查找 LatLng Coord 是否位于其他两个 LatLng Coords 之间

Find if a LatLng Coord is between two other LatLng Coords

Objective

判断一个坐标点是否在另外两个坐标之间。

背景

我正在制作一个 google-maps 应用程序,我需要知道某个点是否在两个 LatLng 点(起点和终点)之间。

我正在寻找如下函数:

var currentCoord = {lat: 51.8732, lng: -118.6346};
var startCoord = {lat: 61.3434, lng: -118.0046};
var endCoord = {lat: 50.5468, lng: -118.5435};

function isBetween(startCoord, endCoord, currentCoord){
    //calculations here
    return "true if currentCoord is between startCoord and endCoord, or false otherwise";
}

我试过的

为了实现这一点,我阅读了几个问题和话题:

代码

无论我尝试什么,我都无法让它工作,但我确实有一个失败实验的最小例子:

"use strict";

/*global google*/

function initialize() {

  let mapOptions = {
    zoom: 3,
    center: new google.maps.LatLng(0, -180),
    mapTypeId: google.maps.MapTypeId.TERRAIN
  };

  let map = new google.maps.Map(document.getElementById('map-canvas'),
    mapOptions);

  let flightPlanCoordinates = [
    new google.maps.LatLng(37.772323, -122.214897),
    new google.maps.LatLng(21.291982, -157.821856),
    new google.maps.LatLng(-18.142599, 178.431),
    new google.maps.LatLng(-27.46758, 153.027892)
  ];

  let flightPath = new google.maps.Polyline({
    path: flightPlanCoordinates,
    geodesic: true,
    strokeColor: '#FF0000',
    strokeOpacity: 1.0,
    strokeWeight: 2
  });

  flightPath.setMap(map);

  google.maps.event.addListener(flightPath, 'mouseover', function(event) {
    console.log("Marker is over the polyline");
  });

  let marker = new google.maps.Marker({
    position: new google.maps.LatLng(37.772323, -122.214897),
    draggable: true,
    map: map,
    title: 'Drag me!'
  });

  marker.addListener('drag', function(event) {

    let startPoint = {
      lat: 37.772323,
      lng: -122.214897
    };
    let endPoint = {
      lat: 21.291982,
      lng: -157.821856
    };
    let currentPoint = {
      lat: marker.getPosition().lat(),
      lng: marker.getPosition().lng()
    };

    if (checkCoordinate(startPoint, endPoint, currentPoint))
      console.log("in line !");
  });
}

google.maps.event.addDomListener(window, 'load', initialize);

function checkCoordinate(start, end, point) {
  var slope = (end.lng - start.lng) / (end.lat - start.lat);
  var newSlope = (end.lng - point.lng) / (end.lat - point.lat);
  return (point.lat > start.lat && point.lat < end.lat && point.lng > start.lng && point.lng < end.lng && slope == newSlope);
}
html,
body,
#map-canvas {
  height: 100%;
  margin: 0px;
  padding: 0px
}
<!DOCTYPE html>
<html>

<head>
  <meta name="viewport" content="initial-scale=1.0, user-scalable=no">
  <meta charset="utf-8">
  <title>Simple Polylines</title>
  <link rel="stylesheet" type="text/css" href="style.css">
  <script src="https://maps.googleapis.com/maps/api/js?v=3"></script>
  <script src="script.js" type="text/javascript"></script>
</head>

<body>
  <div id="map-canvas"></div>
</body>

</html>

问题

一种选择是使用 google.maps.geometry.poly.isLocationOnEdge 方法。

代码片段:

var map;

function initialize() {

  var mapOptions = {
    zoom: 2,
    center: new google.maps.LatLng(0, -180),
    mapTypeId: google.maps.MapTypeId.TERRAIN
  };

  map = new google.maps.Map(document.getElementById('map-canvas'),
    mapOptions);

  var flightPlanCoordinates = [
    new google.maps.LatLng(37.772323, -122.214897),
    new google.maps.LatLng(21.291982, -157.821856),
    new google.maps.LatLng(-18.142599, 178.431),
    new google.maps.LatLng(-27.46758, 153.027892)
  ];

