（ρ/θ）参数化中定义的两条直线的交点 [英] Intersection of two lines defined in (rho/theta ) parameterization

问题描述

创建了用于检测图像中线条的Hough变换的c ++实现。发现的线条用维基百科上描述的rho，theta来表示：

$ b

参数r表示线条与原点之间的距离，而θ是从原点到这个最近点的矢量的角度

如何找到x，y中的交点用r，θ来描述两行的空间？

这里是我目前用于转换进出hough空间的函数：

  //得到'r'（与极点相交的一条线（角，0,0，距离中心的距离）的一条线的长度） （（（float）x）* cosf（theta））给定点和角度（以弧度表示）
内联浮点point2Hough（int x，int y，float theta）{
 return +（（浮点）Y）* SINF（THETA））; 
} 
 
 //在角度theta上得到点y的距离与r的相交点x的距离？糟糕的解释！ > _< 
 inline float hough2Point（int x，int r，float theta）{
 float y; 
 if（theta！= 0）{
 y =（ -  cosf（theta）/ sinf（theta））* x +（（float）r / sinf（theta））; 
} else {
 y =（float）r; // w the theta may == 0 ?! 
} 
 return（y）; 
} 
  

如果这是明显的事情，请提前抱歉..

解决方案

查看维基百科页面<我看到对应于给定的给定的r，θ对的直线的方程为：bb
$ b

 r = xcosθ+ ysinθ pre> 
 
 
因此，如果我理解，给定两对r1，θ1和r2，θ2，找出交点，则必须求解未知数x，y，以下线性2x2系统：
 
 
 x cosθ1+ y sinθ1= r1 
x cosθ2+ y sinθ2= r2 
 
即AX = b，其中 
 
 
 A = [cosθ1sinθ1] b = | r1 | X = | x | 
 [cosθ2sinθ2] | r2 | | Y | 
 
 
Have created a c++ implementation of the Hough transform for detecting lines in images. Found lines are represented using rho, theta, as described on wikipedia:

  
"The parameter r represents the distance between the line and the origin, while θ is the angle of the vector from the origin to this closest point "
How can i find the intersection point in x, y space for two lines described using r, θ?


For reference here are my current functions for converting in and out of hough space:
//get 'r' (length of a line from pole (corner, 0,0, distance from center) perpendicular to a line intersecting point x,y at a given angle) given the point and the angle (in radians)
inline float point2Hough(int x, int y, float theta) {
    return((((float)x)*cosf(theta))+((float)y)*sinf(theta));
}

//get point y for a line at angle theta with a distance from the pole of r intersecting x? bad explanation! >_<
inline float hough2Point(int x, int r, float theta) {
    float y;
    if(theta!=0) {
            y=(-cosf(theta)/sinf(theta))*x+((float)r/sinf(theta));
    } else {
            y=(float)r; //wth theta may == 0?!
    }
    return(y);
}
sorry in advance if this is something obvious..
解决方案
Looking at the Wikipedia page, I see that the equation of a straight line corresponding to a given given r, θ pair is
r = x cosθ + y sinθ 
Thus, if I understand, given two pairs r1, θ1 and r2, θ2, to find the intersection you must solve for the unknowns x,y the following linear 2x2 system:
x cos θ1 + y sin θ1 = r1
x cos θ2 + y sin θ2 = r2
that is AX = b, where
A = [cos θ1  sin θ1]   b = |r1|   X = |x|
    [cos θ2  sin θ2]       |r2|       |y|


                        
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