【发布时间】：2014-09-07 06:38:34
【问题描述】：
我希望得到一条线的距离并开始使用haversine代码。
private static final double _eQuatorialEarthRadius = 6378.1370D;
private static final double _d2r = (Math.PI / 180D);
private static double PRECISION = 0.001;
// Haversine Algorithm
// source: http://stackoverflow.com/questions/365826/calculate-distance-between-2-gps-coordinates
private static double HaversineInM(double lat1, double long1, double lat2, double long2) {
return (1000D * HaversineInKM(lat1, long1, lat2, long2));
}
private static double HaversineInKM(double lat1, double long1, double lat2, double long2) {
double dlong = (long2 - long1) * _d2r;
double dlat = (lat2 - lat1) * _d2r;
double a = Math.pow(Math.sin(dlat / 2D), 2D) + Math.cos(lat1 * _d2r) * Math.cos(lat2 * _d2r)
* Math.pow(Math.sin(dlong / 2D), 2D);
double c = 2D * Math.atan2(Math.sqrt(a), Math.sqrt(1D - a));
double d = _eQuatorialEarthRadius * c;
return d;
}
// Distance between a point and a line
public static double pointLineDistanceTest(double[] aalatlng,double[] bblatlng,double[] ttlatlng) {
double [] a = aalatlng;
double [] b = bblatlng;
double [] c = ttlatlng;
double[] nearestNode = nearestPointGreatCircle(a, b, c);
// System.out.println("nearest node: " + Double.toString(nearestNode[0]) + "," + Double.toString(nearestNode[1]));
double result = HaversineInM(c[0], c[1], nearestNode[0], nearestNode[1]);
// System.out.println("result: " + Double.toString(result));
return (result);
}
// source: http://stackoverflow.com/questions/1299567/how-to-calculate-distance-from-a-point-to-a-line-segment-on-a-sphere
private static double[] nearestPointGreatCircle(double[] a, double[] b, double c[])
{
double[] a_ = toCartsian(a);
double[] b_ = toCartsian(b);
double[] c_ = toCartsian(c);
double[] G = vectorProduct(a_, b_);
double[] F = vectorProduct(c_, G);
double[] t = vectorProduct(G, F);
return fromCartsian(multiplyByScalar(normalize(t), _eQuatorialEarthRadius));
}
@SuppressWarnings("unused")
private static double[] nearestPointSegment (double[] a, double[] b, double[] c)
{
double[] t= nearestPointGreatCircle(a,b,c);
if (onSegment(a,b,t))
return t;
return (HaversineInKM(a[0], a[1], c[0], c[1]) < HaversineInKM(b[0], b[1], c[0], c[1])) ? a : b;
}
private static boolean onSegment (double[] a, double[] b, double[] t)
{
// should be return distance(a,t)+distance(b,t)==distance(a,b),
// but due to rounding errors, we use:
return Math.abs(HaversineInKM(a[0], a[1], b[0], b[1])-HaversineInKM(a[0], a[1], t[0], t[1])-HaversineInKM(b[0], b[1], t[0], t[1])) < PRECISION;
}
// source: http://stackoverflow.com/questions/1185408/converting-from-longitude-latitude-to-cartesian-coordinates
private static double[] toCartsian(double[] coord) {
double[] result = new double[3];
result[0] = _eQuatorialEarthRadius * Math.cos(Math.toRadians(coord[0])) * Math.cos(Math.toRadians(coord[1]));
result[1] = _eQuatorialEarthRadius * Math.cos(Math.toRadians(coord[0])) * Math.sin(Math.toRadians(coord[1]));
result[2] = _eQuatorialEarthRadius * Math.sin(Math.toRadians(coord[0]));
return result;
}
private static double[] fromCartsian(double[] coord){
double[] result = new double[2];
result[0] = Math.toDegrees(Math.asin(coord[2] / _eQuatorialEarthRadius));
result[1] = Math.toDegrees(Math.atan2(coord[1], coord[0]));
return result;
}
// Basic functions
private static double[] vectorProduct (double[] a, double[] b){
double[] result = new double[3];
result[0] = a[1] * b[2] - a[2] * b[1];
result[1] = a[2] * b[0] - a[0] * b[2];
result[2] = a[0] * b[1] - a[1] * b[0];
return result;
}
private static double[] normalize(double[] t) {
double length = Math.sqrt((t[0] * t[0]) + (t[1] * t[1]) + (t[2] * t[2]));
double[] result = new double[3];
result[0] = t[0]/length;
result[1] = t[1]/length;
result[2] = t[2]/length;
return result;
}
private static double[] multiplyByScalar(double[] normalize, double k) {
double[] result = new double[3];
result[0] = normalize[0]*k;
result[1] = normalize[1]*k;
result[2] = normalize[2]*k;
return result;
}
并且有很多错误所以写这个来计算使用方位角得到角度然后使用距离（点 a，目标）计算目标到（点 a，点 b）线。 A点和B点是线，我想要从目标到线的距离。没有大圈。 sin角（方位差）*点a到目标距离=直角三角形的边从AB线上的直角到目标的长度。
public double pointlinedistancetest(){
//set latlng location point a
Location apoint=new Location("");
apoint.setLatitude(lata);
apoint.setLongitude(lona);
//set latlng location point b
Location bpoint=new Location("");
bpoint.setLatitude(latb);
bpoint.setLongitude(lonb);
//set latlng location target to get dis to line
Location tpoint=new Location("");
tpoint.setLatitude(lat);
tpoint.setLongitude(lon);
float pbearingf = apoint.bearingTo(bpoint);
float tbearingf= apoint.bearingTo(tpoint);
double tb=tbearingf;
double ab=pbearingf;
//get angle degree difference
float angle= Math.min((pbearingf-tbearingf)<0?pbearingf-tbearingf+360:pbearingf-tbearingf, (tbearingf-pbearingf)<0?tbearingf-pbearingf+360:tbearingf-pbearingf);
// float angle= Math.min((tbearingf-pbearingf)<0?tbearingf-pbearingf+360:tbearingf-pbearingf, (pbearingf-tbearingf)<0?pbearingf-tbearingf+360:pbearingf-tbearingf);
// min((a1-a2)<0?a1-a2+360:a1-a2, (a2-a1)<0?a2-a1+360:a2-a1)
double aabearing=angle;
float atot=apoint.distanceTo(tpoint);
double atotdis=atot;
//right angle triangle formula
dis=(Math.sin(aabearing))*atotdis;
return (dis);
}
现在两者仍然显示高达 30% 的大量错误。这段代码看起来是否正确，有没有更好的方法可以做到这一点，或者我的错误在哪里。我的 GPS 显示 4 米精度，而我的公式（代码）显示 10 到 20 米的误差，而且似乎不太一样。
【问题讨论】：
-
你试过
distanceTo()或distanceBetween()Location类的方法吗...... -
@MichaelShrestha 我使用 distanceTo() 来获得从点 a 到（目标 ....*sin 角 = 点 a 的对边）的距离。直角三角形。 B 点仅提供一条线的方位角，三角形由目标与 a、b 线上的直角组成，如mathsisfun.com/algebra/trig-finding-side-right-triangle.html。我实际上不知道它与直线相交的位置，我只想知道到它的距离。