Tolerance stackup calculation (ref pmarc example Alex Kurlikovski Fundamentals GD and T book) 1

[greenimi]

I am learning tolerance stackups and I am using Alex Kurlikovski book (Fundamentals of Geometric dimensioning and Tolerancing, 2nd edition). I have a question about the stackup tolerance calculation --fig 9-14 page 265—attached—
Minimum distance X min calculation shown is 4.5mm. Someone around here, who has way more experience than me in those kind of calculations, is claiming that the “real” X min calculation should be 4.1 (not 4.5 shown) because the form error was not included.
In the book: X min is :
69.6 (min length) – 50 (basic) - 10(basic) – 8.6/2 (max size for the hole) - (0.6+1)/2 (hole at the LMC, hole position is at MMC) = 4.5mm
Our expert is claiming the calculation should have started from 69.6 - 0.4 = 69.2 and not from 69.6. Therefore, the result would be 4.1mm and not 4.5mm.
Justification: the size of the feature (length) still has to be ±0.4mm (70.4/69.6 = 70±0.4), if the opposing points meet the size specification and the envelope meet rule#1, then the length meet the requirements. The form error was not taking in consideration for the calculation in the book. (a gage can use all 0.4mm in form error and still be making contact with the datum feature simulator)

I know pmarc had some issues with X min calculation in fig 9-12 page 263 (x min should be 2.7 and not 2.9) and here is that specific thread.

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

And it’s exactly as pmarc stated: “It is weird to disagree with such authority”
Now, going back to our issue (page 265): Ii is our “expert” points us in a right direction or “the unclaimed form error” is not applicable here?

Thank you

 

[pmarc]

Oh, and short addition to my previous post:
- to have a better picture I increased the thickness of the part in my graphic - it has no impact on the stack-up result;
- in cases #2 and #3 datum feature B stays in linear contact with its datum feature simulator (so at least two points of contacts exist).

 

[Belanger]

I thought we had hashed this out before: Doesn't it have something to do with how you define max and min? In other words, is the min distance the same CONSISTENT min all the way through the part? Or just the min at ONE end of the hole? I believe that this makes a difference in how we are all approaching the question.

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

 

[pmarc]

You are right, CH. I attached dimension to wrong end of the right face. But I think that doesn't change anything in my reply. My apologies.

 

[powerhound]

pmarc, I wasn't thinking about it that way. Somehow, your explanation wasn't clear to me either. The graphic worked perfectly though. I see what you're saying now.

Still though, I don't think this detail should be brought up in the context of the OP. It only serves to confuse the issue as evidenced by greenimi's subsequent questions regarding differences in the fundamentals and advanced concepts. Regardless, greenimi's expert was not correct in how he was calculating the distance.

Thanks for clearing it up for me.

John Acosta, GDTP S-0731
Engineering Technician
Inventor 2013
Mastercam X6
Smartcam 11.1
SSG, U.S. Army
Taji, Iraq OIF II

 

[greenimi]

I can recall pmarc gave us an excellent step by step explanation on the calculation of tolerance stack, pls see the link below.

http://www.eng-tips.com/viewthread.cfm?qid=287922

Season,
I know pmarc's stackup calculation. However, I think it does not fit very well my case because pmarc's case HAS an orientation control of the right edge to datum feature A and the datum feature B---perpendicularity 0.1 A primary and B secondary. This perpendicularity also control the form error in pmarc's case. In my case (AK Book) that control (orientation control) does not exist so the form error is controlled by the rule#1, right? And here we got that issue, on how that form error is included in the stackup calculation.

In other words, is the min distance the same CONSISTENT min all the way through the part? Or just the min at ONE end of the hole? I believe that this makes a difference in how we are all approaching the question.

J-P
No, the X min distance is NOT consistent (at leat in our opinion) all the way through the part.
Okay, let's say in AK book is consistent, because is a Fundamentals Book.
But, I would like to ask you guys, how the calculations would be different (exactly as J-P stated above) if the X min distance is NOT cosistent through the part and how the form error (indused by the direct toleranced dimension ± --70±0.4---) shoud be taken in consideration. That I need the help for.
Thank you gentlemen

 

[greenimi]

After re-reading pmarc’s replay for the form error I have to agree that the maximum form error in this case is 0.8.
So, I want to ask pmarc a quick question: if the part is at its minimum length 69.6, why the maximum form error of 0.8 cannot be applied from 69.6 in the opposite direction than the direction you’ve indicated. In other words, what paragraph of Y14.5 standard states that on the top of the part cannot be 69.6 at any cross section but on the bottom cannot be 69.6-0.8= 68.8.

[If left face is perfectly flat and perpendicular to A and the length of the part is 69.6 at the top, the length of the part at the bottom can be 70.4, resulting in 0.8 of both right faces form error and not 0.4.”]

