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On the last lesson, we have seen the properties of the terms contained in it, a little grunge formulation.

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I have I hope that I was able to clarify them for you because it's very important to understand them

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correctly.

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Now in this lesson, we will try to solve a lot of the ground dynamics for a simple two degree of freedom

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plan that a robot manipulator.

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You can ask why planner and to a degree of freedom robot, because otherwise the example can get such

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difficult that you can hardly imagine, not because of understanding, but because of bare hand calculations

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anyway.

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But this example will also encompass what we have learned, and you can apply this to a higher degree

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of freedom robot manipulators.

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So let's get started.

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Here's our robot manipulator.

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A sticker and frame is x-y.

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Excuse me, coronal frame is stuck to the base frame, and two blue dots indicate central mass of two

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links.

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Here is some physical parameters, namely the distance to the central mass of the links link lengths

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and the inertia of the links with respect to the frame attached their center of mass.

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Be careful about with respect to which frame you are doing calculations to start first to be able to

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calculate inertia matrix m because after we calculated Matrix M, we can calculate charismatic save

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from it.

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As you remember, VE got inertia Method X from kinetic energy.

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We will not calculate kinetic energy because we know that inertia mattress corresponds to this part

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of the kinetic energy formulation, so we will do our calculations directly.

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Let's throw the figure again here to do further calculations.

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First, as you can see from the blue box, we have to calculate linear velocity and angular velocity

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jacobean mattresses.

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Let's start with linear velocity Jacobean, as you can see from this formula.

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Linear velocity Jacobean relates linear velocities of the first link with respect to x y z axis with

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angular velocities of the joints.

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Here, the linear velocities belong to the center of mass.

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In order to get this Jacobean, we have to calculate linear velocities first as a function of angular

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velocities, namely to 1.20 to 2.8 central mass of link one will follow circular trajectories while

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joint moving and saw velocity vector will be tangent to this to this circle.

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If we calculate this velocity, then we can find it's X, Y and Z components by projection because we

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know rotation angles, namely Tatooine and T2.

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Here is the calculation of tangential velocity of central mass of link one, which comes from basic

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physics, namely angular velocity times radius.

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In our case, radius is distance to central mass of lynk1.

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Here is the projections of this velocity vector on X and Y or z axis velocity.

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It will surely be zero as there is no motion in z axis.

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We know that from basic geometry angle between, we see one and we see one x vectors.

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View Beta one.

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OK.

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As we know, angle calculating projections is a piece of cake because we will use basic trigonometry

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to find them.

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Here are the equations for b c one x and we see one y.

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We see one x will be a sign of theta one as it waits in front of the angle two to one.

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Be careful.

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Additionally, we have negative sun because it's opposite to the basis frame's x axis.

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However, we see one Y component will be cosine of title one as it's connected to the one, and it will

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not have negative sun as it's in the same direction as Baz's frame's y axis.

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And as we have noted before, Z component will be zero.

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Perfect.

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We have velocity formulas as functions of joint angles from which we can construct Giacobbe Matrix very

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easily.

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Here is the linear velocity Jacobean of the first link.

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OK, now let's calculate the linear velocity Jacobean for the central mass of the second link.

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As you can see angle of the second joint within the respect.

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The Bayes frames x axis is the sum of title one and title two, which is very obvious.

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The relation will be as for the first link.

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Now, let's think a bit about how we can calculate the velocity of the second linked central mass.

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If the account with velocity of tip of the Lynk1, then we can find center of mass of the second link

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by just using the relative velocities formulation.

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Here is the velocity of the first things tip as it will follow circular passed.

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The direction of the velocity will be tangent to the circle.

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We can calculate this velocity as we did in center of mass of the lynk1.

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Here is the formula.

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In this case, just the radius is different.

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Namely, it's now a one length of the lynk1.

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Here is a circular trajectory of the second lensing center of mass, and it's tangential velocity vector

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we can find.

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We see two by relative velocity formation, which is given by the sum of V1 and angular velocity of

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link two times distance to the center of mass.

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Be careful Angular velocity of link tour will not be detached to DOT, but some of the wandered anteater

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to dot because motion of joint want a fixed wing to also.

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OK.

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Now we have to find again projections of velocity vector.

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We see two on X and Y axis, which is indicated in Fig..

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The angle between these two and the C two Y will be some of the first and the second joint angles.

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So we can find the projections using trigonometry.

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If we project vessie to over x axis, we will get we see 2x as this because we see two x will be the

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sum of projection of the one, our x axis and the projection of second term on x axis.

