We have seen some trajectory planning techniques.
However, during the planning of these trajectories, we did not consider one thing which is really
important during trajectory planning.
We have to ensure that activity velocity and acceleration have to be in the limits of the maximal velocity
and acceleration that actuator can output.
Otherwise, actuators will saturate and throughput will not be able to execute the trajectory correctly.
Let's assume that we have planned the trajectory and we noticed that during execution of the trajectory
on realisable velocity and acceleration are required.
What we have to do, we will use kinematics scaling in this case by kinematics scaling.
We will obtain feasible trajectory that can be actually executed by the robot actuators in order to
do kinematic scaling on the trajectory.
It is better to write it in normalized form.
Let's assume that this is the given trajectory we have to execute with length of age and time lengths
of T. Let's ride normalized trajectory now, which is denoted by and uh, in the index.
It changes between zero and one and time lengths is also one.
It's very easy to obtain the trajectory from normalized trajectory.
Indeed, we can ride the given trajectory in this way with respect to the normalized trajectory.
Let's see the logic behind this structure.
This is normalized trajectory.
As you can see, it starts from zero and goes the configuration of one, while the time changes from
zero to one.
However, we know that the lengths of the given trajectory is not always one, but rather h.
So we multiply and normalized trajectory with age and get the desired length of the trajectory.
As you can see, while we get the desired trajectory, length time span stays the same.
Now we have to consider the initial condition of Q zero because trajectory not always starts from zero,
but from Q zero.
So we add Q zero also, OK, we are close to regaining our trajectory.
We have to consider just time span and initial time of two zero.
And these features are encoded inside Q and tilde.
Here is the Q and till the T, as you can see, it is equal to Q and toll where tall is given as below.
As you can see, by subtracting T zero from T v, consider initial condition, namely T zero mem is
slow the trajectory t zero ahead, then by dividing T minus T zero by T T v complete the obtained original
trajectory.
Here is the way we have gone so far.
We started from normalized trajectory and obtained the given trajectory.
OK, this is perfect because this will be very helpful for us.
Now let's try to find the derivative of the trajectory.
As you can see, the derivative of initial condition becomes zero as it is constant and we have to find
derivative of Q.
And till the T, I know that the formulation of Q and tilde t here again, we will use the chain rule
of derivative to find this differentiation by considering this rule.
We cannot differentiate in in this way.
Derivative of tar with respect to T is nothing but one over T.
You can see it easily from the formula of tongue.
In the same way you can find a second derivative of the Q, so we have velocity and acceleration in
our hands.
Let's call this as equation one from equation 1.0.
It's obvious that both velocity and acceleration will get their maximal values when Q and one tau four
and two and two to reach their maximum.
We can also determine at which time instant of toll we will get maximum velocity and acceleration from
equation one point zero.
You can easily observe that we can change maximal velocity and acceleration.
Output during execution of the trajectory can be adjusted by change of T.
Additionally, optimal trajectory in terms of time, namely the trajectory that's executed in minimum
time, can be determined by setting Q1 and Q2 or T to maximum permissible velocity and acceleration
limits and determined.
10.
OK.
Let's see it in the example.
Let's assume that we are given this quick trajectory in normalized form.
So it's zero in time, zero one in time one and velocity of the trajectory is zero at the beginning
and the end.
Is on these boundary conditions, we can ride the system of equations and determine coefficients for
the normalized trajectory after a more efficient have been determined, coefficients have been determined,
normalized trajectory can be constructed and velocity acceleration and other higher order derivatives
can be found if required.
Now let's determine the maximal values of velocity and acceleration for the maximum value of the velocity,
we have to determine the instant total we are.
QM one will be maximum.
We can find it by making its derivative equal to zero, which gives value of zero point five for thought.
If you plug this a volley in velocity formula, we will get this for the maximum value of the normalized
trajectory velocity.
From this, we can easily find the maximum velocity of the original trajectory.
This can be determined from Equation 1.0 zero.
We can do the same for acceleration also.
And here is the value of it.
Now, let's assume that the initial point of the trajectory is 10 and final point is 50, which gives
lengths of h or 40.
Maximum permissible velocity and acceleration values are given a certain age, respectively.
By plugging them into the above formulas we have obtained, we can determine timespan of D that will
help us to execute trajectory while abiding by limits.
As you can see, we have formed two different values for T.
We will surely find their maximum, namely two, because if you find their minimum, then while acceleration
will be in the limits velocity, what sets read which velocity would cause the mothers to saturate actuators
to saturate, which we don't want.
Indeed, you can do this exact calculation for even higher degrees of polynomials, and you will notice
that while the degree of polynomial increases name, smoothness of the trajectory increases.
This time Ti also increases.
So if you want to achieve smoother trajectory, you will have to execute it in larger time span.
Why?
Because your velocity acceleration becomes smoother.
So a robot cannot accelerate instantly, so or decelerate instantly.
So it takes more time to execute smoother trajectory.