robotic path control techniques

7/31/2019 Robotic Path Control Techniques

1/52

Path Control in Robotics

ME 4135

Lecture series 8

Richard R. Lindeke, Ph. D.


2/52


3/52

Path Control and Motion

Types: We will explore the following types of Motion:

Lead Through Path Creation

(Cubic) Polynomial Paths w/ Via Points Minimum Time Trajectory w/ controlled Acceleration

Lower order Path-Poly Control

LSPB Paths

Craigs Method for acceleration smoothing Strict Velocity Control

Joint Interpolated Control

Full Cartesian Control


4/52

Basically this was a technique whereby a skilled operator took arobot arm (for welding or painting) and used it like his/her weldtool or paint sprayer and performed the required process atreasonable speed

The robot is equipped with a position recording device andmemorizes a large number of points during the teaching session

These learned points then would be played back to replicatethe skilled operators motions

Lead Through Path Creation


5/52

Lead Through Path CreationAdvantages:

Simple way to create complex paths

All points are sure to be physically attainable Playback speed can be controlled by an

external device

Disadvantages: Precision placements are required (program

must be replayed at exactly the initialplacement)

Major concern with operator safety: robot ispowered and operator is physically touching it(OSHA rules it unsafe practice!)


6/52

Modern Path Control:

(Lets look at a simple example)

Dr. Ds new Self-powered Automated CoffeeDrinker Robot

It is a simple cantilevered Cartesian deviceequipped with a spherical wrist that responds toeye movement and thoughts to help theoverworked design engineer get coffee while

designing and drafting & typing of Reports It follows a straight line path from the cups

point on a table to the workers mouth in second


7/52

Lets look at a simple example:


8/52

Lets look at a simple example: We see that the Bot must travel a space path

of 16.45 which can be decomposed into amovement of 9.5 along each of the prismatic

joints

For accuracy lets divide each of the these

joint paths into 100 segments

From Physics: Vjointd/ t =(9.5/100)/(.5/100) = 19in/second (a

reasonable speed!)


9/52

Lets look at a simple example: During the 1st Step then:

Joint 1 starts at 0 and moves to 0.095

Moves there in 0.005 seconds How will it do it?

Of course by Accelerating from a stop to 19 in/sec in0.005 seconds

Compute AccreqrV/t Darn this says thatthe acceleration is3800in/sec2 this is 10G!!!!!


10/52

Lets look at a simple example: So this will certainly be difficult to accomplish!

(more likely it will not work)

OSHA would be just as upset as when we had theworker holding on to the powered robot whatshould we do?

I think our approach is too nave!

If we examine the Pos vs. Time, Vel Vs. Time andAcc vs. Time plots we may see why:


11/52

Look at a simple examples

Trajectory Curves:

0.005

Time

POS

Vel

Acc

0.495 0.5

19in/sec

9.5"

-10 g

10 g


12/52

This is Physically Impossible (or

rather very energy intensive) Can we build a reasonable solution that

keeps the acceleration to an achievable

level? What this would mean is we wouldntinstantly in one time step go from

stopped to full speed This can be achieved with a time

polynomial model of motion


13/52

Building a Path Polynomial

Motion Set

2 3

0 1 2 3

21 2 3

2

2 32

2 3

2 6

q t a a t a t a t

dqq t a a t a t dt

d q

q t a a t dt

These are the trajectory

equations for a joint (Position,

Velocity and Acceleration)


14/52

Solving the Path Polynomial is a matter of

finding ais for SPECIFIC PATHS

We would have boundary conditionsfor position and velocity at both ends of

the path We would have the desired total time of

travel

Using these conditions we can solve fora0, a1, a2 and a3 to build a 3

rd order pathpolynomial for the required motion


15/52

Solving the Path Polynomial is a

matter of finding ais for specific paths2 3

0 0 1 0 2 0 3 0

20 1 2 0 3 0

2 3

0 1 2 3

2

1 2 3

2 3

2 3

f f f f

f f f

q a a t a t a t

q a a t a t

q a a t a t a t

q a a t a t

Polys holding at

starting time and

position

Polys holding at

ending time andposition


16/52

Solving the Path Polynomial is amatter of finding a

is for specific paths

Writing these as Matrix Forms:

2 3

000 0 0

2010 0

2 3

2

2

3

10 1 2 3

10 1 2 3

ff f f

ff f

qat t tqat t

qat t tqat t


17/52

Solving the Path Polynomial is amatter of finding a

is for specific paths

If we set t0 = 0 (starting time is when westart counting motion!) then:

00

01

2 3

2

2

3

1 0 0 0

0 1 0 0

10 1 2 3

f f f f

f f f

qa

qa

t t t qat t qa

By examination, a0 = q0 & a1 = q0(dot)


18/52

Solving the Path Polynomial is a

matter of finding ais for specific paths

Completing the solution consists of forming

relationships for: a2 & a3 Done by substitutinga0 &a1 values and solving the

last two equation simultaneously:

0 022

0 0

33

3 2

2

f f f

f

f f f

f

q q t q qa

t

q q t q qa

t

Be Careful andnote the order ofthe positions andvelocities!


