Surface profile vs highest point

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Surface profile vs highest point

    PCDMIS 2018 R2, after iterative alignment to CAD I did scan one of the surfaces. When dimensioning surface profile of the scan and using Xact (not selecting any datums) I am getting max and min of 0.090 - that's the same as legacy form only. Using legacy form and location my max value is 0.595 and min value -0.185, deviation 0.780 . But when I extract the highest point from the scan it comes as 0.666 deviation from nominal. Questions I have are:
    - should there be any correlation between scan's surface profile max or min number and deviation of highest point?
    - don't fully understand why form only allows for only positive tolerance?


  • #2
    The highest deviation is given in the current alignment, which can be different than the "datums" alignment.
    For the form only, it's like a flatness or a circularity, if the form is perfect, then you have zero.

    Comment


    • #3
      Originally posted by JEFMAN View Post
      The highest deviation is given in the current alignment, which can be different than the "datums" alignment.
      For the form only, it's like a flatness or a circularity, if the form is perfect, then you have zero.
      I only have one iterative alignment.

      Comment


      • #4
        Form only releases all degrees of freedom. This creates a best fit to the object measured. As Jefman mentioned.... you are looking at the same feature in two different ways. One is to you alignment one is not and that is causing two different answers.

        Comment


        • #5
          Originally posted by Lord_Warfield View Post
          Form only releases all degrees of freedom. This creates a best fit to the object measured. As Jefman mentioned.... you are looking at the same feature in two different ways. One is to you alignment one is not and that is causing two different answers.
          I am not comparing form only to form and location. I understand the difference. I am comparing max of 0.595 from form and location to 0.666 deviation from nominal of the highest point. Shouldn't they be the same? Or they are not related?

          Comment


          • #6
            Try creating a best fit to the feature releasing all degrees of freedom the n see what it says...

            Comment


            • Lord_Warfield
              Lord_Warfield commented
              Editing a comment
              Hmmm let me read that again...

          • #7
            I simulated and do not see the same issue as you.

            Comment


            • #8
              Here is what I am comparing:
              Code:
              DIM PROF1= PROFILE OF SURFACE OF SET SCN1    FORMANDLOCATION  UNITS=MM ,$
              GRAPH=ON  TEXT=OFF  MULT=100.00  ARROWDENSITY=100  OUTPUT=BOTH
              AX    NOMINAL       +TOL       -TOL       MEAS        DEV     OUTTOL        MAX        MIN
              M        0.000      0.500      0.500      0.780      0.780      0.095      0.595     -0.185 --#-|--->
              
              ASSIGN/HIGHPOINT=MAXINDEX(SCN1.HIT[1..SCN1.NUMHITS].Y)
              DIM HIGHPOINT= LOCATION OF POINT SCN1.HIT[HIGHPOINT]  UNITS=MM ,$
              GRAPH=OFF  TEXT=OFF  MULT=100.00  OUTPUT=BOTH  HALF ANGLE=NO
              AX    NOMINAL       +TOL       -TOL       MEAS        DEV     OUTTOL        MAX        MIN
              Y      744.568      0.500      0.500    743.902     -0.666      0.166    743.902    743.902 <--------
              END OF DIMENSION HIGHPOINT
              Or is the difference because I am checking high point in Y and only half of the surface is in Y, the other half goes on approx. 45 degrees?
              Last edited by marcc; 04-03-2019, 12:14 PM.

              Comment


              • #9
                Change it to T value...

                Comment


                • #10
                  Originally posted by Lord_Warfield View Post
                  Change it to T value...
                  Not sure how to do assignment so it grabs highest T value. When I just change Y to T in my current assignment it does not grab any of the points (when I point my mouse it displays zero where pointing mouse in original assignment displays point # in the scan which has the highest Y deviation). So I only changed to T in dimension part but that's not necessarily point with the highest T value:
                  Code:
                  ASSIGN/HIGHPOINT=MAXINDEX(SCN1.HIT[1..SCN1.NUMHITS].Y)
                  DIM HIGHPOINT= LOCATION OF POINT SCN1.HIT[HIGHPOINT]  UNITS=MM ,$
                  GRAPH=OFF  TEXT=OFF  MULT=100.00  OUTPUT=BOTH  HALF ANGLE=NO
                  AX    NOMINAL       +TOL       -TOL       MEAS        DEV     OUTTOL        MAX        MIN
                  T        0.000      0.500      0.500      0.392      0.392      0.000      0.392      0.392 --------#
                  END OF DIMENSION HIGHPOINT

                  Comment


                  • #11
                    Calculate the max T_Value :
                    ASSIGN/V1=DOT(SCN1.HIT[1..SCN1.NUMHITS].XYZ-SCN1.HIT[1..SCN1.NUMHITS].TXYZ,SCN1.HIT[1..SCN1.NUMHITS].TIJK)
                    Then ASSIGN/V2=MAXINDEX(V1) and dimension the T value of SCN1.HIT[V2]

                    Comment


                    • Lord_Warfield
                      Lord_Warfield commented
                      Editing a comment
                      dang it... i wanted to figure that out and got pulled away from my desk...

                  • #12
                    Originally posted by JEFMAN View Post
                    Calculate the max T_Value :
                    ASSIGN/V1=DOT(SCN1.HIT[1..SCN1.NUMHITS].XYZ-SCN1.HIT[1..SCN1.NUMHITS].TXYZ,SCN1.HIT[1..SCN1.NUMHITS].TIJK)
                    Then ASSIGN/V2=MAXINDEX(V1) and dimension the T value of SCN1.HIT[V2]
                    I am rerunning scan to see if anything changes. When I added first line of your code and pointed mouse on it all the values were zeros. Second line was displaying 1. I entered the way you had it but just curious, can those 2 lines be combined to:
                    Code:
                    ASSIGN/V1=MAXINDEX(DOT(SCN1.HIT[1..SCN1.NUMHITS].XYZ-SCN1.HIT[1..SCN1.NUMHITS].TXYZ,SCN1.HIT[1..SCN1.NUMHITS].TIJK))

                    Comment


                    • #13
                      I did rerun my scan but line
                      Code:
                       
                      ASSIGN/V1=DOT(SCN1.HIT[1..SCN1.NUMHITS].XYZ-SCN1.HIT[1..SCN1.NUMHITS].TXYZ,SCN1.HIT[1..SCN1.NUMHITS].TIJK)

                      displays all zeros so the second line just grabs first point from the scan
                      Last edited by marcc; 04-03-2019, 02:31 PM.

                      Comment


                      • #14
                        Are you running the scan from the CAD values ?
                        Are you running relearn, master, find noms ?
                        If you dimension the scan with text ON, you should see all the t_values on the report, and find out the min and max.
                        Can you post the code, please ?
                        This assignment works fine usually, since I purposed it here, long time ago.

                        Comment


                        • #15
                          I will have to come back to this tomorrow. I am running patch scan, picking path on CAD model and running with find noms. With text ON I can see X,Y,Z,I,J,K and deviation. One of deviations is marked as MIN, one as MAX.

                          Comment

                          Related Topics

                          Collapse

                          Working...
                          X