我有一些经纬度坐标,它们组成了一个多边形区域。我还有一个经纬度坐标来定义车辆的位置。如何检查车辆是否位于多边形区域内?
我有一些经纬度坐标,它们组成了一个多边形区域。我还有一个经纬度坐标来定义车辆的位置。如何检查车辆是否位于多边形区域内?
或者,如果坐标和车辆足够接近,并且不在极地或国际日期变更线附近,您可以假装地球是平的,并将经度和纬度用作简单的x和y坐标。这样,您就可以使用带有简单线段的射线投射算法。如果您不熟悉非欧几何学,则此方法更可取,但是您的多边形边界周围会产生某些畸变,因为弧会被扭曲。
编辑:关于球面几何的更多信息。
一条大圆可以通过垂直于其所在平面的向量(也称为法向量)来确定。class Vector{
double x;
double y;
double z;
};
class GreatCircle{
Vector normal;
}
//arbitrarily defining the north pole as (0,1,0) and (0'N, 0'E) as (1,0,0)
//lattidues should be in [-90, 90] and longitudes in [-180, 180]
//You'll have to convert South lattitudes and East longitudes into their negative North and West counterparts.
Vector lineFromCoordinate(Coordinate c){
Vector ret = new Vector();
//given:
//tan(lat) == y/x
//tan(long) == z/x
//the Vector has magnitude 1, so sqrt(x^2 + y^2 + z^2) == 1
//rearrange some symbols, solving for x first...
ret.x = 1.0 / math.sqrt(tan(c.lattitude)^2 + tan(c.longitude)^2 + 1);
//then for y and z
ret.y = ret.x * tan(c.lattitude);
ret.z = ret.x * tan(c.longitude);
return ret;
}
Vector Vector::CrossProduct(Vector other){
Vector ret = new Vector();
ret.x = this.y * other.z - this.z * other.y;
ret.y = this.z * other.x - this.x * other.z;
ret.z = this.x * other.y - this.y * other.x;
return ret;
}
GreatCircle circleFromCoordinates(Coordinate a, Coordinate b){
Vector a = lineFromCoordinate(a);
Vector b = lineFromCoordinate(b);
GreatCircle ret = new GreatCircle();
ret.normal = a.CrossProdct(b);
return ret;
}
两个大圆在球面上相交于两点。这些圆的叉积形成一个向量,穿过其中一个点。该向量的反极点穿过另一个点。
Vector intersection(GreatCircle a, GreatCircle b){
return a.normal.CrossProduct(b.normal);
}
Vector antipode(Vector v){
Vector ret = new Vector();
ret.x = -v.x;
ret.y = -v.y;
ret.z = -v.z;
return ret;
}
class GreatCircleSegment{
Vector start;
Vector end;
Vector getNormal(){return start.CrossProduct(end);}
GreatCircle getWhole(){return new GreatCircle(this.getNormal());}
};
GreatCircleSegment segmentFromCoordinates(Coordinate a, Coordinate b){
GreatCircleSegment ret = new GreatCircleSegment();
ret.start = lineFromCoordinate(a);
ret.end = lineFromCoordinate(b);
return ret;
}
double Vector::DotProduct(Vector other){
return this.x*other.x + this.y*other.y + this.z*other.z;
}
double Vector::Magnitude(){
return math.sqrt(pow(this.x, 2) + pow(this.y, 2) + pow(this.z, 2));
}
//for any two vectors `a` and `b`,
//a.DotProduct(b) = a.magnitude() * b.magnitude() * cos(theta)
//where theta is the angle between them.
double angleBetween(Vector a, Vector b){
return math.arccos(a.DotProduct(b) / (a.Magnitude() * b.Magnitude()));
}
a
是否与一个大圆 b
相交:
c
,即 a
的整个大圆与 b
的交点。d
,即 c
的反点。c
位于 a.start
和 a.end
之间或者 d
位于 a.start
和 a.end
之间,则 a
与 b
相交。
//returns true if Vector x lies between Vectors a and b.
//note that this function only gives sensical results if the three vectors are coplanar.
boolean liesBetween(Vector x, Vector a, Vector b){
return angleBetween(a,x) + angleBetween(x,b) == angleBetween(a,b);
}
bool GreatCircleSegment::Intersects(GreatCircle b){
Vector c = intersection(this.getWhole(), b);
Vector d = antipode(c);
return liesBetween(c, this.start, this.end) or liesBetween(d, this.start, this.end);
}
如果两个大圆段a
和b
交叉,则满足以下条件:
a
与b
的整个大圆相交b
与a
的整个大圆相交
bool GreatCircleSegment::Intersects(GreatCircleSegment b){
return this.Intersects(b.getWhole()) and b.Intersects(this.getWhole());
}
bool liesWithin(Array<Coordinate> polygon, Coordinate pointNotLyingInsidePolygon, Coordinate vehiclePosition){
GreatCircleSegment referenceLine = segmentFromCoordinates(pointNotLyingInsidePolygon, vehiclePosition);
int intersections = 0;
//iterate through all adjacent polygon vertex pairs
//we iterate i one farther than the size of the array, because we need to test the segment formed by the first and last coordinates in the array
for(int i = 0; i < polygon.size + 1; i++){
int j = (i+1) % polygon.size;
GreatCircleSegment polygonEdge = segmentFromCoordinates(polygon[i], polygon[j]);
if (referenceLine.Intersects(polygonEdge)){
intersections++;
}
}
return intersections % 2 == 1;
}