19/52

Applying it to the Coffee Bot

Start: X = 0; v = 0 @ time = 0

End: X = 9.5; v = 0 @ time = .5 sec a0 = 0 ; a1 = 0

a2 = (3 * 9.5)/(0.52) = 114

a3 = (2 *(- 9.5))/(0.53) = -152


20/52


Here (specifically):

2 3

2

0 0 114 ( 152)

0 2(114) 3( 152)

2(114) 6( 152)

i i i i

i i i

i i

q t t t

q t t

q t


21/52


Simplifying:

2 3

2

114 152

228 456228 912

i i i

i i i

i i

q t t

q t tq t


22/52

Applying it to the Coffee Bot:Position

0

1

2

3

4

5

6

7

8

9

10

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

time (sec)

Pos

ofJoint(in)

Position vs. Time


23/52

Applying it to the Coffee Bot:Velocity

0

5

10

15

20

25

30

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

Time (sec)

Velocity(in/sec)

Joint Velocity Vs. Time


24/52

Applying it to the Coffee Bot:Acceleration

-300

-200

-100

0

100

200

300

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

time (sec)

Acceleration

(in/sec^2)

Acceleration Vs. Time


25/52


Using the Path Polynomial Approach:

Max Velocity is: 28.5 in/sec (compares to 19 in/sec)

Max Acceleration is: 228 in/sec2 (.6 g) compared to 10 g

But, in this method, I require a 100% duty cyclemotor since throughout the entire path, the motor

is accelerating (either with positive or negativeorientation)

Can we make a path solution where we acceleratefor only part of the path? Turns out we can and

we will call it LSPB!


26/52

Studying the LSPB model

In this model, we will carry forward with aparabolic model

In this model, we will determine a time where wewill blend from startup until we reach a constantvelocity (and its greater than 1/100 of the totaltime!)

Here we will see an acceleration followed by aperiod of coasting and then deceleration (oftencalled a trapezoidal velocity model)


27/52

Model Building:

We must define an acceleration constraint (minimum value)such that the acceleration guaranteed to be completedwithin half of the allocated time period of the travel:

min 2 2

2

4

2

12

( )2

2

B A B A

B A

q q q qq

tt

Pos qt

tt half time

q qPos

based on solving:

at:

we want: =

This assures

that there isno overlap forthe BLEND

Regions


28/52

Looking at the motion over thevarious regions:

During Region 1 (while the joint is Accelerating)(time interval 0 to tblend [tb]) the Joint moves:

q = (V/2)*tb tb is the acceleration time

During the region of Constant Velocity the Jointmoves: q = V*(t 2tb)

During Region 3 -- while the joint is deceleratingthe joint moves: q = (V/2)*tb

Total travel distance is qB - qA


29/52

Writing a motion equation:

0

22 2

2

:

b b b bB A t t t t t t t

B A b b b

B A b b b

b

q q q q q

V Vq q t V t t t

q q Vt Vt Vt Vt Vt

BUT

V qt


30/52

Substituting and Isolating theUnknown (blend time):

2

20

B A b b

b b B A

q q qt t qt

qt qt t q q

reforming:

This is a quadratic equation in tb


31/52

Solving for tb:

2

2

4

2

42 2

A B

b

B A

b

qt qt q q qt

q

qt q q qttq

Note:Acceleration issubject to aboveconstraint


32/52


Acceleration constraint:

Blend time:

2

2

2

2

4 9.5 0152 / sec 0.4

.5

' ' 175 / sec

.25 175 4 175 9.5.25

2 175

1006.25.25 .25 0.091

350

.25 0.091 .159sec

b

b

q in g

lets pick q in

t

t


33/52


tlinear=.5 2*0.159 = 0.181 s

Linear Velocity:

Positions: By tb, Joint moves: (27.89/2)*.159 = 2.222 in During linear velocity joint moves: 0.181*27.89 =

5.055 in (thus the pos = 7.277)

During deceleration joint travels 2.222 inAdding them gives full travel distance: 2.222 +

5.055 + 2.222 in 9.5in

175 .159 27.89lin bV q t in/sec


34/52

Plotting the Path trajectory:

Notice: theaccelerator isoff during the

linear travel

segment

.159

Time

POS

Vel

27.89

in/sec

175

in/s^2

Acc

.361 0.5

9.5"

-175

in/s^2

0 in/s^2


35/52

The 2 Previous Path Control Methodsfocused on Start/Stopping Approaches

What can we do if we desire to travel by G12 methods continuing along a path w/o stopping?

