How can I find the center point and radius of an arc segment from quadratic equation generated by a curve fitting software

Question

I am trying to understand how to find the radius and center of an arc segment using a curve fitting software that generates an equation using linear regression analysis. It generates equations such as y= a + bx + cx^2 and states the values of a,b and c. The software nicely graphs the best fit line through all the points of the arc and gives statistics as to the accuracy of the fit.

My problem is that I need to be able to calculate the center of the arc and the radius of the line generated by the software. The original data is a table x and y coordinates digitized from known curved surfaces machined from metal using a computerized milling machine. The goal is to digitize unknown curves and solve for the radius and center of those curves. At this point we are using known arcs to understand the math involved.

Unfortunately when I try to randomly select 3 x,y data points using circle formula or even enter the values into an online calculator, the radius values generated are very inconsistent. I am hoping the curve fitting software can do a better job - if I can just understand how to effectively use the output!

Thanks, Keith

Answer 1

If I understand your question, you are asking for the parameters of what is called the 'osculating circle' at a given point of a curve.

Given that you already have $y=a+bx+cx^2$ as an approximation, then the required center and radius corresponding to the point $(x,y)$ are given by the following procedure:

1. Calculate $p=b+2cx$, $q=2cp/(1+p^2)$, $u=2c-pq$, $v=\sqrt{u^2+q^2}$.
2. The radius is given by $$r=\frac{(1+p^2)^{3/2}}{2c}$$
3. The center has coordinates $$x_0=x-\frac{rq}{v},\quad y_0=y+\frac{ru}{v}$$