查找LatLng Coord是否介于两个其他LatLng Coords之间

时间:2016-07-13 14:03:49

标签: javascript google-maps math coordinates geospatial

目的

要知道坐标点是否在两个其他坐标之间。

背景

我正在制作一个谷歌地图应用程序,我需要知道某个点是否在两个LatLng点之间(开始和结束)。

我正在寻找以下功能:

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";
}

我尝试了什么

为实现这一目标,我读了几个问题和主题:

代码

无论我尝试什么,我都无法让它发挥作用,但我的失败实验确实有一个很小的例子:

"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

  • 如何使用数学方法确定给定坐标是否在两个坐标之间?

3 个答案:

答案 0 :(得分:2)

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

代码段

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>

答案 1 :(得分:0)

通过计算一个点是否属于一条线,我已经找到了解决方案。

在研究了许多数学原理之后,我决定计算两点之间矩阵的行列式,并检查(具有一定的精度)我的给定点是否在点A和B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 http://stackoverflow.com/a/907491/1337392|Distance 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;
}

然而,这个解决方案有一个值得一提的陷阱。该解决方案的问题在于它没有考虑地球的曲率。它只适用于直线。

所以,如果你要检查你镇上的距离,这根本不可能是重要的。但是如果你正在检查穿越太平洋的航班,你应该使用另一种数学方法。

由于这个原因,并且由于折线已经考虑了地球的曲率,我决定使用geocodezip的答案。

PS:我不得不说这很有趣。几个小时前,我回答了一个老问题,我也感谢geocodezip给出的深刻见解,现在他在这里,回答我的问题。有时,我确实相信世界是一个小地方。谢谢你!

答案 2 :(得分:0)

使用上述http://www.movable-type.co.uk/scripts/latlong.html页面中的方位部分,您只需检查

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

这个等式表示所有三个点都位于同一个大圆弧上

(我假设你的“在两个其他坐标之间”具有相同的含义)