Question

我试图找到以度数（纬度，经度）表示的两个linesegments之间的距离。例如，

LINESTRING1 is ((4.42553, -124.159), (4.42553, -124.159)) and
LINESTRING2 is ((-0.061225, -128.428), (-0.059107, -128.428))

然而，找到这个距离的通常方法是由于地球的球形尺寸而给我不正确的距离。有什么方法可以改进下面的代码，使它返回正确的距离：558 km

#include <cstdlib>
#include <map>
#include <vector>
#include <algorithm>
#include <string>
#include <future>
#include <string>
#include <fstream>
#include <sstream>
#include <iostream>

using namespace std;

double dot( double* V1, double* V2 )
{
    return V1[0]*V2[0] + V1[1]*V2[1];
}

void lineLineDist2()
        //Compute the square of the minimum distance between two line segments
{

        double seg1[4];
        seg1[0]=4.42553; seg1[1]=-124.159; seg1[2]=4.42553; seg1[3]=-124.159;
        double seg2[4];
        seg2[0]=-0.061225; seg2[1]=-128.428; seg2[2]=-0.059107; seg2[3]=-128.428;


        double u[2],v[2],w[2];

        //Vector   u = S1.P1 - S1.P0;
        u[0] = seg1[2] - seg1[0];
        u[1] = seg1[3] - seg1[1];

        //Vector   v = S2.P1 - S2.P0;
        v[0] = seg2[2] - seg2[0];
        v[1] = seg2[3] - seg2[1];

        //Vector   w = S1.P0 - S2.P0;
        w[0] = seg1[0] - seg2[0];
        w[1] = seg1[1] - seg2[1];

        double    a = dot(u,u);         // always >= 0
        double    b = dot(u,v);
        double    c = dot(v,v);         // always >= 0
        double    d = dot(u,w);
        double    e = dot(v,w);
        double    D = a*c - b*b;        // always >= 0
        double    sc, sN, sD = D;       // sc = sN / sD, default sD = D >= 0
        double    tc, tN, tD = D;       // tc = tN / tD, default tD = D >= 0                

        // compute the line parameters of the two closest points
        if (D < 0.00000001) {           // the lines are almost parallel
                sN = 0.0;               // force using point P0 on segment S1
                sD = 1.0;               // to prevent possible division by 0.0 later
                tN = e;
                tD = c;
        }
        else {                          // get the closest points on the infinite lines
                sN = (b*e - c*d);
                tN = (a*e - b*d);

                if (sN < 0.0) {         // sc < 0 => the s=0 edge is visible
                        sN = 0.0;
                        tN = e;
                        tD = c;
                }
                else if (sN > sD) {     // sc > 1  => the s=1 edge is visible
                        sN = sD;
                        tN = e + b;
                        tD = c;
                }
        }

        if (tN < 0.0) {                 // tc < 0 => the t=0 edge is visible

                tN = 0.0;

                // recompute sc for this edge
                if (-d < 0.0)
                        sN = 0.0;
                else if (-d > a)
                        sN = sD;
                else {
                        sN = -d;
                        sD = a;
                }
        }
        else if (tN > tD) {             // tc > 1  => the t=1 edge is visible

                tN = tD;

                // recompute sc for this edge
                if ((-d + b) < 0.0)
                        sN = 0;
                else if ((-d + b) > a)
                    sN = sD;
                else {
                        sN = (-d +  b);
                        sD = a;
                }
        }

        // finally do the division to get sc and tc
        sc = (abs(sN) < 0.00000001 ? 0.0 : sN / sD);
        tc = (abs(tN) < 0.00000001 ? 0.0 : tN / tD);

        // get the difference of the two closest points
        double dp[2];

        //Vector   dP = w + (sc * u) - (tc * v);  // =  S1(sc) - S2(tc)
        dp[0] = w[0] + sc *u[0] - tc*v[0];
        dp[1] = w[1] + sc *u[1] - tc*v[1];                

        cout<<"distance="<<dot(dp,dp)<<"\n";
}
//---------------------------------------------------------------------------
int main()
{
    lineLineDist2();
    return 0; 
}

Answer 1

要找到两个段（大圆弧）之间的距离，必须检查它们是否相交（“两个路径的交点”here）。

如果没有，找到从所有段末端到另一段（at the same page）的最小跨轨距离

 dxt = asin( sin(δ13) ⋅ sin(θ13−θ12) ) ⋅ R
 where   δ13 is (angular) distance from start point to third point
 θ13 is (initial) bearing from start point to third point
 θ12 is (initial) bearing from start point to end point
 R is the earth’s radius

找出使用球面坐标表示的线段之间的距离

1 个答案: