OK, after we have seen some preliminary knowledge related dynamics or a little organized dynamics,
let's jump into the derivation of Eyler Lagrange modeling of robotic manipulators.
As we have said before, choice of generalized coordinates during dynamic modeling is very important.
If we don't see a suitable set of generalized coordinates or dynamic equations can become unnecessarily
difficult to obtain.
And we don't want that because dynamic modeling of robotic manipulators is already a complex problem.
We have two choices for generalized coordinates either choose work space coordinates, namely X, Y
and Z, or choose joint angles Q as generalized coordinates.
Indeed, doing calculations with joint angles are easier because we will not use inverse or Jacobean
mattresses, which can cause problems during singularities.
Let's start with the formula of kinetic energy, which we have seen before.
However, as we have two joint angles as generalized coordinates, we need to express linear velocity
and angular velocity terms in terms of joint angles or joint velocities.
Excuse me.
And here we will use Jacoby mattresses in order to convert from joint space velocities to the linear
and angular velocities at work space.
Mainly, we will use velocity kinematics, as you can see from these formulas.
I hope you can remember these formulas because we have seen them during velocity kinematics tutorial
here.
PCI is nothing but the velocity of the center of mass of the link, i.e. here is the velocity and angular
velocity Jacobean.
If you analyze these mattresses, you will see they have interesting structure in this structure.
Take into account the coupling between velocities of the joints, and the structure is very logical
and easy to understand.
Indeed, as you can see, the Jacobins after the link is zero.
However, before the link, i.e. they are non-zero, let's look this example.
Let's assume that we want to find the linear and angular velocity Jacobins falling.
I then I +1 and I +2 are the next links and I am minus one is the previously.
The motion of next means will not affect the velocity of Link I.
You can observe this from your hand.
Do your fingers motion contribute to the velocity of your wrist?
No, because fingers are next things to your wrist.
How are your shoulder?
Is previous linked with respect to your wrist and its motion will contribute to the velocity of the
wrist.
The same logic is valid here.
OK, let's write the equations here again.
If we consider the four equations of VCI and Omega IV, you can plug them into the form of kinetic energy
and write them.
In this way, you can see that both of the terms contain pure dot and its transpose, so we can take
them out as a common multiplier and write the equation in this way.
And finally, let's call the middle matrix as M and write equation in much simpler way.
Does this equation remind you something?
I think so, because this is very similar to the formula of kinetic equation from basic physics, which
was one over two m v squared instead of linear velocity V if you have Q Dot here, which is also velocity
term.
So the middle term has to be must related symptom.
Indeed, your assumption is correct.
M fix is called inertia matrix and it is square symmetric and the positive definite it has to be positive
definite because must always be positive definite.
Additionally, its configuration dependent because, as you can see from the formula, it is function
of Q namely configuration.
Indeed, we can decompose it and write in this way.
This indeed shows us an interesting property of kinetic energy, which is unusual for us because from
the basic physics, we know that the kinetic energy is velocity dependent on it.
However, from this formulation, we can see that it is also configuration dependent as we have expressed
kinetic energy with joint velocities Q Dot.
Then we can find Lagrange Equation because for potential energy, we don't need to express it in terms
of joint angles, I mean, generalized Corness as it is all expressed in joint angles PC II, which
is this with the centre of mass of being IE, is function of just joint angles.
So here is the expression for Lagrangian function.
Here's the formulation for a little Lagrange form that we have seen before.
Let's calculate the first term.
As you can see, it is nothing but the partial derivative of like.
Engine function with respect, joint velocities, and you know that only the kinetic energy is a function
of joint velocities, not potential energy.
So potential.
So partial derivative of potential energy with respect to joint velocities will be zero.
And we will have only the some of the products of mass mattresses with joint velocities at the end.
Do you interested how this last term obtained?
Indeed, it is very easy to see that, but let's take the example of two degrees of freedom manipulator
and calculate partial derivative of kinetic energy with respect to joint velocities when K is equal
to one.
You can see here we have two variables, namely and one two and and two one.
Or as initial matrix is symmetric, they will be the same so we can sum them together and we will get
this final expression.
Now, after we have finished with partial derivative part, let's find its time derivative, which is
very easy.
Don't forget the rule of derivative of summation.
Here we have to find time derivative of inertia matrix also, because as we have said before, it is
configuration dependent, so changes with respect to time.
So it's time derivative will be found in this way.
Here we have used chain rule as inertia matrix, which is a function of configuration, which is also
it serves a function of time.
Here is the chain formula.
I'm sorry for putting it up because there was no free space.
Anyway, at the end of the at the end, the resulting formulation will be like that.
As we have finished with first term, we can continue with the second one.
Namely, partial derivative of Lagrangian function with respect to joint variables.
In order to calculate this term, we have to again write Lagrangian function in terms of kinetic and
potential energies.
Then we can individually find their partial derivatives with respect to joint variables.
The partial derivative of potential energy will be very easy, as it is only a function of configuration,
so we will get only gravity vector.
Now we can combine two terms.
We have calculated and obtained earlier Legrand's dynamic model of the remote manipulator.
If we called the middle term as H K.J IQ, we can ride the formula in simpler way.
As we have said before, C K is nothing but the conservative forces or some of the conservative forces,
meaning the joint actuator towards external or contact forces and joint friction forces.
Don't forget that there there are other forces also, but we cannot model each and every of them.
This is because, firstly, it is impossible to model every one of them in the exact manner.
Secondly, if we even can do that, modelling them will make the dynamic equation complex.
So it is enough for us for now to model just these key components of non-conservative forces.
So finally, we can write a little like large dynamic model of the robot manipulator in this way.
Try to compare this formula because we will need it almost in every future topics.
Before finishing the video, I want to just give short note about Aristotle's symbols.
Let us know the dynamic model here again now focused on this Si matrix, which is called carryall this
matrix and it can be decomposed in this way.
Here, C I J K is nothing but components of this matrix that can be calculated in this way.
This C IG K components are called crystal symbols.
You don't have to memorize them.
Even it would be possible to miss this detail, but I have mentioned it for one important thing that
we will use in future lessons.
We can just see Matrix not only by a combination of C IG K elements, but in other ways.
Also, as long as this equality is valid, then you can ask Why won't we choose see Matrix as combination
of Christoffer symbols?
In order to know this, let's assume that we have calculated the difference between derivative of inertia
matrix and two times Coriolis matrix.
The resultant matrix will be called KN know the important detail is here.
If we are to see as combination of Christoph Waltz symbols, then it will be skewed symmetric.
Namely, its symmetric elements will have different size and diagonal elements will be zero.
Then this equality will hold for any vector of X.
However, if we didn't choose Christoph, all symbols to construct the Coriolis matrix, then end matrix.
Will not be symmetric and previous equality will hold when X is vector of joint velocities only.
Now you can say, OK, so what?
Yes, it can be seen as useless detail for now, but just keep this property in mind and you will see
how useful is it in future lessons?
Well, one more thing.
Don't forget that during this course, we will always assume that sea matrix is constructed by Christopher
symbols so skewed symmetric property of any matrix will always hold.