[GreenVeltlins77]

I failed on a task today and I wanted to ask how you would carry it out...

I wanted to do the following:

Two equal parts are clamped and I defined the reference point with 2 Presets...
I first wanted to process one part, then the other one afterwards. The Tool Call of the particular Preset call (cycl247) and the subprogram call is in the main program. I have a few more labels in the subprogram after M30.

The problem is that it processes the first part but then it can’t get to the preset and the subprogram because of the M30. However, I can’t delete M30 is the subprogram because otherwise the label is not read out anymore.

I have read something about a jump command (FN9)and already tried it out but without success.

Regard
Sven

[schwindl]

Hello
You can solve it with a jump:

Main program:
Fn 9 if. 0 eq 0 goto lbl 99
your lbls

Lbl 99
End PGM


Then you can go back into the program.

[GreenVeltlins77]

Thank you for your quick answer.

I have tried that jump command without success. Where do I have to enter i


My main program:

1 Tool Call ......(neutral point 1)
3 ProgramCall TNCathNC1
4 L Z250 Fmax
5 CyclDef 247 (neutral point 1)
6 ProgramCall TNCathNC1
7 M30

My subprogram NC1:

1 L .......
2 L .......
3 Call LBL 5
4 L ......
5 Call LBL 6
6 M30
LBL 5
LBL 6
END subprogram NC1

Regards
Sven

[GreenVeltlins77]

Do I have to write anything into LBL 99?
Sorry that I ask so dumb questions.. I am only working with HH for q few months now.. There was no training...

Greetings
Sven

[cgTNC]

No, simply LBL 99 as our last block in the subprogram.
Regards

[schwindl]

Nothing comes into LNL 00. Also no lnl 0 at the end.

The lbl 99 is only your jump destination from if 0 eq 0 goto lbl...

If you have to identical parts you can also use the tool orientated process.
There both parts are processed with the first tool, then both with the second tool and so on..
Just have a look at tool orientated processing...