12-17-2007 01:13 PM
12-17-2007 02:04 PM
You have a user control wired to the dt input of the loop. This is very uncommon. The default for the control is 0. So if you run it like that, you'll have something similar to a normal while loop with no wait statement in it. Essentially running as fast as it can eating up all the CPU resources.
Also, you have a 0 in the Processor node. I would leave that at the default of -2 unless you have some very good reason for assigning the thread to a specific processor.
How do you know the inner timed loop is not running?
12-17-2007 02:42 PM
Sorry, I only program occasionally in LabVIEW.
Sorry, I am having trouble understanding concepts of Timed loops. According to the help, dt is the period of the timed loop. I thought I should be able to wire a number to that and that would define how fast the loop goes. I don't see why that should be uncommon.
You say "you have a 0 in the Processor node" . I looked up LabVIEW help and can't find any mention of a Processor Node.
12-17-2007 03:47 PM
12-17-2007 06:58 PM
12-17-2007 07:17 PM
12-18-2007
10:57 AM
- last edited on
01-11-2026
03:46 PM
by
Content Cleaner
12-18-2007 01:39 PM
Thanks Mike,
It was the timed loop name.
I had a main VI which called a subvi with a timed loop. This VI was stopped and the "offending" VI run from the main VI. I had based the offending VI on the first VI so the timed loops had the same name. I changed the name of the one of the timed loops and it all goes.
It's wierd to have named structures, I mean, while loops, case structures etc. don't have names (that I know of). Anyway it's fixed and I can carry on.
Thanks again.