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?