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Tolerance interpretation
Dr. Richard A. Wysk
ISE316
Fall 2010
Agenda
• Introduction to tolerance interpretation
• Tolerance stacks
• Interpretation
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?
TOLERANCE STACKING
What is the expected dimension and tolerances?
D1-4= D1-2 + D2-3 + D3-4
=1.0 + 1.5 + 1.0
t1-4 = ± (.05+.05+.05) = ± 0.15
1.0±.05 1.0±.05
?
1.5±.05
1
2 3
4
Case #1
TOLERANCE STACKING
What is the expected dimension and tolerances?
D3-4= D1-4 - (D1-2 + D2-3 ) = 1.0 t3-4 = (t1-4 + t1-2 + t2-3 )
t3-4 = ± (.05+.05+.05) = ± 0.15
1.0±.05 1.5±.05
1
2 3
4
3.5±.05
Case #2
TOLERANCE STACKING
What is the expected dimension and tolerances?
D2-3= D1-4 - (D1-2 + D3-4 ) = 1.5 t2-3 = t1-4 + t1-2 + t3-4
t2-3 = ± (.05+.05+.05) = ± 0.15
1.0’±.05 ?
1
2 3
4
3.50±0.05
Case #3
1.00’±0.05
From a Manufacturing Point-of-View
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?
1.0±.05 1.0±.05
?
1.0±.05
1
2 3
4
Case #1
Mfg. Process
Let’s try the following (in the same setup)
-cut plane 2
-cut plane 3
Will they be of appropriate quality?
3
2
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
Expectation
• What do we expect when we manufacture something?
PROCESS DIMENSIONAL ACCURACY
POSITIONAL ACCURACY
DRILLING + 0.008 - 0.001
0.010
REAMING + 0.003 (AS PREVIOUS)
SEMI-FINISH BORING
+ 0.005 0.005
FINISH BORING + 0.001 0.0005
COUNTER-BORING (SPOT-FACING)
+ 0.005 0.005
END MILLING + 0.005 0.007
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
What does the Process tolerance chart represent?
• Normally capabilities represent + 3 s
• Is this a good planning metric?
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 3of our parts would be good 99.73% of our dimensions are good.
An Example
We know that (as specified)
D2-3 = 1.5 .05
If one uses a single set up, then
(as produced)
D1-2
and
D1-3
.95 1.0 1.05D1-2
2.45 2.5 2.55
D2-3 = D1-3 - D1-2
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
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.
What about multiple features?
• Mechanical components seldom have 1 feature -- ~ 10 – 100
• Electronic components may have 10,000,000 devices
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
= .99735
= .9865
Success versus number of features
1 feature = 0.9973
5 features = 0.986
50 features = 0.8735
100 features = 0.7631
1000 features = 0.0669
Should this strategy change?