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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

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  • $\begingroup$ From your description, the software produces a best-fit parabolic arc, not a circular one. Unless it’s very short, you’re going to find a good circular approximation to it. $\endgroup$
    – amd
    Nov 26, 2017 at 1:58

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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 $$x_0=x-\frac{rq}{v},\quad y_0=y+\frac{ru}{v}$$
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  • $\begingroup$ In Excel here is the order of values below from cells A to L x y a b c p q u v r Xo Yo Here are raw numbers for first point -3.8 -0.1839 -0.3591 -0.0686 -0.0059 -0.02376 0.000280209811 -0.011793342 0.011796671 84.80556964 -1.785588191 84.59774181 last point -9.9 -0.2552 -0.3591 -0.0686 -0.0059 0.04822 -0.000567676058 -0.011772627 0.011786305 -84.99214447 -13.99356486 84.63830593 q=PRODUCT(2,E2,F2)/SUM(1,PRODUCT(F2,F2)) X0=SUM(A2,-(PRODUCT(J2,G2)/I2) ) $\endgroup$ Nov 26, 2017 at 15:53
  • $\begingroup$ The solution is excellent! Thanks so much. The only problem is that the Xo center value grows over 60 points from -1.7 to -14. Could there be a problem with the Xo or the q calculation? It is amazing how well the radius and Yo numbers seem to be spot on compared to the random values using other methods for radius and center. $\endgroup$ Nov 26, 2017 at 16:01
  • $\begingroup$ It works great for 3 different digitized arc segments so I am happy with this quantum leap forward in this analysis process. However, all 3 suffer from the variation in Xo center position so at least the aberrant values are consistent and not related to the digitization process or the curve fitting software output. $\endgroup$ Nov 26, 2017 at 17:14
  • $\begingroup$ Figured out the problem with calculations being radius was a negative number so I used the abs(r) and all is well with consistent numbers throughout. So glad to get this figured out after weeks of no progress. Thanks again! $\endgroup$ Nov 27, 2017 at 0:41

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