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Until this lesson, we have learned many different things about the representation of both positions

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and orientations.

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Now it's time to combine these two concepts in order to find a way of representing rigid motion of bodies

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and try to formulate this representation.

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Without any further ado, let's continue.

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Rigid motion is nothing but combination of pure translation and rotation because, as you know, we

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can represent in the motion of a rigid body with the translation and rotation.

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So let's specify a coordinated frame of zero.

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And the second cardinal frame of one rotation matrix of R1 zero orientation represents orientation of

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frame or one with respect to frame zero.

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Additionally, suppose that we have a point rigidly attached to the frame of one.

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Its coordinates with respect to frame or one is P1.

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Then we can express point p with respect to frame or zero by multiplying P1 with R1 zero.

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In order to account for the relative orientation and then ADI, which represents distance between the

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centers of O zero and one.

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Now let's consider three coordinate frames, namely O zero 01 and 02, and denote the distance between

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zero and one as D1 and distance between one and or two as the to the point P is attached to the frame

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of 02 and its coordinates on with one or two with the.

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OK, the point is attached to frame 02 and its coordinates on with respect to all too is given as Victor

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P-2.

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Let's try to calculate its corners with respect to frame zero.

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We can do that by first calculating its coordinates with respect to frame one and then frame frame or

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one to frame zero.

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You can see these calculations from the expressions.

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Now, let's substitute Equation one into Equation two, which yields Equation three.

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This represents Point P coordinates with respect to frame or zero, since the relation between zero

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and 02 is also a rigid motion.

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We can write Equations three in this form.

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Let's call this equation four.

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Finally, we can compare Equations three and Equation four to get total relative rotation and vector

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from the center of frame or zero, and to alter from frame from the center of frame or zero to alter,

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which are given by equations five and six.

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From equations of five and six.

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We can see that while we can get total orientation transformation by just multiplying rotation mattresses,

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we can do the same for the translational transformation.

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If you think about the move problems, you can easily see that it would be.

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It would become very difficult to manipulate these rigid body motions by considering above formulas

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because as the number of frames name, the number of rigid bodies increases, the complexity of formulas

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increases and it becomes more difficult to handle another plate them.

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Due to that reason, we will try to find either representation and way of handling of rigid motion.

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Let's try to write equations five and six in matrix form.

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As you can see from this equation, we have separate mattresses which have similar structures.

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By simply multiplying these mattresses, we can manipulate rigid motions between frames or bodies easily.

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These mattresses are called homogeneous transformation mattresses, and they have dimensions of four

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by four.

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This structure can also be represented in this form.

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As you can see, they are composed of rotation matrix, which represents the relative orientation and

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three by one vector, which represents vector from the center of one frame to another.

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Set of all transformation mattresses creates spatial equilibrium group that is called as all three,

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and they have several properties.

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Now let's talk about these properties.

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First of all, as we mentioned earlier, transformation mattresses are included into the Group of Spatial

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Equilibrium Group, and they have dimensional four by four multiplication of homogeneous transformation.

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Mattresses are associative, but not competitive.

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Be careful about that.

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Additionally, I would like to note that the inverse of homogeneous transformation mattresses are not

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equal to their transpose, as in the case of rotational mattresses.

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Here you can see the inverse of homogeneous transformation mattresses, which is also a homogeneous

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transformation mattress and included in the Group of three.

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There are several usages of homogeneous transformation mattresses.

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First, they represent a configuration.

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This means that if we are given to cordon of frames whose relative orientation is represented by R1

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zero and distance between two frames are given by Vector DB, then we can represent their relative configuration

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by homogeneous transformation matrix of H one zero.

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This constitutes both relative orientation and distance between the center of the frames.

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Additionally, we can find homogeneous transformation from frame o zero, two or one by finding inverse

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inverse of h zero.

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Additional homogeneous transformation mattresses can be used to change a reference frame.

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Let's assume that we are given three frames and distance between their centers as d one and the two.

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Then we can find h to zero up by multiplying homogeneous transformation matrix from frame one to zero

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and homogeneous transformation matrix from frame two to one.

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We we can also find h zero two by finding inverse of H two zero.

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We can do the same for Point Pete and change its reference frame from frame one to frame zero by using

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homogeneous transformation metrics.

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You can see that we can do that by multiplying homogeneous transformation metrics with P one simply.

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But it's not correct because dimension of homogeneous transformation matrix is four by four, while

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Vector P one has the dimension of three by one.

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So the multiplication is impossible.

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So what we have to do, the solution is very simple we have to use homogeneous coordinates instead of

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the normal coordinates.

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Doing that is not complicated.

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We just increase the dimensional vector p one to four and add one two below of the vector.

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So now we can successfully do the multiplication and find coordinates of point p in frame zero.

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Finally, let's talk about the composition of homogeneous transformation mattresses.

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But don't worry, because it's exactly the same as in the case of rotation mattresses.

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So let's assume that we have given two frames, namely oh, zero and four one, and the relative configuration

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between them is given by the homogeneous transformation matrix of each one zero.

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We are also given one additional rigid motion of age.

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So if this motion is about the current frame, then the post multiplied these mattresses.

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If it is about the fixed frame, then the press multiply these mattresses.

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So with this, we have finished the theory of this chapter.

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We will have two practical lessons.

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Additionally, the first will be in MATLAB and the second will be applied.

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Ribeau Real Robotics situation in simulation and we will use the techniques VE learned to solve the

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problem.

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See you on the next lesson.