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Stackup calculations (X min / X max.)
3

Stackup calculations (X min / X max.)

RE: Stackup calculations (X min / X max.)

MIN = 2.5 mm
MAX = 6.05 mm

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

RE: Stackup calculations (X min / X max.)

J-P,

If I am not mistaken, your numbers have been calculated with the assumption that LMB size of the datum feature A is 19.0. Are you sure it is the correct value?

Thanks

RE: Stackup calculations (X min / X max.)

What else would the LMB be on an external diameter? There is no datum referenced in the FCF ahead of A.

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

RE: Stackup calculations (X min / X max.)

Pmarc,
Should we use 18.5mm as the LMB size for datum feature A?
Also please correct me/educate me if I am wrong, perfect form at MMC is still needed for the OD and perfect form at LMC is only mandatory for the ID?

What are your x min and x max values?

RE: Stackup calculations (X min / X max.)

I agree with greenimi. Datum feature A is only controlled by a size tolerance, and is subject to Rule #1 (perfect form boundary at MMC). It could have an actual local size of 19, and be bent by 0.5. The LMB, or the maximum inscribed cylinder, would be 18.5. This makes the extreme wall thickness values 2.35 and 6.3.

Evan Janeshewski

Axymetrix Quality Engineering Inc.
www.axymetrix.ca

RE: Stackup calculations (X min / X max.)

I like the way 2.7.1 (c) is worded: Where is no default requirement for a boundary of perfect form at LMC.

RE: Stackup calculations (X min / X max.)

Yes, but the question is the max and min "X" -- which I assumed means the consistent wall thickness. I think you guys are referring to the wall thickness at a cross-section ... would that have something to do with why we get different answers?

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

RE: Stackup calculations (X min / X max.)

greenimi,
My answer to your first two questions is yes.

My numbers in the third question are 2.25 for Xmin and 6.3 for Xmax.

J-P,
For the consistent wall thickness the answer is the same as for the wall thickness at a cross section.

RE: Stackup calculations (X min / X max.)

Sounds good -- I should have sketched it all out before plunking down the numbers!

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

RE: Stackup calculations (X min / X max.)

To get the 6.3 wall thickness value**:

{ LMC OD 19
+ 3 * OD_TOL 0.5
- LMC ID 11
+ LMC Pos Tol 3.0
- 2 * ID_TOL -0.05}
/ 2

To get the 2.25 wall thickness value:

{ LMC OD 19
- OD_TOL 0.5 (<- This is from adding 0.5 to get the MMC condition,
then moving a direct -0.5 from allowable bend/oval into the
diameter-based equation by multiplying by 2.)
- LMC ID 11
- LMC Pos Tol 3.0
+ 0 * ID_TOL -0.05} (<- This is already at LMC; the tolerance moves away from
that boundary and has no contribution.)
/ 2
---------
** That calc is based on figuring how rig the numbers to get the previously mentioned result.

My derivation is as follows:

The diameter of the OD at MMC is 19.5; the LMC diameter is 19.0 so the farthest the datum simulator can travel is 0.5.
The distance from the far outer wall to the offset datum simulator center is 19.5-(19/2) = 10
The distance from the center of the datum simulator to the center of the shifted MMC hole can be (3 + .05)/2
The radius of the hole at MMC is (11 -.05)/2; subtract the shift from the hole -> (8 -.1)/2 = 3.95.
This is the distance from the offset datum simulator center to the far wall of the hole.

Take the difference 10-3.95 = 6.05

or, as above:

{ LMC OD 19
+ 2 * OD_TOL 0.5
- LMC ID 11
+ LMC Pos Tol 3.0
- 2 * ID_TOL -0.05}
/ 2

RE: Stackup calculations (X min / X max.)

3DDave,

The 6.3 for Xmax was calculated with the assumption that the LMB size of the datum feature A (18.5) is not equal to its LMC size (19.0). If you did the same thing, your derivation would be like this:

The diameter of the OD at MMC is 19.5; the LMB diameter is 18.5 so the farthest the datum simulator can travel is 1.0.
The distance from the far outer wall to the offset datum simulator center is 19.5-(18.5/2) = 10.25
The distance from the center of the datum simulator to the center of the shifted MMC hole can be (3 + .05)/2
The radius of the hole at MMC is (11 -.05)/2; subtract the shift from the hole -> (8 -.1)/2 = 3.95.
This is the distance from the offset datum simulator center to the far wall of the hole.