Why the length at the bottom cannot be 68.8? (keep the same form error, but the opposite direction). Reason the measured value of any individual distance at any cross section of a feature of size is still 69.6 (measured at the top) and the rule#1 is still in place and clearly defines an envelope of 70.4 width that the feature of size must pass through.

What am I missing here? What paragraph of the standard is violated by this approach? I know something is not quite right, but At this point and with my limited knowledge I don’t know what?

 

[pmarc]

So, I want to ask pmarc a quick question: if the part is at its minimum length 69.6, why the maximum form error of 0.8 cannot be applied from 69.6 in the opposite direction than the direction you’ve indicated. In other words, what paragraph of Y14.5 standard states that on the top of the part cannot be 69.6 at any cross section but on the bottom cannot be 69.6-0.8= 68.8.

Unless I am reading your question incorrectly, 68.8 is not acceptable because this would simply violate lower limit of size for the width (69.6).

 

[greenimi]

Pmarc,
Our expert’s understanding (the one which is driving the stackup debate) is that there are actually two tolerance zones when relating to planar features together with a direct tolerance dimension ± and both tolerance zones have a width of 0.8.
Each portion of the left and right faces/edges of the part must fall within two parallel planes that are located 69.6 and 70.4 from tangent plane constructed by the highest points of the opposite face of the part.

He thinks that as long as you can measure 69.6 (minimum dimension acceptable) at any cross section of the feature—take subsequent measurements of the top dimension along the width 50.4/49.6 of the part—then the measured value is in tolerance—

So, if in the top (again on each cross section) you are within 70.4 - 69.6 the bottom should then be in the form error tolernace (form error 0.8 then the minimum dimension on the bottom 68.8)



1.3.54 Size, Actual Local
size, actual local: the measured value of any individual
distance at any cross section of a feature of size. See
Fig. 1-1.

 

[greenimi]

By the way, can you recommend me some good threads on this website (or any other sites) where this issue with ± direct tolerance dimension and form error has been discussed?
Thank you pmarc

 

[pmarc]

greenimi,
I think it would be much easier (at least for me), if you prepared a sketch showing the width satisfing 69.6-70.4 actual local size reuirement and in the same having 68.8 somewhere.

 

[greenimi]

The sketch attached has 3 pages
I was trying to make a 3D sketch for better understanding
10mm part thickness does not matter for this form error issue and also, I think does not matter for the final result of the stackup
Please let me know your thoughts

What am I missing ? Where am I wrong or what's not quite right in the whole matter/topic?

Also, please do not forget about some good references (other threads, websites) for this discussion.
Thank you pmarc

 

[powerhound]

No. This is just wrong. You cannot violate the MMC or LMC boundary. If the dimension is 69.6, you can only have form error that moves towards MMC.

John Acosta, GDTP S-0731
Engineering Technician
Inventor 2013
Mastercam X6
Smartcam 11.1
SSG, U.S. Army
Taji, Iraq OIF II

 

[Belanger]

Greenimi, I'm with Powerhound.

Per the ASME rules, every size dimension must pass both the "actual mating envelope" and also the "actual local size." Your expert seems to be forgetting that second part! There is no way that any point across that block can be 68.8 and be acceptable.
(There are exceptions to the ASME rule such as non-rigid parts, but we're not dealing with that here.)

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

 

[pmarc]

greenimi,
I can just repeat after powerhound and J-P. The geometry shown on your pictures simply does not fall within 69.6-70.4 limits of size.

However, I have been thinking and thinking about the statement... :

Our expert’s understanding (the one which is driving the stackup debate) is that there are actually two tolerance zones when relating to planar features together with a direct tolerance dimension ± and both tolerance zones have a width of 0.8.
Each portion of the left and right faces/edges of the part must fall within two parallel planes that are located 69.6 and 70.4 from tangent plane constructed by the highest points of the opposite face of the part.

...and I think I understand what the expert is talking about. This however does not change my standpoint that form tolerance for 69.6-70.4 width has no impact on MIN (and MAX) result of the stackup.

Here is another graphic showing (at least I hope so) why 68.8 should not appear in the calculations:

http://files.engineering.com/getfil...-6de4-439b-897d-189e694493c1&file=stackup.pdf

So concluding all the considerations regarding MIN and MAX calculations shown in Alex's book, my statement is:
- MIN=4.5 - calculated correctly;
- MAX=6.9 - calculated correctly ONLY IF there is a perfect orientation between datum features A and B when the datum feature B is at its MMC=70.4. As I was trying to point out, the drawing does not tell this, thus the 6.9 is not the absolute maximum that can occur in as-produced geometry.

 

[greenimi]

Pmarc,
You are right and you convinced us.
I would like to ask you, if the "X min" dimension would be different if "X" is the distance between datum B and the left hole?
In other words if the feature we are measuring the minimum distance from is a datum feature, is the X min calculated distance different versus the original X min shown in the book?

