20 - PVD3627-August 2011
The power available for the deceleration is equal to:
Deceleration time t
2
“at constant Power” from
Maximal speed
to
Base speed
:
W
P
P
P
t
resis
mot
73776
10776
63000
tan
=
+
=
+
=
So the estimated
s
1.016
73776
²)
1079
²
2513
(
*
02910
.
0
*
5
.
0
²)
²
(
2
1
max
2
=
−
=
Ω
−
Ω
Σ
=
P
J
t
base
duration
t
2
is given by the formula:
s
1.503
016
.
1
487
.
0
2
1
=
+
=
+
=
Λ
t
t
t
Total deceleration time
∆
t
from
Max speed
to 0 will be given by the sum of t
1
and t
2
:
Method #2
t
2
is solution of a nonlinear first order differential equation.
M
resistant
≠
0
s
1.455
)
*
)
9421
(
0275
.
0
*
63000
(
6
02910
.
0
69
.
45
mod
_
)
*
*
(
mod
_
s
0.599
4
.
52
02910
.
0
*
1079
*
0.0275
)
6
63000
2513
(
*
9421
1
*
6
1
)
(
*
1
*
1
-45.69
6
)
9421
(
*
02910
.
0
*
-9421
6
63000
1079
/
2513
/
1079
4
.
52
6
4
.
58
6
63000
kgm²
0.02910
0275
.
0
*
6
2
*
2
0
1
1
0
max
1
0
tan
tan
=
−
−
+
−
=
−
+
=
=
=
=
=
−
−
−
=
−
−
=
=
−
=
=
=
−
=
−
=
=
Ω
=
=
Ω
=
=
−
=
=
−
=
=
=
=
=
=
+
=
Σ
=
−
−
e
t
e
spindle
on
accelerati
e
t
b
c
a
t
e
axis
on
accelerati
d
a
y
t
Ln
c
b
y
Ln
c
t
c
a
c
b
y
s
rad
y
s
rad
y
Nm
M
M
M
d
Nm
M
c
W
P
b
J
J
J
a
x
t
c
x
x
base
t
resis
mot
t
resis
mot
load
motor
λ
µ
λ
λ
µ
λ
Acceleration time t
2
“at constant Power” from
Base speed
to
Maximal speed
:
s
t
t
t
054
.
2
455
.
1
599
.
0
2
1
=
+
=
+
=
Λ
Total acceleration time
∆
t
from 0 to
Maximal speed
is given by the sum of t
1
and t
2
: