okay i have added the following to my sub: P1 L3 M66 Q5 ; O125 IF [#5399 LT 0] (DEBUG, Anzugbolzen nicht in erwartetem zustand!) #5399 = 0 o<on_abort>CALL M2 O125 ENDIF
so it waits for the signal, then gets an timeout (but somehow doesnt show me the message) then resets #5399 to 0 so when i restart it doesnt stop if the toolchange worked (i believe) sometimes when i change the tool it says: duplicate O-word label - already defined in line 15: ' O125 IF [#5399 LT 0]' but its the only one called o125, thats a little strange basically it works, just need to tweak it a little more ps. i once added #5399 to my var file and somehow something resetted my var file but luckily i had a backup Am Samstag, 2. Dezember 2017 19:00:19 UTC+1 schrieb Schooner: > > > On 02/12/17 17:20, Sag ich Dir nich wrote: > > okay thank you, i will try that. > > one more question, does the next M66 reset #5399? > > > You will need to experiment, set a short timeout M66 and don't send a > signal, to set it to -1, then set a longer one and send a signal before > it times out. > #5399 should contain 0 or 1 typically but never -1 > > I don't know if #5399 is set to anything before M66 exits, I would suspect > it would hold the last entered value until M66 exits for whatever reason > and sets the return value accordingly. > You will need to verify, but suspect #5399 is meaningless until M66 does > exit. > > > Am Samstag, 2. Dezember 2017 16:54:35 UTC+1 schrieb Schooner: >> >> It appears that the Q input is of type double, which means it will *only* >> wait for 28,561.6 years before moving again >> >> So set it to a high number and effectively it will never move again >> without an input. >> >> Or you can enter a shorter period and test the value of #5399 to see if >> it is -1 indicating a timeout. >> In that event you can abort. >> >> >> On 02/12/17 15:15, Sag ich Dir nich wrote: >> >> The manual says "M66 wait on an input stops further execution of the >> program, until the selected event (or the programmed timeout) occurs." >> >> so that means if the machine waits for an input but then the timeout >> occurs, the machine continues? Thats not logical to me. I would like the >> machine to stop if the timeout occurs. >> >> In case you were wondering why i bother, its for an automatic toolchanger >> to let the machine know if the pull stud is in the correct position to >> prevent expensiveness, if it is not in the correct position, the machine >> waits until it is... or gets an timeout and stops further execution. >> >> thanks >> -- >> website: http://www.machinekit.io blog: http://blog.machinekit.io >> github: https://github.com/machinekit >> --- >> You received this message because you are subscribed to the Google Groups >> "Machinekit" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to [email protected]. >> Visit this group at https://groups.google.com/group/machinekit. >> For more options, visit https://groups.google.com/d/optout. >> >> >> -- > website: http://www.machinekit.io blog: http://blog.machinekit.io github: > https://github.com/machinekit > --- > You received this message because you are subscribed to the Google Groups > "Machinekit" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected] <javascript:>. > Visit this group at https://groups.google.com/group/machinekit. > For more options, visit https://groups.google.com/d/optout. > > > -- website: http://www.machinekit.io blog: http://blog.machinekit.io github: https://github.com/machinekit --- You received this message because you are subscribed to the Google Groups "Machinekit" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. Visit this group at https://groups.google.com/group/machinekit. For more options, visit https://groups.google.com/d/optout.
