确定:在线段上的点 [英] determinate: is point on line segment
问题描述
我正在尝试编写Java方法，如果point(x，y)位于线段上，则返回boolean真，否则返回false.
I am trying to code an java methods wich returns an boolean true if a point(x,y) is on a line segment and false if not.
我尝试过:
public static boolean OnDistance(MyLocation a, MyLocation b, MyLocation queryPoint) {
double value = java.lang.Math.signum((a.mLongitude - b.mLongitude) * (queryPoint.mLatitude - a.mLatitude)
- (b.mLatitude - a.mLatitude) * (queryPoint.mLongitude - a.mLongitude));
double compare = 1;
if (value == compare) {
return true;
}
return false;
}
但是它不起作用. 帮助和建议，不胜感激.谢谢！
but it doesnt work. help and advice is greatly appreciated. Thank you!
推荐答案
我不是 JAVA 编码器，所以我坚持数学运算...对于初学者，请假设您在飞机上(而不是球面) )
I am not JAVA coder so I stick to math behind ... For starters let assume you are on plane (not sphere surface)
-
我会使用向量数学，所以让:
a,b-是行端点
q-查询点
c=q-a-查询的行方向向量
d=b-a-线方向矢量
a,b - be the line endpoints
q - queried point
c=q-a - queried line direction vector
d=b-a - line direction vector
使用点积进行参数提取
t=dot(c,d)/(|c|*|d|)
t是行参数<0,1>，如果超出范围q不在行内
|c|=sqrt(c.x*c.x+c.y*c.y)向量的大小
dot(c,d)=c.x*d.x+c.y*d.y标量向量相乘
t is line parameter <0,1> if out of range q is not inside line
|c|=sqrt(c.x*c.x+c.y*c.y) size of vector
dot(c,d)=c.x*d.x+c.y*d.y scalar vector multiply
现在计算线上的对应点
e=a+(t*d)
e是ab
计算q和ab
compute perpendicular distance of q and ab
l=|q-e|;
如果(l>treshold)，则q不在行ab上，否则在ab行上.阈值是您仍在接受为线内线的最大距离.不需要l sqrt-ed阈值常数可以用2来代替速度.
if (l>treshold) then q is not on line ab else it is on the line ab. The threshold is the max distance from line you are still accepting as inside line. No need to have l sqrt-ed the threshold constant can be powered by 2 instead for speed.
如果将所有这些都添加到单个方程式中
然后某些事情会自我简化(希望不会犯一些愚蠢的数学错误)
then some things will simplify itself (hope did not make some silly math mistake)
l=|(q-a)-(b-a)*(dot(q-a,b-a)/|b-a|^2)|;
return (l<=treshold);
或
l=|c-(d*dot(c,d)/|d|^2)|;
return (l<=treshold);
如您所见，我们甚至不需要sqrt:)
As you can see we do not even need sqrt for this :)
[注释]
如果需要球形或椭圆形表面，则需要指定更近的半轴.这条线变为弧线/曲线，需要进行一些校正，具体取决于表面的形状.
If you need spherical or ellipsoidal surface instead then you need to specify it closer which it is what are the semi axises. The line become arc/curve and need some corrections which depends on the shape of surface see
，但也可以通过近似来完成，也可以通过对点e的二进制搜索来实现，请参见:
but can be done also by approximation and may be also by binary search of point e see:
使用的矢量数学可以在最后找到:
The vector math used can be found here at the end:
此处3D C ++实现(具有不同的名称):
double distance_point_axis(double *p,double *p0,double *dp)
{
int i;
double l,d,q[3];
for (i=0;i<3;i++) q[i]=p[i]-p0[i]; // q = p-p0
for (l=0.0,i=0;i<3;i++) l+=dp[i]*dp[i]; // l = |dp|^2
for (d=0.0,i=0;i<3;i++) d+=q[i]*dp[i]; // d = dot(q,dp)
if (l<1e-10) d=0.0; else d/=l; // d = dot(q,dp)/|dp|^2
for (i=0;i<3;i++) q[i]-=dp[i]*d; // q=q-dp*dot(q,dp)/|dp|^2
for (l=0.0,i=0;i<3;i++) l+=q[i]*q[i]; l=sqrt(l); // l = |q|
return l;
}
其中p0[3]是轴上的任何点，而dp[3]是轴的方向向量. p[3]是您想要到轴的距离的查询点.
Where p0[3] is any point on axis and dp[3] is direction vector of axis. The p[3] is the queried point you want the distance to axis for.
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