“移动”从 LatLonA 到 LatLonB (WGS84) # 米（地面以上）

发布于 2024-12-10 15:24:54 字数 467 浏览 0 评论 0原文

我需要 C# 中的一个函数来执行以下操作：从 GPS 点 A 向 GPS 点 B 的方向移动 50 米，并计算该点的 GPS 坐标。

例如，我有两个坐标：

LatLon LatLonA = new LatLon(51.83966, 5.04631); // Latitude 51.83966, Longitude 5.04631
LatLon LatLonB = new LatLon(51.84172, 5.01961); // Latitude 51.84172, Longitude 5.01961

我想要的是这样的函数：

function LatLon MoveTowards(LatLon A, LatLon B, double MetersOverGround) 
{
    //code here
}

该函数将返回在 B 方向上距 A x 米的坐标。

原文

I need a function in C# to do the following: move 50 meters from gps-point A in the direction of gps-point B and calculate the GPS-coordinates for that point.

For instance I've got two coordinates:

LatLon LatLonA = new LatLon(51.83966, 5.04631); // Latitude 51.83966, Longitude 5.04631
LatLon LatLonB = new LatLon(51.84172, 5.01961); // Latitude 51.84172, Longitude 5.01961

What I would like is a function like this:

function LatLon MoveTowards(LatLon A, LatLon B, double MetersOverGround) 
{
    //code here
}

That function would return the coordinate which is x meters away from A in the direction of B.

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半世蒼涼 2024-12-17 15:24:54

地球不是球体，甚至也不是椭圆形。在不购买商业图书馆的情况下，您所能期望的最好结果就是一个近似值（对于大多数人来说这已经足够好了）。

您可以首先研究Haversine 公式，以及此页面将会有很大帮助。

或者，如果您想要一个商业库，我已经使用 ProLat 取得了巨大成功

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红ご颜醉 2024-12-17 15:24:54

这就是你想要的。只需使用 Math.Atan2 即可获取 A 到 B 向量的方位角并获取 bearing 参数。

    /// <summary>
    /// Calculates the end-point from a given source at a given range (meters) and bearing (degrees).
    /// This methods uses simple geometry equations to calculate the end-point.
    /// </summary>
    /// <param name="source">Point of origin</param>
    /// <param name="range">Range in meters</param>
    /// <param name="bearing">Bearing in degrees</param>
    /// <returns>End-point from the source given the desired range and bearing.</returns>
    public static PointLatLng CalculateDerivedPosition(PointLatLng source, double range, double bearing)
    {
        const double DEGREES_TO_RADIANS = Math.PI/180;
        const double EARTH_RADIUS_M = 6371000;
        double latA = source.Lat * DEGREES_TO_RADIANS;
        double lonA = source.Lng * DEGREES_TO_RADIANS;
        double angularDistance = range / EARTH_RADIUS_M;
        double trueCourse = bearing * DEGREES_TO_RADIANS;

        double lat = Math.Asin(
            Math.Sin(latA) * Math.Cos(angularDistance) +
            Math.Cos(latA) * Math.Sin(angularDistance) * Math.Cos(trueCourse));

        double dlon = Math.Atan2(
            Math.Sin(trueCourse) * Math.Sin(angularDistance) * Math.Cos(latA),
            Math.Cos(angularDistance) - Math.Sin(latA) * Math.Sin(lat));

        double lon = ((lonA + dlon + Math.PI) % (Math.PI * 2)) - Math.PI;

        return new PointLatLng(
            lat / DEGREES_TO_RADIANS,
            lon / DEGREES_TO_RADIANS);
    }

Here is what you want. Just use Math.Atan2 to obtain the bearing of your A-to-B vector and obtain the bearing parameter.

    /// <summary>
    /// Calculates the end-point from a given source at a given range (meters) and bearing (degrees).
    /// This methods uses simple geometry equations to calculate the end-point.
    /// </summary>
    /// <param name="source">Point of origin</param>
    /// <param name="range">Range in meters</param>
    /// <param name="bearing">Bearing in degrees</param>
    /// <returns>End-point from the source given the desired range and bearing.</returns>
    public static PointLatLng CalculateDerivedPosition(PointLatLng source, double range, double bearing)
    {
        const double DEGREES_TO_RADIANS = Math.PI/180;
        const double EARTH_RADIUS_M = 6371000;
        double latA = source.Lat * DEGREES_TO_RADIANS;
        double lonA = source.Lng * DEGREES_TO_RADIANS;
        double angularDistance = range / EARTH_RADIUS_M;
        double trueCourse = bearing * DEGREES_TO_RADIANS;

        double lat = Math.Asin(
            Math.Sin(latA) * Math.Cos(angularDistance) +
            Math.Cos(latA) * Math.Sin(angularDistance) * Math.Cos(trueCourse));

        double dlon = Math.Atan2(
            Math.Sin(trueCourse) * Math.Sin(angularDistance) * Math.Cos(latA),
            Math.Cos(angularDistance) - Math.Sin(latA) * Math.Sin(lat));

        double lon = ((lonA + dlon + Math.PI) % (Math.PI * 2)) - Math.PI;

        return new PointLatLng(
            lat / DEGREES_TO_RADIANS,
            lon / DEGREES_TO_RADIANS);
    }

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