由网友(过客)分享简介：在导航应用中寻找资源或算法来计算以下内容:Looking for resources or algorithm to calculate the following in a navigation app:如果我当前的 GPS 位置是 (0,0)，并且我以每小时 15 英里的速度前进 32 度，我如何计算我在 10...

在导航应用中寻找资源或算法来计算以下内容:

Looking for resources or algorithm to calculate the following in a navigation app:

如果我当前的 GPS 位置是 (0,0)，并且我以每小时 15 英里的速度前进 32 度，我如何计算我在 10 秒内的位置?

If my current GPS position is (0,0) and I'm heading 32 degrees at 15 miles per hour, how can I calculate what my position will be in 10 seconds?

即:GPSCoordinate predictCoord = GPSCoordinate.FromLatLong(0, 0).AddByMovement(32, 15, TimeSpan.FromSeconds(10));

当前代码基于以下答案:

Current code based on answer below:

public GPSCoordinate AddMovementMilesPerHour(double heading, double speedMph, TimeSpan duration)
{
    double x = speedMph * System.Math.Sin(heading * pi / 180) * duration.TotalSeconds / 3600;
    double y = speedMph * System.Math.Cos(heading * pi / 180) * duration.TotalSeconds / 3600;

    double newLat = this.Latitude + 180 / pi * y / earthRadius;
    double newLong = this.Longitude + 180 / pi / System.Math.Sin(this.Latitude * pi / 180) * x / earthRadius;

    return GPSCoordinate.FromLatLong(newLat, newLong);
}

推荐答案

这是完整的参数答案:

变量:

heading :航向(即从方位角 0° 向后的角度，以度为单位)speed:速度(即速度向量的范数，以英里/小时为单位)lat0, lon0 : 以度为单位的初始坐标dtime : 从起始位置开始的时间间隔，以秒为单位lat, lon : 以度为单位的预测坐标pi : pi 常数 (3.14159...)Rt : 地球半径 miles(6378137.0 米，相当于 3964.037911746 英里) heading : heading (i.e. backwards angle from azimuth 0°, in degrees) speed : velocity (i.e. norm of the speed vector, in miles/hour) lat0, lon0 : initial coordinates in degrees dtime : time interval from the start position, in seconds lat, lon : predicted coordinates in degrees pi : the pi constant (3.14159...) Rt : Earth radius in miles (6378137.0 meters which makes 3964.037911746 miles)

在一个(东、北)局部坐标系中，时间间隔后的位置是:

In an (East, North) local frame, the position after the time interval is :

x = speed * sin(heading*pi/180) * dtime / 3600;
y = speed * cos(heading*pi/180) * dtime / 3600;

(坐标以英里为单位)

您可以从那里计算 WGS84 帧中的新位置(即纬度和经度):

From there you can compute the new position in the WGS84 frame (i.e. latitude and longitude) :

lat = lat0 + 180 / pi * y / Rt;
lon = lon0 + 180 / pi / sin(lat0*pi/180) * x / Rt;

更正最后一行:*sin(phi) 为/sin(phi)

Edit : corrected the last line : *sin(phi) to /sin(phi)