How to properly project distances on a curved cylindrical part with flaws

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

  • How to properly project distances on a curved cylindrical part with flaws

    Hello,

    I have a curved part composed of 2 cylinders joined at an angle of 108°.
    I want to measure 2 distances, from each end of the cylinders to the main centre.

    In theory it is simple, I just have to create 2 lines from the cylinders axes and 3 points to make a plane, then i project the distances on it.
    Capture1.PNG
    But what if I have a flaw making the cylinders axes not coplanar :
    Capture 2.PNG
    Do I have to make a plane with dots 1 & 2 and line next to dot 1, then projecting distances on it ?
    Or should I create a line perpendicular to lines 1 & 2 and measure it ?


    I am lost because we have also a 2D measurement machine (Keyence) and the difference between both machine is 0.3mm.

    Thank you
    Last edited by TheDudeAbides; 04-01-2019, 05:54 AM.

  • #2
    I'm not sure to understand as well, but I think I would do it :
    ASSIGN/V1=CROSS(CYL1.IJK,CYL2.IJK)....................... ..........it's a vector perp to CYL1 and CYL2
    ASSIGN/V2=PT1.XYZ+DOT(PT2.XYZ-PT1.XYZ,CYL1.IJK)......it's the projection of PT2 on the CYL1 axis
    ASSIGN/V3=PT2.XYZ+DOT(PT1.XYZ-PT2.XYZ,CYL2.IJK)......it's the projection of PT1 on the CYL2 axis
    ASSIGN/V4=(V2+V3)/2................................................. ............it's the mid point between both lines
    Then, I would create a generic plane with V4.X,V4.Y and V4.Z as coordinates and V1.I, V1.J and V1.K as vector.

    Comment


    • #3
      Originally posted by JEFMAN View Post
      I'm not sure to understand as well, but I think I would do it :
      ASSIGN/V1=CROSS(CYL1.IJK,CYL2.IJK)....................... ..........it's a vector perp to CYL1 and CYL2
      ASSIGN/V2=PT1.XYZ+DOT(PT2.XYZ-PT1.XYZ,CYL1.IJK)......it's the projection of PT2 on the CYL1 axis
      ASSIGN/V3=PT2.XYZ+DOT(PT1.XYZ-PT2.XYZ,CYL2.IJK)......it's the projection of PT1 on the CYL2 axis
      ASSIGN/V4=(V2+V3)/2................................................. ............it's the mid point between both lines
      Then, I would create a generic plane with V4.X,V4.Y and V4.Z as coordinates and V1.I, V1.J and V1.K as vector.
      Thanks a lot.
      Does it work regardless of alignement ?
      Because V2 & V3 gave me just one number, not coordinates.

      I did what you wrote, but I moved my alignement on each cylinder axis end in order to add the projection only on one axis.
      Not sure if it is the most optimized way to do this, but it seems to work.

      Here is how it looks:
      Capture2.PNG
      Capture.PNG
      Attached Files

      Comment


      • #4
        Originally posted by TheDudeAbides View Post

        Thanks a lot.
        Does it work regardless of alignement ?
        Because V2 & V3 gave me just one number, not coordinates.

        Yep, because I'm wrong !!!!!!!
        ASSIGN/V1=CROSS(CYL1.IJK,CYL2.IJK)....................... ..........it's a vector perp to CYL1 and CYL2
        ASSIGN/V2=PT1.XYZ+DOT(PT2.XYZ-PT1.XYZ,CYL1.IJK)*CYL1.IJK......it's the projection of PT2 on the CYL1 axis
        ASSIGN/V3=PT2.XYZ+DOT(PT1.XYZ-PT2.XYZ,CYL2.IJK)*CYL2.IJK......it's the projection of PT1 on the CYL2 axis
        ASSIGN/V4=(V2+V3)/2................................................. ............it's the mid point between both lines


        DOT gives a number, here a length, and you have from PT1 to add this length along the cyl axis to find a point...

        Sorry for the time lost ...

        Comment


        • #5
          Thanks !
          No problem, no I see how it works.
          I will never be mad at you, you saved me countless times on this forum.

          Comment


          • JEFMAN
            JEFMAN commented
            Editing a comment
            You're welcome !

        Related Topics

        Collapse

        Working...
        X