  var flightPath = new google.maps.Polyline({
    path: flightPlanCoordinates,
    geodesic: false,
    strokeColor: '#FF0000',
    strokeOpacity: 1.0,
    strokeWeight: 2
  });

  flightPath.setMap(map);

  var marker = new google.maps.Marker({
    position: new google.maps.LatLng(37.772323, -122.214897),
    draggable: true,
    map: map,
    title: 'Drag me!'
  });

  marker.addListener('dragend', function(event) {

    var startPoint = {
      lat: 37.772323,
      lng: -122.214897
    };
    var startMarker = new google.maps.Marker({
      position: startPoint,
      map: map
    });
    var endPoint = {
      lat: 21.291982,
      lng: -157.821856
    };
    var endMarker = new google.maps.Marker({
      position: endPoint,
      map: map
    });
    var currentPoint = {
      lat: marker.getPosition().lat(),
      lng: marker.getPosition().lng()
    };

    if (checkCoordinate(startPoint, endPoint, marker.getPosition()))
      console.log("in line !");
  });
}

google.maps.event.addDomListener(window, 'load', initialize);

function checkCoordinate(start, end, point) {
  return google.maps.geometry.poly.isLocationOnEdge(point, new google.maps.Polyline({
    map: map,
    path: [start, end]
  }), 10e-1);
}
html,
body,
#map-canvas {
  height: 100%;
  margin: 0px;
  padding: 0px
}
<!DOCTYPE html>
<html>

<head>
  <meta name="viewport" content="initial-scale=1.0, user-scalable=no">
  <meta charset="utf-8">
  <title>Simple Polylines</title>
  <link rel="stylesheet" type="text/css" href="style.css">
  <script src="https://maps.googleapis.com/maps/api/js?v=3"></script>
  <script src="script.js" type="text/javascript"></script>
</head>

<body>
  <div id="map-canvas"></div>
</body>

</html>

我已经找到了一个解决方案,通过计算一个点是否属于一条线。

研究了很多数学原理后,我决定计算两点之间矩阵的行列式,并(以一定的精度)检查我给定的点是否在A点和B点的直线之间。

/**
 * @const
 * @type        {Number}
 * @description The precision to calculate if a given point is between two other points. Low precisions get precise results but are less forgiving against errors.
 */
const PRECISION = 1;

/**
 * @function    isBetween
 * @description Determines if a point P = (p.x, p.y) lies on the line connecting points S = (S.x, S.y) and E = (E.x, E.y) by calculating the determinant of the matrix. A point is considered to belong to the line if the precision of the calculation is small enough (tests for errors and loss of precision).
 * @param       {Point} start   The start point
 * @param       {Point} end     The end point
 * @param       {Point} point   The point we which to test.
 * @returns     <code>true</code> if the given point belongs to the line, <code>false</code> otherwise.
 * @see         {@link  Matrix Calculation}
 */
function isBetween(start, end, point) {
    return Math.abs((end.lat - start.lat) * (point.lng - start.lng) - (end.lng - start.lng) * (point.lat - start.lat)) < PRECISION;
}

但是,这个解决方案有一个值得一提的陷阱。这个解决方案的问题是它没有考虑地球的曲率。它仅适用于直线。

因此,如果您要检查您所在城镇的距离,这可能根本不重要。但是,如果您正在检查横跨太平洋的航班,您可能应该使用另一种数学方法。

出于这个原因,并且由于多段线已经考虑到地球的曲率,我决定使用 geocodezip 中的答案。


PS:不得不说很搞笑。几个小时前,我回答了一个老问题,我也感谢 geocodezip 提供的富有洞察力的评论,现在他在这里回答了我的问题。有时候,我确实相信世界是一个小地方。谢谢大佬!

使用提到的 http://www.movable-type.co.uk/scripts/latlong.html 页面中的 Bearing 部分,您可以检查

bearing(from currentCoord to startCoord) = 
  bearing(from currentCoord to endCoord) +/- 180 (with some tolerance)

这个方程表示三点都在同一条大圆弧上

(假设你的"is between two other coordinates"是同一个意思)