Kinematics of Robots: Position Analysis
Conventions (‫ ) االتفاقيات‬:
Throughout this book, we will use the following conventions for describing vectors,
frames, transformations, and so on:
Vectors : i, j, k, x,y, z, n,o ,a ,p
Vector components:
Frames:
Transformation:
(transformation of robot relative to Universe, where Universe is a fixed frame)
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Representation of a Point in Space
A point P in space (Figure below) can be represented by its three coordinates
relative to a reference frame as:
where
and
are
the three coordinates of the
point represented in the
reference frame. Obviously,
other coordinate representations
can also be used to describe the
location of a point in space.
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Representation of a Vector in Space
A vector can be represented by three
coordinates of its tail and its head.
If the vector starts at point A and ends at point B,
then it can be represented by
PAB= (Bx-Ax)i + (By-Ay)j+ (Bz-Az)k. Specifically,
if the vector starts at the origin (Figure aside),
then
Where
and
are the three components of the vector in the reference frame.
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Matrix representation
The three components of the vector can also be written in the matrix form as in the
equation (2.5).
This form can be modified to include a scale factor (zooming factor )
as
Where,
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Results of scaling
If 𝜔 > 1, all vector components enlarge
If 𝜔 < 1, all vector components become smaller
If 𝜔 = 1, the size of the vector components remains unchanged
If 𝜔 = 0, vector length is infinity ,this will be used to represent the
direction vectors.
To make the vector into a unit vector, the length will be normalized to
be equal to 1. To do this each component is divided by the magnitude
of this vector (the square root of the sum of the squares of the three
components).
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Example
The directional vector is
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Representation of a Frame at the Origin of a Fixed Reference
Frame
• Axes x,y,z,are used to represent the fixed universe reference frame(RF) Fx,y,z.
• Axes n,o,a are used to represent another moving frame Fn,o,a relative to the RF.
Each direction of each axis of a frame Fn,o,a located at the origin of a
reference frame Fx,y,z (Figure below)
is represented by its three
directional cosines relative
to the RF. Consequently,
the three axes of the
frame can be represented
by three vectors in matrix
form as:
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Representation of a Frame Relative to a Fixed Reference Frame
The location (position) of a movable frame relative to the RF is described by
a vector between the origin of the frame and the origin of the reference
frame (Figure). Similarly, this vector is expressed by its components relative
to the
reference frame. Therefore, the frame can
be expressed by three vectors describing
its directional unit vectors and
a fourth vector describing its location as:
As shown in Equation, the first three vectors are directional vectors with w = 0,
representing the directions of the three unit vectors of the frame Fn,o, a , while the
fourth vector with w = 1 represents the location of the origin of the frame relative
to the reference frame.
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example
The frame F shown in Figure aside is located at
3,5,7 units, with its n-axis parallel to x, its o-axis at
45o relative to the y-axis, and its a-axis at 45o
relative to the z-axis. The frame can be described
by:
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Representation of a Rigid Body
An object can be represented in space by attaching a frame to it and representing the frame as shown
below.
As discussed before, a frame can be represented by a matrix, where the origin of the
frame and the three vectors representing its orientation relative to the reference frame are
expressed. Therefore,
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• As we discussed before, a point in space has only three degrees of freedom; it can
only move along the three reference axes. However, a rigid body in space has six
degrees of freedom, meaning that not only can it move along x-, y-, and z-axes, it
can also rotate about these three axes. Consequently, all that is needed to completely
define an object in space is six pieces of information describing the location of the
origin of the object in the reference frame and its orientation about the three axes.
However, as can be seen in object Equation, twelve pieces of information are given:
nine for orientation, and three for position (this excludes the scale factors on the last
row of the matrix because they do not add to this information). Obviously, there
must be some constraints present in this representation to limit the above to six.
Therefore, we need 6 constraint equations to reduce the above from twelve to six.
The constraints come from the known characteristics of a frame that have not been
used yet, that:
• the three unit vectors n, o, a are mutually perpendicular, and each unit
vector’s length, represented by its directional cosines, must be equal to 1
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These constraints translate into the following six constraint equations:
1. n . o = 0 (the dot-product of n and o vectors must be zero)
2. n . a = 0
3. a . o = 0
4. |n|=1 (the magnitude of the length of the vector must be 1)
5. |o| =1
6. |n| =1
As a result, the values representing a frame in a matrix must be such that the above
equations remain true. Otherwise, the frame will not be correct. Alternatively, the first
three constraints can be replaced by a cross product of the three vectors as:
Since this Equation includes the correct right-hand-rule relationship too, it is
recommended that this equation be used to determine the correct relationship between
the three vectors.
