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■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.
M
Reduction ratio 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.
AP
AP
=
=
135000.0 [μm] × π × 7/53 × 1/3
ΔS
AL × AM
4194304 × 53 × 3
=
135000.0 × π × 7
166723584
=
236250 × π
Here, make calculation on the assumption that is equal to 3.141592654.
AP
AP
=
=
AL × AM
742201.2645075
ΔS
AL has a significant number to first decimal place, round down numbers to two decimal places.
AP
AP
166723584
=
=
AL × AM
742201.2
ΔS
Thus, AP, AL and AM to be set are as follows.
Setting value
Setting item
AP = 166723584
[Pr.2]
AL = 742201.2
[Pr.3]
AM = 1
[Pr.4]
This setting will produce an error for the true machine value, but it cannot be helped.
This error is as follows.
7422012/166723584
2362500π/166723584
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.
8 CONTROL SUB FUNCTIONS
8.3 Functions for Compensating the Control
Belt conveyor
Pulley ratio 1/3
4194304 [pulse]
166723584
166723584(AP)
=
742201.2(AL) × 1(AM)
- 1
× 100 = -8.69 × 10
φ135 mm
-6
[%]
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Table of Contents
Troubleshooting
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