从 kml 文件计算地面覆盖角的纬度/经度

2022-01-14 00:00:00 geometry kml php javascript

我需要在 php 或 javascript 中找到 kml 文件中给出的地面覆盖层的纬度/经度角.

I need to find the corners in lat/lng of a ground overlay given in a kml-file either in php or javascript.

即对于一个具体的例子，我需要从:

I.e. for a specific example I need to get from:

  <LatLonBox>
    <north>60.406505416667</north>
    <south>60.400570555556</south>
    <east>5.3351572222222</east>
    <west>5.3190577777778</west>
    <rotation>3.7088732260919</rotation>
  </LatLonBox>

转角坐标

SW: 60.400316388889;5.3194425
SE: 60.400824722222;5.3355405555556
NE: 60.406759444444;5.3347738888889
NW: 60.406251388889;5.3186730555556

我可以通过

$w=($nw_lng+$sw_lng)/2;
$e=($ne_lng+$se_lng)/2;
$n=($ne_lat+$nw_lat)/2;
$s=($se_lat+$sw_lat)/2;
$rot= rad2deg (atan ( ( $nw_lng - $sw_lng ) / ($sw_lat - $nw_lat ) / 2  ) );

应该很容易回来，但我已经为此花费了几个小时而没有到达那里.有什么建议吗?

Should be easy to get back, but I've used hours for this without getting there. Any tips?

推荐答案

你需要使用球面三角函数，球面几何的一部分，以确保完全准确.但是，由于您只处理球体的一小部分，如果您记住一件事，欧几里德几何就可以了.

You need to use spherical trigonometry, part of spherical geometry for full accuracy. However, since you are dealing with only a small piece of the sphere, euclidian geometry will do if you remember one thing.

随着纬度的增加，经线越来越靠近.例如，在北极附近，纬线几乎是相接的.因此，调节您的纬度差异，通过乘以 cos(纬度)因子来减少它们.这将为您的应用提供足够好的准确性.

As latitude increases, the lines of longitude get closer together. For example, near the North Pole, the latitude lines are almost touching. So condition your latitude differences, diminishing them by mulitlying by a factor of cos(latitude). That will give you good enough accuracy for your app.

 $n = 60.406505416667;
 $s = 60.400570555556;
 $e = 5.3351572222222;
 $w = 5.3190577777778;
 $rotn = 3.7088732260919;

 $a = ($e + $w) / 2.0;
 $b = ($n + $s) / 2.0;
 $squish = cos(deg2rad($b));
 $x = $squish * ($e - $w) / 2.0;
 $y = ($n - $s) / 2.0;

 $ne = array(
   $a + ($x * cos(deg2rad($rotn)) - $y * sin(deg2rad($rotn))) /$squish,
   $b + $x * sin(deg2rad($rotn)) + $y *cos(deg2rad($rotn))
   );
 $nw = array(
   $a - ($x * cos(deg2rad($rotn)) + $y * sin(deg2rad($rotn))) /$squish,
   $b - $x * sin(deg2rad($rotn)) + $y *cos(deg2rad($rotn))
   );
 $sw = array(
   $a - ($x * cos(deg2rad($rotn)) - $y * sin(deg2rad($rotn))) /$squish,
   $b - $x * sin(deg2rad($rotn)) - $y *cos(deg2rad($rotn))
   );
 $se = array(
   $a + ($x * cos(deg2rad($rotn)) + $y * sin(deg2rad($rotn))) /$squish,
   $b + $x * sin(deg2rad($rotn)) - $y *cos(deg2rad($rotn))
   );
 print_r(array(
 'sw'=>$sw,
 'se'=>$se,
 'ne'=>$ne,
 'nw'=>$nw,
 ));

我的 $squish 变量是我提到的 cos(lat).水平长度的相对部分有去挤压功能.正弦表如下所示:

My $squish variable is the cos(lat) I mentioned. There is de-squishing for the relative part of horizontal lengths. The sine table looks like this:

NE: (a + x cos A - y sin A, b + x sin A + y cos A)
NW: (a - x cos A - y sin A, b - x sin A + y cos A)
SW: (a - x cos A + y sin A, b - x sin A - y cos A)
SE: (a + x cos A + y sin A, b + x sin A - y cos A)

也许 tttppp 可以解释与 tttppp 表的差异.

Perhaps tttppp could account for the differences from tttppp's table.

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