Until this lesson, we have learned many different things about the representation of both positions
and orientations.
Now it's time to combine these two concepts in order to find a way of representing rigid motion of bodies
and try to formulate this representation.
Without any further ado, let's continue.
Rigid motion is nothing but combination of pure translation and rotation because, as you know, we
can represent in the motion of a rigid body with the translation and rotation.
So let's specify a coordinated frame of zero.
And the second cardinal frame of one rotation matrix of R1 zero orientation represents orientation of
frame or one with respect to frame zero.
Additionally, suppose that we have a point rigidly attached to the frame of one.
Its coordinates with respect to frame or one is P1.
Then we can express point p with respect to frame or zero by multiplying P1 with R1 zero.
In order to account for the relative orientation and then ADI, which represents distance between the
centers of O zero and one.
Now let's consider three coordinate frames, namely O zero 01 and 02, and denote the distance between
zero and one as D1 and distance between one and or two as the to the point P is attached to the frame
of 02 and its coordinates on with one or two with the.
OK, the point is attached to frame 02 and its coordinates on with respect to all too is given as Victor
P-2.
Let's try to calculate its corners with respect to frame zero.
We can do that by first calculating its coordinates with respect to frame one and then frame frame or
one to frame zero.
You can see these calculations from the expressions.
Now, let's substitute Equation one into Equation two, which yields Equation three.
This represents Point P coordinates with respect to frame or zero, since the relation between zero
and 02 is also a rigid motion.
We can write Equations three in this form.
Let's call this equation four.
Finally, we can compare Equations three and Equation four to get total relative rotation and vector
from the center of frame or zero, and to alter from frame from the center of frame or zero to alter,
which are given by equations five and six.
From equations of five and six.
We can see that while we can get total orientation transformation by just multiplying rotation mattresses,
we can do the same for the translational transformation.
If you think about the move problems, you can easily see that it would be.
It would become very difficult to manipulate these rigid body motions by considering above formulas
because as the number of frames name, the number of rigid bodies increases, the complexity of formulas
increases and it becomes more difficult to handle another plate them.
Due to that reason, we will try to find either representation and way of handling of rigid motion.
Let's try to write equations five and six in matrix form.
As you can see from this equation, we have separate mattresses which have similar structures.
By simply multiplying these mattresses, we can manipulate rigid motions between frames or bodies easily.
These mattresses are called homogeneous transformation mattresses, and they have dimensions of four
by four.
This structure can also be represented in this form.
As you can see, they are composed of rotation matrix, which represents the relative orientation and
three by one vector, which represents vector from the center of one frame to another.
Set of all transformation mattresses creates spatial equilibrium group that is called as all three,
and they have several properties.
Now let's talk about these properties.
First of all, as we mentioned earlier, transformation mattresses are included into the Group of Spatial
Equilibrium Group, and they have dimensional four by four multiplication of homogeneous transformation.
Mattresses are associative, but not competitive.
Be careful about that.
Additionally, I would like to note that the inverse of homogeneous transformation mattresses are not
equal to their transpose, as in the case of rotational mattresses.
Here you can see the inverse of homogeneous transformation mattresses, which is also a homogeneous
transformation mattress and included in the Group of three.
There are several usages of homogeneous transformation mattresses.
First, they represent a configuration.
This means that if we are given to cordon of frames whose relative orientation is represented by R1
zero and distance between two frames are given by Vector DB, then we can represent their relative configuration
by homogeneous transformation matrix of H one zero.
This constitutes both relative orientation and distance between the center of the frames.
Additionally, we can find homogeneous transformation from frame o zero, two or one by finding inverse
inverse of h zero.
Additional homogeneous transformation mattresses can be used to change a reference frame.
Let's assume that we are given three frames and distance between their centers as d one and the two.
Then we can find h to zero up by multiplying homogeneous transformation matrix from frame one to zero
and homogeneous transformation matrix from frame two to one.
We we can also find h zero two by finding inverse of H two zero.
We can do the same for Point Pete and change its reference frame from frame one to frame zero by using
homogeneous transformation metrics.
You can see that we can do that by multiplying homogeneous transformation metrics with P one simply.
But it's not correct because dimension of homogeneous transformation matrix is four by four, while
Vector P one has the dimension of three by one.
So the multiplication is impossible.
So what we have to do, the solution is very simple we have to use homogeneous coordinates instead of
the normal coordinates.
Doing that is not complicated.
We just increase the dimensional vector p one to four and add one two below of the vector.
So now we can successfully do the multiplication and find coordinates of point p in frame zero.
Finally, let's talk about the composition of homogeneous transformation mattresses.
But don't worry, because it's exactly the same as in the case of rotation mattresses.
So let's assume that we have given two frames, namely oh, zero and four one, and the relative configuration
between them is given by the homogeneous transformation matrix of each one zero.
We are also given one additional rigid motion of age.
So if this motion is about the current frame, then the post multiplied these mattresses.
If it is about the fixed frame, then the press multiply these mattresses.
So with this, we have finished the theory of this chapter.
We will have two practical lessons.
Additionally, the first will be in MATLAB and the second will be applied.
Ribeau Real Robotics situation in simulation and we will use the techniques VE learned to solve the
problem.
See you on the next lesson.