Two Datum Features, Single Datum Axis. 4

[ASHWA]

Hi Friends,
Need a help in reg. Two Datum Features, Single Datum Axis.

Is my assumption correct?

Thanks.

 

[axym]

Ashwa,

Bonus tolerance only applies to the feature controlled by the position tolerance.

For datum features, there can be additional tolerance but it isn't called "bonus". The tolerance zone doesn't get larger, but the part can be adjusted relative to the datum. The Y14.5 standard calls it "datum feature shift", and the tables in your calculation call it "mobility". The mobility comes from clearance between the datum feature and its datum feature simulator (gage element).

In Calculation 1, datum feature B is referenced RMB (Regardless of Material Boundary) so the gage element expands to achieve full contact with datum feature B and there is no shifting. So the entries for mobility are zero.

In Calculation 2, datum feature B is referenced MMB (because of the (M) modifier beside the B reference) and the gage element is fixed at the virtual condition size of feature B (in this case, 10 mm). If feature B is produced at a size of 10.4 mm and is perfectly perpendicular to A, the clearance would allow the part to shift by 0.2 in any direction. This would affect the minimum distance, so we see -0.2 in the MIN column.

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[Burunduk]

ASHWA,
The short answer is that datum shift doesn't provide a "bonus" (additional tolerance of position). Bonus means that the tolerance zone size for the axis depends on the feature size, and can grow as the feature size departs from MMC. Datum shift is not that.

But this is all confusing.
In my opinion simplest way to calculate the min and max distance is to analyze it through the "surface interpretation". Calculate the MMB and the Virtual Condition gage sizes. Draw the hard gage which includes the MMB and VC pins 10mm apart. On top of the gage draw the part with holes at their most extreme conditions in terms of the distance: smallest holes as much apart as possible and the largest holes as close to each other as possible. Then do the math and you'll see that the final calculation results, 2.6 and 3.4 for the MMC & MMB case, are correct.

 

[axym]

Hi All,

This next comment has nothing to do with MMB or datum feature shift.

I'm thinking that the MIN distance in Case 1 (with datum feature B referenced RMB) should be 2.6 and not 2.8. Those of you who have done tolerance stackups before - would you agree or disagree?

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[Burunduk]

axym,
I don't have experience with calculating stuck-ups systematically with spreadsheets etc, but when I calculate the distance between the dia. 10.8 and 4 Resultant Conditions located 10 apart, I get 10-10.8/2-4/2=2.6. So I agree.

 

[Kedu]

I'm thinking that the MIN distance in Case 1 (with datum feature B referenced RMB) should be 2.6 and not 2.8

Yes, I agree.
Regardless if MMB or RMB is used, the x min and x max do not change. This is from Alex K. book, isn't it? If yes, this problem has been discussed multiple times here on this forum.

 

[Kedu]

https://www.eng-tips.com/viewthread.cfm?qid=335464

 

[3DDave]

How can they be the same RMB vs MMB?

MMB: If I make a drill fixture with at 10.0 diameter gage pin and the part arrives with a 10.4 diameter hole then I can shift the location of the drilled hole by the amount of slop between the pin and in-process part.

RMB: If I make a drill fixture that has a centering feature then I cannot shift thee location of the drilled hole.

It doesn't seem reasonable that both cases are identical.

All calcs assume + is up on the diagram.
MMB:
For the closest/minimum case:
From the lower possible tangency of the pin to the hole top/upper tangency = 10.4
From the lower possible tangency of the pin to the pin center = 5
From the pin center to the hole true position = 10
From the hole true position to the allowed position -.2/2 = -.1
Additional allowed position at LMC of hole = -(3.6-3.4)/2 = -.2/2 = -.1
From allowed position to lower possible tangency of hole = -3.6/2 = -1.8

So 5 + 10 -.1 -.1 -1.8 = 13 from the lowest tangency of the pin to lowest tangency to hole.
Subract 10.4 from lowest tangency of the large hole to the upper tangency => 2.6

