I'm trying to measure the angle between two needle positions in an analog meter (gauge). such as: image of a angular gauge

After the image processing of several images, I get the processed image below(a image show different needle positions:

demo image to show my question

Here, the black dot points were the feature points of needle area, so I use the line fitting algorithm(usually by least square method, or Hough transform to find the line) to get several line positions. I can get the L1, L2, so the angle between those two lines was theta1. But the needle can located in many different places, so I can finally get many positions such as L3, L4...Ln. Those lines should be ideally intersect in one point, which is the rotation center of the needle. I can calculate the equivalent intersection point of all those lines (the red dot), since it is the rotation constraint of the movement.

So, my question is: How can the equivalent point be used to refine the final angle, I mean I want to get a better value of theta1, theta2, theta3....



2 Answers 2


You could recompute the lines with the constraint that they all go through the red point. If you normally optimize a line


by optimizing both parameters $a$ and $b$, you would now only have 1 degree of freedom for each line. If the red point has coordinates $(x_1,y_1)$ then you get the constrained line equation


with only one parameter $a$. Now you compute $a$ such that the squared distance to the other dots is minimized, and no matter which value of $a$ you obtain, the line will always pass through $(x_1,y_1)$.

  • $\begingroup$ Hi, Matt L, thanks, this is a workable solution, I'm going test it to see whether it did improve the angle measurement. I first need to do some simulation. I put some random point around the idea line, and see whether your method improve the result. $\endgroup$
    – ollydbg23
    Apr 19, 2014 at 14:24
  • $\begingroup$ @ollydbg23 Don't you compute the red point simply by averaging all other intersection points? $\endgroup$
    – Matt L.
    Apr 19, 2014 at 14:29
  • $\begingroup$ No, not average the individual intersection point, but I compute the red point by "total least square", that is to find a point which minimize the distant square sum to all the lines, is this OK? $\endgroup$
    – ollydbg23
    Apr 19, 2014 at 14:43
  • $\begingroup$ To find the equivalent point of several lines, I use an optimization method described in en.wikipedia.org/wiki/…. $\endgroup$
    – ollydbg23
    Apr 19, 2014 at 15:04

Assume, you find the point of intersection as follows:

  1. Let's say your optimum point is $(x,y)$. You could write the distance of this point from each line, for each line. That would read as: $\frac{a_ix+b_iy+c}{\sqrt{a_i^2+b_i^2}}$, where $a_i$, $b_i$, and $c_i$ are line coefficients. We could rewrite this as $a'_i=\frac{a_i}{\sqrt{a_i^2+b_i^2}}$, $b'_i=\frac{b_i}{\sqrt{a_i^2+b_i^2}}$ and $c'_i=\frac{c_i}{\sqrt{a_i^2+b_i^2}}$. Write the distance as $a'_ix+b'_iy+c'_i$. This is a linear equation and you have $N$ equations like that, where $N$ is the number of lines. You can set up an overdetermined linear system and solve for the optimum $a'_i$, $b'_i$ and $c'_i$. Then you could find back the original coefficients. Actually the optimal $(x,y)$ will lie on the nullspace of this matrix and can be obtained by SVD.

  2. Let's now stick to: $\frac{a_ix+b_iy+c}{\sqrt{a_i^2+b_i^2}}$. What you could do is to use non-linear optimization (such as Gauss-Newton) to minimize the sum of these distance. You could also use this step to refine the previous one.

  3. A more involved approach would include some fancy stuff: Intersecting those lines at a single point is actually called Triangulation. It could well be done by setting up a linear system of equations and solving for the optimum point. More detail: http://en.wikipedia.org/wiki/Triangulation_(computer_vision)

After finding the point of intersection you could either update rotation or the translation in order to enforce the line to pass through the optimum point.


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