Robotic Path Control Techniques 5sh3i

Path Control in Robotics ME 4135 Lecture series 8 Richard R. Lindeke, Ph. D.

Motion Types of Interest 

Point – to – Point Motion:  

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All Axes start and end simultaneously All Geometry is computed for targets and relevant t changes which are then forced to be followed during program execution

Path or Trajectory Controller Motion 

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Here the motion is performed through a time sequence of intermediate configurations computed ahead of time (like above but without stop-start operation) or in real time Paths are “Space Curves” for the n-Frame to follow This motion is a continuous scheme to move the T from one location to the next along a desired (straight or curved) line under direct operational control

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 Craig’s Method for acceleration smoothing Strict Velocity Control

t Interpolated Control Full Cartesian Control

Lead Through Path Creation 

Basically this was a technique whereby a skilled operator took a robot arm (for welding or painting) and used it like his/her weld tool or paint sprayer and performed the required process at reasonable speed



The robot is equipped with a position recording device and memorizes a large number of points during the teaching session These learned points then would be “played back” to replicate the skilled operators motions

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Lead Through Path Creation 

Advantages:   

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Simple way to create complex paths All points are sure to be physically attainable Playback speed can be controlled by an external device

Disadvantages: 

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Precision placements are required (program must be replayed at exactly the initial placement) Major concern with operator safety: robot is powered and operator is physically touching it (OSHA rules it unsafe practice!)

Modern Path Control: (Lets look at a simple example) 

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Dr. D’s new ‘Self-powered Automated Coffee Drinker’ Robot It is a simple cantilevered Cartesian device equipped with a spherical wrist that responds to eye movement and thoughts to help the overworked design engineer get coffee while deg and drafting & typing of Reports It follows a straight line path from the cup’s point on a table to the worker’s mouth in ½ second

Lets look at a simple example:

Lets look at a simple example: 

We see that the ‘Bot must travel a space path of 16.45” which can be decomposed into a movement of 9.5” along each of the prismatic ts

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For ‘accuracy’ lets divide each of the these t paths into 100 segments

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From Physics: Vt  d/ t = (9.5/100)/(.5/100) = 19in/second (a reasonable speed!)

Lets look at a simple example: 

During the 1st Step then:    



t 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 in 0.005 seconds Compute Accreqr  V/t Darn – this says that the acceleration is 3800in/sec2 – this is  10G!!!!!

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 the worker holding on to the powered robot – what should we do? I think our approach is too naïve! If we examine the Pos vs. Time, Vel Vs. Time and Acc vs. Time plots we may see why:

Look at a simple example’s Trajectory Curves:

This is Physically Impossible (or rather ‘very energy intensive’) 

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Can we build a reasonable solution that keeps the acceleration to an achievable level? What this would mean is we wouldn’t “instantly” – in one time step – go from stopped to full speed This can be achieved with a “time polynomial” model of motion

Building a ‘Path Polynomial’ Motion Set q  t   a0  a1t  a2t  a3t  L 2

3

dq 2 q& t    a1  2a2t  3a3t  L dt 2 d q & t   2  2a2  6a3t  L q& dt These are the ‘trajectory’ equations for a t (Position, Velocity and Acceleration)

Solving the ‘Path Polynomial’ is a matter of finding ai’s for SPECIFIC PATHS 

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We would have “boundary” conditions for position and velocity at both ends of the path We would have the desired total time of travel Using these conditions we can solve for a0, a1, a2 and a3 to build a 3rd order path polynomial for the required motion

Solving the ‘Path Polynomial’ is a matter of finding ai’s for specific paths

q0  a0  a1t0  a t  a t 2 2 0

3 3 0

q&0  a1  2a2t0  3a t

2 3 0

q f  a0  a1t f  a t  a t 2 2 f

3 3 f

q&f  a1  2a2t f  3a t

2 3 f

‘Poly’s’ holding at starting time and position

‘Poly’s’ holding at ending time and position

Solving the ‘Path Polynomial’ is a matter of finding ai’s for specific paths 

Writing these as Matrix Forms:

 1 t0  0 1   1 tf   0 1

2 0

t 2t0 2 tf 2t f

t  q0  a0   q&   a1 3t  0       q f  a2 t       & q a 3t    3  f 3 0 2 0 3 f 2 f

Solving the ‘Path Polynomial’ is a matter of finding ai’s for specific paths 

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

 1 0  0 1   1 tf   0 1

0 0 2 tf 2t f

0  a0    0 a    1 3 tf  a2   2 3t f  a3



     

 q0  q& 0 

qf   q&f

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

Solving the ‘Path Polynomial’ is a matter of finding ai’s for specific paths 

Completing the solution consists of forming relationships for: a2 & a3

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Done by substituting a0 & a1 values and solving the last two equation simultaneously:

Be Careful and note the order of the positions and velocities!

