blocks|key|5526941|text|好了，我把它弄好了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5526942|问题是一个放错了位置的变量，所以为了我们的未来着想，下面是可用的JAVA实现：|5526943|public+static+final+double+a+=+6378137;
public+static+final+double+f+=+0.0034;
public+static+final+double+b+=+6.3568e6;
public+static+final+double+e+=+Math.sqrt((Math.pow(a,+2)+-+Math.pow(b,+2))+/+Math.pow(a,+2));
public+static+final+double+e2+=+Math.sqrt((Math.pow(a,+2)+-+Math.pow(b,+2))+/+Math.pow(b,+2));

public+static+double[]+ecef2lla(double+x,+double+y,+double+z)+{

++++double[]+lla+=+{+0,+0,+0+};
++++double+lan,+lon,+height,+N+,+theta,+p;

++++p+=+Math.sqrt(Math.pow(x,+2)+%2B+Math.pow(y,+2));

++++theta+=+Math.atan((z+*+a)+/+(p+*+b));

++++lon+=+Math.atan(y+/+x);

++++lat+=+Math.atan(((z+%2B+Math.pow(e2,+2)+*+b+*+Math.pow(Math.sin(theta),+3))+/+((p+-+Math.pow(e,+2)+*+a+*+Math.pow(Math.cos(theta),+3)))));
++++N+=+a+/+(Math.sqrt(1+-+(Math.pow(e,+2)+*+Math.pow(Math.sin(lat),+2))));

++++double+m+=+(p+/+Math.cos(lat));
++++height+=+m+-+N;


++++lon+=+lon+*+180+/+Math.PI;
++++lat+=+lat+*+180+/+Math.PI;+
++++lla[0]+=+lat;
++++lla[1]+=+lon;
++++lla[2]+=+height;
++++return+lla;
}|code-block|syntax|javascript|5526944|注意：ECEF+X+Y+Z的单位是！|offset|length|style|BOLD|5526945|entityMap^0|0|0|0|0|2|3|A|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]|$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

OK so i got this working.

The problem was a misplaced variable, so for the sake of our future here's the working JAVA implementation :

<pre><code>public static final double a = 6378137;
public static final double f = 0.0034;
public static final double b = 6.3568e6;
public static final double e = Math.sqrt((Math.pow(a, 2) - Math.pow(b, 2)) / Math.pow(a, 2));
public static final double e2 = Math.sqrt((Math.pow(a, 2) - Math.pow(b, 2)) / Math.pow(b, 2));

public static double[] ecef2lla(double x, double y, double z) {

 double[] lla = { 0, 0, 0 };
 double lan, lon, height, N , theta, p;

 p = Math.sqrt(Math.pow(x, 2) + Math.pow(y, 2));

 theta = Math.atan((z * a) / (p * b));

 lon = Math.atan(y / x);

 lat = Math.atan(((z + Math.pow(e2, 2) * b * Math.pow(Math.sin(theta), 3)) / ((p - Math.pow(e, 2) * a * Math.pow(Math.cos(theta), 3)))));
 N = a / (Math.sqrt(1 - (Math.pow(e, 2) * Math.pow(Math.sin(lat), 2))));

 double m = (p / Math.cos(lat));
 height = m - N;


 lon = lon * 180 / Math.PI;
 lat = lat * 180 / Math.PI; 
 lla[0] = lat;
 lla[1] = lon;
 lla[2] = height;
 return lla;
}
</code></pre>

Note : The units for the ECEF X Y Z are in Meters !

blocks|key|2207127|text|我找到了这篇文章，并准备在我的应用程序的一部分使用“接受的答案”，但我决定首先运行几个测试来验证算法。我使用在线转换计算器(http://www.sysense.com/products/ecef_lla_converter/index.html)生成的示例数据。我得到的结果不是很好，如下所示。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2207128|-----Test+1---------
Inputs:+++-576793.17,+-5376363.47,+3372298.51
Expected:+32.12345,+-96.12345,+500.0
Actuals:++32.12306332822881,+83.87654999786477,+486.5472474489361
-----Test+2---------
Inputs:+++2297292.91,+1016894.94,+-5843939.62
Expected:+-66.87654,+23.87654,+1000.0
Actuals:++-66.876230479461,+23.87653991401422,+959.6879360172898|code-block|syntax|javascript|2207129|然后，我使用下面的post+(https://gist.github.com/klucar/1536194)中的代码重新运行相同的测试，并获得了更好的结果，如下面的输出所示。|2207130|-----Test+1---------
Inputs:+++-576793.17,+-5376363.47,+3372298.51
Expected:+32.12345,+-96.12345,+500.0
Actuals:++32.12345004807767,+-96.12345000213524,+499.997958839871
-----Test+2---------
Inputs:+++2297292.91,+1016894.94,+-5843939.62
Expected:+-66.87654,+23.87654,+1000.0
Actuals:++-66.87654001741278,+23.87653991401422,+999.9983866894618|2207131|我没有花时间找出“接受的答案”中提供的解决方案中的错误，但我的建议答案是:使用以下代码……|2207132|/*
*
*++ECEF+-+Earth+Centered+Earth+Fixed
*+++
*++LLA+-+Lat+Lon+Alt
*
*++ported+from+matlab+code+at
*++https://gist.github.com/1536054
*+++++and
*++https://gist.github.com/1536056
*/

