Calculating Slot True Position using Midpoint

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  • Calculating Slot True Position using Midpoint

    I have three slots that I need to do true position and width on. To do this, I am taking a point on each side of the slot and using this to output a distance between the two for a width. Then I am constructing a midpoint between the points and using that to output true position. Only thing is, since I'm just using a single point I can't use the MMC bonus on the slot width for the position. It looks like even if I construct a line between the two original points, the true position still won't let me do MMC on that feature. The slot doesn't have a full radius at either end, so the possibility of using a circle is out. Any ideas?

  • #2
    You can construct a generic circle.
    Code:
    DIM DIST1= 2D DISTANCE FROM LINE LIN2 TO LINE LIN3 (CENTER TO CENTER),NO_RADIUS UNITS=IN,$
    GRAPH=OFF TEXT=OFF MULT=500.00 OUTPUT=BOTH
    AX NOMINAL +TOL -TOL MEAS DEV OUTTOL
    M 0.3750 0.0050 0.0050 0.3726 -0.0024 0.0000 --#------
    LIN4 =FEAT/LINE,CARTESIAN,UNBOUNDED,NO
    THEO/<3.8446,0,6.2>,<1,0,0>
    ACTL/<3.8446,-0.001,6.2>,<1,0.0002327,0>
    CONSTR/LINE,MID,LIN2,LIN3
    SLOT1 =GENERIC/CIRCLE,DEPENDENT,CARTESIAN,IN,$
    NOM/XYZ,<LIN4.X.THEO,LIN4.Y.THEO,6.2>,$
    MEAS/XYZ,<LIN4.X,LIN4.Y,LIN4.Z>,$
    NOM/IJK,<0,0,1>,$
    MEAS/IJK,<0,0,1>,$
    DIAMETER/0.375,DIST1.M.MEAS
    Last edited by Schrocknroll; 05-03-2022, 12:07 PM.
    PC-DMIS 2016.0 SP8

    Jeff

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    • #3
      You can also measure planes, construct a 3D width from them, then report position on the width. You can use MMC on a width construction. Are your slots perpendicular/parallel to the datum structure?

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      • dcoss
        dcoss commented
        Editing a comment
        They're perpendicular to the primary datum and parallel to the secondary datum.

    • #4
      Originally posted by Schrocknroll View Post
      You can construct a generic circle.
      Code:
      DIM DIST1= 2D DISTANCE FROM LINE LIN2 TO LINE LIN3 (CENTER TO CENTER),NO_RADIUS UNITS=IN,$
      GRAPH=OFF TEXT=OFF MULT=500.00 OUTPUT=BOTH
      AX NOMINAL +TOL -TOL MEAS DEV OUTTOL
      M 0.3750 0.0050 0.0050 0.3726 -0.0024 0.0000 --#------
      LIN4 =FEAT/LINE,CARTESIAN,UNBOUNDED,NO
      THEO/<3.8446,0,6.2>,<1,0,0>
      ACTL/<3.8446,-0.001,6.2>,<1,0.0002327,0>
      CONSTR/LINE,MID,LIN2,LIN3
      SLOT1 =GENERIC/CIRCLE,DEPENDENT,CARTESIAN,IN,$
      NOM/XYZ,<LIN4.X.THEO,LIN4.Y.THEO,6.2>,$
      MEAS/XYZ,<LIN4.X,LIN4.Y,LIN4.Z>,$
      NOM/IJK,<0,0,1>,$
      MEAS/IJK,<0,0,1>,$
      DIAMETER/0.375,DIST1.M.MEAS
      Just one question about this comment. I follow most of it except for the last line. What are the two values for diameter?

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      • #5
        Originally posted by dcoss View Post

        Just one question about this comment. I follow most of it except for the last line. What are the two values for diameter?
        The first is the nominal and the second creates the measured from the distance.
        PC-DMIS 2016.0 SP8

        Jeff

        Comment


        • #6
          The first value is equal to the nominal slot width or length, whichever one will use the circle for it's position dimension. The second value is the measured width or length of the slot. "DIST1" is the ID of the dimension measuring the slot. "M" is the axis that is being measured. "MEAS" is the measured value.

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          • #7
            this may be a legacy thing, but we have instances where we have to figure the ideal location, move the trihedron there, and then compare actual feature location to it.

            so let's say the TP is to -A- -B-
            1) create alignment using datums -A- & -B-
            2) if there are distances provided create offsets using those otherwise we have to use trig to figure out offsets
            3) now trihedron is located at ideal center of slot
            4) take hits and create midpoints for slot width and length
            5) dimension POSITION of each midpoint
            6) possibly compare location of midpoints of length and width (ideally they'd be at same location)


            Untitled.png

            in this example for slot 2 the trihedron would be offset 0.5 in X+ and 0.8660 in Y+
            Last edited by Jimbaggins1972; 05-03-2022, 07:40 PM.

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