Here we will focus on a method called dog-trackingafter the lead and follow techniques employed in dogracing

Essentially we would have a situation where the path islaid out (as a series of Via Points) and the jointssmoothly maneuver through and between them


36/52

Craigs Dog Tracking Method


37/52


Upon Examination of the motion, we find thatthere are three regimes in the motion

These are: Start up regime Intermediate regimes Stopping regime

Starting and Stopping are similar to LSPB inthe way they compute blend time andacceleration

During an Intermediate regime we computeacceleration by comparing incoming andoutgoing velocities about each point


38/52


Starting regime Equations:

1 2 1

2 2 11 12 12

1

2 112

12 1

12 12 1 2

2

.5

.5

global

d d

d

l d

SGN

t t t

t t

t t t t

Start Acceleration

Start BlendTime

Linear Velocity 12

Time @ linear Velocity


39/52


Stopping Equation:

1

2 1

1 1

1

1

1

11 1

2

.5

.5

n n n global

n n

n d n n d n n

n

n nn n

nd n n

n nl n n d n n

SGN

t t t

t t

t t t t

Stop acceleration

Stop BlendTime

L. Velocity to stop

Time @ L. Velocity


40/52


Intermediate Equations:

.5 .5

k j

jk

djk

k kl jk global

kl jk

k

k

ljk djk j k

t

SGN

t

t t t t

Linear Velocity

Acceleration

Blend time

Time @ L. Velocity


41/52

Upon examination of the set of equation onthe previous 3 slides several point should be

noted: Start and Stop are essentially the same but very

important differencesmust be noted

One cant complete any of the regimes without

looking ahead actually looking ahead to the 2ndpoint beyond to see if a joint is stopping or continuing

Start/Stop require position/time relationships

Intermediate regimes require velocity/time

relationships



42/52


Step 1: Calculate Global usableacceleration (magnitude) constraintbased on LSPB model applied

Pairwise (12; 23; etc)

Step 2: Focus on Start and StopSegments

Step 3: Complete the table ofaccelerations, blend times, linearvelocity and time at linear velocity

2

4i j

dij

qq

t


43/52

Lets Expand on Dr. Ds Coffee

Drinker Bot:PointName

Position(inch)

DeltaPosition

(inch)

ArrivalTime

(sec)

DeltaTime

(sec)A 0 0 --- ----

B 9.5 9.5 0.5 .5

C 11 1.5 3.5 3

D 9.5 -1.5 6.5 3

E 0 -9.5 7.5 1


44/52

Lets Expand on Dr. Ds CoffeeDrinker Bot:

Step 1: Global Acc. Constraint

2

2

2

2

2

2

2

2

4 9.5152 / s

.5

4 1.50.67 /

3

4 1.5.67 / 3

4 9.538 /

1

AB

BC

CD

DE

q in

q in s

q in s

q in s

This is largestshould work globally but lets make sureit doesnt miss so

choose 200ips2

cause its easier to

calculate and is onlyabout .6g


45/52


Next we focus on the start & stop equations:

Starting

Stopping

2

2

(9.5 0) 200 200

2 9.5.5 .5 0.106200

9.5 21.3.5 .5 .106

A

A

AB

q SGN ips

t s

q ips

2

2

9.5 0 200 200

2 0 9.51 1 0.049

200

0 9.59.74

1 .5 0.049

E

E

DE

q SGN ips

t s

q ips

NOTE:cant

compute tlijyet we

lack thedata!


46/52


Considering Intermediate BC:

2

1.50.5

3

200 200

.5 21.25.103

200

.5 .5 .106 .5 .103 .342

C B

BC dBC

B BC AB

BC AB

BB

lAB dAB A B

q qq ips

t

q SGN q q ips

q qt s

q

t t t t s

Now to finish the 1st segment:


47/52


On to CD Segment

2

0.5

200 200

1 .005200

.5 .5 3 .5(.103) .5(.005) 2.946

D C

CD dCD

C CD BC

CD BC

C C

lBC dBC B C

q qq ips

t

q SGN q q ips

q qt s

q

t t t t s

back to finish B--C Segment


48/52


Now for Segment DE:

!

2

9.74

200 9.74 ( .5) 200 200

0.049

3 .5 .5 2.974

1 .5 0.928

earlier

DE

D DE CD

DE CDD

D

lCD C D

lDE D E

ipsq E

q SGN q q SGN ips

q qt s

q

t t t s

t t t s

Completing Seg. C--D

Seg D--E

(is stop point!)


49/52

Summarizing

Pt POST.Time POS time

Ti(blend) Acc L. Vel

[email protected]

A 0 0 ----- ---- .106 +20021.3 .342

B 9.5 .5 9.5 .5 .103 -200.5 2.946

C 11 3.5 1.5 3 .005 -200-.5 2.974

D 9.5 6.5 -1.5 3 .046 -200-9.74 .928

E 0 7.5 -9.5 1 .049 +200


50/52

A Final Thought on Dog-Tracking:

But if we must travel

over a certain point,we can definePseudo-Via pointsthat flank the desiredtarget and force thearm to pass thepseudos and drive

right over the originaldesired target point


51/52

Looking at Velocity Control

AchievablePath

DesiredPath


52/52

With this Velocity Control

The acceleration is set and the new velocity for anupcoming segment is inserted at the appropriatetime (place)

Over the segment, the arm (joints) lags the desiredpath

We extend the next path velocity curve to intersectthe actual achievable path (to a point that is earlier

in time than when we would have expected tochange) this will mean that the joint can catch upto the desired plan for the travel as it blends to thenew velocity

robotic path control techniques

Documents