经纬度点到小弧线段的距离 [英] Distance from Lat/Lng point to Minor Arc segment

问题描述

我需要计算从 lat/lng GPS 点 P 到由其他 2 个 lat/lng GPS 点 A 和 B 描述的线段的最短距离.

'Cross-track distance' 帮助我计算 P 与 A 和 B 描述的大圆之间的最短距离.

然而，这不是我想要的.我需要 P 和 A-B 的 segment 线之间的距离，而不是整个大圆.

我使用了

在第一张图片中，绿色线表示的跨轨道距离是正确的，并且确实是到线段 AB 的最短距离.

在第二张图片中显示了跨轨道距离的问题，在这种情况下，我希望最短距离是简单距离 AP，但跨轨道距离给了我由 红色 行.

如何更改我的算法以考虑到这一点，或检查点 X 是否在 AB 内.是否可以通过计算来做到这一点?还是迭代是唯一可能的(昂贵的)解决方案?(沿 AB 取 N 个点，计算 P 到所有这些点的最小距离)

为简单起见，图像中的所有线条都是直的.实际上，这些是大圆上的小弧

解决方案

首先，一些命名法:
我们的弧是从 p1 到 p2 绘制的.
我们的第三点是p3.
与大圆相交的假想点是 p4.
p1 由 lat1,lon1 定义；p2 by lat2,lon2;等等
dis12 是 p1 到 p2 的距离；等等
Bear12 是从 p1 到 p2 的方位角；等等
dxt 是跨轨道距离.
dxa 是跨弧距离，我们的目标！

请注意，交叉轨道公式依赖于相对方位，bear13-bear12

我们有 3 个案例要处理.

案例1:相对方位为钝角.所以，dxa=dis13.

案例 2.1: 相对方位是尖锐的，并且 p4 落在我们的弧上.所以，dxa=dxt.

案例 2.2: 相对方位是尖锐的，并且 p4 超出了我们的弧线.所以，dxa=dis23

算法:

第1步:如果相对方位是钝角，dxa=dis13
完毕！
第 2 步: 如果相对方位很尖锐:
2.1: 查找 dxt.
2.3:找到dis12.
2.4:找到dis14.
2.4:如果dis14>dis12，dxa=dis23.
完毕！
2.5:如果我们到达这里，dxa=abs(dxt)

MATLAB 代码:

function [dxa] = crossarc(lat1,lon1,lat2,lon2,lat3,lon3)%//CROSSARC 以米为单位计算最短距离%//在圆弧(由 p1 和 p2 定义)和第三点 p3 之间.%//以度为单位输入 lat1,lon1,lat2,lon2,lat3,lon3.lat1=deg2rad(lat1);lat2=deg2rad(lat2);lat3=deg2rad(lat3);lon1=deg2rad(lon1);lon2=deg2rad(lon2);lon3=deg2rad(lon3);R=6371000；%//地球的半径，以米为单位%//公式的先决条件熊12 =熊(lat1，lon1，lat2，lon2)；熊13 =熊(lat1，lon1，lat3，lon3)；dis13 = dis(lat1,lon1,lat3,lon3);差异=绝对(bear13-bear12)；如果差异 >圆周率差异 = 2 * pi - 差异；结尾%//相对方位是否钝?如果差异>(pi/2)dxa=dis13；别的%//求跨轨距离.dxt = asin( sin(dis13/R)* sin(bear13 - Bear12) ) * R;%//p4是否超出了圆弧?dis12 = dis(lat1,lon1,lat2,lon2);dis14 = acos( cos(dis13/R)/cos(dxt/R) ) * R;如果 dis14>dis12dxa=dis(lat2,lon2,lat3,lon3);别的dxa=abs(dxt);结尾结尾结尾函数 [ d ] = dis( latA, lonA, latB, lonB )%DIS 查找两个纬度/经度点之间的距离.R=6371000；d = acos( sin(latA)*sin(latB) + cos(latA)*cos(latB)*cos(lonB-lonA) ) * R;结尾函数 [ b ] = 熊( latA,lonA,latB,lonB )%BEAR 查找从一个纬度/经度点到另一个的方位角.b=atan2( sin(lonB-lonA)*cos(latB) , ...cos(latA)*sin(latB) - sin(latA)*cos(latB)*cos(lonB-lonA) );结尾

示例输出:展示所有案例.请参阅下面的地图.

>>十字弧(-10.1,-55.5,-15.2,-45.1,-10.5,-62.5)答案=7.6709e+05>>交叉弧(40.5,60.5,50.5,80.5,51,69)答案=4.7961e+05>>交叉弧(21.72,35.61,23.65,40.7,25,42)答案=1.9971e+05

地图上的相同输出！:

演示案例1:

演示案例 2.1:

演示案例 2.2:

归功于:http://www.movable-type.co.uk/scripts/latlong.html
对于公式
和:http://www.darrinward.com/lat-long/?id=1788764
用于生成地图图像.

