multiaxis motion - 16.2

motion controller

Motor

motion

commands

Setpoint

Setpoint

Servo

Generator

Scheduler

Drive

																																											setpoint schedule
								motion profile																																						t (s)							setpoint
													ω
						0.4																																					0								0.0
																															t												0.1								0.2
																																											0.2								0.4
																																											0.3								0.4
											0
													0.2												0.6					0.8													0.4								0.4
																																											.....								....

Figure 16.1 A motion controller

The combination of a motion controller, drive and actuator is called an axis. When there is more than one drive and actuator the system is said to have multiple axes. Complex motion control systems such as computer controlled milling machines (CNC) and robots have 3 to 6 axes which must be moved in coordination.

16.2 MOTION PROFILES

16.2.1 Velocity Profiles

A simple example of a velocity profile for a point-to-point motion is shown in Figure 16.2. In this example the motion starts at 20 deg and ends at 100 deg. (Note: in motion controllers it is more common to used encoder pulses, instead of degrees, for positions velocities, etc.) For position control we want a motion that has a velocity of zero at the start and end of the motion, and accelerates and decelerates smoothly.

multiaxis motion - 16.3

θ ( deg)
100
20	t (s)

Figure 16.2 An example of a desired motion (position)

A trapezoidal velocity profile is shown in Figure 16.3. The area under the curve is the total distance moved. The slope of the initial and final ramp are the maximum acceleration and deceleration. The top level of the trapezoid is the maximum velocity. Some controllers allow the user to use the acceleration and deceleration times instead of the maximum acceleration and deceleration. This profile gives a continuous acceleration, but there will be a jerk (third order derivative) at the four sharp corners.

	ω	( deg)
ω	max
		α			–α	max
			max
	0					t (s)
	0	tacc		tmax	tdec	ttotal
	where,		ω max = the maximum velocity
			α max = the maximum acceleration
		tacc, tdec = the acceleration and deceleration times
			tmax =	the times at the maximum velocity
			ttotal =	the total motion time

Figure 16.3 An example of a velocity profile

multiaxis motion - 16.4

The basic relationships for these variables are shown in Figure 16.4. The equations can be used to find the acceleration and deceleration times. These equations can also be used to find the time at the maximum velocity. If this time is negative it indicates that the axis will not reach the maximum velocity, and the acceleration and deceleration times must be decreased. The resulting velocity profile will be a triangle.

tacc

=

tdec

ω

max

(1)

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

α

max

(2)

ttotal

= tacc + tmax + tdec

θ =

1

ω

+ t

ω

1

ω

= ω

tacc

+ t

tdec

(3)

--t

acc

max

max

max

+ --t

dec

max

max

-------

max

+ -------

2

2

2

2

θ

tacc

tdec

(4)

tmax

=

----------- –

----------

–

----------

ω max

2

2

Note: if the time calculated in equation 4 is negative then the axis never reaches maximum velocity, and the velocity profile becomes a triangle.

Figure 16.4 Velocity profile basic relationships

For the example in Figure 16.5 the move starts at 100deg and ends at 20 deg. The acceleration and decelerations are completed in half a second. The system moves for 7.5 seconds at the maximum velocity.

multiaxis motion - 16.5

Given,

θ

start

= 100deg

θ end

= 20deg

deg

α max

deg

ω

= 10

= 20

--------

max

--------

s

s

2

The times can be calculated as,

deg

tacc = tdec

=

ω

max

=

10 s--------

= 0.5s

-----------α

max

20---------------deg

s2

θ

= θ

end – θ start = 20deg – 100deg = –80deg

tmax

=

θ

–

tacc

–

tdec

=

80deg

–

0.5s

–

0.5s

= 7.5s

-----------

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

---------

---------

ω max

2

2

deg

2

2

10--------

s

ttotal

= tacc + tmax + tdec = 0.5s + 7.5s + 0.5s

= 8.5s

Figure 16.5 Velocity profile example

The motion example in Figure 16.6 is so short the axis never reaches the maximum velocity. This is made obvious by the negative time at maximum velocity. In place of this the acceleration and deceleration times can be calculated by using the basic acceleration position relationship. The result in this example is a motion that accelerates for 0.316s and then decelerates for the same time.

