Automating Manufacturing Systems - Process Control and ...

continuous actuators - 24.26
24.7 PRACTICE PROBLEM SOLUTIONS
1.
⎛ ⎞
⎝ ⎠
⎛ ⎞
⎝ ⎠
a) P = 0.125 ------
in R = 0.025---------
in
rot
step
θ =
0.025---------
in
R step
-- = --------------------------
P
0.125 ------
in
rot
= 0.2---------
rot
step
Thus
b) θ = 0.2--------- rot = 72---------
deg
step step
1
------------------
0.2---------
rot = 5 step ---------
rot
step
c)
PPS
3 in ---- s
= ------------------------
0.025---------
in = 120 steps ------------
s
step
2.
a)
pulses ( 3rot) = 5 step ⎞ ⎛ --------- = 15steps
rot ⎝ ⎠
b) pulses
--------------- 25--------
rot ⎛ ⎛ step ⎞ = 5 --------- = 125 steps ------------ = 125 1min ------------ ⎞steps
------------ = 2.08 step
⎛
---------
⎞
s min rot min 60s min s
⎝
⎝ ⎠
⎠
⎝
⎠
3.
stepper motor
vfd
dc motor
speed
very low speeds
limited speed range
wide range
torque
low torque
good at rated speed
decreases at higher speeds
24.8 ASSIGNMENT PROBLEMS
1. A stepper motor is to be used to actuate one joint of a robot arm in a light duty pick and place
application. The step angle of the motor is 10 degrees. For each pulse received from the pulse
train source the motor rotates through a distance of one step angle.
a) What is the resolution of the stepper motor?
b) Relate this value to the definitions of control resolution, spatial resolution, and
accuracy, as discussed in class.
c) For the stepper motor, a pulse train is to be generated by a motion controller.