来自NMEA日志文件的两个经线和长绳之间的c#计算器轴承

本文关键字：之间 计算器 两个 文件 日志 NMEA 来自 | 更新日期: 2023-09-27 17:51:09

从GPS日志中摘录如下:

$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

我将记录的GPS方位与最后和当前位置之间的计算方位进行比较，并使用以下代码循环每行:

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;

的结果如下:

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

是不是我的计算有误，因为它们似乎都没有接近GPS记录的101度左右的方位?

谢谢

来自NMEA日志文件的两个经线和长绳之间的c#计算器轴承

我在代码中发现了一些问题，首先，在解释纬度和经度时，您应该查看地球的哪个象限的位置落入并转换为负的南方或西方位置:

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;

其余的问题源于将度数而不是弧度传递给数学函数，而经度增量的计算似乎是相反的。我引入了一些辅助函数，并将这段代码重写如下:

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轴承尚未确定的第一个无效数据:

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

我注意到从GGA速度，该单位似乎一直是静止或移动非常缓慢。在这种情况下，一些GPS接收器会过滤或保留航向信息，因此可以预期会有一些变化。修改后，我从一辆行驶的车辆上获取了一些GPS数据，结果相差在1度以内。

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