I need to calculate the shortest distance from a lat/lng GPS point P to a line segment described by 2 other lat/lng GPS points A and B.

'Cross-track distance' helps me to calculate the shortest distance between P and the great circle described by A and B.

However, this is not what I want. I need need the distance between P and the line segment of A-B, not the entire great circle.

I have used the following implementation from http://www.movable-type.co.uk/scripts/latlong.html

Formula:    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

The following images hopefully demonstrate the problem I am trying to solve:

In the first image the Cross-Track distance, indicated by the green line is correct and indeed the shortest distance to the line segment AB.

In the second image the problem with cross-track distance is shown, In this case I would want the shortest distance to be the simple distance AP, but Cross-Track distance gives me the distance indicated by the red line.

How do I change my algoritm to take this into account, or check whether or not point X is within AB. Is it possible to do this computationally? Or is iterative the only possible (expensive) solution? (take N points along AB and calculate the min distance from P to all these points)

For simplicity purposes all lines in the images are straight. In reality, these are minor arcs on a great circle

解决方案

First, some nomenclature:
Our arc is drawn from p1 to p2.
Our third point is p3.
The imaginary point that intersects the great circle is p4.
p1 is defined by lat1,lon1; p2 by lat2,lon2; etc.
dis12 is the distance from p1 to p2; etc.
bear12 is the bearing from p1 to p2; etc.
dxt is cross-track distance.
dxa is cross-arc distance, our goal!

Notice that the cross-track formula relies on the relative bearing, bear13-bear12

We have 3 cases to deal with.

Case 1: The relative bearing is obtuse. So, dxa=dis13.

Case 2.1: The relative bearing is acute, AND p4 falls on our arc. So, dxa=dxt.

Case 2.2: The relative bearing is acute,AND p4 falls beyond our arc. So, dxa=dis23

The algorithm:

Step 1: If relative bearing is obtuse, dxa=dis13
Done!
Step 2: If relative bearing is acute:
2.1: Find dxt.
2.3: Find dis12.
2.4: Find dis14.
2.4: If dis14>dis12, dxa=dis23.
Done!
2.5: If we reach here, dxa=abs(dxt)

MATLAB code:

function [ dxa ] = crossarc( lat1,lon1,lat2,lon2,lat3,lon3 )
%// CROSSARC Calculates the shortest distance in meters 
%// between an arc (defined by p1 and p2) and a third point, p3.
%// Input lat1,lon1,lat2,lon2,lat3,lon3 in degrees.
    lat1=deg2rad(lat1); lat2=deg2rad(lat2); lat3=deg2rad(lat3);
    lon1=deg2rad(lon1); lon2=deg2rad(lon2); lon3=deg2rad(lon3);

    R=6371000; %// Earth's radius in meters
    %// Prerequisites for the formulas
    bear12 = bear(lat1,lon1,lat2,lon2);
    bear13 = bear(lat1,lon1,lat3,lon3);
    dis13 = dis(lat1,lon1,lat3,lon3);

    diff = abs(bear13-bear12);
    if diff > pi
        diff = 2 * pi - diff;
    end
    %// Is relative bearing obtuse?
    if diff>(pi/2)
        dxa=dis13;
    else
        %// Find the cross-track distance.
        dxt = asin( sin(dis13/R)* sin(bear13 - bear12) ) * R;

        %// Is p4 beyond the arc?
        dis12 = dis(lat1,lon1,lat2,lon2);
        dis14 = acos( cos(dis13/R) / cos(dxt/R) ) * R;
        if dis14>dis12
            dxa=dis(lat2,lon2,lat3,lon3);
        else
            dxa=abs(dxt);
        end   
    end
end

function [ d ] = dis( latA, lonA, latB, lonB )
%DIS Finds the distance between two lat/lon points.
R=6371000;
d = acos( sin(latA)*sin(latB) + cos(latA)*cos(latB)*cos(lonB-lonA) ) * R;
end

function [ b ] = bear( latA,lonA,latB,lonB )
%BEAR Finds the bearing from one lat/lon point to another.
b=atan2( sin(lonB-lonA)*cos(latB) , ...
    cos(latA)*sin(latB) - sin(latA)*cos(latB)*cos(lonB-lonA) );
end

Sample outputs: Demonstrate all cases. See maps below.

>> crossarc(-10.1,-55.5,-15.2,-45.1,-10.5,-62.5)
ans =
   7.6709e+05
>> crossarc(40.5,60.5,50.5,80.5,51,69)
ans =
   4.7961e+05
>> crossarc(21.72,35.61,23.65,40.7,25,42)
ans =
   1.9971e+05

Those same outputs on the map!:

Demonstrates case 1:

Demonstrates case 2.1:

Demonstrates case 2.2:

Credit to: http://www.movable-type.co.uk/scripts/latlong.html
for the formulas
and: http://www.darrinward.com/lat-long/?id=1788764
for generating the map images.

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