Roundness can be defined by means of different methods. So far I found MRS or MZC, MIC, MCC and LSC.
Complete description in the VI as comment 🙂
So far I managed to calculate LSC because LV provide a VI for it 🙂
Can someone please help me with the other ones?
I can't finad a way to determine the radius and offset of the maximum inscribed circle and minimum circumscribed circle.
Does anyone has the formulas to calculate those circles?
Thanks a lot
Solved! Go to Solution.
For MIC and MCC I took the center as the (x0,y0,z0) that was returned by LSC. Then found the closest and farthest points from that point and used those as the radii for MIC and MCC.
It should be noted that I don't know if this is an official way to do the calculation. It made sense in my head though.
Thanks for the effort Tim but taking the center point from the LSC circle as zero point for MIC & MCC circles is not right 😞
And that's my problem, how to find the middle point of those circles.
I think there is a formula to determine them but I can't find them on the web 😞
I found formulas for area and gravity point but not for MIC & MCC 😞
Anyway, thanks for your work so far.
I can't open your VI, I have V8.5 instead of V8.6
Since (0,0) isnt the center, how about this.
Center x-value = [(max x-value)+(min x-value)] / 2
Center y-value = [(max y-value) + (min y-value)] / 2
Then your center will be (center x, center y)
And to find out how round the circles are:
(largest diameter in y-direction) - (largest diameter in x-direction)
As this difference approaches 0, the circle is more and more circular (less eliptical)
That shouldnt be a problem.
Find your maximum and minimum for both x and y values.
Find the difference between your max-y and min-y. This will be 1 diameter (in y-direction)
Find the difference between your max-x and min-x. This will be the other diameter (in x-direction)
If you subtract one of these differences from the other, you should get a (hopefully small) value.
Edit: Oh wait, you mean if the ellipse is off to an angle..... then you are correct.
Let me see if I can think of a solution to that