16

我知道我当前的位置（{lat:x,lon:y}），我知道我的速度和方向角；如何预测下一次的下一个位置？

4

5 回答 5

14

首先，根据您当前的速度和已知的时间间隔（“下一次”）计算您将行驶的距离：

distance = speed * time

然后你可以使用这个公式来计算你的新位置（lat2/lon2）：

lat2 =asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
dlon=atan2(sin(tc)*sin(d)*cos(lat1),cos(d)-sin(lat1)*sin(lat2))
lon2=mod( lon1-dlon +pi,2*pi )-pi

有关 Javascript 中的实现，请参阅此页面LatLon.prototype.destinationPoint上的函数

对于那些希望更充实地实现上述内容的人的更新，这里是 Javascript：

  /**
  * Returns the destination point from a given point, having travelled the given distance
  * on the given initial bearing.
  *
  * @param   {number} lat - initial latitude in decimal degrees (eg. 50.123)
  * @param   {number} lon - initial longitude in decimal degrees (e.g. -4.321)
  * @param   {number} distance - Distance travelled (metres).
  * @param   {number} bearing - Initial bearing (in degrees from north).
  * @returns {array} destination point as [latitude,longitude] (e.g. [50.123, -4.321])
  *
  * @example
  *     var p = destinationPoint(51.4778, -0.0015, 7794, 300.7); // 51.5135°N, 000.0983°W
  */
  function destinationPoint(lat, lon, distance, bearing) {
     var radius = 6371e3; // (Mean) radius of earth

     var toRadians = function(v) { return v * Math.PI / 180; };
     var toDegrees = function(v) { return v * 180 / Math.PI; };

     // sinφ2 = sinφ1·cosδ + cosφ1·sinδ·cosθ
     // tanΔλ = sinθ·sinδ·cosφ1 / cosδ−sinφ1·sinφ2
     // see mathforum.org/library/drmath/view/52049.html for derivation

     var δ = Number(distance) / radius; // angular distance in radians
     var θ = toRadians(Number(bearing));

     var φ1 = toRadians(Number(lat));
     var λ1 = toRadians(Number(lon));

     var sinφ1 = Math.sin(φ1), cosφ1 = Math.cos(φ1);
     var sinδ = Math.sin(δ), cosδ = Math.cos(δ);
     var sinθ = Math.sin(θ), cosθ = Math.cos(θ);

     var sinφ2 = sinφ1*cosδ + cosφ1*sinδ*cosθ;
     var φ2 = Math.asin(sinφ2);
     var y = sinθ * sinδ * cosφ1;
     var x = cosδ - sinφ1 * sinφ2;
     var λ2 = λ1 + Math.atan2(y, x);

     return [toDegrees(φ2), (toDegrees(λ2)+540)%360-180]; // normalise to −180..+180°
  }
于 2013-10-14T08:35:00.443 回答
7

在 JS 中计算 lat 和 lng 给定的方位和距离：

//lat, lng in degrees. Bearing in degrees. Distance in Km
calculateNewPostionFromBearingDistance = function(lat, lng, bearing, distance) {
  var R = 6371; // Earth Radius in Km

  var lat2 = Math.asin(Math.sin(Math.PI / 180 * lat) * Math.cos(distance / R) + Math.cos(Math.PI / 180 * lat) * Math.sin(distance / R) * Math.cos(Math.PI / 180 * bearing));
  var lon2 = Math.PI / 180 * lng + Math.atan2(Math.sin( Math.PI / 180 * bearing) * Math.sin(distance / R) * Math.cos( Math.PI / 180 * lat ), Math.cos(distance / R) - Math.sin( Math.PI / 180 * lat) * Math.sin(lat2));

  return [180 / Math.PI * lat2 , 180 / Math.PI * lon2];
};

calculateNewPostionFromBearingDistance(60,25,30,1)
[60.007788047871614, 25.008995333937197]
于 2013-10-24T09:03:32.413 回答
4

Java中的相同代码：

    final double r = 6371 * 1000; // Earth Radius in m

    double lat2 = Math.asin(Math.sin(Math.toRadians(lat)) * Math.cos(distance / r)
            + Math.cos(Math.toRadians(lat)) * Math.sin(distance / r) * Math.cos(Math.toRadians(bearing)));
    double lon2 = Math.toRadians(lon)
            + Math.atan2(Math.sin(Math.toRadians(bearing)) * Math.sin(distance / r) * Math.cos(Math.toRadians(lat)), Math.cos(distance / r)
            - Math.sin(Math.toRadians(lat)) * Math.sin(lat2));
    lat2 = Math.toDegrees( lat2);
    lon2 = Math.toDegrees(lon2);
于 2017-10-31T14:04:47.410 回答
1

这段代码对我有用：
1. 我们必须计算距离（速度 * 时间）。
2.代码将距离转换为KM，因为在KM中也使用了earthradius。 

            const double radiusEarthKilometres = 6371.01f;

            kmDistance = kmSpeed * (timer1.Interval / 1000f) / 3600f;

            var distRatio = kmDistance / radiusEarthKilometres;
            var distRatioSine = Math.Sin(distRatio);
            var distRatioCosine = Math.Cos(distRatio);

            var startLatRad = deg2rad(lat0);
            var startLonRad = deg2rad(lon0);

            var startLatCos = Math.Cos(startLatRad);
            var startLatSin = Math.Sin(startLatRad);

            var endLatRads = Math.Asin((startLatSin * distRatioCosine) + (startLatCos * distRatioSine * Math.Cos(angleRadHeading)));

            var endLonRads = startLonRad
                + Math.Atan2(Math.Sin(angleRadHeading) * distRatioSine * startLatCos,
                    distRatioCosine - startLatSin * Math.Sin(endLatRads));

            newLat = rad2deg(endLatRads);
            newLong = rad2deg(endLonRads);
于 2013-12-17T03:02:26.217 回答
1

根据@clody96 和@mike 的回答，这是R中的一个实现，它使用具有速度和时间步长而不是距离的data.frame：

points = data.frame(
  lon = seq(11, 30, 1),
  lat = seq(50, 59.5, 0.5),
  bea = rep(270, 20),
  time = rep(60,20),
  vel = runif(20,1000, 3000)
)

## lat, lng in degrees. Bearing in degrees. Distance in m
calcPosBear = function(df) {
  earthR = 6371000; 

  ## Units meter, seconds and meter/seconds
  df$dist = df$time * df$vel

  lat2 = asin(sin(
     pi / 180 * df$lat) * 
      cos(df$dist / earthR) + 
      cos(pi / 180 * df$lat) * 
      sin(df$dist / earthR) * 
      cos(pi / 180 * df$bea));

  lon2 = pi / 180 * df$lon + 
    atan2(sin( pi / 180 * df$bea) * 
                 sin(df$dist / earthR) * 
                 cos( pi / 180 * df$lat ), 
               cos(df$dist / earthR) - 
                 sin( pi / 180 * df$lat) * 
                 sin(lat2));

  df$latR = (180 * lat2) / pi
  df$lonR = (180 * lon2) / pi

  return(df);
};

df = calcPosBear(points)
plot(df$lon, df$lat)
points(df$lonR, df$latR, col="red")

这带来了与@clody96 相同的结果：

points = data.frame(
  lon = 25,
  lat = 60,
  bea = 30,
  time = 1000,
  vel = 1
)
df = calcPosBear(points)
df

  lon lat bea time vel dist        latR        lonR
1  25  60  30 1000   1 1000 60.00778805 25.00899533
于 2018-06-16T13:42:16.567 回答