//+WGS84+ellipsoid+constants
private+final+double+a+=+6378137;+//+radius
private+final+double+e+=+8.1819190842622e-2;++//+eccentricity

private+final+double+asq+=+Math.pow(a,2);
private+final+double+esq+=+Math.pow(e,2);

private+double[]+ecef2lla(double[]+ecef){
++double+x+=+ecef[0];
++double+y+=+ecef[1];
++double+z+=+ecef[2];

++double+b+=+Math.sqrt(+asq+*+(1-esq)+);
++double+bsq+=+Math.pow(b,2);
++double+ep+=+Math.sqrt(+(asq+-+bsq)/bsq);
++double+p+=+Math.sqrt(+Math.pow(x,2)+%2B+Math.pow(y,2)+);
++double+th+=+Math.atan2(a*z,+b*p);

++double+lon+=+Math.atan2(y,x);
++double+lat+=+Math.atan2(+(z+%2B+Math.pow(ep,2)*b*Math.pow(Math.sin(th),3)+),+(p+-+esq*a*Math.pow(Math.cos(th),3))+);
++double+N+=+a/(+Math.sqrt(1-esq*Math.pow(Math.sin(lat),2))+);
++double+alt+=+p+/+Math.cos(lat)+-+N;

++//+mod+lat+to+0-2pi
++lon+=+lon+%25+(2*Math.PI);

++//+correction+for+altitude+near+poles+left+out.

++double[]+ret+=+{lat,+lon,+alt};

++return+ret;
}|2207133|entityMap|0|LINK|mutability|MUTABLE|url|http://www.sysense.com/products/ecef_lla_converter/index.html|1|https://gist.github.com/klucar/1536194^0|1Q|1P|0|0|0|F|12|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|K|3|L|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1A|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1B|8|@]|9|@]|C|$G|H]]|$1|Q|3|-4|5|6|7|1C|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]]]

I found this post and was ready to use the "accepted answer" in part of my application but I decided to run a couple of tests first to validate the algorithm. I used sample data generated by an online conversion calculator (<a href="http://www.sysense.com/products/ecef_lla_converter/index.html" rel="noreferrer">http://www.sysense.com/products/ecef_lla_converter/index.html</a>). I got less than stellar results as shown by the output below. 

<pre><code>-----Test 1---------
Inputs: -576793.17, -5376363.47, 3372298.51
Expected: 32.12345, -96.12345, 500.0
Actuals: 32.12306332822881, 83.87654999786477, 486.5472474489361
-----Test 2---------
Inputs: 2297292.91, 1016894.94, -5843939.62
Expected: -66.87654, 23.87654, 1000.0
Actuals: -66.876230479461, 23.87653991401422, 959.6879360172898
</code></pre>

Then I reran the same tests using the code from the following post (<a href="https://gist.github.com/klucar/1536194" rel="noreferrer">https://gist.github.com/klucar/1536194</a>) and got much better results as shown by the output below.

<pre><code>-----Test 1---------
Inputs: -576793.17, -5376363.47, 3372298.51
Expected: 32.12345, -96.12345, 500.0
Actuals: 32.12345004807767, -96.12345000213524, 499.997958839871
-----Test 2---------
Inputs: 2297292.91, 1016894.94, -5843939.62
Expected: -66.87654, 23.87654, 1000.0
Actuals: -66.87654001741278, 23.87653991401422, 999.9983866894618
</code></pre>

I didn't take the time to find the error in the solution provided in the "accepted answer" but my suggested answer: Use this code... 

