来自 NMEA 日志文件的两个纬度和经度坐标之间的 c# 计算器轴承？

Question

从 GPS 日志中提取以下内容：

$GPGGA,153500.009,5137.2603,N,00244.8715,W,1,10,0.8,50.6,M,51.4,M,,0000*71
$GPRMC,153500.009,A,5137.2603,N,00244.8715,W,037.7,101.7,300912,,,A*74
$GPGGA,153500.059,5137.2601,N,00244.8706,W,1,10,0.8,50.6,M,51.4,M,,0000*74
$GPRMC,153500.059,A,5137.2601,N,00244.8706,W,038.0,101.8,300912,,,A*76
$GPGGA,153500.109,5137.2600,N,00244.8697,W,1,10,0.8,50.6,M,51.4,M,,0000*78
$GPRMC,153500.109,A,5137.2600,N,00244.8697,W,038.3,101.9,300912,,,A*78
$GPGGA,153500.159,5137.2599,N,00244.8688,W,1,10,0.8,50.5,M,51.4,M,,0000*73
$GPRMC,153500.159,A,5137.2599,N,00244.8688,W,038.6,101.9,300912,,,A*75
$GPGGA,153500.209,5137.2597,N,00244.8679,W,1,10,0.8,50.5,M,51.4,M,,0000*75
$GPRMC,153500.209,A,5137.2597,N,00244.8679,W,038.9,102.0,300912,,,A*76

我正在将记录的 GPS 方位与最后位置和当前位置之间的计算方位与循环遍历每一行的以下代码进行比较：

string[] splitline = line.Split(',');
course = Convert.ToDouble(splitline[8]);
Lat = Convert.ToDouble(splitline[3]);
Long = Convert.ToDouble(splitline[5]);

LatDeg = (Convert.ToInt16(Lat) / 100) + (Lat - (Convert.ToInt16(Lat) / 100) * 100) / 60;
LongDeg = (Convert.ToInt16(Long) / 100) + (Long - (Convert.ToInt16(Long) / 100) * 100) / 60;
lastLatDeg = (Convert.ToInt16(lastLat) / 100) + (lastLat - (Convert.ToInt16(lastLat) / 100) * 100) / 60;
lastLongDeg = (Convert.ToInt16(lastLong) / 100) + (lastLong - (Convert.ToInt16(lastLong) / 100) * 100) / 60;

var dLon = lastLongDeg - LongDeg;
var y = Math.Sin(dLon) * Math.Cos(lastLatDeg);
var x = Math.Cos(lastLatDeg) * Math.Sin(LatDeg) - Math.Sin(lastLatDeg) *           Math.Cos(LatDeg) * Math.Cos(dLon);
Console.WriteLine(DEG_PER_RAD * Math.Atan2(y, x));
Console.WriteLine("> " + course + " <");

lastLat = Lat;
lastLong = Long;
lastcourse = course;

结果如下：

136.131182151555
> 101.8 <
117.480364881602
> 101.9 <
117.480186101881
> 101.9 <
136.130309531745
> 102 <
117.479649572813
> 102 <

我的计算结果是否正确，因为其中 none 似乎接近 101 度左右的 gps 记录方位角？

谢谢

Answer 1

我在代码中发现了一些问题，首先，在解释纬度和经度时，您应该查看位置落入地球的哪个四边形运行t 并将南方转换为负数或西部地区：

Lat = Convert.ToDouble(splitline[3]);
if (splitline[4] == "S")
    Lat = 0.0 - Lat;
Long = Convert.ToDouble(splitline[5]);
if (splitline[6] == "W")
    Long = 0.0 - Long;

其余问题源于将度数而不是弧度传递给数学函数，并且经度增量的计算似乎相反。我引入了一些辅助函数并将该部分代码重写如下：

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

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

double dLon = DegreesToRadians(LongDeg - lastLongDeg);
double y = Math.Sin(dLon) * Math.Cos(DegreesToRadians(lastLatDeg));
double x = Math.Cos(DegreesToRadians(lastLatDeg)) * Math.Sin(DegreesToRadians(LatDeg)) - Math.Sin(DegreesToRadians(lastLatDeg)) * Math.Cos(DegreesToRadians(LatDeg)) * Math.Cos(dLon);
Console.WriteLine((RadiansToDegrees(Math.Atan2(y, x)) + 360.0) % 360);
Console.WriteLine("> " + course + " <");

这给了我以下结果，您的测试数据忽略了尚未确定轴承的第一个无效数据：

109.693614586392
> 101.8 <
100.14641169874
> 101.9 <
100.146411372034
> 101.9 <
109.693611985053
> 102 <

我从 GGA 速度中注意到，该装置似乎一直处于静止状态或移动速度非常慢。在这些情况下，某些 GPS 接收器会过滤或保留航向信息，因此可能会出现一些变化。更改后，我运行通过我从行驶中的车辆获得的一些 GPS 数据，结果彼此相差不到一个度数。

来自 NMEA 日志文件的两个纬度和经度坐标之间的 c# 计算器轴承？

c# calculater bearing between two lat and long coords from NMEA log file?

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gps