On the last lesson, we have seen the properties of the terms contained in it, a little grunge formulation.
I have I hope that I was able to clarify them for you because it's very important to understand them
correctly.
Now in this lesson, we will try to solve a lot of the ground dynamics for a simple two degree of freedom
plan that a robot manipulator.
You can ask why planner and to a degree of freedom robot, because otherwise the example can get such
difficult that you can hardly imagine, not because of understanding, but because of bare hand calculations
anyway.
But this example will also encompass what we have learned, and you can apply this to a higher degree
of freedom robot manipulators.
So let's get started.
Here's our robot manipulator.
A sticker and frame is x-y.
Excuse me, coronal frame is stuck to the base frame, and two blue dots indicate central mass of two
links.
Here is some physical parameters, namely the distance to the central mass of the links link lengths
and the inertia of the links with respect to the frame attached their center of mass.
Be careful about with respect to which frame you are doing calculations to start first to be able to
calculate inertia matrix m because after we calculated Matrix M, we can calculate charismatic save
from it.
As you remember, VE got inertia Method X from kinetic energy.
We will not calculate kinetic energy because we know that inertia mattress corresponds to this part
of the kinetic energy formulation, so we will do our calculations directly.
Let's throw the figure again here to do further calculations.
First, as you can see from the blue box, we have to calculate linear velocity and angular velocity
jacobean mattresses.
Let's start with linear velocity Jacobean, as you can see from this formula.
Linear velocity Jacobean relates linear velocities of the first link with respect to x y z axis with
angular velocities of the joints.
Here, the linear velocities belong to the center of mass.
In order to get this Jacobean, we have to calculate linear velocities first as a function of angular
velocities, namely to 1.20 to 2.8 central mass of link one will follow circular trajectories while
joint moving and saw velocity vector will be tangent to this to this circle.
If we calculate this velocity, then we can find it's X, Y and Z components by projection because we
know rotation angles, namely Tatooine and T2.
Here is the calculation of tangential velocity of central mass of link one, which comes from basic
physics, namely angular velocity times radius.
In our case, radius is distance to central mass of lynk1.
Here is the projections of this velocity vector on X and Y or z axis velocity.
It will surely be zero as there is no motion in z axis.
We know that from basic geometry angle between, we see one and we see one x vectors.
View Beta one.
OK.
As we know, angle calculating projections is a piece of cake because we will use basic trigonometry
to find them.
Here are the equations for b c one x and we see one y.
We see one x will be a sign of theta one as it waits in front of the angle two to one.
Be careful.
Additionally, we have negative sun because it's opposite to the basis frame's x axis.
However, we see one Y component will be cosine of title one as it's connected to the one, and it will
not have negative sun as it's in the same direction as Baz's frame's y axis.
And as we have noted before, Z component will be zero.
Perfect.
We have velocity formulas as functions of joint angles from which we can construct Giacobbe Matrix very
easily.
Here is the linear velocity Jacobean of the first link.
OK, now let's calculate the linear velocity Jacobean for the central mass of the second link.
As you can see angle of the second joint within the respect.
The Bayes frames x axis is the sum of title one and title two, which is very obvious.
The relation will be as for the first link.
Now, let's think a bit about how we can calculate the velocity of the second linked central mass.
If the account with velocity of tip of the Lynk1, then we can find center of mass of the second link
by just using the relative velocities formulation.
Here is the velocity of the first things tip as it will follow circular passed.
The direction of the velocity will be tangent to the circle.
We can calculate this velocity as we did in center of mass of the lynk1.
Here is the formula.
In this case, just the radius is different.
Namely, it's now a one length of the lynk1.
Here is a circular trajectory of the second lensing center of mass, and it's tangential velocity vector
we can find.
We see two by relative velocity formation, which is given by the sum of V1 and angular velocity of
link two times distance to the center of mass.
Be careful Angular velocity of link tour will not be detached to DOT, but some of the wandered anteater
to dot because motion of joint want a fixed wing to also.
OK.
Now we have to find again projections of velocity vector.
We see two on X and Y axis, which is indicated in Fig..
The angle between these two and the C two Y will be some of the first and the second joint angles.
So we can find the projections using trigonometry.
If we project vessie to over x axis, we will get we see 2x as this because we see two x will be the
sum of projection of the one, our x axis and the projection of second term on x axis.
