Page 62 - Electric Drives and Electromechanical Systems

Chapter 2 Analysing a drive system 55

If it is required that moves a load from q 1 to q 2 in a rotary system, and the speed of the
load at the start and finish of the motion are zero, it is possible to define the speed and
acceleration required as a function of time as,
3 2 2 3
qðtÞ¼ q 0 þ 2 ðq 2 q 1 Þt 3 ðq 2 q 1 Þt
t m t m
6 6
_ 2
qðtÞ¼ 2 ðq 2 q 1 Þt 3 ðq 2 q 1 Þt (2.37)
t t
m m
€ qðtÞ¼ 6 ðq 2 q 1 Þ 12 ðq 2 q 1 Þt
t 2 t 3
m m
where t m is the time required to complete the move. As in the case of the triangular and
trapezoidal profile, a polynomial profile can be applied to linear motions, in which case
Eq. (2.26) will be expressed as,
2
xðtÞ¼ a 0 þ a 1 t þ a 2 t þ a 3 t 3 (2.38)

nnn
Example 2.4
Determine the polynomial profile required for the robot joint move, where the joint is initially
at 15 degrees, and is required to move to a position of 75 degrees in 3 s, using the profile defined
by Eq. (2.37).
In making the single smooth motion, four constraints are evident, the initial and final
positions are known, q(0) ¼ 15 degrees and q(3) ¼ 75 degrees, as are the initial and final
velocities which are zero. On substitution the following coefficients are determined:
a 0 ¼ 15:00

a 1 ¼ 0:00

a 2 ¼ 20:00

a 3 ¼ 4:44
when substituted in Eq. (2.37), we obtain:
2
q ¼ 15:0 þ 20:0t 4:44t 3
_
q ¼ 40:0t 13:33t 2
€ q ¼ 40:0 26:66t
Fig. 2.11 shows the position, velocity and acceleration functions for the required profile. It
should be noted that the velocity profile of this movement where distance is specified as a
cubic polynomial is a parabola, and the acceleration is linear.
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