RMB:
For the closest/minimum case:
/not applicable From the lower possible tangency of the pin to the hole top/upper tangency = 10.4
/not applicable From the lower possible tangency of the pin to the pin center = 5
From the pin center to the hole upper tangency = 10.4/2 = 5.2
From the pin center to the hole true position = 10
From the hole true position to the allowed position -.2/2 = -.1
Additional allowed position at LMC of hole = -(3.6-3.4)/2 = -.2/2 = -.1
From allowed position to lower possible tangency of hole = -3.6/2 = -1.8

So 10 -.1 -.1 -1.8 = 8 from the pin center to lowest tangency to hole.
Subtract 5.2 from center of pin to the upper tangency => 2.8

Why would one believe the RMB case should be 2.6? If it depends on resultant condition that answer needs to recognize that resultant condition for perpendicularity on a hole does not apply to the full depth of the hole. The answer, 2.6, can apply only at the surface of the part as measured with a scale. 2.8 is the minimum one would see using a caliper.

The two questions are how close could two pins get to each other when held perpendicular to datum feature A vs what is the shortest crawling distance of a bug on one side?

 

[Kedu]

I guess it is the same discussion as always........
Difference between consistent wall thickness and the max/min wall thickness at any possible cross-section then the answers would be different........

https://www.eng-tips.com/viewthread.cfm?qid=433578

and specially

https://www.eng-tips.com/viewthread.cfm?qid=430008

 

[axym]

Kedu,

Thanks for posting the link to the previous discussion. It looks like the issue was dealt with quite thoroughly in 2012.

3DDave,

I agree that if the possible perpendicularity error on feature B is factored in, then the distance can go down to 2.6 but only at one surface and not through the whole thickness of the part. So a caliper would see 2.8.

However, if the possible form error of feature B is factored in, the distance can go down to 2.6 through the whole thickness. Imagine that feature B is a perfect 10 mm hole, perfectly perpendicular to A. Then cut a 0.4 mm notch on one side, through the whole thickness of the part, in the direction towards the small hole. This will not change the related AME but it will make the largest actual local size of the hole 10.4 (LMC) and bring the hole-to-hole distance down to 2.6. A caliper with knife-edge jaws would see 2.6.

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[3DDave]

Evan, while that makes an interesting example, that part would see the scrap bin for poor workmanship. I think I would say "You are right. <chuck> Now make it right."

 

[axym]

3DDave,

So the quality manager hears that parts are being thrown in the scrap bin, and is wondering what happened because the parts passed the dimensional inspection. The parts get returned to the vendor, who asks "What was it on these parts that didn't meet the drawing?" ;^)

Evan Janeshewski

Axymetrix Quality Engineering Inc.

http://www.axymetrix.ca

 

[3DDave]

So the customer calls the company president and cancels the contract because the parts look like garbage, promising that if he has to eat those parts he'll make sure to mention getting jerked around over parts that look like garbage.

The manufacturing staff had the option to decline to make the parts because they can't produce a part with a uniform finish or mention they are playing dimension games that will make parts that look like junk but skirt on the "rules."

 

[Burunduk]

While the "notch" shaped form error was probably brought up for the simplicity of the example, and in the real world it is likely to be noticed and rejected as a visible defect, there are realistic cases where parts could be approved with a similarly calculated minimum distance resulting from the exact same reasons mentioned by axym. This can be calculated from the Resultant Conditions, which by the way also take Rule#1 form error into account. Instead of the notch-shaped discontinuity, the datum feature B hole could be produced ovoid, with the RAME size still being 10mm, but a cross-section of 10.4 at the direction towards the small hole. The accumulating distance of 5mm(RAME radius) + 0.4(ovoid form error) would be 5.4 from datum axis B. As a result, there could be a uniform wall thickness of 2.6 between the holes possibly without anyone noticing. If the tolerances were tighter there would be even a greater chance that a similar phenomenon would pass undetected during inspection.

 

[ASHWA]

Thankyou Friends...!!

2 Things i have learned from the discussion,

1. There is NO Bonus Tolerance for Datum shift.

2. Have to check before, reg. any similar topics have been discussed.