如果不需要确切地说，这很容易：

在纬度上，1公里是0.009度（从仪表的原始定义开始）。由于您的广场距离中心5公里，因此您只需从中心点加减0.045度即可。

经度方面，它稍微复杂一点：Divide上面的值与纬度的余弦值。

在代码中：

  lat_min = lat_center  -  0.045; 
 lat_max = lat_center + 0.045; 
 long_min = long_center  - （0.045 / Math.cos（lat_center * Math.PI / 180）; 
 long_max = long_center +（0.045 / Math.cos（lat_center * Math.PI / 180）; 
 

（ Math.PI / 180 从度数到弧度）。

注意：不适用于两极。

Lets say I have a lat lng coordinate and I want to place that at the center of a square that is 10km wide and then get the minimum lat/lng and maximum lat/lng.

Is there an easy way to do this that already exists?

If it doesn't need to be exact, it is pretty easy:

For the latitude, 1 km is 0.009 degrees (follows from the original definition of meter). Since your square is 5 km around the center, you just need to add and subtract 0.045 degrees from the center point.

For the longitude, it is slightly more complicated: Divide the above value with the cosine of the latitude.

In code:

lat_min = lat_center - 0.045;
lat_max = lat_center + 0.045;
long_min = long_center - (0.045 / Math.cos(lat_center*Math.PI/180);
long_max = long_center + (0.045 / Math.cos(lat_center*Math.PI/180);

(Math.PI/180 is needed to convert from degrees to radians).

Beware: Does not work around the poles.