a2

3 q  

a3

2 q  

f

0

 q0   t f  2q&0  q&f 

 q f   t f  q&f  q&0 





t 2f

t 3f

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

Applying it to the Coffee ‘Bot 

Here (specifically):

qi  0  0ti  114t  (152)t 2 i

q&i  0  2(114)ti  3(152)t & q&  2(114)  6(  152) t i i

2 i

3 i

Applying it to the Coffee ‘Bot 

Simplifying:

qi  114t  152t 2 i

3 i

q&i  228ti  456t & q&  228  912 t i i

2 i

Applying it to the Coffee ‘Bot: Position Position vs. Time

Applying it to the Coffee ‘Bot: Velocity

t Velocity Vs. Time

Applying it to the Coffee ‘Bot: Acceleration

Acceleration Vs. Time

Applying it to the Coffee ‘Bot  

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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 cycle motor since throughout the entire path, the motor is accelerating (either with positive or negative orientation) Can we make a path solution where we accelerate Turns out we can and for only part of the path? we will call it LSPB!

Studying the LSPB model 

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In this model, we will carry forward with a ‘parabolic model’ In this model, we will determine a time where we will blend from startup until we reach a constant velocity – (and its greater than 1/100 of the total time!) Here we will see an acceleration followed by a period of “coasting” and then deceleration (often called a trapezoidal velocity model)

Model Building: 

We must define an acceleration constraint (minimum value) such that the acceleration guaranteed to be completed within half of the allocated time period of the travel: qB  q A  4  qB  q A   & & qmin   2 2 t t 2 based on solving: This assures that there is & &2 Pos  1 qt 2 no overlap for the ‘BLEND at: t  t (half time) 2

 

qB  q A   we want: Pos = 2

Regions’

Looking at the motion over the various regions: 

During Region 1 (while the t is Accelerating) (time interval 0 to tblend [tb]) the t moves: q = (V/2)*tb 

tb is the acceleration time

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During the region of ‘Constant Velocity’ the t moves: q = V*(t – 2tb)

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

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Total travel distance is qB - qA

Writing a motion equation: qB  q A  q0tb  qtb  t tb   q t tb  t qB  q A  V tb  V  t  2tb   V tb 2 2 qB  q A  Vt  2Vtb  Vtb  Vt  Vtb BUT : & &b V  qt

Substituting and Isolating the Unknown (blend time):

& & tb  qt & & qB  q A  qt

2 b

reforming:

& &  qt & & tb   qB  q A   0 qt 2 b

This is a quadratic equation in tb

Solving for tb: & &    qt  t 

& &  qt

b

tb  t  2

& &  qt

2

& q A  qB   4q&

& 2q& 2

& qB  q A   4q& & 2q&



Note: Acceleration is subject to above constraint

Applying it to the Coffee ‘Bot 

Acceleration constraint: gg 4   9.5  0  2 q  152 in / sec  0.4 g 2 .5 gg

lets ' pick ' q  175in / sec 2 

Blend time:tb  .25 

 .25 175

2

 4  175  9.5

2  175

1006.25  .25   .25  0.091 350  tb  .25  0.091  .159sec

Applying it to the Coffee ‘Bot 

tlinear=.5 – 2*0.159 = 0.181 s

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Linear Velocity:

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Positions:  By t , t moves: (27.89/2)*.159 = 2.222 in b  During linear velocity t moves: 0.181*27.89 = 5.055 in (thus the pos = 7.277”)  During deceleration t travels 2.222 in  Adding them gives full travel distance: 2.222 + 5.055 + 2.222 in  9.5in

& tb  175  .159  27.89 in/sec Vlin  q&

Plotting the Path trajectory:

Notice: the accelerator is ‘off’ during the linear travel segment

The 2 Previous Path Control Methods focused on Start/Stopping Approaches 

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

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Here we will focus on a method called ‘dog-tracking’ after the lead and follow techniques employed in dog racing

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Essentially we would have a situation where the path is laid out (as a series of Via Points) and the ts smoothly maneuver through and between them

Craig’s Dog Tracking Method

Craig’s Dog Tracking Method 

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Upon Examination of the motion, we find that there are three ‘regimes’ in the motion These are:   

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Start up regime Intermediate regimes Stopping regime

Starting and Stopping are similar to LSPB in the way they compute blend time and acceleration During an Intermediate regime we compute acceleration by comparing incoming and outgoing velocities about each point

Craig’s Dog Tracking Method 

Starting regime Equations:

& SGN      & & & 1 2 1 global t1  td 12  t

2 d 12

 2   2  1     & &  1 

 2  1 td 12  .5  t1 tl12  td 12  t1  .5  t2

& 12 

Start Acceleration

Start Blend Time

Linear Velocity 1→2 Time @ linear Velocity

Craig’s Dog Tracking Method Stopping Equation: & SGN      & & & n n 1 n global



tn  td  n1 n  t

&  n 1 n 

2 d  n 1 n

2   n   n1   & & n

  n   n1  td  n1 n  .5  tn

tl  n1 n  td  n1 n  tn  .5  tn1

Stop ‘acceleration’

