# Thread: In LMC application the X distance is always 2.7

1. ## In LMC application the X distance is always 2.7 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.

thilag  Reply With Quote

2. 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

+ 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)
___________________

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.  Reply With Quote

3. Hello Kelly

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  Reply With Quote

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