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example
• Find the missing elements in the object equation aside.
• Solution:
• Using the constraint equations:
•
simplifying these equations yields
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Completing the exam
• Solving these six equations gives
• The reason for multiple solutions is that with the given parameters, it is possible to have two sets of
mutually perpendicular vectors in opposite directions. The final matrix will be:
• The same problem may be solved using
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Completing the exam
which replace the three equations for the dot products.
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Homogeneous Transformation Matrices
• A matrix Aij is called a square matrix if the number of rows (m) is equal to the
number of columns (n) so that the dimensions of the matrix are m×n=m×m. Matrix which
represents the direction alone is a 3×3 matrix. If the matrix represents both direction and
position(location), a scale factor should be added to make the matrix square
(4×4).Matrices of this form are called homogeneous matrices and we refer to them as:
• Matrices multiplication has no commutative
property . So, A×B≠B×A even for square matrices.
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Representation of Transformations
A transformation is defined as making a movement in space.
A transformation may be in one of the following forms:
-A pure translation: the moving frame (MF),
(which may represents an object) moves in space without any change in its
direction.
-A pure rotation about an axis: the MF moves in space (rotates) without any
change in its location (position)
-A combination of translations and/or rotations : the MF(Fn,o,a )moves in
space and changes its origin(location) and orientation(rotation)
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Representation of a Pure Translation
• In the pure translation, the only thing that changes
Is the location of the origin of the moving frame (MF)relative to the RF
as shown in the figure aside. In matrix form, the new
frame representation may be found by pre-multiplying
the frame with a matrix representing the transformation.
This equation is symbolically written as:
Note that the new location of the frame relative to the fixed reference frame can be found by
adding the vector representing the translation to the vector representing the original location of the
origin of the frame (P+d).
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Example
• A frame F has been moved 10 units along the
y-axis and 5 units along the z-axis of the
reference frame. Find the new location
of the frame.
Solution:
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Representation of a Pure Rotation about an Axis
Let’s assume that a MF(Fnoa), located at the origin of the
RF(Fxyz), rotates an angle of
about the x-axis of the RF.
Let’s also assume that attached to the rotating frame Fnoa, is
a point or vector p, with coordinates px, py, and pz relative
to the RF and pn, po, and pa relative to the MF. As the MF
rotates about the x-axis, point p attached to the frame will
also rotate with it. Before rotation, the coordinates of the
point in both frames are the same (remember that the two
frames are at the same location and are parallel to each
other). After rotation, the pn, po, and pa coordinates of the
point remain the same in the rotating frame Fnoa, but px, py,
and pz will be different in the Fxyz frame.
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Now let’s look at the same coordinates in 2D as if we were standing on the x-axis.
The coordinates of point p are shown
before and after rotation in the figure aside.
The
coordinates of point p relative to the RF are
px, py, and pz, while its coordinates
relative to the rotating frame (to which the
point is attached) remain as pn, po, and pa.
from the figure we can write
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The previous equations can be written in matrix form as:
The obtained results show that: in order to get the
coordinates of point p (or vector p) in the RF, the coordinates
in the MF must be pre-multiplied by the rotation matrix. The
results in symbolical form can be written as:
In conventional form:
Where,
transformation of frame R relative to universe (U)
p relative to frame R
p relative to the universe U
The same results for the rotation of a frame about the y- and z-axes of the reference frame
can be obtained as:
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Example
• A point p(2,3,4)T is attached to a rotating frame. The frame rotates 90o about the
x-axis of the reference frame. Find the coordinates of the point relative to the
reference frame after the rotation, and verify the result graphically.
Solution:
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Representation of Combined Transformations
• Combined transformations consist of a number of successive translations and
rotations about the fixed reference frame axes or the moving current frame axes.
To see how combined transformations are handled, let’s assume that a point pnoa
is attached to a rotating frame Fnoa at the origin of the reference frame and the
current frame (MF)is subjected to the following three successive transformations
relative to the reference frame Fxyz:
1. Rotation of α degrees about the x-axis,
2. Followed by a translation of [l1,l2,l3] (relative to the x-, y-, and z-axes
respectively),
3. Followed by a rotation of β degrees about the y-axis.
Solution:
- After the first transformation, the coordinates of the point relative to the RF are
- After the second transformation, the coordinates of the point relative to the RF
are
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after the third transformation, the coordinates of the point relative to the
reference frame will be:
As you can see, the coordinates of the point relative to the reference frame at
the conclusion (‫) ختام‬of each transformation are found by pre-multiplying the
coordinates of the point by each transformation matrix.
Consequently, the order of matrices written is the opposite of the order of
transformations performed.