multiaxis motion - 16.6

Given,

θ start

= 20deg

θ end

= 22deg

deg

α max

deg

ω

= 10

= 20

--------

max

--------

s

s

2

The times can be calculated as,

deg

tacc = tdec

=

ω

max

=

10 s

= 0.5s

-----------α

max

20---------------deg

s2

θ

= θ

end – θ start = 22deg – 20deg = 2deg

tmax =

θ

–

tacc

–

tdec

=

2deg

–

0.5s

–

0.5s

= –0.3s

-----------

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

---------

---------

ω max

2

2

deg

2

2

10--------

s

The time was negative so the acceleration and deceleration times become,

θ

1

α

2

-- =

--

t

2

2

max

acc

θ

2deg

2

tacc

=

=

= 0.1s

= 0.316s

-----------α max

---------------20deg

tmax

= 0s

s2

Figure 16.6 Velocity profile example without reaching maximum velocity

Given the parameters calculated for the motion, the setpoints for motion can be calculated with the equations in Figure 16.7.

multiaxis motion - 16.7

Assuming the motion starts at 0s,

0s ≤

t <

tacc

1

2

θ ( t)

α

+ θ

=

--

max

t

start

2

tacc ≤

t <

tacc + tmax

1

2

θ ( t)

=

α

t

+ ω

( t – t ) +θ

--

2

max

acc

max

acc

start

tacc + tmax ≤ t < tacc + tmax + tdec

1

2

+ ω

1

( t – t

)

2

+ θ

θ ( t)

=

--α

t

t

+ --

α

– t

tacc + tmax + tdec ≤

t

2

max

acc

max

max

2

max

max

acc

start

θ ( t) = θ end

Figure 16.7 Generating points given motion parameters

A subroutine that implements these is shown in Figure 16.8. In this subroutine the time is looped with fixed time steps. The position setpoint values are put into the setpoint array, which will then be used elsewhere to guide the mechanism.

multiaxis motion - 16.8

void generate_setpoint_table(

double t_acc, double t_max, double t_step, double vel_max, double acc_max,

double theta_start, double theta_end, double setpoint[], int *count){

double t, t_1, t_2, t_total; t_1 = t_acc;

t_2 = t_acc + t_max;

t_total = t_acc + t_max + t_acc; *count = 0;

for(t = 0.0; t <= t_total; t += t_step){ if( t < t_1){

setpoint[*count] = 0.5*acc_max*t*t + theta_start; } else if ( (t >= t_1) && (t < t_2)){

setpoint[*count] = 0.5*acc_max*t_acc*t_acc + vel_max*(t - t_1) + theta_start;

} else if ( (t >= t_2) && (t < t_total)){

setpoint[*count] = 0.5*acc_max*t_acc*t_acc + vel_max*(t_max)

+ 0.5*acc_max*(t-t_2)*(t-t_2) + theta_start;

} else {

setpoint[*count] = theta_end;

}

*count++;

}

setpoint[*count] = theta_end; *count++;

}

Figure 16.8 Subroutine for calculating motion setpoints

In some cases the jerk should be minimized. This can be achieved by replacing the acceleration ramps with a smooth polynomial, as shown in Figure 16.9. In this case two quadratic polynomials will be used for the acceleration, and another two for the deceleration.

multiaxis motion - 16.9

ω ( deg)	ω ( t) = At2	+ Bt + C

ω max

α max					–α max

t (s)

0

0

tacc

tmax

tdec

ttotal

where,

ω

max = the maximum velocity

α max = the maximum acceleration

tacc, tdec = the acceleration and deceleration times

tmax = the times at the maximum velocity ttotal = the total motion time

Figure 16.9 A smooth velocity profile

An example of calculating the polynomial coefficients is given in Figure 16.10. The curve found is for the first half of the acceleration. It can then be used for the three other required curves.

multiaxis motion - 16.10

Given,		θ start	θ end					ω max
The constraints for the polynomial are,
ω ( 0) = 0			ω	tacc		=	ω	max
				2				2
				--------			-----------
d	ω	( 0) = 0	d		tacc		=	α max
----			---- ω		-------
dt			dt		2

These can be used to calculate the polynomial coefficients,

0 = A02 + B0 + C

0 = 2A0 + B

ω				2				ω max
	max	= Atacc					A =	-----------
								2
								tacc
α		= 2Atacc					A =	α max
	max							-----------
								2tacc
	ω	max	=	α	max
A = -----------		2		2-----------tacc
	tacc
	α	max	=		α max		α	max2
A = -----------				------------------------	2ω	max	= ---------------
	2tacc			2			4ω	max
					---------------α	max

The equation for the first segment is,

α max

C = 0

B = 0

tacc =	2ω	max
	---------------
	α	max

ω ( t)		α		max2	2	0 ≤ t <	tacc
	=	---------------					-------
		4	ω	max t			2

The equation for the second segment can be found using the first segment,

ω ( t) = ω

ω ( t) = ω

	α		max2			2
max	– ---------------( tacc – t)
	4	ω	max
	α		max2	2		2
max	– ---------------( t				– 2tacct + tacc)
	4	ω	max

tacc	≤ t < tacc
-------
2

Figure 16.10 A smooth velocity profile example