Table of Contents
Advertisement
APPENDICES
A
A
P
P
=
=
∆
S
A
A
360.00000(degree) 3/11
L
M
=
360.00000(degree) 3
=
1080.00000
Here, reduce the above result since the A
65535.
A
11264
P
=
A
A
67.50000
L
M
Next, since the A
setting range is up to 0.65535, set 0.06750 as A
L
by 1000 with A
.
M
A
11264(A
P
=
A
A
0.06750(A
L
M
Thus, A
, A
and A
P
L
A
=11264
P
A
=0.06750
L
A
=1000
M
(4) When mm is set as the control unit for conveyor drive (calculation including π)
When the belt conveyor drive is used, the conveyor diameter is 135mm, the
pulley ratio is 1/3, the motor is HC-SF (16384PLS/rev) and the reduction ratio of
the reduction gear is 7/53
Motor
Reduction
gear 7/53
As the travel value of the conveyor is used to exercise control, set mm as the
control unit.
First, find how many millimeters the load (machine) will travel (∆S) when the
motor turns one revolution (A
A
(number of pulses per motor revolution)
P
∆S (machine travel value per motor revolution) = 135000.0(µm)×π×reduction
Substitute this for the above relational expression.
At this time, make calculation with the reduction ratio 7/53×1/3 remaining as a
fraction.
A
A
P
P
=
=
∆
S
A
A
L
M
=
Here, make calculation on the assumption that π is equal to 3.14159.
A
A
P
P
=
=
∆
S
A
A
L
M
16384(PLS)
16384(PLS) 11
180224
)
P
) 1000(A
)
L
M
to be set are as follows.
M
Pulley
ratio 1/3
).
P
16384(PLS)
135000.0( m)
7/53 1/3
16384(PLS) 53 3
135000.0( m)
2605056
2968802.6
APP − 70
setting must be made not more than
P
Belt conveyor
= 16384 (PLS)
= 135000.0(µm)×π×7/53×1/3
7
and multiply it
L
135mm
ratio
Hide quick links:
Advertisement
Table of Contents
