Proper Tolerance Stack Methodology 2

[tz101]

Anyone with good suggestions for how to treat overlapping entities within a linear tolerance analysis? Specifically, I am interested in how to allow for cases where similar components fit into the same exact linear space within a given assembly. For purposes of doing RSS or Monte Carlo analyses, do I add this "element" in twice to my total stackup, or simply ignore one of the overlapping components?

 

[GregLocock]

Um, if you get no helpful replies (and this isn't) it may be because your question needs re-writing. Can you expand a bit, and maybe give a similar example?

Cheers

Greg Locock

 

[GregLocock]

Ah, do you mean like fitting a handful of pencils into a pencil case where only the longest one matters?

If so I am 90% certain that that will break any simple statistical analysis, Monte Carlo is the obvious solution.

The reason is closely allied to the following problem: what computation time is required to find the longest pencil out of 3 pencils? and 10 pencils? There is a practical technique for which the time is equal, yet obviously more computation is required for the second case.

Cheers

Greg Locock

 

[tz101]

Greg,

Somewhat difficult to describe. If, for instance, two separate but identical spacers run parallel within the same linear space of an assembly I am conducting a linear tolerance stack up on, then do I take into consideration both overall length tolerance ranges when calculating my RSS and/or Monte Carlo results, or do I simply ignore one of the spacers altogether?

Thanks for the input.

 

[GregLocock]

Again with 90% confidence:

(a) you can't use RSS

(b) you'll need to put some logic in your Monte Carlo Sim to pick the longest of the two.

I think we posted simultaneously, does my pencils in a case analogy hold? I think it does.

Are there only two spacers in your actual case? Do you know the statistical distribution of their lengths?

Alternatively if you are being crude and just doing a min-max stackup then you can just use the tolerance of one spacer, I think.





Cheers

Greg Locock

 

[tz101]

Greg,

Don't know that the pencils example fits in the manner you have described. You mention only the "longest" piece mattering. My concern is with identical length pieces, both having identical tolerance range on overall length. Say, for instance, two instances of a part number 12345 with an overall length of 4.25±.005". My question is whether to only use tolerance range from one of these parallel instances (.010), or to factor both overall length tolerances (.020) into my Monte Carlo results. Am I making matters harder than they have to be?

Thanks, and God bless!

 

[GregLocock]

Do a mini Monte carlo to analyse the resulting distribution.

As a crude example

Lets say each one is equally likely to be high or low but twice as likely as either of those to be medium

Spacer 1 lo lo lo lo m m m m hi hi hi hi

Spacer 2 lo m m hi lo m m hi lo m m hi

result lo m m hi m m m hi hi hi hi hi

so 1/2 of the pairs are high, 5/12 are medium, and only 1/12 is low

That's fairly interesting, if you had 5 or so spacers they'd pretty much sit at the top limit.



Cheers

Greg Locock

 

[tz101]

Perhaps I need to do some more explaining. Not being a SPC person, or even having a QC background, I have been asked to perform tolerance stackup analyses on furniture assemblies for my employer. My background has been exclusively in mechanical design. My sole purpose in the original question posed was to ascertain whether or not I need to include ALL variables that contribute to the particular stack in question, or can I ignore "duplicate components"? I have been using a spreadsheet-based analysis tool purchased off the web to analyze (RSS) all assembly stacks to this point. The spreadsheet is pretty straightforward, asking for user inputs for each feature dimension and its associated upper and lower limits. Where I tend to get confused is whether or not to "double (triple, quadruple, etc.) enter assembly components into the stackup equation when dealing with multiple "overlapping" instances of the same part number. While your mini Monte Carlo is educational, I still am left wondering about the answer to my question.

Thanks, and God bless!

 

[GregLocock]

well, the tolerance range is the same as a single spacer, but the distribution of tolerances is very much biased towards the high side.

Cheers

Greg Locock

 

[ivymike]

The spreadsheet is pretty straightforward, asking for user inputs for each feature dimension and its associated upper and lower limits. Where I tend to get confused is whether or not to "double (triple, quadruple, etc.) enter assembly components into the stackup equation when dealing with multiple "overlapping" instances of the same part number.

Greg has shown the answer...

Your spreadsheet is apparently doing a stackup of dimensions in SERIES. Your parts are fitted in parallel. The effect on the stackup is as Greg describes- the range of variation doesn't change, but the distribution shifts markedly to the high end of the spec limits (because you're taking the max value of 5 tries). You can probably look up a formula for adjusting your distribution accordingly.

 

[ivymike]

...which the spreadsheet probably can't account for - so you would not want to double-count. (finishing the thought)

By stacking these parts in parallel, you're reducing the chances that you'll ever hit the lower end of the spec, but you're not eliminating the chance. If the spreadsheet doesn't let you input details of the part distribution within the limits, then the limits to enter are the ones you have for each part, and you enter all the parallel parts as though they're one.

 

[unclesyd]

While looking for another book I ran across this:

http://www.omega.com/bobi/productpage.asp?id=ME-1718