Variable tolerance of a chamfer

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  • Variable tolerance of a chamfer

    PCD CAD 2020 R1 SP7.....

    Am measuring a chamfer diameter and angle. Per my customer's spec, the tolerance of the angle varies and is driven by the measured size of the cone. How can I make PCD look at the diameter and then automatically change the tolerance it uses for the chamfer angle's dimensional output?


    Cone nominal Ø 0.6880-0.7030 (Ø0.6955 ± 0.0075)

    At min CHF DIA, ang can be 119.500°-120.500°
    At nom CHF DIA, ang can be 124.458°-115-668°
    at max CHF DIA, ang can be 127.958°-119.500°


    Thanks in advance.
    Last edited by DAN_M; 10-27-2020, 12:01 PM.
    SF7107(PCD), SF454(PCD), 152614(Quindos), 9159(Quindos), 7107(Quindos), B&S Manual, M&M Gear Checker

  • #2
    Hi Dan,
    First,I would say that values are strange...( max < min ?) and it seems that nominal dia is missing.

    I would :
    Code:
    ASSIGN/TOL_ARRAY=(119.5,120.5,115.668,124.458,119.5,127.958)
    ASSIGN/DELTA_D=(0.7030-0.6880)/3
    IF CHF_DIA>=0.6880 AND CHF_DIA<=(0.6880+DELTA_D)
    ASSIGN/TOL_MIN=TOL_ARRAY[1]
    ASSIGN/TOL_MAX=TOL_ARRAY[2]
    END IF
    IF CHF_DIA>(0.6880+DELTA_D) AND CHF_DIA<=(0.6880+2*DELTA_D)
    ASSIGN/TOL_MIN=TOL_ARRAY[3]
    ASSIGN/TOL_MAX=TOL_ARRAY[4]
    END IF
    IF CHF_DIA>(0.6880+2*DELTA_D) AND CHF_DIA<=(0.6880+3*DELTA_D)
    ASSIGN/TOL_MIN=TOL_ARRAY[5]
    ASSIGN/TOL_MAX=TOL_ARRAY[6]
    END IF
    IF CHF_DIA<0.6880 OR CHF_DIA>(0.6880+3*DELTA_D)
    COMMENT/OPER
    "Part is OOT !"
    END IF
    Then use tol_min and tol_max in tol values.

    Comment


    • DAN_M
      DAN_M commented
      Editing a comment
      Hi Jeff, this is great, unfortunately not quite what I am looking for. You have this broken down by RANGE (which is about as far as I got). My engineers shot me down & would like it to come up with EXACT tolerance. So if the CHF was Ø0.7015, they'd like the EXACT calculated amount of allowable tolerance for the angle, not a "high limit range" for the angle. Any ideas? This is above any beyond my math skills to come up with a formula for this and then translate that into PC DMIS. =/

  • #3
    Ok, I understand.
    The problem is that the tol is not really linear.
    Do they know it ?
    (I changed you values to
    0,688 119,5 120,5
    0,6955 124,458 125,668
    0,703 127,958 129,5
    Which seems to be more logical (maybe I'm wrong !)
    tolerance.JPG

    Comment


    • DAN_M
      DAN_M commented
      Editing a comment
      I triple checked my numbers and they are correct. I know this is less logical and makes it more difficult but this is what I've got =/

    • JEFMAN
      JEFMAN commented
      Editing a comment
      DAN_M : I don't have any doubt about your checking, just about the customer values !!!!!
      How can the max be lower than the min ?
      How can you have a tol of 1° at the min, and close to 9° at the nom or the max dia ???????
      I would call the customer to its own check !!!!!

    • vpt.se
      vpt.se commented
      Editing a comment
      Gotta agree with JEFMAN , those numbers makes no sense.

  • #4
    Let say you have 2 segments for each tolerance.
    You have for each one 2 points (min dia , max dia, min angle, max angle)
    You can write
    min angle=a* min dia + b
    max angle=a*max dia + b
    A and b are unknown, but can be calculated :
    a=(max angle - min angle)/(max dia - min dia)
    b= min angle- a * min dia.

    The tol angle for meas dia is a* meas dia + b.