Take the difference 10.25-3.95 = 6.3

RE: Stackup calculations (X min / X max.)

At least I know what is required to get you to show your work.

What's interesting is that there is no example in the standard of an LMB datum reference that has both an explanation and no other related FCF. Figure 7-17, which is where the question came from, contains no 'means this' section. That's an odd omission considering it's a featured example.

RE: Stackup calculations (X min / X max.)

pmarc

Its easy to understand MMB, but I am confused at LMB here, may I know how do you get the LMB figure 18.5? Thanks lot.

Season

RE: Stackup calculations (X min / X max.)

Hi All,

After all that effort, I mistyped one of the values in my previous post - I meant to type 2.25 and not 2.35. No, really, I had it right and didn't have to look at pmarc's calculation ;^).

I used a different method to calculate the result. I sketched the boundaries and an extreme as-produced feature, and then measured the wall thicknesses.



The black lines are the boundaries and the brown lines are the as-produced part.

3DDave,

I think that they apply a "perfect form at LMC" mindset to the wording in Y14.5. If there is no explanation provided, as in Fig. 7-17, then there can't be any errors. If they say a little bit, as in 2.7.1 (c), then there can be one error as you pointed out. If they say a lot more ...

Evan Janeshewski

Axymetrix Quality Engineering Inc.
www.axymetrix.ca

RE: Stackup calculations (X min / X max.)

Seasonlee,

Draw a banana shaped cylinder with section dimension of 19 inside a 19.5 boundary. You lose .5 on each side.

RE: Stackup calculations (X min / X max.)

Quote (axym)

I meant to type 2.25 and not 2.35. No, really, I had it right and didn't have to look at pmarc's calculation ;^).

Evan,

We believe you…..
For at least two reasons:
1. We know you know this stuff.
2. Pmarc calculations did not include X min (2.25 calculations), but only 6.30 (X max). He included only the final result for X min. (2.25);

RE: Stackup calculations (X min / X max.)

greenimi,

Thanks. I actually didn't even read pmarc's or 3DDave's derivations in detail. I have a hard time thinking of these things in terms of center geometry and radii - I do much better with diameters, and the surface interpretation. I also need the picture - I can see things in my mind, but only up to a certain level of complexity.

I agree with 3DDave that the lack of a "means this" for Fig. 7-17 is a concern. The only explanation in the text is "allowable displacement results when the datum feature departs from LMB". It appears that Y14.5 didn't think through the implications of a datum feature referenced at LMB, but without a tolerance referencing it at LMC (that would switch the perfect form boundary to LMC, according to the special rule). I don't think that this figure (or its explanation) has been changed in the new draft, so there must not have been many (or perhaps any) comments on it during the review process.

Evan Janeshewski

Axymetrix Quality Engineering Inc.
www.axymetrix.ca

RE: Stackup calculations (X min / X max.)

2
Just to be clear, I have not shown any of my derivations yet - I merely corrected 3DDave's derivation for Xmax and, to be honest, some pieces of it are still a mistery to me smile

For what it is worth, my preferred way of doing stacks is to use A.Krulikowski's method, but I noticed that the method isn't easy when it comes to explaining its logic/mechanics (especially to those who are not familiar with it). That is why in this forum I have always tried to explain my derivations by using additional sketch(es) and bringing the numbers down to simple equations. Here I will make no exception, although in this case the sketch looks pretty similar to Evan's graphic.

http://files.engineering.com/getfile.aspx?folder=8...

Side note #1:
I agree with everything that Evan said about lack of sufficient explanation on how to calculate LMB size for unrelated features of size in current Y14.5. From what I see the figure has been completely removed from the draft of new Y14.5, however there is one statement in the draft that changes the story a little bit, in my opinion. In para. 7.11.8 they say: "When an LMB equal to LMC is the design requirement for a given datum feature, a zero geometric tolerance at LMC is specified to the datum feature [...]". They explain it using an example with secondary and tertiary datum features controlled with 0@LMC geometric tolerances, but that statement stays true for primary (unrelated) datum features too. In other words, if in OP sketch the datum feature A had had a zero straigthness at LMC applied (as a DML control), the LMB of A would have been 19.0. Without it the LMB is 18.5.

