英文:
Translation to plot line with hough transform when using polar coordinates
问题
我有点难以理解使用极坐标进行Hough变换后如何绘制线条。
在OpenCV示例中，他们这样做：
a = math.cos(theta)
b = math.sin(theta)
x0 = a * rho
y0 = b * rho
pt1 = (int(x0 + 1000*(-b)), int(y0 + 1000*(a)))
pt2 = (int(x0 - 1000*(-b)), int(y0 - 1000*(a)))
我理解他们在同一条线上取两个远离的点，但我难以理解x点的“投影”/乘以sin以及y点的乘以cos。
有人能为我解释一下吗？
英文:
I have a bit of trouble understanding how lines are plotted after doing a Hough Transform using polar coordinates.
On the OpenCV example they do it like this:
a = math.cos(theta)
b = math.sin(theta)
x0 = a * rho
y0 = b * rho
pt1 = (int(x0 + 1000*(-b)), int(y0 + 1000*(a)))
pt2 = (int(x0 - 1000*(-b)), int(y0 - 1000*(a)))
I understand that they are taking two points far apart on the same line, but I have trouble understanding the "projection"/ multiplication by the sin for the x point and the cos for the y point.
Can anyone enlighten me?
答案1
得分: 0
“r”和“theta”的定义如下。
因此，蓝线方向单位向量V为(cos(theta), sin(theta))，并且该线上的一个点可表示为r * **V**（在您的描述中为(x0, y0)）。
沿着该点移动可以得到该线上的其他点。沿着垂直于V的方向单位向量U，它可以表示为(-sin(theta), cos(theta))或(sin(theta), -cos(theta))。
因此，该线上的点为：
r * **V** + k * **U**
其中，标量k的值是任意的移动量。（在您的描述中，使用的k值为1000和-1000）
因此，对于
x点的sin和y点的cos
答案是，U：沿着该线的单位方向向量。
英文:
Definitions of r and theta are as shown.
So, the blue line directional unit vector V is ( cos(theta), sin(theta) ),
and one point on the line is given as <code>r * V</code> (In your description, this is (x0,y0) ).
Points on the line can be obtained by moving along the line from this point.
The along directional unit vector U (which is perpendicular to V) , is ( -sin(theta), cos(theta) ) or ( sin(theta), -cos(theta) ).
So, points on the line is :
<code>r * V + k * U</code>
where, value of the scalar k is arbitrary amount of movement. (In your description, used k values are 1000 and -1000)
Therefore, short answer for
> the sin for the x point and the cos for the y
is, U : the unit directional vector along the line.
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