Harmonic Drive FHA Series, Manual, Page: 32 / 51

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2-4-6 Effective torque and average speed
Additionally to the former studies, the effective torque and the average speed should be studied.
(1) The effective torque should be less than allowable continuous torque specified by the driver.
(2) The average speed should be less than allowable continuous speed of the actuator.
Calculate the effective torque and the average speed of an operating cycle as shown in the former figure.
Tm: effective torque (N•m)
Ta: maximum torque (N•m )
Tm =
Ta2 x (ta + td) + Tr2 x tn
Tr: load torque (N•m)
√
t ta:
acceleration time (s)
td: deceleration time (s)
tr: running time at constant speed (s)
Nav =
N / 2 x ta + N x tr + N / 2 x td
t: time for one duty cycle (s)
t Nav:
average speed (r/min)
N: driving speed (r/min)
If the result is greater than the value in the table below, calculate once again after reducing the duty cycle.
Model FHA-17C FHA-25C FHA-32C FHA-40C
Items 50 100 160 50 100 160 50 100 160 50 100 160
Reduction Ratio 1:50 1:100 1:160 1:50 1:100 1:160 :50 1:100 1:160 :50 1:100 1:160
Continuous torque N•m 15 24 24 35 75 85 60 130 200 85 190 300
Continuous speed r/min 70 35 22 70 35 22 60 30 19 50 25 16
•
Example 3: getting effective torque and average speed
The parameters are the same as the example 1 and 2 for an FHA-25C-50.
(1) Effective torque
From the parameters of T
a
=T
d
=150 N
•
m,Tr =0 N
•
m, t
a
=0.097 s, tr=0.243 s, t
d
=0.085 s, t=2 s,
Tm =
150
2
x (0.097+0.085)
= 45 N
•
m
√
2.0
As the value of Tm (45N
•
m) exceeds its allowable continuous torque (35N
•
m), it is impossible to drive the actuator
continuously on the duty cycle. The following equation is introduced by converting the equation for effective torque.
The limited time for one duty cycle can be obtained by substituting the continuous torque for the Tm of the
following equation.
T =
Ta
2
x (ta + td) + Tr
2 x
Tr
Tm
2
Substituting 150 N
•
m for Ta, 150 N
•
m for Td, 0 N
•
m for Tr, 35 N
•
m Tm, 0.097 s for ta,
0.243 s for tr, and 0.085 s for td:
T =
150
2
x (0.097+0.085)
= 3.34
35
2
Namely, when the time for one duty cycle is set more than 3.4s, the effective torque [Tm] becomes less
than 34.0 N
•
m, and the actuator can drive the load with lower torque than the continuous torque continuously.
(2) Average speed
From the parameters of N =60 r/min, ta=0.097 s, tr=0.243 s, td=0.085 s, t=3.5 s,
Nav =
60/2 x 0.097+60 x 0.242+60/2 x 0.085)
= 5.88 r/min
3.4
As the speed is less than the continuous speed of the FHA-25C-50,
it is possible to drive it continuously on new duty cycle.
Chapter 2 Guidelines for sizing