
++++ecef.add(y);
++++ecef.add(z);

++++return+ecef;


}|code-block|syntax|javascript|1279405|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can do it this way on Java.

<pre><code>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;


}
</code></pre>

I have some earth-centered coordinate points given as latitude and longitude (<a href="http://en.wikipedia.org/wiki/World_Geodetic_System" rel="noreferrer">WGS-84</a>).

How can i convert them to Cartesian coordinates (x,y,z) with the origin at the center of the earth?

Converting from longitude\latitude to Cartesian coordinates

Python

Java

我有一些以纬度和经度()表示的地心坐标点。如何将它们转换为原点位于地球中心的笛卡尔坐标(x，y，z)？

问从经纬度到笛卡尔坐标的转换
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回答 6

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