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Tolerance interpretation

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

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Tolerance interpretation

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

  2. Agenda • Introduction to tolerance interpretation • Tolerance stacks • Interpretation

  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. Case #1 1 4 2 3 1.0±.05 1.5±.05 1.0±.05 ? 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

  5. Case #2 1 4 2 3 1.0±.05 1.5±.05 3.5±.05 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

  6. Case #3 1 4 2 3 ? 1.0’±.05 1.00’±0.05 3.50±0.05 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

  7. Case #1 1 4 2 3 1.0±.05 1.0±.05 1.0±.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?

  8. Let’s try the following (in the same setup) -cut plane 2 -cut plane 3 3 2 Mfg. Process 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. 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 Expectation • What do we expect when we manufacture something?

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

  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. .95 1.0 1.05 2.45 2.5 2.55 D1-2 We know that (as specified) D2-3 = 1.5  .05 If one uses a single set up, then (as produced) and D1-2 D1-3 D2-3 = D1-3 - D1-2

  15. 1.4 1.5 1.6 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

  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 = .99735 = .9865

  19. 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

  20. Should this strategy change?

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