1. Tolerance interpretation
Dr. Richard A. Wysk
IE550
Fall 2008

2. Agenda Introduction to tolerance interpretation Tolerance stacks

3. Tolerance interpretation
Frequently a drawing has more than one datum
How do you interpret features in secondary or tertiary drawing planes?
How do you produce these?
Can a single set-up be used?

4. TOLERANCE STACKING Case #1
1.0±.05 ?
1.5±.05 1 2 3 4
Case #1
What is the expected dimension and tolerances?
D1-4= D1-2 + D2-3 + D3-4 =
t1-4 = ± ( ) = ± 0.15

5. TOLERANCE STACKING Case #2
1.0±.05
1.5±.05 1 2 3 4
3.5±.05
Case #2
What is the expected dimension and tolerances?
D3-4= D (D1-2 + D2-3 ) = 1.0
t3-4 =  (t1-4 + t1-2 + t2-3 )
t3-4 = ± ( ) = ± 0.15

6. TOLERANCE STACKING Case #3
1.0’±.05 ? 1 2 3 4
3.50±0.05
Case #3
1.00’±0.05
What is the expected dimension and tolerances?
D2-3= D (D1-2 + D3-4 ) = 1.5
t2-3 = t1-4 + t1-2 + t3-4
t2-3 = ± ( ) = ± 0.15

7. From a Manufacturing Point-of-View How will the part be produced?
1.0±.05 ? 1 2 3 4
Case #1
Let’s suppose we have a wooden part and we need to saw.
Let’s further assume that we can achieve  .05 accuracy per cut.
How will the part be produced?

8. Mfg. Process Will they be of appropriate quality?3 2
Let’s try the following (in the same setup)
-cut plane 2
-cut plane 3
Will they be of appropriate quality?

9. So far we’ve used Min/Max Planning
We have taken the worse or best case
Planning for the worse case can produce some bad results – cost

10. Expectation What do we expect when we manufacture something? PROCESS
DIMENSIONAL ACCURACY
POSITIONAL ACCURACY
DRILLING
0.010
REAMING
(AS PREVIOUS)
SEMI-FINISH BORING
0.005
FINISH BORING
0.0005
COUNTER-BORING (SPOT-FACING)
END MILLING
0.007

11. Size, location and orientation are random variables
For symmetric distributions, the most likely size, location, etc. is the mean
2.45
2.5
2.55

12. What does the Process tolerance chart represent?
Normally capabilities represent + 3 s
Is this a good planning metric?

13. An Example
Let’s suggest that the cutting process produces  (, 2) dimension where (this simplifies things)
=mean value, set by a location
2=process variance
Let’s further assume that we set = D1-2 and that =.05/3 or 3=.05
For plane 2, we would surmise the 3of our parts would be good 99.73% of our dimensions are good.

14. We know that (as specified)
If one uses a single set up, then
(as produced)
.95
1.0
1.05
D1-2
2.45
2.5
2.55
and
D1-2
D1-3
D2-3 = D D1-2

15. Sums of i.i.d. N(,) are normal
What is the probability that D2-3 is bad?
P{X1-3- X1-2>1.55} + P{X1-3- X1-2<1.45}
Sums of i.i.d. N(,) are normal
N(2.5, (.05/3)2) +[(-)N(1.0, (.05/3)2)]= N (1.5, (.10/3)2)
So D2-3
1.4
1.5
1.6

16. The likelihood of a bad part is
P {X2-3 > 1.55}-1 P {X2-3 < 1.45}
(1-.933) + (1-.933) = .137
As a homework, calculate the likelihood that
D1-4 will be “out of tolerance” given the same logic.

17. What about multiple features?
Mechanical components seldom have 1 feature -- ~ 10 – 100
Electronic components may have 10,000,000 devices

18. Suppose we have a part with 5 holes
Let’s assume that we plan for + 3 s for each hole
If we assume that each hole is i.i.d., the
P{bad part} = [1.0 – P{bad feature}]5 =
= .9865

19. Success versus number of features

20. Should this strategy change?