Fabios

Hi, for reset integral error in a application with PID, i read status word and when it is = 7 or = 8 or = -4, i reset integral error. Is there other condition that require reset integral error? With best regards Fabio

speed test Ok Damian, i have tested with a V280, it work fine! i have used 30000, and result is exact! Thank you very much! Fabio

300000? why? 1000ms*60sec/2bolts = 30000

Software test Damian, please look image in the link "software test". I have only a doubt: MF2 (TimeBetweenPulses) remain ever > 0, also when the wheel is stop. Second your opinion is a good way to insert a timer (TD 1 sec) and ceck if for 1 sec no arrive impulse from I47 then MF2 = 0? Regards Fabio

> Fabios, the previous posts were deleted because they contained a wrong approach, according to you. Ok, no problem. >I know you asked for a software solution, but it looks like the sensor is pointing to the nuts on the outside, and there are 8 of them. >This should help accuracy? Ziwi, you are right. If i use 8 bolts i have more precision: 10(pulse in one second)*60(sec)/8(bolts)*2,34(ratio 433/1011) = 175,5 rpm one pulse more: 11(pulse in one second)*60(sec)/8(bolts)*2,34(ratio 433/1011) = 193,05 rpm It's no the best but it's better. There is only one problem.....it's no possible! Damian your solution i like it! I study it and test! If i need help i will contact you, according to you! Regards Fabio

> Also, as you shown before, if you take larger intervals the differences will be smaller (At the expense of accuracy). Sure now is not accuracy. Why now i don't see precedent post? Regards Fabio

Hi, i must measure the speed of a turbine. There is pulse sensor on a main wheel with 2 bolts. Every one rotary i have 2 pulse. I use an Hight Speed Input(I47/MI0). The main wheel transmit the rotary to a more little secondary wheel. When the main wheel go to 433 rpm, secondary wheel go to 1011 rpm. Please see the two attachments. As you can see i have used Frequency mesurement 1000 ms, so in MI0 i have F = impulse number in one second. If i multiply this number for 60 sec i obtain number of pulse in one minute. But now i must ti divide this number for 2 (number of Bolts(bulloni in italian). So as you can see in MI183 i have Main wheel RPM. If now i linearize from 433 to 1011, i obtain secondary wheel rpm. The system work fine, but i have the following problem: for example if F(MI0) = 10 then speed of secondary wheel = 702 rpm; if F = 11 then speed of secondary wheel = 772 rpm; if F = 12 then speed of secondary wheel = 842 rpm; There is no middle way So for my application is necessary more precision. Is there another more accuracy sistem via software? Fabio

Hi, my error! excuse me for the disturb. Regards Fabio

Hi, i know that "Activate Autotune has any sense only in a real closed loop system, on real working conditions", but for test usually i check all on my desk. Now it happen that autotune does not start, status word remain = 0 and nothing happen. I have used many time autotune pid and it work fine. I'd like to know: is there some condition that not permit to autotune to start? Regards Fabio

New product and old problem....

Hi, i must start up 2 I/O LC3. On the manual http://www.unitronics.com/Data/Uploads/IO_expansions/IO-LC1%20IO-LC3.pdf on page 10 there is chart for Effective Resolution. Can you explain me this chart please? Ciao Fabio