How to determine the specification?

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    The diameter specification for large ring  is:20mm+/-0.04mm, the small ring one is: 18.5mm+/-0.04mm.  If the small ring is placed into the large ring,  what is the specification for the space?



    The specification for the space between large and small should be: 1.5mm+/-0.08mm.
    SUL: 20.04mm-18.46mm=1.58mm;
    SLL: 19.96mm-18.54mm=1.42mm;
    According to this calculation, space diameter should be 1.42mm-1.58mm. It’s 1.5 mm with +/-0.08mm tolerance.
    I hope my answer will meet your expectations.



    When you have an assembly of N items, with a tolerance of +/-t on each item, then the total tolerance is given by +/- (N x t)/SQRT(N).
    For example, if 100 plates of thickness 1+/-0.01 mm are stacked one over the other, the total thickness shall have a thickness of 100 +/- 0.1 and NOT 100+/-1.0.
    Reverse of it goes like this: If a customer has specified an assembly to have a total dimension of 10 +/- 0.1 mm, with the assembly made of 10 items of 1mm thk. each, the tolerance for the items shall be 0.1/SQRT(10) = 0.0316.
    But, your question is regarding the difference and not the sum. Since, it is the dimension of the rings that are specified, I would go by what Marcin has written.


    Tim Wygant

    Wouldn’t it depend on the type of fit desired (clearance, transitional, interference)?



    Sri, your equation of +/- (N x t)/SQRT(N) is one of many forms of statistical tolerance summing.
    However, in my experience, and many of my peers, along with years of data collection on tolerance means, even with high volume automotive industry production, I have never seen statistical tolerancing to be consistant nor accurate on assemblies of two to four parts; whether the parts are made by ourselves or suppliers under very tight SPC control, the resulting actual average tolerance mean value tends to be away from the central mean value of the given tolerance, most often the actual measured results tend to fall 30% off central mean towards the maximum material condition.
    In other words if the part has an inside diameter of 20.00mm +/- .04mm.  The overal average of hundereds of parts would tend to have an actual mean value of  [20.00mm – (.04 x .30)] = 19.988 mm.
    I may have swayed slightly away from the intended question by Cactus; and I apologize.  But I believe Marcin’s answer to be the true answer for this particular case.



    Let me pick an example that should be near and dear to your heart. Who does statistical toleranceing work for – Honda or Harley?
    The answer is Honda not because of the statistics, but because of the operating philosophy. The focus of what we call SMED is not how long it takes to change over, but instead on how quickly we can bring the process on target. The Honda folks worry about processes being on target, not meeting specs.
    PS – Does that mean I would ride a Goldwing? No way.


    Tim F

    The question is 3 weeks old, so I hope “Cactus” is still around!Sri’s equation works well to find the total tolerance of a stack of parts, assuminmg that you have measured the mean & tolerance of the parts. It doesn’t work so well the other way around – to give specs that will meet a given total.Using Sri’s example of stacked , suppose you told the shop to make a bunch of parts 1 +/- 0.03 mm thick that you were planning to stack. They give you a bunch of parts that are 0.98 +/- 0.01 mm thick. That’s 10 Sigma quality, but a stack of ten is almost guaranteed to be too small.In other words, +/- (N x t)/SQRT(N) works well as a PREDICTION when adding tolerances, but it doesn’t work well as a SPECIFICATION of tolerances.Finally, if the cylinders are concentric, then the radii, not diameters determine the gap between the two parts. Then the simple answer is 0.75 +/- 0.04 mm for the gap all around, rather than 1.50 +/- 0.08 mm. Tim F


    Peter C

    Whether Marcin or Sri is correct depends on the interpretation of the original question. I.e., if you feed the two rings with the said (guaranteed) diamters into an assembly the distance of the (concentric) rings can range betwwen 0.75-0.04mm and +0.04mm.
    On the other hand, if we assume a certain process capability z (for example 6 sigma) for the respective rings, than we can ask with which specification limits the resulting assembly is again at z sigma (e.g., 6 sigma). In this case Sri is right with 0.75+/-0.028mm (of course only regarding variance due to the variances of the rings :-)
    Another way to look at it: If we accept Marcin’s specification limits we can ask what sigma level the assembly will be at: Sqrt(2)*z (i.e., if the rings were at 6 sigma, the assembly with 0.75+/-0.04mm was at 8.5 sigma).

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