将经度和纬度值转换为自定义大小的网格

问题描述

我正在制作一个Java程序，该程序将一组经/纬度坐标分类为具有自定义大小的特定矩形，因此实际上，将地球表面映射到自定义网格中并能够识别出哪个矩形/多边形关键所在.

I am making a java program that classifies a set of lat/lng coordinates to a specific rectangle of a custom size, so in effect, map the surface of the earth into a custom grid and be able to identify what rectangle/ polygon a point lies in.

我正在研究的方法是使用地图投影(可能是Mercator).

The way to do this I am looking into is by using a map projection (possibly Mercator).

例如，假设我要将长形/经度分类为100m x 100m的正方形"，

For example, assuming I want to classify a long/lat into 'squares' of 100m x 100m,

44.727549、10.419704和44.727572、10.420460将分类为X区域

44.727549, 10.419704 and 44.727572, 10.420460 would classify to area X

和

44.732496、10.528092和44.732999、10.529465将分类为区域Y，因为它们相距100m以内.(假设它们位于相同的边界内)

44.732496, 10.528092 and 44.732999, 10.529465 would classify to area Y as they are within 100m apart.(this assumes they lie within the same boundary of course)

我不太担心变形，因为我不需要显示地图，但是我确实需要能够知道一组坐标属于哪个多边形.

Im not too worried about distortion as I will not need to display the map, but I do need to be able to tell what polygon a set of coordinates belong to.

这可能吗?任何建议欢迎.谢谢.

Is this possible? Any suggestions welcome. Thanks.

修改

省略两极的投影也是可以接受的损耗

Omitting projection of the poles is also an acceptable loss

推荐答案

这是我的最终解决方案(在PHP中)，为每个100m的正方形创建一个bin:

Here is my final solution (in PHP), creates a bin for every square 100m :

function get_static_pointer_table_id($lat, $lng)
{
    $earth_circumference = 40000; // km

    $lat_bin = round($lat / 0.0009);
    $lng_length = $earth_circumference * cos(deg2rad($lat));
    $number_of_bins_on_lng = $lng_length * 10;
    $lng_bin = round($number_of_bins_on_lng * $lng / 360);
    //the 'bin' unique identifier
    return $lat_bin . "_" . $lng_bin;

}

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