Created Date: September 26,2006

Created by: al.kaufman

I 've had to do the same recently on a PC based control system, but the logic is the same.

Use a 1-Shot on an immediate input of your pulse. Use the 1-Shot to grab the accumulated value of a timer. Divide 4000 by the accumulated value in milli-seconds to get the RPM, assuming 4 pulses per rev. Also put the 1-Shot as an N.C. contact as the condition to run / reset the timer. That's it.

Adjust the dividend according to the number of pulses per rev and resolution of your timer. ie: Use 600 for six pulses per rev and .01 sec ticks, etc.

Created Date: September 26,2006

Created by: Rich1955

Sp8------------udc 176

spo Nc---------udc 176

X3 incoderx4---udc 176

K9999999

X3 NC-----------Ld cta 176

out v2000

That is the fastest way I know

Created Date: September 28,2006

Created by: twbrock

Sorry I took so long to get this posted by I have been busy. This works from 1 rpm to 1000 rpm and more as long as the scan rate is less than 60 mS. Only need 1 pulse per rev. Theta can be 1 to 99 but I am using 90 on a couple of units and this seams to work very well.

Rung 1

STR SP0

LD K6000

OUT V2104

LD K90

OUT V2105

LD K100

OUT V2106

Rung 2

STR X0

PD C0

Rung 3

STR C0

ANDN C1

STR C1

ANDN C0

ORSTR

OUT C1

Rung 4

STR C1

AND X1

STR C5

TMRAF T0 K30000

Rung 5

STRN C1

AND X1

STR C6

TMRAF T2 K30000

Rung 6

STRN C1

PD C3

Rung 7

STR C3

LD V2104

DIV TA0

OUT V2100

OUT C5

Rung 8

STR C3

MATHBCD V2102 "(V2102 * V2105) + ((V2106 - V2105) * V2100) "

DIV V2106

OUT V2102

OUT V2103

Rung 9

STR C1

PD C4

Rung 10

STR C4

LD V2104

DIV TA2

OUT V2101

OUT C6

Rung 11

STR C4

MATHBCD V2102 "(V2102 * V2105) + ((V2106 - V2105) * V2101) "

DIV V2106

OUT V2102

OUT V2103

Rung 12

STRN X1

LD K0

OUT V2100

OUT V2101

OUT V2102

OUT V2103

Rung 13

END

Created Date: September 28,2006

Created by: al.kaufman

Originally posted by al.kaufman:

I 've had to do the same recently on a PC based control system, but the logic is the same.

Use a 1-Shot on an immediate input of your pulse. Use the 1-Shot to grab the accumulated value of a timer.

The general equation is:

Speed (rpm)=60/(n*t) where: n=pulses/rev

t=time between pulses (sec)

Example:

for n=4, t=.030 sec Speed=60/(4*.030)=500 rpm

Note: If you use a standard input for the pulses, you want 2+ scans per on-off pulse.

This sets the max speed you can reliably measure at:

Speed=60/(n*tmin) where tmin= 2*scan time

Example:

for n=15, and scan time =.037 sec

Speed (max) = 60/(15*2*.037) = 54 rpm

Above that speed you need faster inputs, a faster scan, or a high speed counter.

I am doing this with a PC based control system which gives very fast scan times.

A standard prox on a regular input is fast enough for the rpm range I need.

Created Date: September 12,2006

Created by: trevorstripling

I have downloaded and got the calc RPM example working for my application, but the input sensor I have only pulses 4 times per revolution and does not give me enough resolution at low rpms. I have a shaft that starts at 0 rpm and ramps up to 8000. I need to sample 100-200 times per second to ensure it is not increasing in speed faster than a predetermined acceleration rate.

What I need to do (or at least think) is instead of count the pulses per time period to determine the shaft rpm, is to trap the amount of time between pulses instead.

Can anyone give me a good example of how to do this with a DL06 using the high speed inputs?

Trevor