Be careful about signs which come from directions of the vectors we can find v one x vector because
it is the same as V.
Could we see one x on the previous page after we found the One X. We can plug it into the place and
do some manipulations.
We can do exactly the same for B C two y, which is some of the projections of B one and the second.
Term over y axis, and surely we see tools that will be zero as there is no motion on that axis as we
have velocity formulations, we can find Jacoby in Matrix four link to which is given as this Whoa,
this was a lot of calculations.
Now the concept becomes much, much simpler.
We have to calculate now the angle Jacobins for each link, which is very easy, especially for this
example, angular velocity of Link one will be affected only due to the motion of joint one.
And as the rotation is over z axis, it will have on the Z component, which is indicated within its
vector of K had
power linked to angular velocity will be also affected by the motion of joint one and joint two.
And as it also rotates on the opposite axis, it will have angular velocity on K had direction only.
OK.
They have angular velocities in our hands so we can get, lynk1, angular velocity Jacobean and a link
to angular velocity.
Jacobean is a right.
Another simple note in the equation of inertia matrix.
There was also our eye rotation matrix.
What about that?
It will be identity in our case.
Why?
Because the robot is planning the frames of central mass of lynk1 and linked to it will be in the same
direction as bases frame, so there will not be in the rotation.
Be careful.
Ah, I was the rotation matrix from frame in center of mass of the link to the Baz's frame because we
wrote inertia matrix with respect to the frame in center of mass.
In order to make it constant and one another, not here, we calculated Jacobins by bare hands, but
for higher degrees of freedom.
You had to use formulation that we have seen before in which we have calculated homogeneous transformation
mattresses.
Otherwise, you cannot do them by hand.
Here are the Jacobins we have calculated this is the formula of the inertia matrix again, as our robot
manipulator in this case is to a degree of freedom.
Inertia Matrix will have dimensions of two by two.
As shown here, with its elements given below, we have three components.
As the inertia matrix is always symmetric, I will not do the calculations.
I have done this calculations with butler script and you can do the same.
I will provide the Matlow script for you, but you can analyze, OK, we have found inertia matrix.
Now we have to find Coriolis mattress, which will be derived from Enoshima tricks.
So I have written the components of inertia matrix here again.
Diesel Compute Coriolis Matrix by Christopher Symbol's.
As we have said before, in order to get skewed, symmetric and matrix, then we can use them to obtain
C matrix components.
As you can see, a calculation of crystal ball symbols is nothing but partial derivatives of inertia
matrix components with respect to joint angles.
You can easily calculate them with MATLAB.
Here is the crystal ball symbols and in MATLAB file.
I have provided the calculation of these two.
After we have calculated Christoffer symbols, we can get Coriolis Matrix C Perfect.
We have done 90 percent of the work.
Now we have to calculate the potential energy of the robot in order to get gravity vector, which is
really very easy.
Here's the figure in which we will do calculations.
First, let's indicate gravity accelerations, which are gravitational explorations, which are opposite
to the y axis.
Let's start with the calculation of potential energy of link one, which is given by this lot that we
are familiar from, from basic physics.
Each one is the heart of the central mass of the link, one from Baz's frame, which we can obtain easily
by a right triangle and some trigonometry.
As we have found each one, we can plug it and obtain some formula for potential energy of lynk1.
Sure, the central for central mass.
Let's do the same for linked to a center of mass.
We can calculate the hate of lean to central mass from Baz's frame easily by just summing these two
individual hates me and H two, one and H two two, we can get these two heads from take our metric
relations again.
Be careful about angles now as we have.
Contrary to the hate speech to one and edge two to be can plug and obtain the potential energy for the
center of mass.
The second link total potential energy is nothing but the sum of these two potential energies.
Now we can easily calculate gravity vector one by just finding partial derivative of the total potential
energy with respect to Joint Angle one and Gravity Vector two by just finding partial derivative with
respect the second joint angle.
OK, we have calculated all the terms we need now.
It's remained to just plug them into this formulation and obtain joint talks to one and two.
And here it is.
We calculated like an initial formulation for two degrees of freedom plan and a robot manipulator.
You can see the amount of calculations, even in this simple case.
Now imagine what will happen for six degrees of freedom robot manipulators in three dimensions.
Surely you have to do it with MATLAB.