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multiaxis motion - 16.15

16.3.1 Slew Motion

When the individual axis of a machine are not coordinated this is known as slew motion. Each of the axes will start moving at the same time, but finish at separate times. Consider the example in Figure 16.15. A three axis motion is required from the starting angles of (40, 80, -40)deg, and must end at (120, 0, 0)deg. The maximum absolute accelerations and decelerations are (50, 100, 150) degrees/sec/sec, and the maximum velocities are (20, 40, 50) degrees/sec.

Joint angle (degrees)

180

90

 

 

 

 

 

 

 

 

 

 

time(sec)

θ

3

θ

2

 

 

-90

 

θ

1

 

 

 

Joint velocity (degrees/sec)

ω

max

 

α

max

 

tacc

tmax

tdec

 

Figure 16.15 Multi-axis slew motion

The calculations for the motion parameters are shown in Figure 16.16. These are done in vector format for simplicity. All of the joints reach the maximum acceleration. The fastest motion is complete in 1.13s, while the longest motion takes 4.4s.

multiaxis motion - 16.16

The area under the velocity curve is the distance (angle in this case) travelled. First we can determine the distance covered during acceleration, and deceleration and the time during acceleration, and deceleration.

 

tacc = tdec =

ω

max

20

40

50

= ( 0.4, 0.4, 0.333) sec.

 

 

 

 

----------- =

-----, --------, --------

 

 

 

α

max

50

100

150

 

 

 

 

 

 

 

 

tacc ω

max.vel.

 

 

0.4( 20) 0.4( 40) 0.333( 50)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

θ acc. = θ dec. =

---------------------------

=

 

------------------, ------------------, ------------------------

 

=

( 4, 8, 8.33) deg.

 

 

 

 

 

 

2

 

2

2

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The next step is to examine the moves specified,

θ move = θ end θ start = ( 120 – 40, 0 – 80, 0 – ( –40) ) = ( 80, –80, 40) deg.

Remove the angles covered during accel./deccel., and find the travel time at maximum velocity.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

θ move

– 2θ acc

 

 

80 – 2( 4)

80 – 2( 8)

40 – 2( 8.333)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

tmax

=

 

---------------------------------- =

 

 

----------------------, ----------------------, ---------------------------------

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ω max

 

20

 

 

 

40

 

 

50

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

tmax

=

( 3.6, 1.6, 0.46668) sec.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ttotal

= tacc + tmax + tdec

= ( 4.4, 2.4, 1.13) s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 16.16 Calculated times for the slew motion

16.3.1.1 - Interpolated Motion

In interpolated motion the faster joints are slowed so that they finish in coordination with the slowest. This is essential in devices such as CNC milling machines. If this did not occur a straight line cut in the x-y plane would actually be two straight lines. The slew motion example can be extended to be slew motion where all joints finish their motion at 4.4s. This can be done by accelerating at the maximum acceleration, but setting a new maximum velocity. This is shown in the example in Figure 16.17 using the results from the example in Figure 16.16.