如果你想要驾驶距离/路线(张贴在这里，因为这是谷歌上两点之间距离的第一个结果，但对大多数人来说，驾驶距离更有用)，你可以使用谷歌地图距离矩阵服务:

getDrivingDistanceBetweenTwoLatLong(origin, destination) {

 return new Observable(subscriber => {
  let service = new google.maps.DistanceMatrixService();
  service.getDistanceMatrix(
    {
      origins: [new google.maps.LatLng(origin.lat, origin.long)],
      destinations: [new google.maps.LatLng(destination.lat, destination.long)],
      travelMode: 'DRIVING'
    }, (response, status) => {
      if (status !== google.maps.DistanceMatrixStatus.OK) {
        console.log('Error:', status);
        subscriber.error({error: status, status: status});
      } else {
        console.log(response);
        try {
          let valueInMeters = response.rows[0].elements[0].distance.value;
          let valueInKms = valueInMeters / 1000;
          subscriber.next(valueInKms);
          subscriber.complete();
        }
       catch(error) {
        subscriber.error({error: error, status: status});
       }
      }
    });
});
}

计算距离——尤其是大距离——的主要挑战之一是解释地球的曲率。如果地球是平的，计算两点之间的距离就会像计算直线一样简单!哈弗辛公式包括一个常数(下面是R变量)，它表示地球的半径。根据你是用英里还是公里来测量，它分别等于3956英里或6367公里。 基本公式是:

Dlon = lon2 - lon1 dat = lat2 - lat1 = (sin (dlat / 2)) ^ 2 + cos (lat1) * cos (lat2) * (sin (dlon / 2)) ^ 2 C = 2 * atan2(√(a)，√(1-a)) distance = R * c(其中R为地球半径) R = 6367公里OR 3956英里

     lat1, lon1: The Latitude and Longitude of point 1 (in decimal degrees)
     lat2, lon2: The Latitude and Longitude of point 2 (in decimal degrees)
     unit: The unit of measurement in which to calculate the results where:
     'M' is statute miles (default)
     'K' is kilometers
     'N' is nautical miles

样本

function distance(lat1, lon1, lat2, lon2, unit) {
    try {
        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);
        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
    } catch (err) {
        console.log(err);
    }
}

function getDistanceFromLatLonInKm(position1, position2) {
    "use strict";
    var deg2rad = function (deg) { return deg * (Math.PI / 180); },
        R = 6371,
        dLat = deg2rad(position2.lat - position1.lat),
        dLng = deg2rad(position2.lng - position1.lng),
        a = Math.sin(dLat / 2) * Math.sin(dLat / 2)
            + Math.cos(deg2rad(position1.lat))
            * Math.cos(deg2rad(position2.lat))
            * Math.sin(dLng / 2) * Math.sin(dLng / 2),
        c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    return R * c;
}

console.log(getDistanceFromLatLonInKm(
    {lat: 48.7931459, lng: 1.9483572},
    {lat: 48.827167, lng: 2.2459745}
));

非常感谢这一切。我在Objective-C iPhone应用程序中使用了以下代码:

const double PIx = 3.141592653589793;
const double RADIO = 6371; // Mean radius of Earth in Km

double convertToRadians(double val) {

   return val * PIx / 180;
}

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

        double dlon = convertToRadians(place2.longitude - place1.longitude);
        double dlat = convertToRadians(place2.latitude - place1.latitude);

        double a = ( pow(sin(dlat / 2), 2) + cos(convertToRadians(place1.latitude))) * cos(convertToRadians(place2.latitude)) * pow(sin(dlon / 2), 2);
        double angle = 2 * asin(sqrt(a));

        return angle * RADIO;
}

纬度和经度是十进制的。我没有在asin()调用中使用min()，因为我使用的距离非常小，以至于它们不需要min()。

它给出了错误的答案，直到我传入弧度的值-现在它几乎与从苹果地图应用程序中获得的值相同:-)

额外的更新:

如果你使用的是iOS4或更高版本，那么苹果会提供一些方法来实现相同的功能:

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

    MKMapPoint  start, finish;


    start = MKMapPointForCoordinate(place1);
    finish = MKMapPointForCoordinate(place2);

    return MKMetersBetweenMapPoints(start, finish) / 1000;
}

这是一个简单的PHP函数，它将给出一个非常合理的近似值(误差小于+/-1%)。

<?php
function distance($lat1, $lon1, $lat2, $lon2) {

    $pi80 = M_PI / 180;
    $lat1 *= $pi80;
    $lon1 *= $pi80;
    $lat2 *= $pi80;
    $lon2 *= $pi80;

    $r = 6372.797; // mean radius of Earth in km
    $dlat = $lat2 - $lat1;
    $dlon = $lon2 - $lon1;
    $a = sin($dlat / 2) * sin($dlat / 2) + cos($lat1) * cos($lat2) * sin($dlon / 2) * sin($dlon / 2);
    $c = 2 * atan2(sqrt($a), sqrt(1 - $a));
    $km = $r * $c;

    //echo '<br/>'.$km;
    return $km;
}
?>

如前所述;地球不是一个球体。它就像马克·麦奎尔决定用来练习的一个很旧很旧的棒球——到处都是凹痕和凸起。简单的计算(像这样)把它当作一个球体。

不同的方法或多或少的精确取决于你在这个不规则的卵形上的位置以及你的点之间的距离(它们越近，绝对误差范围就越小)。你的期望越精确，计算就越复杂。

更多信息:维基百科地理距离

下面是Haversine公式的typescript实现

static getDistanceFromLatLonInKm(lat1: number, lon1: number, lat2: number, lon2: number): number {
    var deg2Rad = deg => {
        return deg * Math.PI / 180;
    }

    var r = 6371; // Radius of the earth in km
    var dLat = deg2Rad(lat2 - lat1);   
    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;
}