Question

0 .995-1.000 0 1.010-1.025 .006M) 1.000 1 500ーイーー 3 0001 6.00

.XX   = ± .01

.XXX = ± .005

ANGLES = ± 1°

1.     Calculate the Virtual Condition of the PIN.               Calculate the Virtual Condition of the HOLE.

V.C._P=                               V.C._P/2=                                       V.C._H=                           V.C._H/2=                            

2.     Calculate the Resultant Condition of the PIN.           Calculate the Resultant Condition of the HOLE.

R.C._P=                               R.C._P/2=                                       R.C._H=                            R.C._H/2=                            

Calculate the maximum and minimum distances indicated on the drawing above:

3.     Maximum X =

                                                                                                                                                                                 

4.     Minimum X =

                                                                                                                                                                                 

5.     Maximum Y =

                                                                                                                                                                                 

6.     Minimum Y =

                                                                                                                                                                                 

Answer 1

1. VCP = 1.000+0.006 = 1.006 (MMC + GT) VCP/2 = 0.503

VCH = 1.010 - 0.006 = 1.004 (MMC - GT) VCH = 0.502

2. RCP = 0.995-0.006-0.005 = 0.984 (LMC - GT - BT) RCP/2=0.492

RCH = 1.025+0.006+0.015 = 1.046 (LMC + GT + BT) RCH/2=0.523

{MMC = MAXIMUM MATERIAL CONDITION, LMC = LEAST MATERIAL CONDITION, GT = GEOMETRIC TOLERANCE, BT = BONUS TOLERANCE}

3.MAXIMUM X = 6 - 4.999 = 1.001

(Maximum deviation of the hole to the left and least dimension of the hole)

4.MINIMUM X = 6- 5.0185 = 0.9815

(Maximum deviation of the hole to the right and biggest dimension of the hole)

5.MAXIMUM Y = 3.012 - 0.4975 - 0.505 = 2.0095

(Maximum deviation of the hole and pin away from the middle of the base and least dimension of the hole and pin)

6.MINIMUM Y = 2.988 - 0.5 - 0.5125 = 1.9755

(Maximum deviation of the hole and pin to the middle of the base and biggest dimension of the hole and pin)