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Thread: lbl problem

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  1. #1
    Dear community
    I have a weird problem! I have a program where I have to mill an outer diameter in helix with different feed depths. That looks like this:

    x250 y0 z20 ro fmax
    z0 fauto
    x200 y0 rl fauto
    cc x0y0
    lbl10
    cpi 360 iz-2 dr+
    call lbl 10/rep2
    lbl11
    cpi 360 iz-1 dr+
    call lbl 11/rep2
    cc x0y0
    cpi 360 dr+
    z20

    Now I have the problem that the machine sometimes mills 1 mm too deep!!! ( from 1000parts one will be a reject)
    What could be the raison???
    Is it because I haven’t programmed a Inl 0 ??
    Controller Heidenhain itnc530

    Regards

  2. #2
    Join Date
    Sep 2008
    Posts
    2313
    Hello

    if you want to be sure then you have to work with if-inquiries.
    Then you can request the final depth with the parameters.
    Gruß<br /><!--fonto:Comic Sans MS--><span style="font-family:Comic Sans MS"><!--/fonto-->Schwindl<!--fontc--></span><!--/fontc-->

  3. #3
    Hello
    What is an if-inquiry?
    Regards

  4. #4
    Join Date
    Dec 2001
    Posts
    5151
    Hello
    What is an if-inquiry?
    Regards

    @ appenzeller

    You will find the if-inquiries behind the soft key “JUMPS”
    Here a few examples:

    FN 9: IF SAME, JUMP
    e.g. FN 9: IF +Q1 EQU +Q3 GOTO LBL 12
    If both values or parameter is the same, jump to the stated label

    FN 10: IF NOT EQUAL, JUMP
    e.g.. FN 10: IF +10 NE –Q5 GOTO LBL 1
    If both values or parameter not equal, jump to the stated label


    FN 11: IF BIGGER, JUMP
    e.g. FN 11: IF+Q1 GT+10 GOTO LBL 5
    If first value or parameter bigger than second value/parameter, jump to the stated label

    FN 12: IF SMALLER, JUMP
    e.g. FN 12: IF+Q5 LT+0 GOTO LBL “EXAMPLELBL“
    If first value/parameter smaller than second value/parameter, jump to the stated label
    <b>Gruß / Regards</b><br /><i><b>CNCFanatiker</b></i><br />AdminCrew<br /><br />Wenn wir immer das tun was wir können, dann bleiben wir immer das was wir sind.<br />If we always this act what we are able to do, then we always remain what we are.

  5. #5
    e.g.
    q1 = -6 ;1.depth absolute(infeed2)
    q2 = -8 ;2.depth absolute(infeed 1)
    q3 = 0 ;counter
    x250 y0 z20 ro fmax
    z0 fauto
    x200 y0 rl fauto
    cc x0y0
    lbl10
    q3 = g3 - 2
    cpi 360 iz-2 dr+
    if q3 ne q1 goto lbl10
    lbl11
    q3 = g3 - 1
    cpi 360 iz-1 dr+
    if q3 ne q1 goto lbl11
    cc x0y0
    cpi 360 dr+
    z20
    Pr

  6. #6
    Hello everybody
    I know that I can do it with parameters but I know it’s also possible with a cycle!
    I would prefer not using parameter because otherwise my shift partner will struggle...
    Of course many ways lead to the goal....wink.gif)

    But that doesn’t answer my question about the missing Ibl 0!!!!!!
    Is here anybody that is familiar with this problem??

    Regards

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