In LMC application the X distance is always 2.7

[thilag]

Hello Friends

Kindly refer the attached image. In the attached image figure the X distance is 2.7, but i do know how to calculate it for different hole sizes (4.2 to 4.6) and for Datum B 50.4 - 50.8.

Anybody please guideme.

thanks in advance

thilag

[Kelly_Bramble]

Homework? This is a poor example on how LMC might be used in the real world of engineering design...

#1 ignore the perpendicular MMC on the OD.

OD (datum B) LMC size= 50.4

Hole LMC size = dia. 4.6

Hole position tolerance allowed at LMC = dia. .4
Datum Bonus Tolerance allowed at LMC = dia. 0

Outer tangent edge of hole on radius.

For Hole size at LMC

40/2 (Basic or TED radius)
+ 4.6/2 (hole Outer Tangent Edge)
+ 0.4/2 (hole position tolerance start radius)
+ 0/2 (hole position bonus tolerance at radius)
+ 0/2 (Datum position bonus tolerance radius)
___________________
radius = 22.5


50.4/2 = 25.2 Radius of datum dia.


25.2-22.5 = 2.7 edge distance



Do the calculation on radius for each size don't forget bonus tolerance for hole and datum feature as they diverge.

Your book should show method.

[thilag]

Hello Kelly

Thanks for your reply.

With your help i did / find the minimum distance = 2.7 for #2 and #3 and plotting the procedure calculated under for future reference.

Ignore the Outer diameter 50.8 - 50.4 Perpendicularity at MMC at any condition (i.e., OD = 50.4/2 always).
Now for the # 1 Hole Ø4.6 OD Ø50.4
Hole dia = 4.6+0.4+(4.6-4.6) = 4.6+0.4 = 5
Minimum distance = (50.4/2)-(40/2)-(5/2) = 25.2-20-2.5 = 25.2-22.5 = 2.7


Now for the # 2 Hole Ø4.4 OD Ø50.6
Hole dia = 4.4 + 0.4 + (4.6-4.4) + (50.8-50.6) = 4.4 + 0.4 + 0.2 + 0.2 = 5.2
Minimum distance = (50.6/2)-(40/2)-(5.2/2) = 25.3-20-2.6 = 25.3-22.6 = 2.7


Now for the # 3 Hole Ø4.2 OD Ø50.8
Hole dia = 4.2+0.4+(4.6-4.2)+(50.8-50.4) = 4.2+0.4+0.4+0.4 = 5.4
Minimum distance = (50.8/2)-(40/2)-(5.4/2) = 25.4-20-2.7 = 25.4-22.7 = 2.7

thanks

thilag