Google地图两个圆形交点 [英] Google Maps Two Circles Intersection Points

问题描述

是否有一种简单的方法可以获得两个交点（如果可用）的 lat / lng

  var R = 6371; // km 
 var dLat =（lat2-lat1）.toRad（）; 
 var dLon =（lon2-lon1）.toRad（）; 
 var lat1 = lat1.toRad（）; 
 var lat2 = lat2.toRad（）; 
 
 var a = Math.sin（dLat / 2）* Math.sin（dLat / 2）+ 
 Math.sin（dLon / 2）* Math.sin（dLon / 2） * Math.cos（lat1）* Math.cos（lat2）; 
 var c = 2 * Math.atan2（Math.sqrt（a），Math.sqrt（1-a））; 
  

和我们的

  AC = c / 2 
  

如果给出圆半径Rd是公里， / p>

  AB = Rd / R = Rd / 6371 
  

现在我们可以找到角度

$ $ $ $ $ $ $ c $ A = arccos（tg（AC）* ctg （AB））

起始方位（AF方向）：

  var y = Math.sin（dLon）* Math.cos（lat2）; 
 var x = Math.cos（lat1）* Math.sin（lat2） -  
 Math.sin（lat1）* Math.cos（lat2）* Math.cos（dLon）; 
 var brng = Math.atan2（y，x）; 
  

相交点的方位：

  B_bearing = brng  -  A 
 D_bearing = brng + A 
  

相交点的坐标：

pre $ var latB = Math.asin（Math.sin（lat1）* Math.cos（Rd / R）+
Math.cos（lat1）* Math.sin（Rd / R）* Math.cos（B_bearing））;
var lonB = lon1.toRad（）+ Math.atan2（Math.sin（B_bearing）* Math.sin（Rd / R）* Math.cos（lat1），
Math.cos（Rd / R）-Math.sin（LAT1）* Math.sin（LAT2））;


与D_bearing相同

latB lonB以弧度为单位

Is there an easy way to get the lat/lng of the intersection points (if available) of two circles in Google Maps API V3? Or should I go with the hard way?

EDIT : In my problem, circles always have the same radius, in case that makes the solution easier.

解决方案

Yes, for equal circles rather simple solution could be elaborated:
Let's first circle center is A point, second circle center is F, midpoint is C, and intersection points are B,D. ABC is right-angle spherical triangle with right angle C.

We want to find angle A - this is deviation angle from A-F direction. Spherical trigonometry (Napier's rules for right spherical triangles) gives us formula:

cos(A)= tg(AC) * ctg(AB) where one symbol denote spherical angle, double symbols denote great circle arcs' angles (AB, AC). We can see that AB = circle radius (in radians, of course), AC = half-distance between A and F on the great circle arc. To find AC (and other values) - I'll use code from this excellent page

var R = 6371; // km
var dLat = (lat2-lat1).toRad();
var dLon = (lon2-lon1).toRad();
var lat1 = lat1.toRad();
var lat2 = lat2.toRad();

var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
        Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2); 
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 

and our

AC = c/2

If circle radius Rd is given is kilometers, then

AB = Rd / R = Rd / 6371

Now we can find angle

A = arccos(tg(AC) * ctg(AB))

Starting bearing (AF direction):

var y = Math.sin(dLon) * Math.cos(lat2);
var x = Math.cos(lat1)*Math.sin(lat2) -
        Math.sin(lat1)*Math.cos(lat2)*Math.cos(dLon);
var brng = Math.atan2(y, x);

Intersection points' bearings:

B_bearing = brng - A
D_bearing = brng + A

Intersection points' coordinates:

var latB = Math.asin( Math.sin(lat1)*Math.cos(Rd/R) + 
              Math.cos(lat1)*Math.sin(Rd/R)*Math.cos(B_bearing) );
var lonB = lon1.toRad() + Math.atan2(Math.sin(B_bearing)*Math.sin(Rd/R)*Math.cos(lat1), 
                     Math.cos(Rd/R)-Math.sin(lat1)*Math.sin(lat2));

and the same for D_bearing

latB, lonB are in radians

这篇关于Google地图两个圆形交点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！