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? D 1-4 = D D D 3-4 = t 1-4 = ± ( ) = ± ±.05 ? 1.5± Case #1

TOLERANCE STACKING What is the expected dimension and tolerances? D 3-4 = D (D D 2-3 ) = 1.0 t 3-4 =  (t t t 2-3 ) t 3-4 = ± ( ) = ± ±.051.5± ±.05 Case #2

TOLERANCE STACKING What is the expected dimension and tolerances? D 2-3 = D (D D 3-4 ) = 1.5 t 2-3 = t t t 3-4 t 2-3 = ± ( ) = ± ’±.05 ? ±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 ? 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? PROCESSDIMENSIONAL ACCURACY POSITIONAL ACCURACY DRILLING REAMING (AS PREVIOUS) SEMI-FINISH BORING FINISH BORING COUNTER-BORING (SPOT-FACING) END MILLING

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

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  = D 1-2 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. An Example

We know that (as specified) D 2-3 = 1.5 .05 If one uses a single set up, then (as produced) D 1-2 and D D D 2-3 = D D 1-2

What is the probability that D 2-3 is bad? P{X X 1-2 >1.55} + P{X X 1-2 <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 D

The likelihood of a bad part is P {X 2-3 > 1.55}-1 P {X 2-3 < 1.45} (1-.933) + (1-.933) =.137 As a homework, calculate the likelihood that D 1-4 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 = =.9865

Success versus number of features 1 feature = features = features = features = features =

Should this strategy change?