Outliers !

[[Da...]]

I run very very close tolerance parts. They're either ground or honed to tolerances at or below .0001(inch) surface finish is typically below 10 RA and 90% of my parts are bores. So, with the background covered, here is my question.

We check cylindricity to usually .0000050 Inch. I collect large numbers of points for each circle then, recall feature points into a cylinder. It works best this way, and allows me to separate individual circle sizes and roundnesses. Dirt , or even the hint of dirt will blow the numbers pretty quickly. I used to use 2-3 additional point and 2-3 iterations for outlier reduction. Then after reading the cookbook changed to 5 additional points and only 1 iteration. This doesn't seem to work quite as well.

Does a second iteration look for the same 3 sigma error or does it refigure the sigma based on the remaining points AFTER the 1st points are removed? I am really hoping that the later is the case. I would feel better about using multiple iterations if each used a smaller factor for elimination. Hope this makes sense. Oh , and Happy Friday !

[[To...]]

Hi question Dave. I don't really know the answer to this but if I was on "Who Wants to be a Millionaire?", I would say "YES! That's my final answer!" da-da-donnnn.

I mean, yes it would reduce the sigma, thus the limit range. Again, this is a guess.

[[Da...]]

I guessed that way also. It makes sense. I guess I was looking for a supporting opinion, so there it is ! 😃

We run lots of parts, and stopping to RE-clean a part and rerun the cycle is a PITA !! 🤣

[[De...]]

The first time through it calculates the standard deviation from all the points.

The next iteration it calculates it from the remaining points.

By nature it has to work this way, if it did not recalculate the standard deviation based on the remaining points, the standard deviation would be the same as the first iteration. Since the first iteration already used that standard deviation to remove all of those points, if it did not recalculate subsequent iterations would never remove any more points.

[[Na...]]

It does in fact re-calculate the sigma once the first iteration of data has been removed.

[[No...]]

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Damn, too slow! 🤣
But your explanation is far better than mine 😉

[[To...]]

Does that mean I won a million dollars?

[[Öz...]]

Hello Dave,

According to standard deviation %99.73 (give a number with 2 digits after coma and make yourself more believable 🙂) ) of the points are in -3s/+3s. If the production gets more stable most of the points will remain in -2s/+2s then more and more stable it will go to just 1s range. General recommended use of outliers is -3s/+3s and 5 adjacent points. But you want to use it with iteration. If there is an outlier (this can be a dirt or vibration or etc.) out of -3s/+3s range in the first iteration it will be deleted. Then calypso will recalculate the sigma value and make and outlier elimination again. But in a high precision part like yours the points will be in -2s/+2s . so then iteration will not have an effect. But if you use it with -2s/+2s and iteration it will have an effect. But then the important think is how much point did you removed after outlier elimination. The limit is max %5 of the points. If you removed more than %5 of the points with outlier elimination that means the outlier elimination is done wrong!

[[Ma...]]

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No, but you will receive a participation award. We are all winners here.

Mark