geolocation - 如何使用方向角和速度计算下一次的经纬度

Question

我知道我现在的位置({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？

Answer 1

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，请参阅函数 LatLon.prototype.destinationPoint在 this page

更新 对于那些希望对上述内容进行更充实的实现的人，这里是在 Javascript 中:

  /**
  * 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°
  }