計算終點給定距離，方位，起點

Question

我試圖找到目的地點，給定起點lat/long，距離爲&。下面這個網站的計算器給了我想要的結果。

http://www.movable-type.co.uk/scripts/latlong.html

當我試圖通過代碼來實現一樣的，我沒有得到正確的結果。

下面是我的代碼 -

private GLatLng pointRadialDistance(double lat1, double lon1, 
       double radianBearing, double radialDistance) 
    { 
     double rEarth = 6371.01; 
     lat1 = DegreeToRadian(lat1); 
     lon1 = DegreeToRadian(lon1); 
     radianBearing = DegreeToRadian(radianBearing); 
     radialDistance = radialDistance/rEarth; 
     double lat = Math.Asin(Math.Sin(lat1) * Math.Cos(radialDistance) + Math.Cos(lat1) 
         * Math.Sin(radialDistance) * Math.Cos(radianBearing)); 
     double lon; 
     if (Math.Cos(lat) == 0) 
     { // Endpoint a pole 
      lon = lon1; 
     } 
     else 
     { 
      lon = ((lon1 - Math.Asin(Math.Sin(radianBearing) * Math.Sin(radialDistance)/Math.Cos(lat)) 
          + Math.PI) % (2 * Math.PI)) - Math.PI; 
     } 
     lat = RadianToDegree(lat); 
     lon = RadianToDegree(lon); 
     GLatLng newLatLng = new GLatLng(lat, lon); 
     return newLatLng; 
    } 

    public double Bearing(double lat1, double long1, double lat2, double long2) 
    { 
     //Convert input values to radians 
     lat1 = DegreeToRadian(lat1); 
     long1 = DegreeToRadian(long1); 
     lat2 = DegreeToRadian(lat2); 
     long2 = DegreeToRadian(long2); 

     double deltaLong = long2 - long1; 

     double y = Math.Sin(deltaLong) * Math.Cos(lat2); 
     double x = Math.Cos(lat1) * Math.Sin(lat2) - 
       Math.Sin(lat1) * Math.Cos(lat2) * Math.Cos(deltaLong); 
     double bearing = Math.Atan2(y, x); 
     return bearing; 
    } 

    public double DegreeToRadian(double angle) 
    { 
    return Math.PI * angle/180.0; 
    } 

    public double RadianToDegree(double angle) 
    { 
     return 180.0 * angle/Math.PI; 
    } 

在主程序中，調用子程序如下 -

double bearing = Bearing(-41.294444, 174.814444, -40.90521, 175.6604); 
GLatLng endLatLng = pointRadialDistance(-41.294444, 174.814444, bearing, 80); 

我得到以下結果 -

Bearing=1.02749621782165 
endLatLng=-40.5751022737927,174.797458881699 

答案我預計是-40.939722,175.646389（從上面的網站鏈接）。

任何人都可以提示我在代碼中犯了什麼錯誤嗎？

Answer 1

這裏有一些代碼可以實現你想要做的事情。

public static GeoLocation FindPointAtDistanceFrom(GeoLocation startPoint, double initialBearingRadians, double distanceKilometres) 
{ 
    const double radiusEarthKilometres = 6371.01; 
    var distRatio = distanceKilometres/radiusEarthKilometres; 
    var distRatioSine = Math.Sin(distRatio); 
    var distRatioCosine = Math.Cos(distRatio); 

    var startLatRad = DegreesToRadians(startPoint.Latitude); 
    var startLonRad = DegreesToRadians(startPoint.Longitude); 

    var startLatCos = Math.Cos(startLatRad); 
    var startLatSin = Math.Sin(startLatRad); 

    var endLatRads = Math.Asin((startLatSin * distRatioCosine) + (startLatCos * distRatioSine * Math.Cos(initialBearingRadians))); 

    var endLonRads = startLonRad 
     + Math.Atan2(
      Math.Sin(initialBearingRadians) * distRatioSine * startLatCos, 
      distRatioCosine - startLatSin * Math.Sin(endLatRads)); 

    return new GeoLocation 
    { 
     Latitude = RadiansToDegrees(endLatRads), 
     Longitude = RadiansToDegrees(endLonRads) 
    }; 
} 

public struct GeoLocation 
{ 
    public double Latitude { get; set; } 
    public double Longitude { get; set; } 
} 

public static double DegreesToRadians(double degrees) 
{ 
    const double degToRadFactor = Math.PI/180; 
    return degrees * degToRadFactor; 
} 

public static double RadiansToDegrees(double radians) 
{ 
    const double radToDegFactor = 180/Math.PI; 
    return radians * radToDegFactor; 
} 

Answer 2

幾何庫（V3）很簡單的解決方案，如果你沒有使用谷歌地圖API V3一problema（取決於應用程序 - 實時資產跟蹤，例如 - 免費的許可證是不適用或者你可能不想從V2重構到V3）。

1：

<script type="text/javascript" src="http://maps.google.com/maps/api/js?libraries=geometry&sensor=false"></script> 

第二：建立起點，航向和距離

var nyc = new google.maps.LatLng(40.715, -74.002); 
var distance = 5576673; 
var heading = 51.2145; 

3：去那裏

var endPoint = google.maps.geometry.spherical.computeOffset(nyc, distance, heading); 
var london = new google.maps.Marker({ 
    position: endPoint, 
    map: map 
}); 

與您當前的一起聲明宣佈額外的庫做完了，你現在在倫敦小鎮。瞭解有關computeDistance，computeHeading和computeArea更多信息：

http://www.svennerberg.com/2011/04/calculating-distances-and-areas-in-google-maps-api-3/

http://code.google.com/intl/en/apis/maps/documentation/javascript/geometry.html

Answer 3

下面是JavaScript代碼在http://www.movable-type.co.uk/scripts/latlong.html執行當中我寫了我自己，在我自己的項目中使用。如果您願意，可以將其實施到您的項目中。

注意：座標是一個具有X（經度），Y（緯度），Z（高度）屬性的類。 ToDegree（）和ToRadian（）是Double類型的擴展。最後，GetTarget（）是Coordinate實例的擴展。

/// <summary>Calculates the destination coordinate by given angle and distance.</summary> 
/// <param name="origin">Origin.</param> 
/// <param name="bearing">Azimuth.</param> 
/// <param name="distance">Distance (km).</param> 
/// <returns>Coordinate.</returns> 
public static Coordinate GetTarget(
this Coordinate origin, double bearing, double distance, double altitude = 0) 
{ 
    var d = distance/6371; 
    var rlat = origin.Y.ToRadian(); 
    var rlon = origin.X.ToRadian(); 
    var rbearing = bearing.ToRadian(); 
    var lat2 = rlat + (d * Math.Cos(rbearing)); 
    var dlat = lat2 - rlat; 
    var dphi = Math.Log((Math.Tan((lat2/2) + (Math.PI/4)))/(Math.Tan((rlat/2) + (Math.PI/4)))); 
    var q = 
     Math.Abs(dlat) > 0.0000000001 
     ? dlat/dphi 
     : Math.Cos(rlat); 
    var dlon = (d * Math.Sin(rbearing))/q; 

    if (Math.Abs(lat2) > Math.PI/2) 
    { 
     lat2 = lat2 > 0 ? Math.PI : Math.PI - lat2; 
    } 

    var lon2 = (rlon + dlon + Math.PI) % (2 * Math.PI) - Math.PI; 

    return new Coordinate 
    { 
     X = lon2.ToDegree(), 
     Y = lat2.ToDegree(), 
     Z = origin.Z 
    }; 
}