 

[pmarc]

If I understood you correctly, you would face with the same problem like for Xmax in the original stackup - that is, lack of orientation tolerance on datum feature B wrt A would make the stackup impossible to complete in 100%. If, however, you had this orientation tolerance specified on the print, it would be similar to the scenario I created in the presentation already mentioned by SeasonLee.

http://files.engineering.com/getfil...ae-9e71-85710eb4d765&file=Tolerance_stack.pdf

Did I get your question right?

 

[greenimi]

Pmarc,

Yes, I forgot about that stackup calculation you provided awhile ago and was mentioned by SeasonLee. In that calculation you have an orientation control and the tolerance for the orientation is included in the stackup. Now, this perpendicularity control (in your example) is 0.1, so is smaller than the form error of 60.3-59.7 = 0.6. A warm-up question for you: the perpendicularity tolerance must be smaller than the form error or not? In other words can the perpendicularity tolerance be bigger than 0.6 (again in your stackup example)

And to expand a little bit:

I am questioning what’s happened if no orientation control is provided (as it is in AK book example). Why you said the stackup is impossible to complete 100% (“lack of orientation tolerance on datum feature B wrt A would make the stackup impossible to complete in 100%.”) Why we cannot use the maximum form error of 0.8 (again in AK example) to be included in the stackup ?

Also, can you verify a little bit that 7.986 (page 3 and Case#3 in your sketch) dimension. I did the same construction on my Cad system and I got for X max 7.7 (and not 7.986). But, I have used 0.8 max form error to get to my X max of 7.7. And by the way, if you said it’s impossible to calculate/complete 100%, then how you get to 7.986 –your result---what assumptions did you use—since the lack of orientation control?
The link to your stackup (for X max) is below, just for your reference.

Thank you pmarc

 

[pmarc]

In that calculation you have an orientation control and the tolerance for the orientation is included in the stackup. Now, this perpendicularity control (in your example) is 0.1, so is smaller than the form error of 60.3-59.7 = 0.6. A warm-up question for you: the perpendicularity tolerance must be smaller than the form error or not? In other words can the perpendicularity tolerance be bigger than 0.6 (again in your stackup example)

The perpendicularity tolerance value can be greater than size tolerance (form error).

Let's focus on my example for a moment and assume that perpendicularity tolerance wrt A|B on datum feature C is 1.0, and not 0.1. Imagine first that datum feature C is perfectly flat. In such case whole allowable form error = 0.6 is "moved" to the opposite (top) face and the maximum orientation error of datum feature C to A|B can be 1.0. Now, since nothing is perfect, picture that actual form error of datum feature C is for example 0.2. What does it mean? It just means that maximum allowable perpendicularity error of the datum feature C to A|B is reduced to 0.8 (1.0-0.2). And by analogy, for actual form error of datum feature C = 0.6 (maximum possible), the maximum perpendicularity error will be reduced to 0.4 = 1.0-0.6.

So, to conclude shortly, the drawing can specify the perpendiculariy tolerance greater than the size tolerance. It is just that actual allowable errors of form and perpendicularity for this type of geometry are dependent on each other.
----

I am questioning what’s happened if no orientation control is provided (as it is in AK book example). Why you said the stackup is impossible to complete 100% (“lack of orientation tolerance on datum feature B wrt A would make the stackup impossible to complete in 100%.”) Why we cannot use the maximum form error of 0.8 (again in AK example) to be included in the stackup?

I think what I said above, explains this. Doesn't it?
----

Also, can you verify a little bit that 7.986 (page 3 and Case#3 in your sketch) dimension. I did the same construction on my Cad system and I got for X max 7.7 (and not 7.986). But, I have used 0.8 max form error to get to my X max of 7.7.

Again, you are sticked to the belief that the maximum orientation error is limited by the maximum form error, which is not correct. Additionally, notice that in my graphic I played with angles (90, 85, 80 degress). Most likely, if I played with linear distances and assumed 0.8 in one of my sketches, I would get the 7.7. But that does not mean I could not assume 1.0 or 2.0 for instance in other scenarios.
----

And by the way, if you said it’s impossible to calculate/complete 100%, then how you get to 7.986 –your result---what assumptions did you use—since the lack of orientation control?

This is not the absolute maximum that can be obtained. This number is for 80 degrees angle between datum feature B and datum plane A. If I showed additional page with the geometry at 75 deg angle, the result would be greater than 7.986. Actually, in the absence of the orientation tolerance between A and B, my graphic could be showing 10 deg angle between A and B and it would not be violating the print in Alex's book in any way.

Did it help?

 

[pmarc]

greenimi,
After re-reading first part of my last reply about relationship between size tolerance and perpendicularity tolerance, I am more than sure that it caused a serious confusion. Therefore I am attaching another graphic in which I am trying to explain why the perpendicularity tolerance can be greater than size (and form) tolerance.

http://files.engineering.com/getfil...f7-8b68-80fa33a82129&file=another_graphic.pdf

 

[CheckerHater]

Shouldn't your perpendicularity be refinement of size tolerance?