我目前正在使用下面的功能，它不能正常工作。根据谷歌Maps，这些坐标之间的距离(从59.3293371,13.4877472到59.3225525,13.4619422)是2.2公里，而函数返回1.6公里。我怎样才能使这个函数返回正确的距离?
function getDistanceFromLatLonInKm(lat1, lon1, lat2, lon2) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2-lat1); // deg2rad below
var dLon = deg2rad(lon2-lon1);
var a =
Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon/2) * Math.sin(dLon/2)
;
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI/180)
}
jsFiddle: http://jsfiddle.net/edgren/gAHJB/
我在typescript和ES6中实现了这个算法
export type Coordinate = {
lat: number;
lon: number;
};
求两点之间的距离:
function getDistanceBetweenTwoPoints(cord1: Coordinate, cord2: Coordinate) {
if (cord1.lat == cord2.lat && cord1.lon == cord2.lon) {
return 0;
}
const radlat1 = (Math.PI * cord1.lat) / 180;
const radlat2 = (Math.PI * cord2.lat) / 180;
const theta = cord1.lon - cord2.lon;
const radtheta = (Math.PI * theta) / 180;
let dist =
Math.sin(radlat1) * Math.sin(radlat2) +
Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
if (dist > 1) {
dist = 1;
}
dist = Math.acos(dist);
dist = (dist * 180) / Math.PI;
dist = dist * 60 * 1.1515;
dist = dist * 1.609344; //convert miles to km
return dist;
}
获取坐标数组之间的距离
export function getTotalDistance(coordinates: Coordinate[]) {
coordinates = coordinates.filter((cord) => {
if (cord.lat && cord.lon) {
return true;
}
});
let totalDistance = 0;
if (!coordinates) {
return 0;
}
if (coordinates.length < 2) {
return 0;
}
for (let i = 0; i < coordinates.length - 2; i++) {
if (
!coordinates[i].lon ||
!coordinates[i].lat ||
!coordinates[i + 1].lon ||
!coordinates[i + 1].lat
) {
totalDistance = totalDistance;
}
totalDistance =
totalDistance +
getDistanceBetweenTwoPoints(coordinates[i], coordinates[i + 1]);
}
return totalDistance.toFixed(2);
}
使用Haversine公式，源代码:
//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
//::: :::
//::: This routine calculates the distance between two points (given the :::
//::: latitude/longitude of those points). It is being used to calculate :::
//::: the distance between two locations using GeoDataSource (TM) prodducts :::
//::: :::
//::: Definitions: :::
//::: South latitudes are negative, east longitudes are positive :::
//::: :::
//::: Passed to function: :::
//::: lat1, lon1 = Latitude and Longitude of point 1 (in decimal degrees) :::
//::: lat2, lon2 = Latitude and Longitude of point 2 (in decimal degrees) :::
//::: unit = the unit you desire for results :::
//::: where: 'M' is statute miles (default) :::
//::: 'K' is kilometers :::
//::: 'N' is nautical miles :::
//::: :::
//::: Worldwide cities and other features databases with latitude longitude :::
//::: are available at https://www.geodatasource.com :::
//::: :::
//::: For enquiries, please contact sales@geodatasource.com :::
//::: :::
//::: Official Web site: https://www.geodatasource.com :::
//::: :::
//::: GeoDataSource.com (C) All Rights Reserved 2018 :::
//::: :::
//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
function distance(lat1, lon1, lat2, lon2, unit) {
if ((lat1 == lat2) && (lon1 == lon2)) {
return 0;
}
else {
var radlat1 = Math.PI * lat1/180;
var radlat2 = Math.PI * lat2/180;
var theta = lon1-lon2;
var radtheta = Math.PI * theta/180;
var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
if (dist > 1) {
dist = 1;
}
dist = Math.acos(dist);
dist = dist * 180/Math.PI;
dist = dist * 60 * 1.1515;
if (unit=="K") { dist = dist * 1.609344 }
if (unit=="N") { dist = dist * 0.8684 }
return dist;
}
}
样例代码是根据LGPLv3许可的。
我试着通过命名变量让代码更容易理解，
我希望这能有所帮助
function getDistanceFromLatLonInKm(point1, point2) {
const [lat1, lon1] = point1;
const [lat2, lon2] = point2;
const earthRadius = 6371;
const dLat = convertDegToRad(lat2 - lat1);
const dLon = convertDegToRad(lon2 - lon1);
const squarehalfChordLength =
Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(convertDegToRad(lat1)) * Math.cos(convertDegToRad(lat2)) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
const angularDistance = 2 * Math.atan2(Math.sqrt(squarehalfChordLength), Math.sqrt(1 - squarehalfChordLength));
const distance = earthRadius * angularDistance;
return distance;
}
谷歌的答案
https://cloud.google.com/blog/products/maps-platform/how-calculate-distances-map-maps-javascript-api
支票由三部分组成。
https://www.distancefromto.net/
static distance({ x: x1, y: y1 }, { x: x2, y: y2 }) {
function toRadians(value) {
return value * Math.PI / 180
}
var R = 6371.0710
var rlat1 = toRadians(x1) // Convert degrees to radians
var rlat2 = toRadians(x2) // Convert degrees to radians
var difflat = rlat2 - rlat1 // Radian difference (latitudes)
var difflon = toRadians(y2 - y1) // Radian difference (longitudes)
return 2 * R * Math.asin(Math.sqrt(Math.sin(difflat / 2) * Math.sin(difflat / 2) + Math.cos(rlat1) * Math.cos(rlat2) * Math.sin(difflon / 2) * Math.sin(difflon / 2)))
}
你使用的是哈弗辛公式，它计算了当乌鸦飞行时球面上两点之间的距离。您提供的谷歌Maps链接显示距离为2.2公里，因为它不是一条直线。
Wolfram Alpha是进行地理计算的一个很好的资源，它还显示了这两点之间的距离为1.652公里。
如果您正在寻找直线距离(如crow文件)，则您的函数工作正常。如果你想要的是开车距离(或骑自行车距离、公共交通距离或步行距离)，你必须使用一个映射API(谷歌或Bing是最流行的)来获得适当的路线，其中将包括距离。
顺便提一下，谷歌Maps API在其Google . Maps .geometry.spherical命名空间(查找computeDistanceBetween)中提供了一个打包的球面距离方法。这可能比你自己卷更好(首先，它使用了更精确的地球半径值)。
对于挑剔的我们来说，我说的“直线距离”，指的是“球面上的直线”，实际上当然是一条曲线(即大圆距离)。
