MMB - Indirect Modification

[Jacob Cheverie]

I have attached a very crude drawing of a simple part. There are two datum planes (A and D) and two datum cylinders (B and C). Let's assume that I wish to assign a positional tolerance on datum C, with datum references A and B (applied at MMB). The question is this: What is the size of the MMB of datum B?

Now, the MMB size will equal the MMC size if there is no geometric tolerance assigned to B. Y14.5 states that, if there is a geometric tolerance assigned to B with respect to the preceding datum structure (in this case, A), then the MMB size will be equal to the MMC size of datum feature B plus the allowed geometric tolerance on B with respect to datum feature A. In this case, it would be MMC size plus .005.

Here's where I'd like some feedback - If, contrary to the dimensioning style of ASME, datum feature B were to have a positional tolerance with respect to datum feature D (and no tolerance w.r.t. A), could some proportion of the geometric tolerance be used to increase the MMB size of datum feature B? The proportion would certainly depend on the perpendicularity between datum features D and A. As the two datum features depart from perfect perpendicularity, less of the positional tolerance zone (1.25 from D) can be used to shift datum B toward, or away from, datum C.

My question ultimately comes down to: How rigid is the requirement that the boundary increasing tolerance be related to datums of higher precedence, exclusively speaking?

 

[Burunduk]

Jacob Cheverie,
First, since datum feature B is a hole, not an external cylinder, the MMB is the MMC size minus the geometric tolerance with reference to the higher precedence datum, not plus.

Secondly, datum feature D does not influence the MMB size of B for the position tolerance on C. When this position requirement is inspected, the part is fixtured against a planar datum feature simulator A and mated with a cylindrical datum feature simulator B perpendicular to A. The size of datum feature simulator B is its MMB, and it takes into account the smallest allowed actual local size of the actual datum hole B and the variation of that datum hole relative to datum plane A - that is, the .005 maximum perpendicularity error. Since the part is not fixtured against D for the .005 at (M) position tolerance inspection, the position error of B relative to D has no influence on the MMB of B, that is - on the size of the datum feature simulator B for the considered control.

 

[Jacob Cheverie]

Burunduk,

Thanks for catching that, I’m used to dealing with pins and I didn’t think too much about the plus/minus.

I’m sure I’m off base with this method of thinking, it was just a curiosity. I was thinking more along the lines of an assembly where there is no geometric control on B with respect to A (I put one on the drawing for case #1). In case #2, I assume that there is only a control on B with respect to D and I’m wondering if, in an assembly, accounting for the position tolerance on B with respect to D would actually yield more datum shift/displacement on the control of C. If so, it would act as a substitute for the lack of control on B with respect to A. The added datum shift would then be dependent on the Perpendicularity error of D wrt A.

I guess the thing that would end this way of thinking would be if Y14.5 requires an orientation control of 0 at MMC on B wrt A if no control is clearly stated.

Again, I may be wrong. I’m trying to think of a method that may assist poorly drawn prints.

 

[Burunduk]

You are not off base.
Position with reference to |A|B(M)| means there must be consideration of the allowed variation of datum feature B relative to A in deciding the size of the B datum feature simulator, so I guess if their relationship is not controlled directly in the print you have no choice but considering the indirectly controlled allowed orientation error. With that said I'm not sure that simple addition of the position tolerance of B with reference to D and the orientation tolerance of D relative to A will give an accurate value for X in the calculation of MMC-X=MMB for B.