火星坐标系 (GCJ-02) 与百度坐标系 (BD-09) 的转换算法

水深无声 2022-07-14 23:24 227阅读 0赞

关于 GCJ-02 和 BD-09 ，请参考 [http://developer.baidu.com/map/question.htm\#qa0043][http_developer.baidu.com_map_question.htm_qa0043] 。

[算法][Link 1]代码如下，其中 `bd_encrypt` 将 GCJ-02 坐标转换成 BD-09 坐标， `bd_decrypt` 反之。

**\[cpp\]** [view plain][] [copy][view plain]

[print][view plain] [?][view plain]

1.  \#include <math.h>
2.  
3.  **const****double** x\_pi = 3.14159265358979324 \* 3000.0 / 180.0; 
4.  
5.  **void** bd\_encrypt(**double** gg\_lat, **double** gg\_lon, **double** &bd\_lat, **double** &bd\_lon) 
6.  \{ 
7.  **double** x = gg\_lon, y = gg\_lat; 
8.  **double** z = sqrt(x \* x + y \* y) + 0.00002 \* sin(y \* x\_pi); 
9.  **double** theta = atan2(y, x) + 0.000003 \* cos(x \* x\_pi); 
10.  bd\_lon = z \* cos(theta) + 0.0065; 
11.  bd\_lat = z \* sin(theta) + 0.006; 
12. \} 
13. 
14. **void** bd\_decrypt(**double** bd\_lat, **double** bd\_lon, **double** &gg\_lat, **double** &gg\_lon) 
15. \{ 
16. **double** x = bd\_lon - 0.0065, y = bd\_lat - 0.006; 
17. **double** z = sqrt(x \* x + y \* y) - 0.00002 \* sin(y \* x\_pi); 
18. **double** theta = atan2(y, x) - 0.000003 \* cos(x \* x\_pi); 
19.  gg\_lon = z \* cos(theta); 
20.  gg\_lat = z \* sin(theta); 
21. \}

http://blog.csdn.net/coolypf/article/details/8569813

[http_developer.baidu.com_map_question.htm_qa0043]: http://developer.baidu.com/map/question.htm#qa0043
[Link 1]: http://lib.csdn.net/base/datastructure
[view plain]: http://blog.csdn.net/coolypf/article/details/8569813#