Bi directional positional tolerancing

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  • Bi directional positional tolerancing

    I have a elongated hole in one of our products. There are two FCF's assigned to this hole. They control position-- and they are perpendicular to each other. In two dimensional terms, the tolerance zone can be the same as rectangular.
    No diameter symbol is given in the discription of the FCF, but maximum material conditions are given in both FCF's for this elongated hole.

    My question is, does one FCF control the dimension in "X" and the other control the dimension in "Y"? If this is correct, do you take the square root of "X squared" and multiply it by 2? Do I get bonus in both directions or in one direction of the elongated hole for calculating "X"?

    Every good flight ends with a great landing!
    PCDMIS CAD ++ 2014, PH10MQ

  • #2
    Yes, each is for 1 axis, one for the width (get the bonus from the width) and one for the length (bonus from the length). Use the true pos function and select the appropriate axis.

    sigpicHave a homebrew


    • #3
      Position the DF in 'X' and the LF in '"Y" or vise versa. Do the following to get bonus on the LF

      Position to the Length of a Slot
      1. Add slot to program SLOT1 =AUTO/ROUND SLOT,SHOWALLPARAMS = YES,SHOWHITS = NO THEO/0,-35,0.001,-1,0,0,11,16,0,-1,0.00018 ACTL/0,-35,0.001,-1,0,0,11,16,0,-1,0.00018 TARG/0,-35,0.001,-1,0,0,0,-1,0.00018 THEO_THICKNESS = 0,RECT,IN,ONERROR = YES,$ AUTO MOVE = BOTH,DISTANCE = 10,RMEAS = None,None,None,$ READ POS = YES,FIND HOLE = NO,REMEASURE = NO,$ INIT = 3,PERM = 3,SPACER = 3,DEPTH = 1,HITS = 18,MEAS ANGLE = 160 2. After slot, hit enter and type “GEN” and hit Tab once. Change name to SLOT1L. Hit Tab once and F7 until feature changes to CIRCLE. Hit Tab three times and change OUT to IN. F9 the Generice feature and change from RADIUS to DIAMETER. Hit OK. Change underlined areas to what is below. T = Theoretical, LF = Length of Feature SLOT1L =GENERIC/CIRCLE,DEPENDENT,RECT,IN,$ NOM/XYZ,SLOT1.TX,SLOT1.TY,SLOT1.TZ,$ MEAS/XYZ,SLOT1.X,SLOT1.Y,SLOT1.Z,$ NOM/IJK,SLOT1.TI,SLOT1.TJ,SLOT1.TK,$ MEAS/IJK,SLOT1.I,SLOT1.J,SLOT1.K,$ DIAMETER/SLOT1.TLF,SLOT1.LF 3. Position the slot as below: a) Position to the width of a slot DIM LOC1= TRUE POSITION OF SLOT SLOT1 UNITS=MM ,$ GRAPH=OFF TEXT=OFF MULT=10.00 OUTPUT=BOTH DEV PERPEN CENTERLINE=OFF DISPLAY=DIAMETER AX MEAS NOMINAL +TOL -TOL BONUS DEV OUTTOL Z -9.057 -9.000 0.057 DF 11.000 11.000 0.350 0.200 0.200 0.000 0.000 -#- TP MMC 2.000 0.200 0.000 0.104 #-- END OF DIMENSION LOC1 b) Position to the length of a slot DIM LOC2= TRUE POSITION OF CIRCLE SLOT1L UNITS=MM ,$ GRAPH=OFF TEXT=OFF MULT=10.00 OUTPUT=BOTH DEV PERPEN CENTERLINE=OFF DISPLAY=DIAMETER AX MEAS NOMINAL +TOL -TOL BONUS DEV OUTTOL Y -33.807 -33.800 0.007 DF 16.000 16.000 0.350 0.200 0.200 0.000 0.000 -#- TP MMC 4.000 0.200 0.000 0.014 #-- END OF DIMENSION LOC2
      sigpic GDTPS - 0584


      • #4
        Insp 212 assumes you can do an Auto Round Slot. If you don't have CAD++ you cannot. You will need to create two circles (one of the diameter equal to length and the other of diameter equal to width) and use them respectively to get bonus tolerance IF you do not have CAD++
        Bill Jarrells
        A lie can travel half way around the world while the truth is putting on its shoes. - Mark Twain


        • #5
          I have same case here; it looks both of FCFs takes care of their own dimension; a bit of help here :


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