如何在C#中计算距离一点但朝向另一点的坐标X米

Question

假设我有两个相距约222.33米的坐标：

A: 49.25818, -123.20626
B: 49.25813, -123.2032

这两个点构成一个线段。

如何计算距离A或BX米但朝向另一点的Z点的坐标？

我已经使用System.Device.Location库知道了我的两个点之间的距离。

GeoCoordinate A = new GeoCoordinate(49.25818, -123.20626);
GeoCoordinate B = new GeoCoordinate(49.25813, -123.2032);
var distanceInMeters = A.GetDistanceTo(B);
// distanceInMeters = 222.33039783713738

我在找这样的东西：

GeoCoordinate GetPointTowards(GeoCoordinate fromPoint, GeoCoordinate towardPoint, double distanceInMeter) {
[???]
}

我想我可能需要轴承或其他东西才能获得新的点位置。

我发现的大多数例子都是针对iOS、Android或带有特定库的GMap的。。

Answer 1

以下是我将如何做到这一点的概述。使用这种方法，无需明确处理坐标和距离之间的单位差异，因为采用目标与总距离的比率可以消除单位。

totalDistance = distance in meters between point A and point B.
targetDistance = distance in meters to travel from point A to point B
ratio = targetDistance / totalDistance
diffX = B.X - A.X
diffY = B.Y - A.Y
targetX = A.X + (ratio * diffX)
targetY = A.Y + (ratio * diffY)

但这不能处理边缘情况，比如在经度179度加上3度，就会使你处于经度-178度。

Answer 2

这是我从http://www.movable-type.co.uk/scripts/latlong.html.分数从0到1，是从第一个点到输出位置的第二个点的距离上的分数。您可以随时修改它，使其取直线距离值。

public static (double Lat, double Lon) IntermediatePoint((double Lat, double Lon) StartPoint, (double Lat, double Lon) EndPoint, double fraction)
{
if (fraction < 0 || fraction > 1)
throw new ArgumentOutOfRangeException();
double angDist = Distance(StartPoint, EndPoint) / radius;
double lat1 = StartPoint.Lat * (Math.PI / 180);
double lon1 = StartPoint.Lon * (Math.PI / 180);
double lat2 = EndPoint.Lat * (Math.PI / 180);
double lon2 = EndPoint.Lon * (Math.PI / 180);
double a = Math.Sin((1 - fraction) * angDist) / Math.Sin(angDist);
double b = Math.Sin(fraction * angDist) / Math.Sin(angDist);
double x = a * Math.Cos(lat1) * Math.Cos(lon1) + b * Math.Cos(lat2) * Math.Cos(lon2);
double y = a * Math.Cos(lat1) * Math.Sin(lon1) + b * Math.Cos(lat2) * Math.Sin(lon2);
double z = a * Math.Sin(lat1) + b * Math.Sin(lat2);
double lat3 = Math.Atan2(z, Math.Sqrt(Math.Pow(x, 2) + Math.Pow(y, 2)));
double lon3 = Math.Atan2(y, x);
return (lat3 * (180 / Math.PI), lon3 * (180 / Math.PI));
}
public static double Distance((double Lat, double Lon) point1, (double Lat, double Lon) point2)
{
double φ1 = point1.Lat * (Math.PI / 180.0);
double φ2 = point2.Lat * (Math.PI / 180.0);
double Δφ = (point2.Lat - point1.Lat) * (Math.PI / 180.0);
double Δλ = (point2.Lon - point1.Lon) * (Math.PI / 180.0);
double a = Math.Sin(Δφ / 2) * Math.Sin(Δφ / 2) + Math.Cos(φ1) * Math.Cos(φ2) * Math.Sin(Δλ / 2) * Math.Sin(Δλ / 2);
double c = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
return radius * c;
}

radius是以米为单位表示地球半径的常数。