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Example
• A point p(7,3,1)T is attached to a frame Fnoa and is subjected to the following
transformations. Find the coordinates of the point relative to the reference frame
at the conclusion of transformations.
1. Rotation of 90o about the z-axis,
2. Followed by a rotation of 90o about the y-axis,
3. Followed by a translation of [4,- 3,7].
Solution: The matrix equation representing the transformation is:
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Completing the example
• Graphically can be done as shown in the figures below
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Combined Transformations Relative to the Rotating Frame
• Transformations can be made relative to the current or moving frame instead of RF. To
calculate the changes in coordinates of a point attached to the MF relative to the RF, the
transformation matrix is post multiplied instead.
• Example: point p(7,3,1)T, attached to Fnoa, is subjected to the
transformations relative to the MF as listed below, Find the coordinates of the point
relative to the reference frame after transformations are completed.
1. A rotation of 90o about the a-axis,
2. Then a translation of [4,- 3,7] along n-, o-, a-axes
3. Followed by a rotation of 90o about the o-axis.
Solution:
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Example
A frame F was rotated about the y-axis 90o, followed by a rotation about the o-axis of 30o, followed by a
translation of 5 units along the n-axis, and finally, a translation of 4 units along the x-axis. Find the total
transformation matrix.
Solution: The following set of matrices, written in the proper order to represent transformations relative to
the reference frame or the current frame describes the total transformation:
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Inverse of Transformation Matrices(‫)معكوس مصفوفات النقل‬
Inverse matrices are needed in the robot analysis. One
situation where transformation matrices may be involved
can be seen in the following analysis. Suppose the robot
in Figure aside is to be move toward part P in order to drill
a hole in the part. The end effector(bit) frame (E) has to be at
the hole where the drilling to be done. The location of the
point where the hole will be drilled can be related to the
reference frame U through two independent paths: one
through the part, one through the robot.
To calculate (UTE ) using the forward path, all transformations of this path should be found. In fact, all
the transformation are known except the transformation of frame H relative to frame R (RTH ) . To
calculate this matrix (RTH ) , unlike in an algebraic equation, we cannot simply divide the right side by the left
side of the equation. We need to pre- or post-multiply by inverses of inverse matrices to eliminate them. As a
result, we will have:
So, we get
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Inverse matrix of homogeneous transformation matrix
For a homogenous 4 ×4 transformation matrix, it can be shown that the matrix inverse can be written
by dividing the matrix into two portions; the rotation portion of the matrix can be simply transposed, as it
is still unitary. The position portion of the homogeneous matrix is the negative of the dot product of the
p-vector with each of the n-, o-, and a-vectors, as follows:
𝑛𝑥
𝑜
= 𝑥
𝑎𝑥
0
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𝑛𝑦
𝑜𝑦
𝑎𝑦
0
𝑛𝑧
𝑜𝑧
𝑎𝑧
0
−(𝑝𝑥 𝑛𝑥 + 𝑝𝑦 𝑛𝑦 + 𝑝𝑧 𝑛𝑧 )
−(𝑝𝑥 𝑜𝑥 + 𝑝𝑦 𝑜𝑦 + 𝑝𝑧 𝑛𝑧 )
−(𝑝𝑥 𝑎𝑥 + 𝑝𝑦 𝑎𝑦 + 𝑝𝑧 𝑎𝑧 )
1
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Example 1
A frame F has the same directions as the universe frame and located at 1,5,4 units. If the
frame rotates 40o about the x-axis, Calculate the inverse transformation matrix
Solution: The matrix after rotation of 40o about the x-axis is:
1
0
0
1
0 0.766 −0.643 5
0 0.643 0.766 4
0
0
0
1
The inverse of this matrix is:
𝟏
𝟎
𝟎
𝟎
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𝟎
𝟎
−𝟏
𝟎. 𝟕𝟔𝟔 𝟎. 𝟔𝟒𝟑 −𝟔. 𝟒𝟎𝟐
−𝟎. 𝟔𝟒𝟑 𝟎. 𝟕𝟔𝟔 𝟎. 𝟏𝟓𝟏
𝟎
𝟎
𝟏
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Example 2
In a robotic set-up, a camera is attached to the fifth link of a 6-DOF robot. It observes an object and
determines its frame relative to the camera’s frame. Using the following information, determine the
necessary motion the end effector must make to get to the object:
Solution: Referring to the previous Equation of robot for drilling, we can write a similar equation
that relates the different transformations and frames together as:
Since RT5 Appears on both sides of the equation, we can simply neglect it, then ETobj can be
found as
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Completing the example
Substituting the matrices and the inverses in the above equation will result:
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