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Thread: Angle/Degree

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  1. #1
    Join Date
    Mar 2014
    Posts
    19

    Angle/Degree




    I learn CNC code and I have a question.

    I try to make 15 Degree and 0.75 (center)
    My code is:

    G97 M08;
    G54 T S;
    G0 Z2.;
    X80. (How to calculate start point?)
    G01 Z0.06 F0.2;
    A105. X82. F0.05;

    noseR is 0.4

  2. #2
    Join Date
    Mar 2014
    Posts
    19

    Re: Angle/Degree


  3. #3
    Join Date
    Jan 2009
    Posts
    103

    Re: Angle/Degree

    Tangent(15) X (.5) = .133975

  4. #4
    Join Date
    Mar 2014
    Posts
    19

    Re: Angle/Degree

    How did you get .133975
    I got 2.7060
    Explain please.

    And what is X?

  5. #5
    Join Date
    Jan 2009
    Posts
    103

    Re: Angle/Degree

    Your going to have to give more information than what your showing.

  6. #6
    Join Date
    Mar 2014
    Posts
    19

    Re: Angle/Degree

    Draft:


    Program
    G97 M08;
    G54 T S;
    G0 Z2.;
    X80. (How to calculate start point?)
    G01 Z0.06 F0.2; OR Z0. ?
    A105. X82. F0.05;

    noseR is 0.4

    Or what is the correct and accurate program?

  7. #7
    Join Date
    Sep 2010
    Posts
    1230

    Re: Angle/Degree

    Quote Originally Posted by Readme00 View Post
    How did you get .133975
    I got 2.7060
    Explain please.

    And what is X?
    If you refer to the attached diagram, you will see that the radial value for X can't possibly be greater than the Z value of the chamfer. The two known components are the angle of 15deg and the side adjacent to the angle. What you don't know, and need, is the side opposite the angle. As the Tan of an angle equals the ratio of the opposite and adjacent sides of a Right Triangle, you can use the Tan trig function to solve the length of the opposite side X in the diagram.

    Click image for larger version. 

Name:	Angle1.jpg 
Views:	6 
Size:	12.1 KB 
ID:	242580

    Tan 15 = X / 0.5
    Tan 15 x 0.5 = X
    X = Tan 15 x 0,5
    X = 0.133975

    X solved in the above is a radial value, therefore, multiply by 2 to obtain the diameter value of 0.26795

    To get the X value of the start of the chamfer:

    X = 82.0 + 0.26795
    X = 82.26795

    Regards,

    Bill

  8. #8
    Join Date
    Jan 2009
    Posts
    103

    Re: Angle/Degree

    Thanks for the help Bill,
    Looking at the Print and the code was a little confusing. It looks like a cross section of a 82mm dia.
    plate with a .5 +.5/-0 chamfer and the code he shows looks like milling code. He also is showing a tool Rad.
    of .4, which confused me as to what he wanted to calculate for. He's showing in his code A105 X82. which
    looks like a polar move to the O.D. which would make the start point X80.5 (82 - (2 x mean chamfer size of .75)).
    I'm not sure if he's looking to calculate for the tool radius either. If it's a mill, His code is showing X82. which seems
    like the print is not oriented correctly. It may help if the OP provided more info as to machine type and what he is
    trying to do. Thanks again, and by the way I've enjoyed reading many of your solutions you've posted.

    Mark

  9. #9
    Join Date
    Sep 2010
    Posts
    1230

    Re: Angle/Degree

    Hi Mark,
    Having another look at the OP's code I see what you mean.

    To the OP
    Following is how to calculate the start and finish coordinates for a 0.4 rad Tool Nose Radius.

    Referring to the attached picture:

    Attachment 242582

    1. Solve Z1
    Z1 = Tan 15 x 0.5
    Z1 = 0.133975

    2. Solve X2
    X2 = Tan 7.5 x .4
    X2 = 0.052661

    3. Solve Start X Pos of Tool

    X Start = 81 + X2 x 2 - 0.8
    X Start = 81 + 0.052661 x 2 - 0.8
    X Start = 80.305322

    4. Solve Z3
    Tan 52.5 = Z3 / 0.4
    Z3 = 0.4 / Tan 52.5
    Z3 = 0.306931

    5. Solve Finish Z Pos of Tool

    Z Finish = (-Z1 + Z3) - 0.4
    Z Finish = (-0.133975 + 0.306931) - 0.4
    Z Finish = -0.227044


    N1 G00 X78.000 Z1.000
    G01 Z0.000
    G01 X80.305
    G01 X82.000 Z-0.227

    Regards,

    Bill

  10. #10
    Join Date
    Mar 2014
    Posts
    19

    Re: Angle/Degree



    How to calculate degrees, if:
    X1 50.9
    X2 48

    The correct degree is 2.51592 (Mastercam) but I want to know how to calculate it.

  11. #11
    Join Date
    Sep 2002
    Posts
    1949
    tan(phi) = 1.45 / 33 = 0.043939...
    phi = atan(0.043939...) = 2.51592 deg

    1.45: difference of the two diameters 50,9 and 48 divided by 2 (i.e. difference of the radii)

  12. #12
    Join Date
    Mar 2014
    Posts
    19

    Re: Angle/Degree

    Thanks

  13. #13
    Join Date
    Jun 2019
    Posts
    1
    Quote Originally Posted by CNCFr View Post
    tan(phi) = 1.45 / 33 = 0.043939...
    phi = atan(0.043939...) = 2.51592 deg

    1.45: difference of the two diameters 50,9 and 48 divided by 2 (i.e. difference of the radii)
    How to You Get 33

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