Mark W. Spong - Robot Modeling and Control

(2.58)

Now consider three coordinate frames o 0 x 0 y 0 z 0, o 1 x 1 y 1 z 1, and o 2 x 2 y 2 z 2. Let d 1be the vector from the origin of o 0 x 0 y 0 z 0to the origin of o 1 x 1 y 1 z 1and d 2be the vector from the origin of o 1 x 1 y 1 z 1to the origin of o 2 x 2 y 2 z 2. If the point p is attached to frame o 2 x 2 y 2 z 2with local coordinates , we can compute its coordinates relative to frame o 0 x 0 y 0 z 0using

(2.59)

and

(2.60)

The composition of these two equations defines a third rigid motion, which we can describe by substituting the expression for from Equation ( 2.59) into Equation ( 2.60)

(2.61)

Since the relationship between and is also a rigid motion, we can equally describe it as

(2.62)

Comparing Equations ( 2.61) and ( 2.62) we have the relationships

(2.63)

(2.64)

Equation ( 2.63) shows that the orientation transformations can simply be multiplied together and Equation ( 2.64) shows that the vector from the origin o 0to the origin o 2has coordinates given by the sum of (the vector from o 0to o 1expressed with respect to o 0 x 0 y 0 z 0) and (the vector from o 1to o 2, expressed in the orientation of the coordinate frame o 0 x 0 y 0 z 0).

One can easily see that the calculation leading to Equation ( 2.61) would quickly become intractable if a long sequence of rigid motions were considered. In this section we show how rigid motions can be represented in matrix form so that composition of rigid motions can be reduced to matrix multiplication as was the case for composition of rotations.

In fact, a comparison of Equations ( 2.63) and ( 2.64) with the matrix identity

(2.65)

where 0 denotes the row vector (0, 0, 0), shows that the rigid motions can be represented by the set of matrices of the form

(2.66)

Transformation matrices of the form given in Equation ( 2.66) are called homogeneous transformations. A homogeneous transformation is therefore nothing more than a matrix representation of a rigid motion and we will use SE (3) interchangeably to represent both the set of rigid motions and the set of all 4 × 4 matrices of the form given in Equation ( 2.66).

Using the fact that is orthogonal it is an easy exercise to show that the inverse transformation is given by

(2.67)

In order to represent the transformation given in Equation ( 2.58) by a matrix multiplication, we must augment the vectors and by the addition of a fourth component of 1 as follows,

(2.68)

(2.69)

The vectors and are known as homogeneous representationsof the vectors and , respectively. It can now be seen directly that the transformation given in Equation ( 2.58) is equivalent to the (homogeneous) matrix equation

(2.70)