Side note #2:
Evan, my sincere apologies for not waiting with my reply until you corrected the typo in your post smile

RE: Stackup calculations (X min / X max.)

pmarc
Excellent, your sketch illustration always worth a star.

Season

RE: Stackup calculations (X min / X max.)

pmarc,

I checked the public review draft, and the figure hasn't been removed - it's still the same as before. I agree with you that adding a straightness control of zero at LMC would make the LMB 19 instead of 18.5. Here's a question - would this override the Rule #1 boundary, so that perfect form at MMC would not be required?

Evan Janeshewski

Axymetrix Quality Engineering Inc.
www.axymetrix.ca

RE: Stackup calculations (X min / X max.)

Ah yes, you are right, Evan. The figure is in the draft indeed - it is now fig. 10-18. For some reason (most likely because it is in chapter 7 in 2009 standard) I also searched in chapter 7 in the draft. My mistake.

As for your question, perfect form at MMC would not be required in this case. Para. 2.7.1(d) in 2009 standard does not explicitly say that, but fortunately in the draft it was clarified. Para. 5.7.1(d) says: "Where a geometric tolerance is specified to apply at LMC, perfect form at LMC is required and there is no requirement for perfect form at MMC."

RE: Stackup calculations (X min / X max.)

pmarc,

What parts are a mystery? It must be easy enough to show where it's unclear. It was clear enough to correct it.

RE: Stackup calculations (X min / X max.)

Quote (axym)

I agree with you that adding a straightness control of zero at LMC would make the LMB 19 instead of 18.5. Here's a question - would this override the Rule #1 boundary, so that perfect form at MMC would not be required?

Quote (pmarc)

As for your question, perfect form at MMC would not be required in this case. Para. 2.7.1(d) in 2009 standard does not explicitly say that, but fortunately in the draft it was clarified.

ASME Y14.5-2009 para. 2.7.1(d) may not say it, but doesn't para. 2.7.1(a) cover this?

Quote (ASME Y14.5-2009 para. 2.7.1(a))

No variation in form is permitted if the regular feature of size is produced at its MMC limit of size unless a straightness or flatness tolerance is associated with the size dimension or the Independency symbol is applied per para. 2.7.3.

pylfrm

RE: Stackup calculations (X min / X max.)

Quote (3DDave)

subtract the shift from the hole -> (8 -.1)/2 = 3.95

This is unclear but luckily I did not have to touch it to get to the correct number.

pylfrm,
Personally I do not like the statement from 2.7.1(a) because to me it implies that the straightness or flatness of zero at MMC would not be considered by the committee as legal callout.

But if we take that out of equation then I guess 2.7.1(a) could be used to prove that when produced at MMC size the feature controlled with straightness tolerance of zero at LMC does not have to have perfect form.

RE: Stackup calculations (X min / X max.)

pmarc - that's because you didn't read the line above it. (11 - 0.05)/2 - (3 + 0.05)/2 = (8 - 0.1)/2

I call 2.7.1 half-baked. It has interlocking requirements that conflict. I bet they have escaped notice due to the rarity of the LMC callout and the lack of good examples in the standard.

RE: Stackup calculations (X min / X max.)

Quote (3DDave)

pmarc - that's because you didn't read the line above it. (11 - 0.05)/2 - (3 + 0.05)/2 = (8 - 0.1)/2

Yes, no other reasonable explanation comes to my mind smile

RE: Stackup calculations (X min / X max.)

To all who posted:

I have not studied and LMC datum concepts like it appears everyone who is posting has. So I am trying to follow the posts as a expanding-my-understating exercise. The "dialoig" is amazing! But I have one observation I would like confirmed.

The min / max calculations have issues because Y14.5-2009 and the Nov 2105 DRAFT "avoid" clarifying text that would define that "there is perfect from at LMC for LMC datums" (or not).

Certified Sr. GD&T Professional

RE: Stackup calculations (X min / X max.)

mkcski -- No version of the standard has said that an LMB reference automatically means perfect form at LMC. I haven't read the new draft completely, but from the snippets given by pmarc and Evan, it says that if you wish for LMB to equal LMC, then you need to put a geo tol of zero at LMC on that datum feature.

I don't think any of that makes for an issue on this calculation, because the designer apparently didn't want that perfect-form requirement. (Maybe perfect form at LMB was desired, thus it can be confusing, witnessed by my flub on the initial answer!)

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

RE: Stackup calculations (X min / X max.)

