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Minimum Circumscribed Circle


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Hi,

Just wandering, does anybody know how many degree of scanning of a circular element is required to compute the diameter with Evaluation method "Minimum Circumscribed Circle".

I kind of feel, as long as we measure the circular element less than 360 degree, it will report the LSQ diameter does not matter what we put in the Evaluation method option. But I have no proof. I tried to search with the F1 help document, but didn't find anything related to this calculation.

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Create a test, copy the circle feature (LSQ), change the copied circles
evaluation (MCC) and run both. Now see what the difference is, if any?

Another words, create two identical circles, one evaluated at LSQ the
other at MCC. Run them both. Is there a difference?
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Also I would put in 500 on number of points. This will give you more points after the software drops of points at the first of the feature when it is run. When I went to class my teacher said to have minimum of 500 points on features for this reason.
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Hi, Clarke and Kyle,

Thanks for the reply,

The hole I got problem with is a pivot hole which only has 150 degree around. I have tried to measure the LSQ and Max Inscribed Circle as two different features, but I get no different result. I also tried to increased the point density to be 1000 points, but still makes no difference. (Please see the report attached)

On the same part, there is another bolt hole which is a full cylinder, and I scan 380 degree around. I did a test on this bolt hole, and I found If I manually masked more than 177 degree of angle out of the measurement, there will be no difference between LSQ, Min and Max Circles. Is that possible that the Min/Max best fit requires at least 3 opposite elements exist to calculate the best fit boundary? Otherwise, it will just report the LSQ value anyways?

Thank you very much.

222.jpg82.0154 Picture.JPG

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If you look at all of the results you can see that they are different. Look at your X, Y, and Form results. I'd be hesitant to use MIC/MCC on an incomplete circle either way.
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I think Kyle is right with the number of points but have you tried increasing the number of decimal places? Maybe the difference is only a few tenths.
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Thank you for your reply, the X, Y, Z, D and A1, A2 looks identical between LSQ and Maximum Inscribed Element if I scan less than 180 degree. Could you please confirm this on your CMM with any scan of full circle if you mask out more than 180 degree or with a scan of less than half circle? Maybe there are some setting I need to change on my CMM, I am not sure. I want to see how other people's CMM is handling this situation. Thank you.

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With less than 360 degrees of a circle I usually resort to constraining
the two axis of whatever the work-plane is. (For you it looks like XY).

Constraining location returns a more accurate size, constraining size
(Radius) returns a more accurate location.
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Thank you for your reply. As Kyle suggested, I have increased the number of points by one order (from 177 per circle to 1000 per circle), but I didn't see any difference up to 0.01um (5 decimal places). On the contrast, a full scan circle with almost same the diameter, coined by the same die on the same part will report about 10um difference on diameter between LSQ and Max Inscribed Circle.

I also tried to copy the LSQ circle as a new feature and change the assignment for evaluation method to "Maximum Inscribed Elements". So basically, I measured the same circle twice and two features. The difference on diameter is 0.03um (See the diameter on the attached picture). I do not think the 0.03um is the difference between the two different best fit method (LSQ and Max Inscribed Elements), it is just from the repeat-ability of the CMM.

Could you please confirm this on your CMM? Will your CMM actually report any difference between Max/Min/LSQ with a circle scanned less than 180 degree?

4444.jpg

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I have to agree with Clarke on this. With less than 180° sweep you need to constrain the X,Y,Z coordinates for an accurate result for the size of the radius. If a position call out is also necessary, I would go to features and create another circle and do recall feature points. I would constrain the size of the Ø/Radius and this will give you the proper coordinates for position.
I have successfully used this on arcs with 60° sweeps. Our source inspector has yet to prove me wrong with the results, and he has tried.

Greg
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