Tolerance stack question 1

[pmarc]

I believe this is quite interesting tolerance stack.
Question is - what is maximum and minimum distance X as shown in attached picture?

http://files.engineering.com/getfile.aspx?folder=63724088-1f5f-4368-98e5-009be626f065&file=stack.PNG

Any feedback will be appreciated.

 

[ctopher]

Curious...
First, what do you think it is?

Chris, CSWA
SolidWorks 14
SolidWorks Legion

 

[pmarc]

It does not actually matter what this is. Let's say it is a very simplified geometry of a real thing that does not even look like this part. What is important is the dimensioning and tolerancing scheme applied to this part. Let's also say the scheme properly reflects function of this part.

 

[pmarc]

I have just noticed small, but significant, typo in the original picture. Size limits for the pattern of two holes on the right are 1.216-1.224, not 1.116-1.224. My apologies for that.

I am attaching corrected picture.

http://files.engineering.com/getfil...42eb-b06b-720c324467d9&file=stack_correct.PNG

 

[pmarc]

My numbers are:
MIN = 7.154
MAX = 7.218

 

[greenimi]

Late night: 7.168/7.202
I will redo it tomorrow morning.

 

[3DDave]

It's a trick question.

Because the length of B is not defined, the amount of rotation available relative to A|B(M) is not defined. All that is possible to know is that .010inch/unknown depth of notch is allowed when orienting the part to A|B(M) to locate the .804 max and 2.40 max size holes. This is important because C is held to A|B(s) and datum feature C can rotate within limits established by A|B(M)

Because the limits of C are not defined, the amount datum feature C is allowed to be rotated from C in the A|B(M) frame of reference is not defined.

Only A|D(M)|C forces datum feature C to be parallel to datum simulator C, but this is not the same orientation limitation as A|B(M)|C where B(M) is the control.

I guess this is easy to calculate if the datum scheme is largely ignored, but since the example is designed to depend on separate orientation controls it seems like they should be included. There just isn't enough information to do so.

The sensible reference would be to A|C|B(m) but that would take most of the fun out of the analysis.

 

[pmarc]

Dave,
Center hole and both left holes are located relative to the same datum features, referenced in the same order of precedence, and at the same material boundary conditions. Do you still think the limits that you mentioned are needed to calculate this stack?

 

[3DDave]

The question is about the distance between the right and left holes. They have different reference frames.

 

[pmarc]

Dave,
You are correct, there is not enough information on the drawing to be able to calculate the stack. A star for you for bringing this up.

I designed another version of the drawing where I think the missing size limits are not needed to find the MIN and MAX values of X. But like you said, in this case most of the fun is gone.

http://files.engineering.com/getfil...-4348-424a-a815-8ea9057baa6e&file=stack_2.PNG

 

[greenimi]

Pmarc,

Now , I am confused.

Q1: How did YOU calculate the stack –you numbers 7.154/7.218 on the original datum scheme--(if the info was not enough)?

Q2: Are those numbers still applicable/valid to the new datum scheme (A primary, C secondary and B(MMB) or D(MMB) tertiary)?

Thank you

 

[greenimi]

And one more question: why you went back to 1.224/1.116 size hole the holes.

You said:
"I have just noticed small, but significant, typo in the original picture. Size limits for the pattern of two holes on the right are 1.216-1.224, not 1.116-1.224. My apologies for that."

You lastest datum scheme has 1.224/1.116 hole size.

 

[pmarc]

greenimi,

Q1: I calculated my numbers -- 7.154-7.218 -- because I thought the drawing contained all necessary info to calculate them. But it is not the case, thus my numbers are wrong.
Q2: No, these numbers are not applicable to the new datum scheme. I did not post new numbers, because I did not want to spoil the fun (assuming that anyone is still interested to calculate this stack, after all the confusion I made).
Q3: This is typo again. The correct hole size is 1.216-1.224.

 

[greenimi]

X min.: 7.168
X Max.: 7.204

How far off am I?

.804/.796:
VC: .784,
RC: .824;

1.224/1.216:
VC: 1.208,
RC: 1.240;

2.400/2.396:
VC: 2.386
RC: 2.414

X min.: 3.9 + 4.3 -(.824/2) - (1.240/2) = 7.168
X Max.: 3.9 + 4.3 - (.784/2) - (1.208/2) = 7.204

That's all I am capable of doing. sorry I do have a limited GD and T knowledge. More to learn.

Do I need to consider .005 perpendicularity to A? Does it affect the stack?
Also .014 datum shift from D(MMB)? How this plays a role?

Thank you

 

[pmarc]

greenimi,
Your numbers are correct.

Perpendicularity tolerance of C wrt A does not influence the stack. It would influence Xmin, if X was a distance between one of the holes and the surface C.

Also, no datum feature shift (neither B nor D) is included in the stack, mainly because both shifts do not act in the direction of the stack. They would have to be taken into account, if the goal of the stack was to find a vertical distance between holes of different patterns.

 

[greenimi]

Thank you pmarc,
So, those are the valid numbers/calculations for the "no fun" scenario/case.
Would you mind, if I am asking you to add the missing size limits and lets do the math for the "fun" scenario/case (if all possible...I mean without software VisVSA, CeTol, etc as I do not have those available).
I might need more explanations on how to do it correctly from you or 3DDave.
Thank you

 

[pmarc]

greenimi,
I am afraid that if I add those missing size limits to my initial drawing, it won't be easy to figure out Xmin and Xmax without a software.

 

[greenimi]

pmarc,

Fair enough.

Safe to say that this kind of calculations won't show up on the certification exam (at least not the ones from your initial drawing + missing size limits)?

 

[pmarc]

You can sleep easy, greenimi ;-)
This kind of calculations won't show up on the certification exam.

 

[bxbzq]

Another approach. I used two sets of cartoon gages to do the stack.
On first cartoon gage which establish DRF A|B(M)|C, V is the distance between small holes to datam C. So
Vmin = 10.038
-.804
+.784(VC)/2
+10.45
------------
= 10.038

Vmax = 10.058
-.784/2
+10.45
----------
= 10.058

On the second cartoon gage which establish DRF A|D(M)|C, W is the distance between big holes to datum C. So
Wmin = 2.854
+1.208(VC)/2
+2.25
-----------
= 2.854

Wmax = 2.87
+1.224
-1.208/2
+2.25
-----------
= 2.87

Datum shift from Datum D does not play a role in horizontal direction as the right edge of the part needs to rest against datum C simulator.

Xmin = Vmin - Wmax = 7.168
Xmax = Vmax - Wmin = 7.204