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SMD17E2 User Manual
CALCULATING MOVE PROFILES
51
S-Curve Acceleration Equations (continued)
Determining Waveforms by Values (continued)
Example 2, Jerk = 400
Because a
f
is greater than the programmed acceleration of 58,000 steps/sec
2
, the resulting acceleration is a
trapezoidal S-curve. As shown in figure
R3.7, two additional calculations must be made. The first is the time
(t
1
) it takes to jerk to the programmed acceleration value. The second is the time (t
2
) it takes to accelerate to
half of the required change in speed (S
m
).
The time for this acceleration phase is 2(t1 + t2), which equals 2(0.2500 sec + 0.1336 sec) or 0.7672 seconds.
Time spent in the constant acceleration period is (2(0.1336))/0.7672) or 34.8% of the entire phase.
S
m
30,000 steps/sec
2
--------------------------------------- 15,000 steps/sec== S
m
= midpoint of change in speed
J
100j
a
-----------
j
Ja
100
---------==
J Acceleration Jerk parameter=
j physical jerk property=
a
f
calculated final acceleration=
j
400 58,000 steps/sec²
100
-------------------------------------------------------=
j 232,000 steps/sec³=
Just as displacement
1
2
---
at
2
, speed
1
2
---
jt
2
==
15,000 steps/sec
232,000 steps/sec³ t
2
2
-----------------------------------------------------=
t
2
15,000 steps/sec
116,000 steps/sec³
--------------------------------------------=
t 0.3596 seconds=
Just as speed = at, acceleration = jt
a
f
232,000 steps/sec³ 0.3596 sec=
a
f
83,427 steps/sec²=
Figure R3.7 Calculating Trapezoidal S-Curve
Acceleration
a
Time
2
1
232,000 steps/sec³ t
1
58,000 steps/sec²= jt a=
t
1
0.25 seconds =
Determine speed at t
1
: Speed
1
2
---
jt
2
=
S
1
232,000 steps/sec³ 0.25
2
2
--------------------------------------------------------------=
S
1
7,250 steps/sec=
Determine remaining change in speed and required time
based on programmed acceleration
S
2
S
m
S–
1
15,000 7,250– steps/sec==
S
2
7,750 steps/sec=
S
2
a
c
t
2
t
2
S
2
a
c
==
t
2
7,750 steps/sec
58,000 steps/sec²
-----------------------------------------=
t
2
0.1336 seconds=