LLA（经纬高）坐标转换成ENU（东北天）坐标的详细推导

这是一篇经纬高（LLA）坐标转东北天坐标（ENU）的详细推导，并给出近似转换的过程和结果

参考资料：
https://blog.csdn.net/qq_34213260/article/details/109133847

ECEF坐标系和ENU坐标系之间的关系如上图所示，各坐标系的定义在此不再赘述。

LLA坐标：Lat、Lon、Alt坐标，表示在ECEF坐标系下，其中Lat和Lon用角度（${}^{。}$）或弧度（rad）表示，Alt用米（m）表示；
ENU坐标：东北天坐标系下的坐标，该坐标系的原点需要指定，原点向东为x轴，原点向北为y轴，原点指向天为z轴，构成右手直角坐标系，该系下的坐标表示都是米制单位；

LLA坐标转换成ENU坐标的过程主要分为两步：LLA->ECEF->ENU

其中主要参数的定义如下：

$$b=6356752.3142$$
$$e^2=f(2-f)$$
$$N=\frac{a}{\sqrt{1-e^{2}si{n}^2(lat)}}$$
其中b为椭球短半径，e为椭球偏心率

（1）LLA->ECEF:（无需给定起点，一对一转换）
从最上面一幅图中可以轻易得到同一点在这两个坐标系之间的转换关系，如下：
$$X_{ecef}=(N+alt)cos(lat)cos(lon)$$
$$Y_{ecef}=(N+alt)cos(lat)sin(lon)$$
$$Z_{ecef}=(N(1-e^2)+alt)sin(lat)$$
（2）ECEF->ENU:（需要给定起点）
假设ENU坐标系的坐标原点对应的ECEF坐标和LLA坐标分别为：
$$O_{ecef}=(X_0,Y_0,Z_0)$$
$$O_{lla}=(la{t}_0,lo{n}_0,al{t}_0)$$
假设空间中有另一点P，且：
$$P_{ecef}=(X,Y,Z)$$
则P在ENU系下的坐标为：
$$P_{enu}=R_{ECEF}^{ENU}(P_{ecef}-O_{ecef})\text{\hspace{1em}(1)}$$
其中只有$R_{ECEF}^{ENU}$是未知的，下面对其进行推导：

$R_{ECEF}^{ENU}$可以分解为如下三个旋转过程：ECEF–>绕$Z_{ecef}$轴逆时针旋转$\lambda$（经度）–>绕$Y_{ecef}^{{}^{\prime }}$轴逆时针旋转（$90-\varphi$）（$\varphi :$纬度）–>绕$Z_{ecef}^{{}^{\prime \prime }}$逆时针旋转90–>ENU

上述问题是典型的空间中点不动，坐标系转动的问题，按顺序记三次旋转分别为$R_1、R_2、R_3$，则：
$$R_1=\left[\begin{array}{ccc}cos(\lambda )& sin(\lambda )& 0\\ -sin(\lambda )& cos(\lambda )& 0\\ 0& 0& 1\end{array}\right]$$
$$R_2=\left[\begin{array}{ccc}sin(\varphi )& 0& -cos(\varphi )\\ 0& 1& 0\\ cos(\varphi )& 0& sin(\varphi )\end{array}\right]$$
$$R_3=\left[\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 1\end{array}\right]$$
则：
$$R_{ECEF}^{ENU}=R_3\ast R_2\ast R_1=\left[\begin{array}{ccc}-sin(\lambda )& cos(\lambda )& 0\\ -sin(\varphi )cos(\lambda )& -sin(\varphi )sin(\lambda )& cos(\varphi )\\ cos(\varphi )cos(\lambda )& cos(\varphi )sin(\lambda )& sin(\varphi )\end{array}\right]$$
将上式带入到公式（1）中可以得到：
$$P_{enu}=R_{ECEF}^{ENU}(P_{ecef}-O_{ecef})=\left[\begin{array}{ccc}-sin(lo{n}_0)& cos(lo{n}_0)& 0\\ -sin(la{t}_0)cos(lo{n}_0)& -sin(la{t}_0)sin(lo{n}_0)& cos(la{t}_0)\\ cos(la{t}_0)cos(lo{n}_0)& cos(la{t}_0)sin(lo{n}_0)& sin(la{t}_0)\end{array}\right]\ast \left[\begin{array}{c}X-X_0\\ Y-Y_0\\ Z-Z_0\end{array}\right]\text{\hspace{1em}(2)}$$
至此完成了LLA坐标到ENU坐标的转换，且无近似，下面对公式（2）进行近似推导

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由于
$$X=(N+alt)cos(lat)cos(lon)$$
$$Y=(N+alt)cos(lat)sin(lon)$$
$$Z=(N(1-e^2)+alt)sin(lat)$$
$$X_0=(N_0+al{t}_0)cos(la{t}_0)cos(lo{n}_0)$$
$$Y_0=(N_0+al{t}_0)cos(la{t}_0)sin(lo{n}_0)$$
$$Z_0=(N_0(1-e^2)+al{t}_0)sin(la{t}_0)$$
所以
$$\begin{array}{rl}X-X_0& =(N+alt)cos(lat)cos(lon)-(N_0+al{t}_0)cos(la{t}_0)cos(lo{n}_0)\\ & =(N+alt)cos(lat)cos(lon)-(N+al{t}_0)cos(la{t}_0)cos(lo{n}_0)\\ & =(N+al{t}_0+\delta alt)cos(la{t}_0+\delta lat)cos(lo{n}_0+\delta lon)-(N+al{t}_0)cos(la{t}_0)cos(lo{n}_0)\\ & =(N+al{t}_0)cos(la{t}_0+\delta lat)cos(lo{n}_0+\delta lon)-(N+al{t}_0)cos(la{t}_0)cos(lo{n}_0)+\delta altcos(la{t}_0+\delta lat)cos(lo{n}_0+\delta lon)\\ & =(N+al{t}_0)\left[cos(la{t}_0+\delta lat)cos(lo{n}_0+\delta lon)-cos(la{t}_0)cos(lo{n}_0)\right]+\delta altcos(la{t}_0+\delta lat)cos(lo{n}_0+\delta lon)\\ & =(N+al{t}_0)\left[(cos(la{t}_0)cos(\delta lat)-sin(la{t}_0)sin(\delta lat))(cos(lo{n}_0)cos(\delta lon)-sin(lo{n}_0)sin(\delta lon))-cos(la{t}_0)cos(lo{n}_0)\right]+\delta altcos(la{t}_0+\delta lat)cos(lo{n}_0+\delta lon)\\ & =(N+al{t}_0)(-cos(la{t}_0)sin(lo{n}_0)cos(\delta lat)sin(\delta lon)-sin(la{t}_0)cos(lo{n}_0)sin(\delta lat)cos(\delta lon))+\delta altcos(la{t}_0+\delta lat)cos(lo{n}_0+\delta lon)\\ & =(N+al{t}_0)(-cos(la{t}_0)sin(lo{n}_0)\ast \delta lon-sin(la{t}_0)cos(lo{n}_0)\ast \delta lat)+\delta altcos(la{t}_0+\delta lat)cos(lo{n}_0+\delta lon)\\ & =(N+al{t}_0)(-cos(la{t}_0)sin(lo{n}_0)\ast \delta lon-sin(la{t}_0)cos(lo{n}_0)\ast \delta lat)+\delta alt\left[(cos(la{t}_0)-sin(la{t}_0)\ast \delta lat)(cos(lo{n}_0)-sin(lo{n}_0)\ast \delta lon)\right]\\ & =(N+al{t}_0)(-cos(la{t}_0)sin(lo{n}_0)\ast \delta lon-sin(la{t}_0)cos(lo{n}_0)\ast \delta lat)+\delta alt\left[cos(la{t}_0)cos(lo{n}_0)-cos(la{t}_0)sin(lo{n}_0)\ast \delta lon-cos(lo{n}_0)sin(la{t}_0)\ast \delta lat\right]\\ & =(N+al{t}_0)(-cos(la{t}_0)sin(lo{n}_0)\ast \delta lon-sin(la{t}_0)cos(lo{n}_0)\ast \delta lat)+\delta altcos(la{t}_0)cos(lo{n}_0)\end{array}$$
同理，有
$$Y-Y_0=(N+al{t}_0)(\delta loncos(la{t}_0)cos(lo{n}_0)-\delta latsin(la{t}_0)sin(lo{n}_0))+\delta altcos(la{t}_0)sin(lo{n}_0)$$
将以上两式带入到（2）中可以得到点P在ENU系下的东向坐标：
$$\begin{array}{rl}e& =-sin(lo{n}_0)\ast (X-X_0)+cos(lo{n}_0)\ast (Y-Y_0)\\ & =-sin(lo{n}_0)((N+al{t}_0)(-cos(la{t}_0)sin(lo{n}_0)\ast \delta lon-sin(la{t}_0)cos(lo{n}_0)\ast \delta lat)+\delta altcos(la{t}_0)cos(lo{n}_0))+cos(lo{n}_0)((N+al{t}_0)(\delta loncos(la{t}_0)cos(lo{n}_0)-\delta latsin(la{t}_0)sin(lo{n}_0))+\delta altcos(la{t}_0)sin(lo{n}_0))\\ & =(N+al{t}_0)(sin(lo{n}_0)cos(la{t}_0)sin(lo{n}_0)\ast \delta lon)+(N+al{t}_0)(cos(la{t}_0)co{s}^2(lo{n}_0)\ast \delta lon)\\ & =(N+al{t}_0)\ast \delta loncos(la{t}_0)(si{n}^2(lo{n}_0)+co{s}^2(lo{n}_0))\\ & =(N+al{t}_0)\ast \delta loncos(la{t}_0)\\ & \approx a\ast \delta loncos(la{t}_0)\end{array}$$
类似的，可以得到n、u坐标，最终得到如下的近似结果：
$$P_{enu}=\left[\begin{array}{c}e\\ n\\ u\end{array}\right]=\left[\begin{array}{c}a\ast \delta loncos(la{t}_0)\\ a\ast \delta lat\\ \delta alt\end{array}\right]$$