两个纬度和经度之间的中点

Question

我正在尝试将http://www.movable-type.co.uk/scripts/latlong.html 中给出的代码 sn-p 转换为 java。但我没有得到与网站相同的结果。这是我的代码，用于查找给定纬度和经度的两点之间的中点

midPoint(12.870672,77.658964,12.974831,77.60935);
    public static void midPoint(double lat1,double lon1,double lat2,double lon2)
    {
   double dLon = Math.toRadians(lon2-lon1);
        double Bx = Math.cos(lat2) * Math.cos(dLon);
        double By = Math.cos(lat2) * Math.sin(dLon);
        double lat3 = Math.atan2(Math.sin(lat1)+Math.sin(lat2),Math.sqrt( (Math.cos(lat1)+Bx)*(Math.cos(lat1)+Bx) + By*By) );
        double lon3 = lon1 + Math.atan2(By, Math.cos(lat1) + Bx);
        System.out.print(lat3 +" " + lon3 );
    }

我不确定 dLon 是否正确。所以请大家帮我弄清楚。 P.S.我需要找到中点的经纬度

Answer 1

您需要转换为弧度。将其更改为以下内容：

public static void midPoint(double lat1,double lon1,double lat2,double lon2){

    double dLon = Math.toRadians(lon2 - lon1);

    //convert to radians
    lat1 = Math.toRadians(lat1);
    lat2 = Math.toRadians(lat2);
    lon1 = Math.toRadians(lon1);

    double Bx = Math.cos(lat2) * Math.cos(dLon);
    double By = Math.cos(lat2) * Math.sin(dLon);
    double lat3 = Math.atan2(Math.sin(lat1) + Math.sin(lat2), Math.sqrt((Math.cos(lat1) + Bx) * (Math.cos(lat1) + Bx) + By * By));
    double lon3 = lon1 + Math.atan2(By, Math.cos(lat1) + Bx);

    //print out in degrees
    System.out.println(Math.toDegrees(lat3) + " " + Math.toDegrees(lon3));
}

Answer 2

Android Google Maps Utilities 更容易：

LatLngBounds bounds = new LatLngBounds(start, dest);
bounds.getCenter();

更新： 更好地使用构建器（为什么参见Bad Losers Answer）：

LatLngBounds.builder().include(start).include(dest).build().getCenter();

Answer 3

如果您想正确处理反子午线（经度 +/-180 ）。

这是说明问题的测试：

LatLng mp = midPoint(new LatLng(-43.95139,-176.56111),new LatLng(-36.397816,174.663496));
public static LatLng midPoint (LatLng SW, LatLng NE) {
    LatLngBounds bounds = new LatLngBounds(SW, NE);
    Log.d("BAD!", bounds.toString() + " CENTRE: " + bounds.getCenter().toString());
    bounds = LatLngBounds.builder().include(SW).include(NE).build();
    Log.d("GOOD", bounds.toString() + " CENTRE: " + bounds.getCenter().toString());
    return bounds.getCenter();
}

实际结果：

BAD!: LatLngBounds{southwest=lat/lng: (-43.95139,-176.56111), northeast=lat/lng: (-36.397816,174.663496)} CENTRE: lat/lng: (-40.174603,-0.948807)
GOOD: LatLngBounds{southwest=lat/lng: (-43.95139,174.663496), northeast=lat/lng: (-36.397816,-176.56111)} CENTRE: lat/lng: (-40.174603,179.051193)

构造函数技术产生一个中心经度向外 180 度！

Answer 4

您还需要将其他公式中使用的纬度和经度值转换为弧度。您可以在页面下方约 3/5 的代码中看到这一点。余弦距离公式的球面定律最后给出了线索​​：

（请注意，此处和所有后续代码片段中，为简单起见，我没有显示从度数到弧度的转换；完整版本请参见下文）。

Answer 5

以下是@dogbane 的java 代码转换为Kotlin：

private fun midPoint(lat1: Double, lon1: Double, lat2: Double, lon2: Double) : String {
    var lat1 = lat1
    var lon1 = lon1
    var lat2 = lat2
    val dLon: Double = Math.toRadians(lon2 - lon1)
    //convert to radians
    lat1 = Math.toRadians(lat1)
    lat2 = Math.toRadians(lat2)
    lon1 = Math.toRadians(lon1)
    val Bx: Double = Math.cos(lat2) * Math.cos(dLon)
    val By: Double = Math.cos(lat2) * Math.sin(dLon)
    val lat3: Double = Math.atan2(Math.sin(lat1) + Math.sin(lat2), Math.sqrt((Math.cos(lat1) + Bx) * (Math.cos(lat1) + Bx) + By * By))
    val lon3: Double = lon1 + Math.atan2(By, Math.cos(lat1) + Bx)
    var result: String = ""
    result = Math.toDegrees(lat3).toString() + "," + Math.toDegrees(lon3).toString()
    return result;
}

Answer 6

这是@dogbane 的Java 代码转换为TypeScript。

type LatLng = {
  lat: number;
  lng: number;
};

function calculateMidPoint(latLngA: LatLng, latLngB: LatLng) {
  function toRadians(degress: number): number {
    return degress * (Math.PI / 180);
  }

  function toDegrees(radians: number): string {
    return (radians * (180 / Math.PI)).toFixed(4);
  }

  const lngDiff = toRadians(latLngB.lng - latLngA.lng);
  const latA = toRadians(latLngA.lat);
  const latB = toRadians(latLngB.lat);
  const lngA = toRadians(latLngA.lng);

  const bx = Math.cos(latB) * Math.cos(lngDiff);
  const by = Math.cos(latB) * Math.sin(lngDiff);

  const latMidway = toDegrees(
    Math.atan2(
      Math.sin(latA) + Math.sin(latB),
      Math.sqrt((Math.cos(latA) + bx) * (Math.cos(latA) + bx) + by * by)
    )
  );
  const lngMidway = toDegrees(lngA + Math.atan2(by, Math.cos(latA) + bx));

  console.log(
    `Midway point between ${latLngA} and ${latLngB} is: Lat: ${latMidway}, lng: ${lngMidway}`
  );
}

Answer 7

我的上一份工作是制作一个跟踪模块，我使用这个公式来计算 2 个坐标之间的距离。

//Location lat and lon
double locLat = -23.548333;
double locLon = -46.636111;

//Destination lat and lon
double dstLat = -22.902778;
double dstLon = -43.206667;

double arcoAB = 90 - (dstLat);
double arcoAC = 90 - (locLat);

double difLon = locLon - (dstLon);

double cosA = Math.cos(Math.toRadians(arcoAC)) * Math.cos(Math.toRadians(arcoAB)) + Math.sin(Math.toRadians(arcoAC)) * Math.sin(Math.toRadians(arcoAB)) * Math.cos(Math.toRadians(difLon));
double acosCosA = Math.toDegrees(Math.acos(cosA));

double raio = 2 * Math.PI * 6371;
double distance = (raio * acosCosA) / 360;

return distance; //Distance in KM, convert to anything else (miles, meters..) if you need..

你可以得到中点除以距离2。

啊，这另一个公式也有效：

double dLat = Math.toRadians(dstLat - locLat);
double dLon = Math.toRadians(dstLon - locLon);

double a = Math.sin(dLat / 2) * Math.sin(dLat / 2)
           + Math.cos(Math.toRadians(locLat)) * Math.cos(Math.toRadians(dstLat))
           * Math.sin(dLon / 2) * Math.sin(dLon / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
double d = 6371 * c;

return d; //Distance in KM