2.SELECTION METHODS CONFIGURATIONS AND SELECTION B-65382EN/02
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2.1.3 Root Mean Square Force
The root mean square force means the root mean value of the force
required in one duty cycle. The root mean square force must not be
greater than the rated continuous force of the motor. If the root mean
square force exceeds the rated continuous force, the motor may
overheat. It is desirable to allow a margin of about 20% by taking a
load variation into account.
The rated continuous force of the motor varies depending on the type
of cooling method used for the motor (no cooling, air cooling, or
water cooling). It also varies depending on the heat dissipation and
other characteristics of the machine. Be extremely careful of force for
supporting frictional load and weight along a vertical axis because it
sometimes fluctuates largely. Even when the motor is at rest, it keeps
producing force to prevent drifting.
An example of calculation is shown below.
For explanation, a simple duty cycle as shown below is assumed.
After 0.2-second positioning for a 150-mm stroke, cutting is
performed for 5 seconds in the constant feed mode. It is assumed that
a cutting load (cutting reaction force) of 1,000 N is applied in the
direction of thrust during cutting. After cutting, the machine stops for
0.5 seconds. Other conditions are the same as those listed in
Subsections 2.1.1 and 2.1.2.
0.1s 0.1s 0.5s5.0s
1 cycle (5.8 seconds)
Rate [m/s]
Time
[seconds]
Cutting in the constant feed mode, cutting
load: 1,000 N
The machine stops.
0.05s
0.05s
Positioning in the rapid traverse mode
According to the above conditions, create a force distribution chart for
one cycle. The force distribution chart is made as follows:
5
5
,
,
5
5
1
1
5
5
N
N
d
d
u
u
r
r
i
i
n
n
g
g
a
a
c
c
c
c
e
e
l
l
e
e
a
a
t
t
i
i
o
o
n
n
4
4
,
,
3
3
0
0
5
5
N
N
d
d
u
u
r
r
i
i
n
n
g
g
d
d
e
e
c
c
e
e
l
l
e
e
a
a
t
t
i
i
o
o
n
n
6
6
0
0
5
5
N
N
d
d
u
u
r
r
i
i
n
n
g
g
s
s
t
t
o
o
p
p
1 cycle (5.8 seconds)
0.1s 0.1s 0.5s
1
1
,
,
6
6
0
0
5
5
N
N
d
d
u
u
r
i
i
n
n
g
g
c
c
u
u
t
t
t
t
i
i
n
n
g
g
a
a
t
t
a
a
c
c
o
o
n
n
s
s
t
t
a
a
n
n
t
t
f
f
e
e
e
e
d
d
r
r
a
a
t
t
e
e
Force [N]
Time
[seconds]
5
5
,
,
5
5
1
1
5
5
N
N
d
d
u
u
r
r
i
i
n
n
g
g
a
a
c
c
c
c
e
e
l
l
e
e
r
r
a
a
t
t
i
i
o
o
n
n
4
4
,
,
3
3
0
0
5
5
N
N
d
d
u
u
r
r
i
i
n
n
g
g
d
d
e
e
c
c
e
e
l
l
e
e
r
r
a
a
t
t
i
i
o
o
n
n
5.0s
0.05s
0.05s
5,515 N during acceleration is the required maximum force calculated
in Subsection 2.1.2. 4,305 N during deceleration is obtained as
follows:
Force required during deceleration=(300+34)×9.8×1.5-605
4305[N]
because the load force contributes to deceleration.