    So with some assignments :

    Code:
    ASSIGN/V1min=(119.5,124.458,,127.9 58)
    ASSIGN/V1max=(120.5,125.668,129.5)
    ASSIGN/V2=(0.688,0.6955,0.703)
    ASSIGN/A1MIN=(V1MIN[2]-V1MIN[1])/(V2[2]-V2[1])
    ASSIGN/B1MIN=V1MIN[1]-A1MIN*V2[1]
    ASSIGN/A1MAX=(V1MAX[2]-V1MAX[1])/(V2[2]-V2[1])
    ASSIGN/B1MAX=V1MAX[1]-A1MAX*V2[1]
    ASSIGN/A2MIN=(V1MIN[3]-V1MIN[2])/(V2[3]-V2[2])
    ASSIGN/B2MIN=V1MIN[2]-A2MIN*V2[2]
    ASSIGN/A2MAX=(V1MAX[3]-V1MAX[2])/(V2[3]-V2[2])
    ASSIGN/B2MAX=V1MAX[2]-A2MAX*V2[2]
    IF CHF_DIA>=0.6880 AND CHF_DIA<=0.6955
    ASSIGN/TOL_MIN=A1MIN*CHF_DIA+B1MIN
    ASSIGN/TOL_MAX=A1MAX*CHF_DIA+B1MAX
    END IF
    IF CHF_DIA>0.6955 AND CHF_DIA<=0.703
    ASSIGN/TOL_MIN=A2MIN*CHF_DIA+B2MIN
    ASSIGN/TOL_MAX=A2MAX*CHF_DIA+B2MAX
    END IF
    To be checked because I'm not sure of values, and here, it's time to go home ...
    Hope this helps.


    Comment


    • #5
      This may or may not be the most efficient way to do this. But it does work.


      Code:
      CON1 =FEAT/CONTACT/CONE/DEFAULT,CARTESIAN,IN
      THEO/<0,0,0>,<0,0,1>,120,-0.1,0.6955
      ACTL/<0,0,0>,<0,0,1>,120,-0.1,0.6955
      TARG/<0,0,0>,<0,0,1>
      START ANG=0,END ANG=360
      ANGLE VEC=<1,0,0>
      SHOW FEATURE PARAMETERS=NO
      SHOW CONTACT PARAMETERS=NO
      DIM LOC1= LOCATION OF CONE CON1 UNITS=IN ,$
      GRAPH=OFF TEXT=OFF MULT=10.00 OUTPUT=NONE HALF ANGLE=NO
      AX NOMINAL MEAS +TOL -TOL DEV OUTTOL
      D 0.6880 0.6955 0.0150 0.0000 0.0075 0.0000 --------#---------
      END OF DIMENSION LOC1
      ASSIGN/V1=LOC1.D.DEV/.0075
      ASSIGN/V2=(LOC1.D.DEV/.0075)-1
      IF/V1<1
      ASSIGN/LOWEROUTPUT1="REPORT"
      ASSIGN/LOWERMIN=-((((115.668-119.50)*V1)+119.5)-120)
      ASSIGN/LOWERMAX=((((124.458-120.5)*V1)+120.5)-120)
      END_IF/
      IF/V1>1
      ASSIGN/LOWEROUTPUT1="NONE"
      END_IF/
      DIM LOC2= LOCATION OF CONE CON1 UNITS=IN ,$
      GRAPH=OFF TEXT=OFF MULT=10.00 OUTPUT=LOWEROUTPUT1 HALF ANGLE=NO
      AX NOMINAL MEAS +TOL -TOL DEV OUTTOL
      A 120.0000 120.0000 LOWERMAX LOWERMIN 0.0000 0.0000 -#----------------
      END OF DIMENSION LOC2
      IF/V1>1
      ASSIGN/UPPEROUTPUT1="REPORT"
      ASSIGN/UPPERMIN=-((((119.5-115.688)*V2)+115.688)-120)
      ASSIGN/UPPERMAX=((((127.958-124.458)*V2)+124.458)-120)
      END_IF/
      IF/V1<1
      ASSIGN/UPPEROUTPUT1="NONE"
      END_IF/
      DIM LOC3= LOCATION OF CONE CON1 UNITS=IN ,$
      GRAPH=OFF TEXT=OFF MULT=10.00 OUTPUT=UPPEROUTPUT1 HALF ANGLE=NO
      AX NOMINAL MEAS +TOL -TOL DEV OUTTOL
      A 120.0000 120.0000 UPPERMAX UPPERMIN 0.0000 0.0000 -#----------------
      END OF DIMENSION LOC3
      Attached Files

      Comment


      • #6
        JEFMAN what do you think?

        Comment


        • #7
          Originally posted by LaserJay View Post
          JEFMAN what do you think?
          You're right, your calculation gives the right tolerance, mine gives the boundaries !!!!!
          I didn't thought subtracting the nominal !!!!!!!

          Comment

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