They mostly have problems since LMC/LMB references can not typically (ever?) be verified by go/no-go gauge as they lie on the surface only at LMC or are embedded in the part. Lack of general applicability doesn't help either - most interest in tolerances that affect fit and LMC/LMB can not typically (ever?) affect fit in a way that is any but a side effect. Since there is no gauge there's no physical datum simulator.

The main thing that's missing is a diagram of what an LMC datum reference means in the context of an unmodified feature of size - that the LMB is based on a banana (or similar) shape that forms the limit for the unrelated actual minimum material
envelope, which is not a shape that is often seen in machined parts.

Since the example was from Y14.5 and all it was used for was a cartoon without explanation I'd say that's a good place for improvement.

RE: Stackup calculations (X min / X max.)

3DDave:

I totally agree with your thinking. The modern world runs runs on interchangeable parts, which supports application of MMC assembly concepts. I have only seen one justified use of LMC for features (not datums) - controlling the ligament "thickness" between holes in a heat-exchange tube sheet.

Certified Sr. GD&T Professional

RE: Stackup calculations (X min / X max.)

This (the OP excercise) could be a good question / problem for the GDTP-2009 certification exam and it is from the standard ( from the "suppose to know material").

RE: Stackup calculations (X min / X max.)

Nah - it's what I call a sucker question. If you know the trick it's easy; if you don't it's disproportionately difficult. If there was a 'means this' for that diagram then the trick is obvious. There are a lot of MMC examples. I suspect if this went through there would be a 90-95% fail rate.

Edit - Except for those who add covering this example to their GDTP test prep courses.

RE: Stackup calculations (X min / X max.)

Quote (mkcski)

I have only seen one justified use of LMC for features (not datums) - controlling the ligament "thickness" between holes in a heat-exchange tube sheet.

Why not directly tolerance the ligament thickness? Seems like that would be a much more accurate encoding of the functional requirement.

pylfrm

RE: Stackup calculations (X min / X max.)

I'm confused about how we're getting 18.5 from "19 +0.5/-0"?

RE: Stackup calculations (X min / X max.)

The maximum diameter is 19.5 so that limits the form. At 19 the outer part can bend to wedge into that 19.5. In the bent condition the largest perfectly formed cylinder that can fit is 18.5.

If you draw a picture of these steps - 19.5 diameter, bent 19.0 local size minimum, and then fit a cylinder that lies fully within the bent 19. local size bent version you'll see that you loose .5 from one side of the 19.5 and an additional .5 from the other side.

RE: Stackup calculations (X min / X max.)

Lee

I am with you about using 18.5 Here is an exact example from Alex book where he uses the given limits of size.

RE: Stackup calculations (X min / X max.)

A.K book does not use the form error for the datum feature inside diameter A (RFS in the positional callout)
Not sure why the book does not use it................., but it is a difference between thin wall and min wall.
If 14.2-13.8 = 0.4 form error is taken into the calculation for the minimum wall calculation then you find find the "correct" answer to be 0.45.
Again: thin wall (probabbly constant accross the section) 0.65, but minimum wall = minimum possible distance x min. that can ever happen in a single cross section is 0.45.




RE: Stackup calculations (X min / X max.)

That's not the same example really, because that doesn't have LMB on the datum reference. Having LMB means that there is no Rule #1 at the least material size, which is the reason the other example used 18.5.

edit: sorry greenimi -- I didn't notice your comment is saying the same thing.

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

RE: Stackup calculations (X min / X max.)

Quote (Belanger)

That's not the same example really, because that doesn't have LMB on the datum reference.
Right. Unfortunately, there is no similar example using LMB. If you want to use Alex approach, the trick is to use 18.5 as LMC. Where the VC=MMC size (19.5) or min datum shift=0 (|MMC rad-VC rad) and the max datum shift is 1.0 (|LMC rad - VC rad|)

RE: Stackup calculations (X min / X max.)

I mean, what's the definition of "perfect form" here? If the hole tolerance tells me that it applies only when a Ø19+0.5/-0 shaft is at LMC (and therefore "perfect form"), shouldn't that mean only when you've got a perfectly straight Ø19 shaft in which the upper limit doesn't come into play?

RE: Stackup calculations (X min / X max.)

The perfect form limitation only applies at MMC unless the feature itself has a geometric tolerance that applies at LMC. Since the outer diameter has no geometric tolerance applied, then its perfect form is at MMC.

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