纬度，经度之间的点与线之间的最小距离 [英] Minimum distance between a point and a line in latitude, longitude

问题描述

我有两行经纬线

A：3.222895，101.719751

B：3.227511，101.724318

和1点

C：3.224972，101.722932

如何计算C点和一行之间的最小距离，由A点和B'
如果您还可以提供计算和目标代码，这将是方便的。距离约为89米（使用Google地球中的标尺）。

解决方案

感谢mimi和这篇伟大的文章 http://www.movable-type.co.uk/scripts/latlong.html 但他们并没有给出整体情况。这是一个细节。使用Google地球使用地标收集所有这些积分，以标记位置。确保在偏好设置中lat / long设置为十进制度。

  lat A = 3.222895 
 lon A = 101.719751 
 lat B = 3.222895 
 lon B = 101.719751 
 lat C = 3.224972 
 lon C = 101.722932 
地球半径，R = 6371 
  

1。首先，您必须找到从A到C和A到B的方位。

轴承公式

  bearingAC = atan2（sin（Δλ）* cos（φ2），cos（φ1）* sin（φ2）-sin（φ1）* cos（φ2）* cos（Δλ））
 bearingAB = atan2 sin（φλ）* cos（φ2），cos（φ1）* sin（φ2）-sin（φ1）* cos（φ2）* cos（Δλ））
  $ p> 
 
 
φ是纬度，λ是经度，R是地球半径
 
 
 
  2。使用球面定律求出A到C距离
  distanceAC = acos（sin（φ1）* sin（φ2 ）+ cos（φ1）* cos（φ2）* cos（Δλ））* R 
  
  3。查找跨轨距离
  distance = asin（sin（distanceAC / R）* sin（bearingAC-bearing AB ））* R 
  
  Objective-C代码 
  double lat1 = 3.227511; 
双lon1 = 101.724318; 
 double lat2 = 3.222895; 
 double lon2 = 101.719751; 
 double lat3 = 3.224972; 
 double lon3 = 101.722932; 
 
 double y = sin（lon3  -  lon1）* cos（lat3）; 
 double x = cos（lat1）* sin（lat3） -  sin（lat1）* cos（lat3）* cos（lat3-lat1）; 
 double bearing1 = radiansToDegrees（atan2（y，x））; 
 bearing1 = 360  - （bearing1 + 360％360）; 
 
 double y2 = sin（lon2  -  lon1）* cos（lat2）; 
 double x2 = cos（lat1）* sin（lat2） -  sin（lat1）* cos（lat2）* cos（lat2  -  lat1）; 
 double bearing2 = radiansToDegrees（atan2（y2，x2））; 
 bearing2 = 360  - （轴承2 + 360％360）; 
 
 double lat1Rads = degreesToRadians（lat1）; 
 double lat3Rads = degreesToRadians（lat3）; 
 double dLon = degreesToRadians（lon3  -  lon1）; 
 
 double distanceAC = acos（sin（lat1Rads）* sin（lat3Rads）+ cos（lat1Rads）* cos（lat3Rads）* cos（dLon））* 6371; 
 double min_distance = fabs（asin（sin（distanceAC / 6371）* sin（degreesToRadians（bearing1）-degreesToRadians（bearing2）））* 6371）; 
 
 NSLog（@bearing 1：％g，bearing1）; 
 NSLog（@bearing 2：％g，bearing2）; 
 NSLog（@distance AC：％g，distanceAC）; 
 NSLog（@min distance：％g，min_distance）; 
  
其实有一个图书馆。您可以在这里找到 https://github.com/100grams/CoreLocationUtils  
 
I have a line with two points in latitude and longitude

A: 3.222895, 101.719751

B: 3.227511, 101.724318  

and 1 point

C:  3.224972, 101.722932

How can I calculate minimum distance between point C and a line consists of point A and B? 
It will be convenient if you can provide the calculation and objective-c code too. The distance is around 89 meters (using ruler in Google Earth).
解决方案
Thanks to mimi and this great article http://www.movable-type.co.uk/scripts/latlong.html but they don't give the whole picture. Here is a detail one. All this points are collected using Google Earth using Placemark to mark the locations. Make sure lat/long are set to decimal degrees in Preferences.
lat A = 3.222895  
lon A = 101.719751  
lat B = 3.222895  
lon B = 101.719751  
lat C = 3.224972  
lon C = 101.722932  
Earth radius, R = 6371
1. First you have to find the bearing from A to C and A to B.

Bearing formula  
bearingAC = atan2( sin(Δλ)*cos(φ₂), cos(φ₁)*sin(φ₂) − sin(φ₁)*cos(φ₂)*cos(Δλ) )  
bearingAB = atan2( sin(Δλ)*cos(φ₂), cos(φ₁)*sin(φ₂) − sin(φ₁)*cos(φ₂)*cos(Δλ) ) 
φ is latitude, λ is longitude, R is earth radius

2. Find A to C distance using spherical law of cosines  
distanceAC = acos( sin(φ₁)*sin(φ₂) + cos(φ₁)*cos(φ₂)*cos(Δλ) )*R
3. Find cross-track distance  
distance = asin(sin(distanceAC/ R) * sin(bearingAC − bearing AB)) * R
Objective-C code  
double lat1 = 3.227511;
double lon1 = 101.724318;
double lat2 = 3.222895;
double lon2 = 101.719751;
double lat3 = 3.224972;
double lon3 = 101.722932;

double y = sin(lon3 - lon1) * cos(lat3);
double x = cos(lat1) * sin(lat3) - sin(lat1) * cos(lat3) * cos(lat3 - lat1);
double bearing1 = radiansToDegrees(atan2(y, x));
bearing1 = 360 - (bearing1 + 360 % 360);

double y2 = sin(lon2 - lon1) * cos(lat2);
double x2 = cos(lat1) * sin(lat2) - sin(lat1) * cos(lat2) * cos(lat2 - lat1);
double bearing2 = radiansToDegrees(atan2(y2, x2));
bearing2 = 360 - (bearing2 + 360 % 360);

double lat1Rads = degreesToRadians(lat1);
double lat3Rads = degreesToRadians(lat3);
double dLon = degreesToRadians(lon3 - lon1);

double distanceAC = acos(sin(lat1Rads) * sin(lat3Rads)+cos(lat1Rads)*cos(lat3Rads)*cos(dLon)) * 6371;  
double min_distance = fabs(asin(sin(distanceAC/6371)*sin(degreesToRadians(bearing1)-degreesToRadians(bearing2))) * 6371);

NSLog(@"bearing 1: %g", bearing1);  
NSLog(@"bearing 2: %g", bearing2);  
NSLog(@"distance AC: %g", distanceAC);  
NSLog(@"min distance: %g", min_distance);
Actually there's a library for this. You can find it here https://github.com/100grams/CoreLocationUtils

                        
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