问题
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
回答1:
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.
回答2:
For 100 - 1000m spherical problems, it is easy to just convert to
cartesian space, using a equirectangular projection.
Then it continues with school mathematics:
Use the function "distance from line segment" which is easy to find ready implemented.
This fucntion uses (and sometimes returns) a relative forward/backward position for the projected point X on the line A,B. The value is
- in the interval [0,1] if the projected point is inside the line segment.
- it is negative if X is outside before A,
- it is >1 if outside after B.
If the relative position is between 0,1 the normal distance is taken, if outside the shorter distance of the both start and line-end points, A,B.
An example of such / or very similar an cartesian implementaion is Shortest distance between a point and a line segment
回答3:
/**
* Calculates the euclidean distance from a point to a line segment.
*
* @param v the point
* @param a start of line segment
* @param b end of line segment
* @return an array of 2 doubles:
* [0] distance from v to the closest point of line segment [a,b],
* [1] segment coeficient of the closest point of the segment.
* Coeficient values < 0 mean the closest point is a.
* Coeficient values > 1 mean the closest point is b.
* Coeficient values between 0 and 1 mean how far along the segment the closest point is.
*
* @author Afonso Santos
*/
public static
double[]
distanceToSegment( final R3 v, final R3 a, final R3 b )
{
double[] results = new double[2] ;
final R3 ab_ = b.sub( a ) ;
final double ab = ab_.modulus( ) ;
final R3 av_ = v.sub( a ) ;
final double av = av_.modulus( ) ;
if (ab == 0.0) // a and b coincide
{
results[0] = av ; // Distance
results[1] = 0.0 ; // Segment coeficient.
}
else
{
final double avScaProjAb = av_.dot(ab_) / ab ;
final double abCoeficient = results[1] = avScaProjAb / ab ;
if (abCoeficient <= 0.0) // Point is before start of the segment ?
results[0] = av ; // Use distance to start of segment.
else if (abCoeficient >= 1.0) // Point is past the end of the segment ?
results[0] = v.sub( b ).modulus() ; // Use distance to end of segment.
else // Point is within the segment's start/end perpendicular boundaries.
{
if (avScaProjAb >= av) // Test to avoid machine float representation epsilon rounding errors that would result in expection on sqrt.
results[0] = 0.0 ; // a, b and v are colinear.
else
results[0] = Math.sqrt( av * av - avScaProjAb * avScaProjAb ) ; // Perpendicular distance from point to segment.
}
}
return results ;
}
the above method requires cartesian 3D space arguments and you asked to use lat/lon arguments. To do the conversion use
/**
* Calculate 3D vector (from center of earth).
*
* @param latDeg latitude (degrees)
* @param lonDeg longitude (degrees)
* @param eleMtr elevation (meters)
* @return 3D cartesian vector (from center of earth).
*
* @author Afonso Santos
*/
public static
R3
cartesian( final double latDeg, final double lonDeg, final double eleMtr )
{
return versor( latDeg, lonDeg ).scalar( EARTHMEANRADIUS_MTR + eleMtr ) ;
}
For the rest of the 3D/R3 code or how to calculate distance to a path/route/track check https://sourceforge.net/projects/geokarambola/
回答4:
Adding a Java version to wdickerson answer:
public static double pointToLineDistance(double lon1, double lat1, double lon2, double lat2, double lon3, double lat3) {
lat1 = Math.toRadians(lat1);
lat2 = Math.toRadians(lat2);
lat3 = Math.toRadians(lat3);
lon1 = Math.toRadians(lon1);
lon2 = Math.toRadians(lon2);
lon3 = Math.toRadians(lon3);
// Earth's radius in meters
double R = 6371000;
// Prerequisites for the formulas
double bear12 = bear(lat1, lon1, lat2, lon2);
double bear13 = bear(lat1, lon1, lat3, lon3);
double dis13 = dis(lat1, lon1, lat3, lon3);
// Is relative bearing obtuse?
if (Math.abs(bear13 - bear12) > (Math.PI / 2))
return dis13;
// Find the cross-track distance.
double dxt = Math.asin(Math.sin(dis13 / R) * Math.sin(bear13 - bear12)) * R;
// Is p4 beyond the arc?
double dis12 = dis(lat1, lon1, lat2, lon2);
double dis14 = Math.acos(Math.cos(dis13 / R) / Math.cos(dxt / R)) * R;
if (dis14 > dis12)
return dis(lat2, lon2, lat3, lon3);
return Math.abs(dxt);
}
private static double dis(double latA, double lonA, double latB, double lonB) {
double R = 6371000;
return Math.acos(Math.sin(latA) * Math.sin(latB) + Math.cos(latA) * Math.cos(latB) * Math.cos(lonB - lonA)) * R;
}
private static double bear(double latA, double lonA, double latB, double lonB) {
// BEAR Finds the bearing from one lat / lon point to another.
return Math.atan2(Math.sin(lonB - lonA) * Math.cos(latB), Math.cos(latA) * Math.sin(latB) - Math.sin(latA) * Math.cos(latB) * Math.cos(lonB - lonA));
}
来源：https://stackoverflow.com/questions/32771458/distance-from-lat-lng-point-to-minor-arc-segment