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Thread: In LMC application the X distance is always 2.7

  1. #1
    Associate Engineer
    Join Date
    Nov 2015
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    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.

    Anybody please guideme.

    thanks in advance

    thilag
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  2. #2
    Administrator Kelly Bramble's Avatar
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    Bold Springs, GA
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    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.
    Last edited by Kelly Bramble; 11-20-2015 at 08:06 AM.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  3. #3
    Associate Engineer
    Join Date
    Nov 2015
    Posts
    6
    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 4.6 OD 50.4
    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 4.4 OD 50.6
    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 4.2 OD 50.8
    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

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