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Roundness calculations

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Hi there,

 

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

Alain

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Message 1 of 24
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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.

Message Edited by elset191 on 06-02-2009 09:26 AM
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Tim Elsey
Certified LabVIEW Architect
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Message 2 of 24
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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

 

Regards

Alain

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Well, if you find a way to find the proper center, you just have to switch two wires.  I've attached my VI in 8.5 so you can see it.
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Tim Elsey
Certified LabVIEW Architect
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Half of the job is done 🙂

I'll keep looking for the other half Smiley Wink

 

Thanks for the V8.5 vi.

 

Regards

Alain

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Message 5 of 24
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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)

Cory K
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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)

Cory K
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The ellipse doesn't necessarily have it's axes aligned with the X and Y axes.
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Tim Elsey
Certified LabVIEW Architect
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Message 8 of 24
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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

Message Edited by Cory K on 06-02-2009 10:13 AM
Cory K
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You can just do a rotation of all the points (or the axes themselves.  I've done something like this before. ) if you can figure out the angle.  Then treat it like it's aligned properly.
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Tim Elsey
Certified LabVIEW Architect
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Message 10 of 24
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