Specifications

33
Motion Profiles
Triangular Motion Profile
In this motion profile the actuator is
linearly accelerated and decelerated to
provide the desired incremental move
time. All of the movement is either in
acceleration or deceleration with no
steady run speed.
Move distance (x) in time (t)
Distance x/2 is moved in time t/2
(accelerate)
Distance x/2 is moved in time t/2
(decelerate)
Example of Triangular Motion Profile
Given:
Move distance ( x ) = 12 inches
Time to move (t) = 8 seconds
Trapezoidal Motion Profile
In this type motion profile the actuator is
linearly accelerated to and decelerated
from the desired running speed to
provide the desired incremental move
time.
Example of Trapezoidal Motion Profile
Given: Move distance ( x ) = 18 inches
Time (t) = 8 seconds
Maximum acceleration rate = 2 inches /
second / second
Calculate:
Vavg =
x
/t =
12
/8 = 1.5 inches / second
V
max = 2 x Vavg = 2 x 1.5 = 3 inches /
second
t
acc = tdec = (
1
/2) t = 4 seconds
V
max 3 in/sec
a = ——— = ———— = .75 in/sec
2
tacc 4 sec
Therefore when programming set
actuator speed to 3 inches / second and
acceleration to .75 inches / second /
second
V
avg =
x
/t
Vmax = 2 Vavg = 2
x
/t
tacc = tdec =
1
/2t
V
max
a = ———
tacc
a = Acceleration (in/sec
2
)
x = Total distance (in)
t = Total move time (sec)
t
aac = Time to accelerate (sec)
t
dec = Time to decelerate (sec)
V
max = Maximum velocity (in/sec)
V
avg = Average velocity (in/sec)
a = Acceleration (in/sec
2
)
d = Deceleration (in/sec
2
)
x = Total distance (in)
t = Total move time (sec)
t
aac = Time to accelerate (sec)
t
dec = Time to decelerate (sec)
V
max = Maximum velocity (in/sec)
V
avg = Average velocity (in/sec)
t
c = Time at constant velocity
Calculate:
Vavg =
x
/ t =
18
/ 8 = 2.25 inches / second
Assume t
c = 0.75 x t = 0.75 x 8 = 6 seconds
t
acc = tdec =
(t - tc)
/2 =
(8-6)
/2 = 1
V
max = Vavg x t / (
1
/2 x tacc + tc +
1
/2 x tdec)
V
max = 2.25 x
8
/ (
1
/2 x 1 + 6 +
1
/2. 1) = 2.25 x
8
/ 7 = 2.57
V
max 2.57
a = ——— = ———— = 2.57 in/sec
2
tacc 1
Since this is greater than the maximum allowable acceleration rate of 2 inches /
second / second some parameters must be modified to reduce the acceleration. The
acceleration rate required can be reduced by increasing the time allowed for
acceleration. This is done by decreasing the time at constant speed. Doing this
increases the value of maximum speed (V
max).
Assume t
c = 5 seconds
then t
acc =
( 8- 5 )
/ 2 = 1.5 seconds
V
max = 2.25 x
8
/ (
1
/2 x 1.5 + 5 +
1
/2 x 1.5) = 2.25 x
8
/ 6.5 = 2.77 inches / second
V
max 2.77
a = ——— = ———— = 1.85 in/sec
2
tacc 1.5
Therefore when programming set actuator speed to 2.77 inches / second and
acceleration to 1.85 inches / second / second
V
max
V
avg
=
t
acc
Accelerate
(time in seconds)
t
t
dec
Decelerate
(time in seconds)
Velocity
(in/sec)
V
max
2
V
max
t
acc
Accelerate
(time in seconds)
t
t
dec
Decelerate
(time in seconds)
t
c
Constant velocity
(time in seconds)
Velocity
(in/sec)