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(OP)
When calculating a RSS is there a minimum number of tolerances than can be used? I have seen RSS results with 2 tolerances(the very minimum) used for the calculation and have been told that an RSS should not be used unless there are 5 or more tolerances used in the calculation. Nobody where I work can tell me where the 5 or more rule originated. All my research into this has not come up with a reason why this is enforced.

Any input/comments would be greatly appreciated!

Thanks,

L. Jones

ljones,

I don't have my tolerance stack book with me.

Are you working out the mean dimension values, or the standard deviations? There is some number below which standard deviation has no meaning, and I would think that number was very much greater than five.

--
JHG

(OP)
We are working out the standard deviation of nominal dims with equal bilateral tolerances. The RSS sums the variances of the pooled dimensions and the square root of the is the standard dev. of the pooled(assembly. You most likely know that already. My stats books do not state anything about having a minimum number of inputs.

L. Jones

I've used it with as little as 2 dimensions although the benefit is bigger with more dimensions in the stack. I don't see anything in the theory that invalidates just 2 dimensions. You do need to make sure that statistical process control is being applied to the dimensions. If they are not mean centered and capable you are kidding yourself.

----------------------------------------

The Help for this program was created in Windows Help format, which depends on a feature that isn't included in this version of Windows.

(OP)
Thanks dgallup,

From researching I was inclined to accept that 2 bilaterally toleranced dims would be acceptable. Stats usually generate more reliable results with increased number of reasonable data but one does not always have a lot data to use.

There is no rule about this, even from statistical theory. But simply realize that the more variables you have in the calculation, then the more confidence you can have in the RSS answer.
It's like rolling dice -- a histogram tracking the result of many rolls of one die will be a flat distribution. Two dice will show a bump in the center (think of the game of craps). But if you roll ten dice, then the histogram will be much closer to a Gaussian bell-curve distribution.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

You need to look at the confidence for the assumption that the resulting distribution will be normal based on the inputs.

My guess is that the confidence interval for a small quantity is so large that it is not useful to use RSS.

This looks like a promising paper, in which the RSS is multiplied by 1.5 as a correction factor. I need to read the rest of the paper to see where that came from, but the first few pages covers the essential problems involved with the analysis.

Dave, you may be referring to Arthur Bender's SAE paper from years ago for the 1.5 correction factor. That number was derived empirically, not from any theory. He did a study of actual production parts compared to the RSS predictions, and there seemed to be a 1.5 factor between them. Over the years that's been proven somewhat accurate, although many people will fudge that number to 1.4 or 1.6.

And you make a good point about the initial assumptions. Even if the OP has enough inputs for everyone to say "yeah, let's do an RSS" there are still many caveats that need to be verified. Here are a few of them:
-- it needs to be a linear stack-up (no sines/cosines etc.)
-- each input should be a controlled process, with a normal distribution
-- the process average should match the design mean
-- the tolerances being stacked should be independent

People often run with the RSS answer without checking into these caveats first.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

I was referring to a Boeing paper that referenced Bender.

It goes into details about other correction factors that depend on the actual distributions.

The problem with most tolerance analysis is that it can be difficult, without a captive and interested manufacturing and inspection group, to determine the actual manufacturing variation distribution. Without that, the rest is a paper exercise.

This is one thing that became obvious when using VSA for tolerance analysis - it was easy enough to calculate what percentage of paper parts would be acceptable, but impossible to predict manufacturing yield. Another interesting factor was the contribution of fixturing, which should reduce any allowed range to a tiny one.

(OP)
Thanks for the valuable feedback and comments!

Belanger makes a really good point about caveats. People will often do a RSS calculation without meeting or understanding the assumptions
(I prefer the term "rules") for the analysis. I have observed people changing tolerances from equal bilateral to +0/-value or +value/-0 to get a RSS that they feel is more functional without knowing they have violated the assumptions/caveats(rules).
For a RSS calculation, 0(zero) is more like a condition of no tolerance than a tolerance. The caveats listed would also require that
unequal bilateral be converted to an equal bilateral for input. Many people calculate RSS without ever knowing there are caveats/assumptions that need to be followed.

(OP)
Lets say we have a 2 dimension stack to evaluate. Each tolerance in the stack is the RSS for the dim. That would mean 99.7% of the measured values for the dim would fall with the RSS tolerance. So now we complete a RSS calculation with the 2 dims. The confidence for the result would be 0.997*0.997=0.994. If more dims are in the stack the confidence in the result would keep going down. A 5 dim stack would be 0.985. Is this reasoning wrong?

L. Jones

How do you know that 99.7% of the measured values for the dim fall within that RSS tolerance? I don't think anything in RSS theory tells us that.

Recall that RSS gives an answer, but one that is only valid if the capability of the dimension/process is controlled. And the better the capability, the more confidence you have (i.e., closer to 100%). So I think the missing variable in your scenario is the standard deviation (or Cpk).

Along the lines of your question, however, is another approach put forth by a guy named Gilson in the 1970s. He proposed a graduated scale on the worst-case answer, based on the number of dims in a stack.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

(OP)
John-Paul,
You are correct. RSS Theory does not tell us the percentage of the measured values one needs. Knowing the Cpk would allow one to figure out the percentage of measurements. In the example I proposed, with a little quick research after you posting, the 2 dims would require a Cpk of 1 so that 99.7% of the values would fall within a specific range. Higher Cpks would increase the confidence in a 2 dim stack when the Cpk are known for each of the dims. Unfortunately, we do not always know the Cpks when designing new products or processes.

Note. While investigating this I found several references to Gilson. Most stated that his work was done in the 50s and one stated that it was later shown to be inaccurate.

Thanks,
Les

Good to know that about Gilson (not sure why I thought he was from the 1970s). Your initial scenario from this morning is still an interesting one -- although my gut says that if you have more dims(such as 5) then the percentage would go up, not down.

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

(OP)
Gilson is referenced in the link that 3DDave provided.
Going back to the 2 dim stack. Assume both dims are produced with a process capability Cpk of 1.33. At this Cpk 99.4% of the dims will come within their tolerance. This would mean that the confidence in the 2 stack would be higher that 5 dims with a Cpk of 1.33. This Cpk was picked because it is the lowest one deemed capable. When calculating a RSS for s single measurement more measurements are better. I think everyone working with stacks has been told/trained that. Every additional dim in a RSS stack would slightly lower the confidence level in the result. I have done RSS assembly stacks with over 30 dims. Simply because there are alot of dims does not mean the confidence in the result is better that a stack with 2 dims. It makes sense for the level to decrease when there are several dims being input.

?

The confidence that the distribution is normal, regardless of the underlying distribution shape, increases with increasing numbers of contributors and the RSS tolerance width increases much more slowly than the arithmetic/worst case.

In other words, the confidence that the results of an RSS calculation are represented by the normal distribution increases with the number of contributors.

Multiplying the probabilities doesn't seem correct. You are looking for the probability that the sum of two (or more) values to be within a certain range. Multiplication only tells the probability that each contributor is within a given range at the same time.

Instead, if one adds the 3 sigma variance of one to the 3 sigma variance of the other you get the 3 sigma variance of the stack. The square root of this summation is the 3 sigma deviation of the distribution.

ljones -- are you sure that "every additional dim in a RSS stack would slightly lower the confidence level in the result"? Statistics theory has an important concept called the central limit theorem; this tells us that the more variables (dimensions) are in play, the more the overall answer tends toward a bell-curve shape.

So I'd have way more confidence in the RSS method when 30 dims are involved rather than 2! (Granted, I'm still leaving out the Cpk aspect; I'm just talking about using RSS versus not using RSS.)

John-Paul Belanger
Certified Sr. GD&T Professional
Geometric Learning Systems

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