Omron SYSMAC CJ Operation Manual page 303

How-to's
5-1-3-7 Example 5
.
m motor_revolution = n machine_cycle = n 12.24 motor_revolution
The evident solution is: n = 100 and m = 1224. Or, when we simplify the
factors: n = 25 and m = 306. Therefore: Pn205 = m – 1 = 305. With these
settings, executing MOVE(180) moves the moving part 180 tenths of an angle
degree or 18 angle degrees in forward direction.
M
17-bit absolute
1:10 Gear
encoder
Pulley: 12 teeth
50mm between teeth
The mechanical system uses a servo motor with a 17-bit absolute encoder.
The mechanical gear ratio of the gearbox is 1:10. The pulley has got 12 teeth,
and each two are 50 mm apart. One complete turn of the pulley equals 144
stations on the main wheel. The distance between two stations is 50 mm. The
mechanical measurement units must mm. Total repeat distance must be the
distance between two stations, 50mm.
With the same procedure as in example 1, we have:
Pn202
UNITS =
Pn203
17
.
2 encoder_counts
.
1 motor_revolution
17
.
2 10 encoder_counts
=
.
12 50 mm
Therefore, if we use the mechanical system to set the electronic gear ratio, we
have:
One possible solution is:
17
Because 2 /50 is a number with an infinite number of decimal digits, we can
choose the following:
Pn202
UNITS
Pn203
Therefore, the parameters are:
.
.
10 motor_revolution 1 pulley_revolution
. .
.
1 pulley_revolution
17
Pn202
2
UNITS =
Pn203 50
17
2
UNITS =
50
Pn202 = 5
Pn203 = 6
Pn205 = 4
10 10
17 17 17
= 2 = 2 = 2
.
50 12 600
Section 5-1
. .
Main Wheel: 144 stations
50 mm between stations
. .
1 station
.
12 station 50mm
10
12
1 1 1
17 15
= 2 = 2
2
.
60 2 15 15
=
291