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Be careful about signs which come from directions of the vectors we can find v one x vector because

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it is the same as V.

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Could we see one x on the previous page after we found the One X. We can plug it into the place and

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do some manipulations.

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We can do exactly the same for B C two y, which is some of the projections of B one and the second.

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Term over y axis, and surely we see tools that will be zero as there is no motion on that axis as we

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have velocity formulations, we can find Jacoby in Matrix four link to which is given as this Whoa,

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this was a lot of calculations.

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Now the concept becomes much, much simpler.

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We have to calculate now the angle Jacobins for each link, which is very easy, especially for this

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example, angular velocity of Link one will be affected only due to the motion of joint one.

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And as the rotation is over z axis, it will have on the Z component, which is indicated within its

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vector of K had

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power linked to angular velocity will be also affected by the motion of joint one and joint two.

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And as it also rotates on the opposite axis, it will have angular velocity on K had direction only.

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OK.

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They have angular velocities in our hands so we can get, lynk1, angular velocity Jacobean and a link

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to angular velocity.

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Jacobean is a right.

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Another simple note in the equation of inertia matrix.

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There was also our eye rotation matrix.

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What about that?

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It will be identity in our case.

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Why?

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Because the robot is planning the frames of central mass of lynk1 and linked to it will be in the same

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direction as bases frame, so there will not be in the rotation.

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Be careful.

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Ah, I was the rotation matrix from frame in center of mass of the link to the Baz's frame because we

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wrote inertia matrix with respect to the frame in center of mass.

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In order to make it constant and one another, not here, we calculated Jacobins by bare hands, but

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for higher degrees of freedom.

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You had to use formulation that we have seen before in which we have calculated homogeneous transformation

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mattresses.

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Otherwise, you cannot do them by hand.

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Here are the Jacobins we have calculated this is the formula of the inertia matrix again, as our robot

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manipulator in this case is to a degree of freedom.

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Inertia Matrix will have dimensions of two by two.

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As shown here, with its elements given below, we have three components.

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As the inertia matrix is always symmetric, I will not do the calculations.

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I have done this calculations with butler script and you can do the same.

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I will provide the Matlow script for you, but you can analyze, OK, we have found inertia matrix.

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Now we have to find Coriolis mattress, which will be derived from Enoshima tricks.

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So I have written the components of inertia matrix here again.

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Diesel Compute Coriolis Matrix by Christopher Symbol's.

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As we have said before, in order to get skewed, symmetric and matrix, then we can use them to obtain

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C matrix components.

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As you can see, a calculation of crystal ball symbols is nothing but partial derivatives of inertia

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matrix components with respect to joint angles.

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You can easily calculate them with MATLAB.

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Here is the crystal ball symbols and in MATLAB file.

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I have provided the calculation of these two.

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After we have calculated Christoffer symbols, we can get Coriolis Matrix C Perfect.

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We have done 90 percent of the work.

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Now we have to calculate the potential energy of the robot in order to get gravity vector, which is

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really very easy.

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Here's the figure in which we will do calculations.

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First, let's indicate gravity accelerations, which are gravitational explorations, which are opposite

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to the y axis.

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Let's start with the calculation of potential energy of link one, which is given by this lot that we

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are familiar from, from basic physics.

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Each one is the heart of the central mass of the link, one from Baz's frame, which we can obtain easily

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by a right triangle and some trigonometry.

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As we have found each one, we can plug it and obtain some formula for potential energy of lynk1.

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Sure, the central for central mass.

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Let's do the same for linked to a center of mass.

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We can calculate the hate of lean to central mass from Baz's frame easily by just summing these two

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individual hates me and H two, one and H two two, we can get these two heads from take our metric

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relations again.

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Be careful about angles now as we have.

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Contrary to the hate speech to one and edge two to be can plug and obtain the potential energy for the

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center of mass.

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The second link total potential energy is nothing but the sum of these two potential energies.

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Now we can easily calculate gravity vector one by just finding partial derivative of the total potential

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energy with respect to Joint Angle one and Gravity Vector two by just finding partial derivative with

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respect the second joint angle.

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OK, we have calculated all the terms we need now.

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It's remained to just plug them into this formulation and obtain joint talks to one and two.

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And here it is.

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We calculated like an initial formulation for two degrees of freedom plan and a robot manipulator.

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You can see the amount of calculations, even in this simple case.

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Now imagine what will happen for six degrees of freedom robot manipulators in three dimensions.

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Surely you have to do it with MATLAB.