Rotation About an Arbitrary Axis - Kennesaw State University
Chapter 9
Rotation About an
Arbitrary Axis
9.1
Quick Review
Given a point P = (x; y; z; 1) in homogeneous coordinates, let P 0 = (x0 ; y 0 ; z 0 ; 1)
be the corresponding point after a rotation around one of the coordinate axis has
been applied. You will recall the following from our studies of transformations:
1. Rotation about the x-axis by an angle x , counterclockwise (looking along
the x-axis towards the origin). Then P 0 = Rx P where the rotation matrix,
Rx ,is given by:
3
2
1
0
0
0
6 0 cos x
sin x 0 7
7
Rx = 6
4 0 sin x
cos x 0 5
0
0
0
1
2. Rotation about the y-axis by an angle y , counterclockwise (looking along
the y-axis towards the origin). Then P 0 = Ry P where the rotation matrix,
Ry ,is given by:
2
3
cos y 0 sin y 0
6
0
1
0
0 7
7
Ry = 6
4 sin y 0 cos y 0 5
0
0
0
1
3. Rotation about the z-axis by an angle z , counterclockwise (looking along
the z-axis towards the origin). Then P 0 = Rz P where the rotation matrix,
Rz ,is given by:
2
3
cos z
sin z 0 0
6 sin z
cos z 0 0 7
7
Rz = 6
4 0
0
1 0 5
0
0
0 1
117
118
CHAPTER 9. ROTATION ABOUT AN ARBITRARY AXIS
Note that the above transformations also apply to vectors.
You will also recall that
Rx 1
= RxT
Ry 1
= RyT
Rz 1
= RzT
This means in particular that these matrices are orthogonal. It can also be
proven that the product of two orthogonal matrices is itself an orthogonal matrix
(see problems at the end of the chapter). So, if we combine several rotations
about the coordinate axis, the matrix of the resulting transformation is itself an
orthogonal matrix.
One way of implementing a rotation about an arbitrary axis through the
origin is to combine rotations about the z, y, and x axes. The matrix of the
resulting transformation, Rxyz , is
2
3
Cy Cz
Cy Sz
Sy
Sx Sy Sz + Cx Cz
Sx Cy 5 (9.1)
Rxyz = Rx Ry Rz = 4 Sx Sy Cz + Cx Sz
Cx Sy Cz + Sx Sz Cx Sy Sz + Sx Cz
Cx Cy
where
Ci = cos
i
and Si = sin
i
for i = x; y; z
From what we noticed above, Rxyz is an orthogonal matrix. This means that
its inverse is its transpose.
9.2
Rotation About an Arbitrary Axis Through
the Origin
Goal: Rotate a vector v = (x; y; z) about a general axis with direction vector rb
(assume rb is a unit vector, if not, normalize it) by an angle (see ¡gure 9.1).
Because it is clear we are talking about vectors, and vectors only, we will omit
the arrow used with vector notation.
We begin by decomposing v into two components: one parallel to rb and one
perpendicular to rb. Let us denote vk the component parallel to rb and v? the
component perpendicular to rb. You will recall from our study of vectors that
vk
Similarly,
= comprbv
v rb
b
=
2r
kb
rk
= (v rb) rb since rb is a unit vector
v?
=
=
orthrbv
v (v rb) rb
9.2. ROTATION ABOUT AN ARBITRARY AXIS THROUGH THE ORIGIN119
And we have
v = v k + v?
Figure 9.1: Rotation about a general axis through the origin, showing the axis
of rotation and the plane of rotation (see [VB1])
Let T denote the rotation we are studying. We need to compute T (v).
T (v)
= T vk + v?
= T vk + T (v? )
since T is a linear transformation. Also,
T vk = vk
120
CHAPTER 9. ROTATION ABOUT AN ARBITRARY AXIS
Figure 9.2: Rotation about a general axis through the origin, showing vectors
on the plane of rotation (see [VB1])
9.2. ROTATION ABOUT AN ARBITRARY AXIS THROUGH THE ORIGIN121
since vk has the same direction as rb and we are rotating around an axis with
direction vector rb(see ¡gure 9.1). Therefore,
T (v) = vk + T (v? )
So, T (v? ) is the only quantity we need to compute. For this, we create a two
dimensional basis in the plane of rotation (see ¡gure 9.1 and 9.2). We will use
v? as our ¡rst basis vector. For our second, we can use
w
=
=
Looking at ¡gure 9.2, we see that
T (v? )
=
rb
rb
(9.2)
v?
v
cos v? + sin w
=
cos v? + sin (b
r
v)
and therefore
T (v)
= vk + T (v? )
=
(v rb) rb + cos v? + sin (b
r
=
(1
=
=
v)
(v rb) rb + cos [v (v rb) rb] + sin (b
r
v)
(v rb) rb + cos v cos (v rb) rb + sin (b
r
cos ) (v rb) rb + cos v + sin (b
r
v)
v)
We would like to express this as a matrix transformation, in other words, we
want to ¡nd the matrix R such that T (v) = Rv. For this, we ¡rst establish
some intermediary results.
Lemma 97 If u = (ux ; uy ; uz ) and v = (vx ; vy ; vz ) then
2
3
0
uz uy
0
ux 5 v
u v = 4 uz
uy ux
0
Proof.
u
v
=
2
i
ux
vx
j
uy
vy
uy v z
= 4 u z vx
ux vy
2
0
= 4 uz
uy
k
uz
vz
3
uz vy
u x vz 5
uy v x
uz
0
ux
3
uy
ux 5 v
0
................
................
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