<pre><code>/*
*
* ECEF - Earth Centered Earth Fixed
* 
* LLA - Lat Lon Alt
*
* ported from matlab code at
* https://gist.github.com/1536054
* and
* https://gist.github.com/1536056
*/

// WGS84 ellipsoid constants
private final double a = 6378137; // radius
private final double e = 8.1819190842622e-2; // eccentricity

private final double asq = Math.pow(a,2);
private final double esq = Math.pow(e,2);

private double[] ecef2lla(double[] ecef){
 double x = ecef[0];
 double y = ecef[1];
 double z = ecef[2];

 double b = Math.sqrt( asq * (1-esq) );
 double bsq = Math.pow(b,2);
 double ep = Math.sqrt( (asq - bsq)/bsq);
 double p = Math.sqrt( Math.pow(x,2) + Math.pow(y,2) );
 double th = Math.atan2(a*z, b*p);

 double lon = Math.atan2(y,x);
 double lat = Math.atan2( (z + Math.pow(ep,2)*b*Math.pow(Math.sin(th),3) ), (p - esq*a*Math.pow(Math.cos(th),3)) );
 double N = a/( Math.sqrt(1-esq*Math.pow(Math.sin(lat),2)) );
 double alt = p / Math.cos(lat) - N;

 // mod lat to 0-2pi
 lon = lon % (2*Math.PI);

 // correction for altitude near poles left out.

 double[] ret = {lat, lon, alt};

 return ret;
}
</code></pre>

blocks|key|2201346|text|如果您对使用已发布的java库在笛卡尔坐标系和大地坐标系之间进行转换感兴趣，您可以使用java轨道传播库orekit。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2201347|虽然Orekit提供了一些高级功能，但它并不是很容易深入研究。为了帮助你入门，我写了一个样例代码来做一些转换。我的代码是用Scala编写的，但它使用的是java库，所以它应该还是有帮助的。请参阅示例代码here。|2201348|entityMap|0|LINK|mutability|MUTABLE|url|https://www.orekit.org/|1|https://github.com/gastoneb/Vector3D_LLA_conversion_example^0|1G|6|0|0|2T|4|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|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 are interested in using a published java library to convert between cartesian and geodetic coordinate systems, you can use the java orbit propagation library <a href="https://www.orekit.org/" rel="nofollow">orekit</a>. 

While Orekit provides some advanced features, it is not very easy to dive into. To help get you started I've written a sample code does some conversions. My code is in Scala but it uses the java library, so it should still be helpful. See the example code <a href="https://github.com/gastoneb/Vector3D_LLA_conversion_example" rel="nofollow">here</a>.

Ive looked through the posts in the site and haven't found my problem ...
as the head line says im trying to convert from ecef to lla .

I'm using this document : <a href="http://www.google.co.il/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=4&amp;ved=0CE0QFjAD&amp;url=http://www.nalresearch.com/files/Standard%2520Modems/A3LA-XG/A3LA-XG%2520SW%2520Version%25201.0.0/GPS%2520Technical%2520Documents/GPS.G1-X-00006%2520%28Datum%2520Transformations%29.pdf&amp;ei=t9MMUum3KpH6sgbE04DoDg&amp;usg=AFQjCNFddElyrN6ssKPyU5JTNh_q4uD3LQ&amp;sig2=73dM3EjUmDVXC4jU0SIFMw&amp;bvm=bv.50768961,d.Yms" rel="nofollow">Conversion article</a>
in the direct formula , not the iterate formula
and this site for result comparison : <a href="http://www.oc.nps.edu/oc2902w/coord/llhxyz.htm" rel="nofollow">ECEF2LLA</a>

Im developing in java so my code is as follows :

<pre><code>public static final double a = 6378137;
public static final double f = 1/298.257223563;
public static final double b = a*(1-f);
public static final double e = Math.sqrt((Math.pow(a, 2)-Math.pow(b, 2))/Math.pow(a, 2));
public static final double e2 = Math.sqrt((Math.pow(a, 2)-Math.pow(b, 2))/Math.pow(b, 2));
public static double[] ecef2lla(double x , double y , double z){
 double[] lla = {0,0,0};
 double lon,lat,height,N;
 double p = Math.sqrt(Math.pow(x, 2)+Math.pow(y, 2));
 double theta = Math.atan((z*a)/(p*b));
 lon = Math.atan(y/x);
 lon = lon*180/Math.PI;

 lat = Math.atan(((z+Math.pow(e2, 2)*b*Math.pow(Math.sin(theta), 3))/((p-Math.pow(e,2)*a*Math.pow(Math.cos(theta), 3)))));
 lat = (lat*180)/Math.PI;

 N= a/(Math.sqrt(1-Math.pow(e*Math.sin(lat), 2)));
 height = (p/Math.cos(theta)) - N;
 lla[0] = lon;
 lla[1] = lat;
 lla[2] = height;
 return lla;

}
</code></pre>

I'm getting wrong height data.
I've tried to move to radians and degrees and what not .

Thank you in advance !

ECEF to lla (lat,lon,alt) in java

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我已经浏览了网站上的帖子，但没有发现我的问题……正如标题所说的，我正在尝试从ecef转换到lla。我在直接公式中使用这个文档：，而不是迭代公式，这个站点用于结果比较：我是用java开发的，所以我的代码如下：public static final double a = 6378137;public static fina...

问java中将ECEF转换为lla (lat，lon，alt)
EN

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