Tolerance stack help needed. 2

[greenimi]

I need some help here to confirm if I am on the right track or not. I would like to find out if the X min/ X max dimensions (see sketch attached) are correct calculated.
I realize none of the fine gentlemen on this form do not work for me, therefore is more like a favor to ask for someone who has strong knowledge in those kind of calculations (or even maybe some kind of reliable tolerance stackup software Vis-VSA, 3DCS , CeTol or other).
I am trying to understand the concept (how the calculations should be done CORRECTLY) otherwise I will be only a “number generator” and my calculations are worthless sheet of paper.
Note: do not consider the form error on the datum feature A. I know does affect the calculation, but I am not there yet. (I have to walk before I can run)
I got
X min.: .057
and
X max.: .076
(should be no surprise if I am telling you that not everybody around here agrees with those numbers).
Thank you

 

[pmarc]

I got the same numbers.

And I don't think that form error of datum feature will make any difference in results.

 

[CheckerHater]

I am thinking of .057 and .068

To greenimi: did anyone "around there" suggested those numbers?

 

[greenimi]

CH,
As a matter of fact: YES. (two of .068’s):
X max. [(.505+.002) – (.372-.001)]/2 = .068

Other numbers for X max: .071 and .064. Go figure………
(X min. is more in agreement: .057)

I used virtual boundaries and resultant condition boundaries:
X max. .517-.365 = .152; .152/2 = .076
(.517 = .505 +.012 pos at LMC)
(.365 = .372-.007 pos at LMC)

I feel somewhat safe (99.999999%, confidence level) now that pmarc got my numbers !!!!

 

[pmarc]

I am honored, greenimi.

Question to CH:
Did you take datum feature shift into account?

 

[CheckerHater]

Counter-question:
Did you take simultaneous requirement into account?

 

[pmarc]

No, I did not.

 

[pmarc]

Ah, sorry. I thought you asked: "Did you take datum feature shift into account"?
Yes, I did simultaneous requirement into account.

 

[CheckerHater]

If you imagine "umbrella-shaped" gage designed to check two positions at once, it's datum-simulator part will never touch the datum feature.

There was a tip on Tec-Ease named "Datum shift is not bonus". Unfortunately free tips are not free anymore. :-(

 

[pmarc]

It is not that the only case when datum feature shift is not taken into account is when datum feature contacts its simulator. Datum features shift is also not taken into account when two or more features, between which the stack is calculated, are gaged simultaneously. This is exactly what we have got here, and this is the key to this stack.

Besides, according to the drawing datum feature simulator pin is [Ø].248, right? So, simply picture the datum feature A having its actual mating envelope = exactly [Ø].248. And here you have the contact between the datum feature and its simulator.

 

[Tarator]

Here you go...

Note: No datum shift because of the default SIM. REQ.!

 

[Belanger]

I agree on the minimum, but I am getting a different number for X maximum: .0735.

I get this from
-0.186 (radius of inner diameter)
+0.0005 (half of inner diameter's stated position tol)
+0.003 (half of inner diameter's max bonus tol)
+0.001 (half of outer diameter's stated position tol)
+0.0025 (half of outer diameter's max bonus tol)
+0.2525 (radius of outer diameter)

I know this may be a different approach than what some of you used. And no, I didn't use any datum shift. But since some of you get the max being .076, and some get .068, I thought I'd throw my vote in; let me know if I've goofed somewhere


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

 

[Tarator]

Belanger,

When you use dia .378 (which is MMC), you can't have bonus! Same applies for dia .495 (which is MMC). Btw, you should use dia .495 for min. calculation.

 

[3DDave]

Both the pin and hole share the same axis, as mentioned, due to the simultaneous requirement.

The pin (.372 to .378) has an MMC virtual diameter of .379; Since the smallest pin diameter is confined to the virtual diameter, if the pin is the smallest size (.372) one side can be offset by .379/2 = .1895; the amount remaining on the other side is .1825.

The MMC virtual condition of the hole is (.495-.002) which has a radius of .2465. This is the closest the hole can approach the axis. The largest hole is .505. The farthest radius will be .505 -.2465 = .2585

The difference is .076, which is the largest gap.

The smallest gap is 0.057 is the difference in MMC virtual condition radii.

 

[Belanger]

Tarator -- I think you missed that my calculation was for the maximum of X, and that occurs when the smaller diameter is at .372 and when the larger diameter is at .505. Thus, both of these numbers are LMC, and that means both get the full bonus tolerance. (For the min of X, yes...we'd use the MMC values, and no bonus.)

Dave -- I think I follow that, but let me sleep on it tonight. I'm just trying to reconcile my "Krulikowski" method with your VC method.

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

 

[pmarc]

J-P,
How about this?

-0.186 (radius of inner diameter)
+0.0005 (half of inner diameter's stated position tol)
+0.003 (half of inner diameter's max bonus tol)
+0.001 (half of outer diameter's stated position tol)
+0.005 (half of outer diameter's max bonus tol)
+0.2525 (radius of outer diameter)

You simply used wrong number for the half of outer diameter's max bonus tol (5th line in the stack).

 

[Belanger]

Aha! Thanks, pmarc. The total size difference was .010, so half of that is .005. For some reason I saw .005 and cut that in half.

That's what happens when I do simple math late at night



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

 

[waqasmalik]

Hi all,i would like it to proceed hurtful if greenimi dont mind.
If i take form error into account both for smaller diameter and larger diameter
( obviously for maximum value of x) then what my results will be?

 

[waqasmalik]

Hi all,i would like it to proceed hurtful if greenimi dont mind.
If i take form error into account both for smaller diameter and larger diameter
( obviously for maximum value of x) then what my results will be?

 

[greenimi]

waqasmalik,

As pmarc stated (and personally I believe him -- he has a loooooooooooog record of being right)
“And I don't think that form error of datum feature will make any difference in results. “24 Mar 14 13:19

Pmarc,
Please don’t let me down.....