Stop Blend Time

L. Velocity to stop

Time @ L. Velocity

Craig’s Dog Tracking Method 

Intermediate Equations:

  &  jk

k

 j 

tdjk

& SGN  &  &  & & & k kl jk global tk

&  &    kl

jk

& & k tljk  tdjk  .5  t j  .5  tk

Linear Velocity

Acceleration

Blend time

Time @ L. Velocity

Craig’s Dog Tracking Method 

Upon examination of the set of equation on the previous 3 slides several point should be noted: 

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 

Start and Stop are essentially the same but very important differences must be noted One can’t complete any of the regimes without looking ahead – actually looking ahead to the 2 nd point beyond to see if a t is stopping or continuing Start/Stop require position/time relationships Intermediate regimes require velocity/time relationships

Craig’s Dog Tracking Method 

Step 1: Calculate Global ‘usable’ acceleration (magnitude) constraint based on LSPB model applied Pairwise (1→2; 2→3; etc)

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Step 2: Focus on Start and Stop Segments

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Step 3: Complete the table of accelerations, blend times, linear velocity and time at linear velocity

4  q   & q& i j

t

2 dij

Lets Expand on Dr. D’s Coffee Drinker ‘Bot: Point Name

Position Delta (inch) Position (inch)

Arrival Time (sec)

Delta Time (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

Lets Expand on Dr. D’s Coffee Drinker ‘Bot: 

Step 1: Global Acc. Constraint 4  9.5 2 &AB  q&  152 in / s .52 4  1.5 2 & q&   0.67 in / s BC 32 4  1.5 2 & & qCD    .67 in / s 32 4  9.5 2 & q&   38 in / s DE 12

This is largest – should work globally – but lets make sure it doesn’t miss so choose 200ips2 ‘ cause it’s easier to calculate and is only about .6g

Lets Expand on Dr. D’s Coffee Drinker ‘Bot: 

Next we focus on the start & stop equations: q&&A  SGN (9.5  0)  200  200ips 2 

Starting

t A  .5  .52  2  9.5 q&AB  9.5

200

 .5  .5  .106 

 0.106 s

 21.3ips

2 & q& E  SGN  9.5  0  200  200ips



Stopping

t E  1  12 

 q&DE 

2   0  9.5 

0  9.5 

200

 1  .5  0.049 

 0.049 s

 9.74ips

NOTE: can’t compute tlij yet – we lack the data!

Lets Expand on Dr. D’s Coffee Drinker ‘Bot: 

Considering Intermediate B→C: qC  qB   

1.5 q&BC   0.5ips tdBC 3 g g   2 & q&  SGN q  q 200   200 ips  BC AB B   g  g  q  q  BC AB .5  21.25     tB    .103s & q&  200 B Now to finish the 1st segment:

tlAB  tdAB  t A  .5  t B  .5  .106  .5  .103  .342 s

Lets Expand on Dr. D’s Coffee Drinker ‘Bot: 

q&CD

On to “C→D” Segment

qD  qC   

tdCD

 0.5ips

2 & & & q&  SGN q  q 200   200 ips   C CD BC

tC

q&CD  q&BC   

1   .005 s & q& 200 C

back to finish B--C Segment

tlBC  tdBC  .5t B  .5tC  3  .5(.103)  .5(.005)  2.946 s

Lets Expand on Dr. D’s Coffee Drinker ‘Bot: 

Now for Segment D→E:

q&DE   earlier !

E (is stop point!)  9.74ips

2 & & & q&  SGN q  q 200  SGN  9.74  (  .5)  200   200 ips     D DE CD

tD

q&DE  q&CD   

 0.049 s & q& D

Completing Seg. C--D

tlCD  3  .5tC  .5t D  2.974s Seg D--E

tlDE  1  .5t D  t E  0.928s

Summarizing Pt

POS

T. Time POS

A

0

0

B C

9.5 11

.5 3.5

----9.5 1.5

time

Ti (blend )

----

.106

.5 3

.103 .005

Acc

L. Vel

Time @L. Vel

+20 0

21.3

.342

.5

2.94 6

-.5

2.97 4

-200 -200

D

9.5

6.5

-1.5

3

.046

-200 -9.74 .928

E

0

7.5

-9.5

1

.049

+20 0

A Final Thought on DogTracking: But if we must travel over a certain point, we can define ‘Pseudo-Via’ points that flank the desired target and force the arm to the ‘pseudos’ and drive right over the original desired target point

Looking at Velocity Control

Desired Path Achievable Path

With this “Velocity Control” 

The acceleration is set and the new velocity for an segment is inserted at the appropriate time (place)

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Over the segment, the arm (ts) lags the desired path

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We extend the next path velocity curve to intersect the actual achievable path (to a point that is earlier in time than when we would have expected to change) this will mean that the t can ‘catch up’ to the desired plan for the travel as it ‘blends’ to the new velocity