Instruction manual

App - 6
A
PPENDI
X
3.2 Selection example including regenerative
Mechanical conditions
Load mass
Feed speed
Efficiency of mechanical system
Feed distance
Friction
Feed time
1cycle time
: 20 kg
: 2m/s
: 0.9
: 1600 mm
: 0.01
: 1s
: 2s
(1) Temporary selection and calculation of load power
When the operation pattern is decided by setting the acceleration time as t1=0.2s, the constant speed time
will be t2=0.6s.
Acceleration: a=2[m/s]/0.2[s]=10[m/s
2
]
Select temporarily the linear servo motor LM-H2P2B-24M, which mass ratio is 9.6 times when the load
mass is 20kg.
Mass
Magnetic suction
Continuous thrust
Maximum thrust
: 2.5 kg
: 1900N
: 240N
: 600N
(2) Calculation of acceleration time and deceleration time thrust (Counting only friction)
Ff =
(M 9.8+1900) = 0.01 {(20+2.5) 9.8 + 1900} = 21.2 [N]
Fma = M
a + Ff = (20 + 2.5) 10m/s
2
+ 21.2 = 246.2 [N]
Fmd = –M
a + Ff = –(20 + 2.5) 10m/s
2
+ 21.2 = –203.8 [N]
(3) Calculation of continuous effective load thrust
Frms =
(Fma
2
t1
+Ff
2
t2
+ Fmd
2
t3
) / to
(246.2
2
0.2 + 21.2
2
0.6 + (-203.8)
2
0.2) / 2 = 101.7 [N]
=
Frms/
= 101.7/0.9 =113 [N]
As a result of this, the continuous thrust is less than 240N, and therefore the linear servo motor is available.
Frms/
= 246.2/0.9 =273.6 [N]
As a result of this, the maximum thrust is less than 600N, and therefore the linear servo motor is available.