Cpk – Multiple Tolerances Applied to a Single Dimension
February 24, 2020 at 2:19 pm #246343
HV1234Participant@HV1234 Include @HV1234 in your post and this person will
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I have a feature that is measured at multiple locations on a part. Easiest way to think of it is the space (web) between multiple holes in a grid. So as the hole diameters vary, and the true position of the holes vary, the web in-between them vary.
The requirements are:
Cpk of 1.0 or greater.
Web greater than or equal to 1.0 on 80 % of the webs.
Web 0.5 to 1.0 on 20% of the webs.
To make the numbers easy let’s say 100 parts are measured and there are 100 webs on each part.
If all 100 webs on all 100 parts are greater than 1.0 then I would just use the >/= 1.0 tolerance, and calculate my Cpk as usual. All webs >1.0 is actually most desirable condition. I know I would actually be underestimating the Cpk, but at least I would understand the math.
However, if I had 80% >1.0 and 20% 0.5-1.0 on a given part that part is still acceptable per the tolerance. I could have anywhere from 1 of the 100 parts to 100 of the 100 parts with anywhere from an 80/20 split to a 99/1 split within those two tolerances and still have conforming parts. And that is where I lose any idea how to calculate Cpk and I’ve not found anything that helps.0February 27, 2020 at 1:39 pm #246387
Mike CarnellParticipant@Mike-Carnell Include @Mike-Carnell in your post and this person will
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@HV1234 I am not sure I understand this completely but let me see if I can help. It has been a very long time since I have used GD&T. When you say that >/= 1.0 is good and 0.5-1.0 is also then in terms of a Cpk they are both within specification. That is about the extent of what you get from Cpk. It really doesn’t explain degrees of goodness. It explains how much in inside and how much is outside the specification and the distance (measured in std dev’s) the mean is from the specification. Basically it sounds like when you create the two categories you are taking variable data and turning it into attribute data.
As I said I may not completely understand what you are trying to do.0
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