距Lat / Lng点到小弧段的距离 [英] Distance from Lat/Lng point to Minor Arc segment

问题描述

我需要计算从纬度/经度GPS点P到其他纬度/经度GPS点A和B描述的线段的最短距离。

'交叉距离'帮助我计算P和A和B所描述的大圆之间的最短距离。然而，这不是我想要的。我需要AB和AB的段之间的距离，而不是整个大圆圈。

我使用了< a href =http://www.movable-type.co.uk/scripts/latlong.html =noreferrer> http://www.movable-type.co.uk/scripts/latlong.html

 公式：dxt = asin（sin（δ13）⋅sin（θ13-θ12））⋅R 
其中：
δ13是从起点到第三点的（角度）距离
θ13是从起点到第三点的（初始）方位
θ12是从起点到终点的（初始）方位点
 R是地球半径
  

下面的图片有希望证明我正在尝试的问题解决：

案例2.1 ：相对方位是尖锐的，AND p4落在我们的弧上。
因此，dxa = dxt。

案例2.2：相对方位是尖锐的，AND p4超出我们的弧度。
因此，dxa = dis23

算法：

第1步：如果相对方位是钝角，则dxa = dis13

完成！

第2步：如果相对方位是尖锐的：

2.1：查找dxt。

2.3：查找dis12。

2.4：查找dis14。

2.4：如果dis14> dis12，dxa = dis23。

完成！

2.5：如果我们到达这里，dxa = abs（dxt）

MATLAB代码：
$ b $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $' // CROSSARC计算弧（由p1和p2定义）和第三个点p3之间的最短距离（以米为单位
％//）。
％//以度数输入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
bear12 = bear（lat1，lon1，lat2，lon2）;
bear13 =熊（lat1，lon1，lat3，lon3）;
dis13 = dis（lat1，lon1，lat3，lon3）;

％//相对轴承钝角？
if abs（bear13-bear12）>（pi / 2）
dxa = dis13;
else
％//找到跨轨距离。
dxt = asin（sin（dis13 / R）* sin（bear13 - bear12））* R;

％// p4是否超出了弧度？
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

函数[d] = dis（latA，lonA，latB，lonB）
％DIS查找两个经纬度之间的距离/ lon点。
R = 6371000;
d = acos（sin（latA）* sin（latB）+ cos（latA）* cos（latB）* cos（lonB-lonA））* R;
end

函数[b] = bear（latA，lonA，latB，lonB）
％BEAR查找从一个纬度点到另一个纬度的方位。
b = atan2（sin（lonB-lonA）* cos（latB），...
cos（latA）* sin（latB） - sin（latA）* cos（latB）* cos（lonB- lonA））;
end

示例输出：演示所有情况。见下面的地图。

 >> 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 
  

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

为公式

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);

    %// Is relative bearing obtuse?
    if abs(bear13-bear12)>(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.

这篇关于距Lat / Lng点到小弧段的距离的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！