In the last video we learned to express the robot's joint force and torque limits as constraints
on the feasible accelerations s-double-dot along the path theta-of-s, as a function of
the state (s, s-dot).
In this video, we'll express those constraints graphically and gain some insight into the
time-optimal time-scaling problem.
Let's draw the plane (s, s-dot).
At the beginning of the motion, s and s-dot are both equal to zero.
At the end of the motion, s is equal to one and s-dot is equal to zero.
The start and end states are shown as dots.
Since we require the motion along the path to be monotonic, that is, the robot always
moves forward along the path, s-dot must always be positive.
So we only need to draw the top-right quadrant of the (s, s-dot) plane.
If the robot moves very slowly along the path, then the motion of the robot is essentially
along the s-axis from the start to the end.
A more typical time scaling might look something like this.
s-dot starts out at zero, increases to a maximum value, then drops back to zero at the end
of the motion.
Until now we have been expressing a time scaling as s as a function of time, but here we're
plotting it as s-dot as a function of s.
The same time scaling can be represented either way, but in this time-optimal problem it's
more convenient to express a time scaling as s-dot-of-s.
Now let's look at a particular state of the robot on its trajectory, indicated by this
point (s, s-dot).
We can draw the tangent vector to the time scaling, as shown here.
Now let's get rid of the time scaling so we can focus on this tangent vector.
The tangent vector consists of a horizontal component and a vertical component.
The horizontal component expresses the rate of change of s, so it is just s-dot, which
can be drawn as proportional to the height of the point along the s-dot axis.
The vertical component expresses the rate of change of s-dot, in other words, the acceleration
s-double-dot.
If we assumed that s-double-dot is always zero, then the tangent vectors at states (s,
s-dot) would look like this.
The horizontal component of a vector is determined by the s-dot value of the point.
Now let's focus on one particular tangent vector at the state (s, s-dot).
The horizontal component is s-dot.
But now let's assume that the vertical component, the acceleration s-double-dot, can be any
value in the range from L of (s, s-dot) to U of (s, s-dot), the range of feasible accelerations
according to the dynamics.
Summing these vertical vectors with the horizonal vector, we get the vectors shown here.
These vectors form a cone called the feasible motion cone.
At this state (s, s-dot), the tangent vector to the time scaling must be inside this cone
to satisfy the actuator limits.
Therefore, a time scaling like this would be OK at this state, as the tangent vector
lies inside the feasible motion cone.
If, instead, our feasible motion cone looked like this, the tangent vector is outside the
cone, and this time scaling is not possible according to the robot's actuator limits.
You could imagine drawing the motion cone at every point in the plane, and the problem
is to get from the start state to the goal state as quickly as possible while keeping
the tangent to the time scaling inside all motion cones along the curve.
Since we want to go as fast as possible, the time scaling should always be as high as possible,
and to find such a time scaling we could forward integrate the upper edge of the motion cones,
starting from the initial state.
Here you see a curve that is found by numerically integrating the maximum possible accelerations.
Because this curve travels along the edge of the motion cones, at least one actuator
is always operating at a limit.
This time scaling causes the robot to follow the path as fast as possible, but it does
not bring the robot to a stop at the end of the path, as required.
So now imagine backward integrating from the end state along the lower edge of the feasible
motion cones.
This integral curve intersects the other integral curve at a switch point s-star.
The time scaling you see here, represented by the two segments obtained by numerical
integration, is the time-optimal time scaling.
During the first segment, the robot maximally accelerates along the path, and during the
second segment the robot maximally decelerates along the path.
It is clear why this is time optimal: in the first segment, the robot cannot go any faster,
and in the second segment, if the speed s-dot were any higher at any given s, the robot
would not be able to come to a stop.
This time scaling keeps the speed s-dot as high as possible at all times, and therefore
the duration of the motion is as short as possible.
This kind of trajectory is called a "bang-bang" trajectory, because one or more of the actuators
"bangs" against a limit during the first segment, then one or more of the actuators "bangs"
against a limit during the second segment.
Compare this to a non-optimal trajectory, where the tangent to the time scaling is in
the interior of the motion cones, not on the edges.
The speed of the robot at any given position s is lower than what it is for the time-optimal
time scaling.
So this is the basic idea behind the time-scaling algorithm, except for one hitch, which I'll
describe now.
Let's plot the motion cone at a particular state (s, s-dot).
If you keep s constant but increase s-dot, you get a different motion cone.
If you increase s-dot further, then the motion cone may reduce to a single vector, where
the lower acceleration limit is the same as the upper acceleration limit.
If you increase s-dot further, then no motions are feasible, and this means the robot is
traveling too fast for the actuators to keep the robot on the path.
In general, we could plot a velocity limit curve: at states on this curve, only a single
acceleration is possible, and at states above this curve, the robot leaves the path immediately.
At states below the curve, there is a cone of possible tangent vectors.
Now, considering the existence of a speed limit, we might end up with a situation as
illustrated here: the maximum acceleration curve and the maximum deceleration curve do
not intersect, but instead run into the velocity limit curve.
Therefore, bang-bang control is not possible.
What to do in this case is the subject of the next, and final, video of Chapter 9.