On 12/02/2017 12:00 PM, [email protected] wrote:
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
Oops, fired off that last message too soon. Mick's right: M66 L[1-4]
is for digital inputs only; after execution, it will hold the value of
the digital input specified by the P flag, or -1 in case of timeout.
Also, the P flag can range from 0 to 63 (IIRC) if you set `loadrt motmod
num_dio=64` in the HAL file; the M66 manual isn't quite right on that point.
I haven't been following your project, but I hope I get to see a video
of your tool changer in operation once you're done with it!
John
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
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