How to locate imaginary ball gage inside a cone

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  • How to locate imaginary ball gage inside a cone

    Hello, I'm hoping to get some help on what seems to be the most challenging measurement I've come across.
    My goal is to measure the distance from a planar datum (perpendicular to the axis of the cone) to the top of an imaginary gage ball that would rest inside of a cone.

    I have tried a few things:
    • Construct a circle using a cone and picked each of the 3 options I see.
    I'm uncertain if I understand correctly what these constructions are doing but this is what I believe they do. I would appreciate feedback if I'm wrong about how they work.
    1. I believe the first option was I input at what diameter of the cone I want to read height at.
    2. The second option was to determine a "distance" which I presumed to be the distance I'm trying to confirm.
    3. The third one looks as if it's asking me what size of ball I want to place into the cone and then it "should" feed back a height. This appears to feed back a diameter instead though.

    I've included an image of my problem below:

    cmmgageproblem.jpg

    Attached Files

  • #2
    some answers here :

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    • #3
      Code:
      con1       =feat/contact/cone/default,cartesian,out
                  theo/<0,0,0>,<0,0,1>,60,20,25
                  actl/<0,0,0>,<0,0,1>,60,20,25
                  targ/<0,0,0>,<0,0,1>
                  start ang=360,end ang=360
                  angle vec=<-1,0,0>
                  show feature parameters=no
                  show contact parameters=yes
                    numhits=9,numlevels=3,depth=0,end offset=0
                    sample hits=0,spacer=0.381
                    avoidance move=no,distance=25
                    onerror=no,read pos=no
                  show hits=no
      con2       =feat/cone,cartesian,in,ang,no
                  theo/<0,0,-21.651>,<0,0,1>,60,25,48.094
                  actl/<0,0,-21.651>,<0,0,1>,60,25,48.094
                  constr/cone,cast,con1,dependent
      a1         =alignment/start,recall:startup,list=yes
                  alignment/level,zplus,con2
                  alignment/trans,xaxis,con2
                  alignment/trans,yaxis,con2
                  alignment/trans,zaxis,con2
                  alignment/end
                  assign/halfang=con1.a/2
                  assign/opp=12.7/2
                  assign/myhyp=opp/sin(deg2rad(halfang))
      f1         =generic/sphere,dependent,cartesian,out,$
                  nom/xyz,<0,0,myhyp>,$
                  meas/xyz,<0,0,myhyp>,$
                  nom/ijk,<0,0,1>,$
                  meas/ijk,<0,0,1>,$
                  diameter/12.7,12.7
      dim loc1= location of sphere f1  units=mm ,$
      graph=off  text=off  mult=10.00  output=both  half angle=no
      ax    nominal       +tol       -tol       meas        dev     outtol
      x        0.000      0.050     -0.050      0.000      0.000      0.000 ----#----
      y        0.000      0.050     -0.050      0.000      0.000      0.000 ----#----
      z       12.700      0.050     -0.050     12.700      0.000      0.000 ----#----
      end of dimension loc1

      Steps...

      1) Measure cone
      2) Create a cast cone from your cone (this is the same cone, but the origin is from the point of the cone not the centroid)
      3) Align to the cone (Level to axis and origin on the point
      4) from here the trig is simple, your sphere radius is the Opposite side of a right angled triangle, half the cone angle is your angle.
      5) Calculate the hypotenuse
      6) Construct a generic sphere at 0,0,Hypotenuse with the gauge diameter.
      Last edited by NinjaBadger; 04-16-2019, 03:29 AM.
      Applications Engineer
      Hexagon UK

      Comment


      • #4
        I would simply measure the bottom. smaller cone and then the cylinder just above it. Now create a circle where these two intersect and looking at it from a side view x or y, call the distance from A to the circle in the z axis. (If your drawing is correct then it seems that this is the dimension you are looking for.)

        Comment


        • #5
          Hi. I know that you can construct a circle of a certain diameter from a measured cone and that the constructed circle will be placed where it exists at the
          "This is my word... and as such is beyond contestation."

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