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Centcir

I'm using CENTCIR to find the location of the pin within my contour. I can find the center of the circle, but how can I evaluate the distance to the end of the pin?

Attached is a print snippet. How can I evaluate the 2.7753 and 1.5175 dimensions?





  • Attached is the WDB and the plot fi it helps. I had to rename the extension WDB to docx to be able to upload.


  • Hello
    Evaluate the distance from the upper planen to centerpoint and add half the center circle diameter.
    Evaluate the distance from the rigt plane to P2.

    Hope this helps.


  • Hello
    Evaluate the distance from the upper planen to centerpoint and add half the center circle diameter.
    Evaluate the distance from the rigt plane to P2.

    Hope this helps.




    I was thinking there might be a command for this such as 3McdCirCir2, except one element would be an axis. There is a command MCDCIAX. but this is to the center of the circle.


  • I was thinking there might be a command for this such as 3McdCirCir2, except one element would be an axis. There is a command MCDCIAX. but this is to the center of the circle.


    Also, The P2 tangency point and the maximum point on the circle are dfferent.
  • Hello.
    What i can see on the drawing the 1.5175 distance is to the tangency point.
    I did a test om a Cad model. You must insert your own values and element names.


    CENTCIR (NAM=PROFIL, PLA=XY, DIA=100, ANG=180, SPT=4, PNT=P(1))
    CENTCIR_P (NAM=PROFIL, RES=(CENT, PNT1, PNT2), PLA=XY, DIA=100)

    !Calculate the distance to the centerpoint
    DIPNTAXI (NAM=DIS(1), CSY=CSY(1), EL1=CENT, EL2=AXI(1), CPY=DEF$DIS1)

    !Calculate the distance to the tangentpoint
    ​DIPNTAXI (NAM=DIS(2), CSY=CSY(1), EL1=PNT2, EL2=AXI(2), CPY=DEF$DIS1)

    !Add the radius off the centering ball
    XY=DIS(1).$A+50

    !Run DFNELE just one time
    DFNELE (NAM=DIS(3), TYP=POI, CSY=CSY(1))

    !Add the calculated realvalue to the element
    PUTVAL (OBJ=DIS(3), DSC=X, TYP=ELE, VAL=XY)

    !Evaluate
    AddEva (NAM=DIS(3), EDT=N)


    Hope this helps
  • Hello.
    What i can see on the drawing the 1.5175 distance is to the tangency point.
    I did a test om a Cad model. You must insert your own values and element names.


    CENTCIR (NAM=PROFIL, PLA=XY, DIA=100, ANG=180, SPT=4, PNT=P(1))
    CENTCIR_P (NAM=PROFIL, RES=(CENT, PNT1, PNT2), PLA=XY, DIA=100)

    !Calculate the distance to the centerpoint
    DIPNTAXI (NAM=DIS(1), CSY=CSY(1), EL1=CENT, EL2=AXI(1), CPY=DEF$DIS1)

    !Calculate the distance to the tangentpoint
    ​DIPNTAXI (NAM=DIS(2), CSY=CSY(1), EL1=PNT2, EL2=AXI(2), CPY=DEF$DIS1)

    !Add the radius off the centering ball
    XY=DIS(1).$A+50

    !Run DFNELE just one time
    DFNELE (NAM=DIS(3), TYP=POI, CSY=CSY(1))

    !Add the calculated realvalue to the element
    PUTVAL (OBJ=DIS(3), DSC=X, TYP=ELE, VAL=XY)

    !Evaluate
    AddEva (NAM=DIS(3), EDT=N)


    Hope this helps


    That does help to find the distance to the point of tangency. Thank you.

    The distance from the edge to the POT is 1.5136. The distance to the end of the pin is 1.5175. I'm trying to figure out the best way to add the pin radius (.175) to the center distance then evaluate.
  • Hello
    I make another attempt to explain.

    I explain in XY plane

    First you calculate the distance to the center of the centering circle.
    Make sure that the distance in Z betveen the axis and profil i close to zero, otherwise you have to prjekt the points (PRJPTS)

    DIPNTAXI (NAM=DIS(1), CSY=CSY(1), EL1=CENT, EL2=AXI(1), CPY=DEF$DIS1)
    DIPNTAXI (NAM=DIS(2), CSY=CSY(1), EL1=CENT, EL2=AXI(2), CPY=DEF$DIS1)

    Get the result as an real value. It´s the A value från the element and it´s the same as DXYZ value from the evaluation above.

    GETVAL (NAM=XY_DIS(1), OBJ=DIS(1), DSC=A, TYP=ELE)
    GETVAL (NAM=XY_DIS(2), OBJ=DIS(2), DSC=A, TYP=ELE)

    Take this values and add the radius for the centering cicle.

    XY_DIS_RAD(1)=XY_DIS(1)+(0.35/2)
    XY_DIS_RAD(2)=XY_DIS(2)+(0.35/2)

    Define an element to put the calculated value in. That needs to be done only one time.

    DFNELE (NAM=DIS(3), TYP=POI, CSY=CSY(1))

    Put in the values that has been calculated as X and Y coordinates

    PUTVAL (OBJ=DIS(3), DSC=X, TYP=ELE, VAL=XY_DIS_RAD(1))
    PUTVAL (OBJ=DIS(3), DSC=Y, TYP=ELE, VAL=XY_DIS_RAD(2))

    Evaluate the result in X and Y

    AddEva (NAM=DIS(3), EDT=N)​



  • Hello
    I make another attempt to explain.

    I explain in XY plane

    First you calculate the distance to the center of the centering circle.
    Make sure that the distance in Z betveen the axis and profil i close to zero, otherwise you have to prjekt the points (PRJPTS)

    DIPNTAXI (NAM=DIS(1), CSY=CSY(1), EL1=CENT, EL2=AXI(1), CPY=DEF$DIS1)
    DIPNTAXI (NAM=DIS(2), CSY=CSY(1), EL1=CENT, EL2=AXI(2), CPY=DEF$DIS1)

    Get the result as an real value. It´s the A value från the element and it´s the same as DXYZ value from the evaluation above.

    GETVAL (NAM=XY_DIS(1), OBJ=DIS(1), DSC=A, TYP=ELE)
    GETVAL (NAM=XY_DIS(2), OBJ=DIS(2), DSC=A, TYP=ELE)

    Take this values and add the radius for the centering cicle.

    XY_DIS_RAD(1)=XY_DIS(1)+(0.35/2)
    XY_DIS_RAD(2)=XY_DIS(2)+(0.35/2)

    Define an element to put the calculated value in. That needs to be done only one time.

    DFNELE (NAM=DIS(3), TYP=POI, CSY=CSY(1))

    Put in the values that has been calculated as X and Y coordinates

    PUTVAL (OBJ=DIS(3), DSC=X, TYP=ELE, VAL=XY_DIS_RAD(1))
    PUTVAL (OBJ=DIS(3), DSC=Y, TYP=ELE, VAL=XY_DIS_RAD(2))

    Evaluate the result in X and Y

    AddEva (NAM=DIS(3), EDT=N)​





    Now THAT is what I was looking for. Actually, the previous example will also be very useful. I have several variations of that profile with different pin diameters that I need to write programs for. Thank you for taking the time with a detailed solution !