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8 CONTROL SUB FUNCTIONS
8.3 Functions for Compensating the Control
■When "mm" is set as the control unit for conveyor drive (calculation including )
When the belt conveyor drive is used, the conveyor diameter is 135 mm, the pulley ratio is 1/3, the motor is HG-KR
(4194304 pulses/rev) and the reduction ratio of the reduction gear is 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 (AP).
AP (Number of pulses per rotation) = 4194304 [pulse]
S (Movement amount per rotation)
= 135000.0 [m] Reduction ratio
= 135000.0 [m] 7/53 1/3
Substitute this for the above expression (1).
At this time, make calculation with the reduction ratio 7/53 1/3 remaining as a fraction.
Here, make calculation on the assumption that is equal to 3.141592654.
AL has a significant number to first decimal place, round down numbers to two decimal places.
Thus, AP, AL and AM to be set are as follows.
This setting will produce an error for the true machine value, but it cannot be helped.
This error is as follows.
AP (Number of pulses per rotation) = 4194304 [pulse]
S (Movement amount per rotation)
= 135000.0 [m] Reduction ratio
= 135000.0 [m] 7/53 1/3
It is equivalent to an about 86.9 [m] error in continuous 1 km feed.
Setting value Setting item
AP = 166723584 [Pr.2]
AL = 742201.2 [Pr.3]
AM = 1 [Pr.4]
M
Reduction ratio 7/53 Pulley ratio 1/3
Belt conveyor
φ135 mm
AP
AL × AM
=
4194304 [pulse]
135000.0 [μm] × π × 7/53 × 1/3
=
4194304 × 53 × 3
135000.0 × π × 7
=
166723584
236250 × π
AP
ΔS
=
AP
ΔS
=
AP
AL × AM
=
166723584
742201.2645075
AP
ΔS
=
AP
AL × AM
=
166723584
742201.2
=
166723584(AP)
742201.2(AL) × 1(AM)
7422012/166723584
2362500π/166723584
- 1
× 100 = -8.69 × 10
-6
[%]
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