Rotation About an Arbitrary Axis

Rotation About an Arbitrary Axis

9.1 Quick Review

Given a pointP = (x; y; z;1)in homogeneous coordinates, letP⁰= (x⁰; y⁰; z⁰;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 thex-axis by an angle x, counterclockwise (looking along thex-axis towards the origin). ThenP⁰=RxPwhere the rotation matrix, Rx,is given by:

Rx= 2 66 4

1 0 0 0

0 cos x sin x 0 0 sin x cos x 0

0 0 0 1

3 77 5

2. Rotation about they-axis by an angle y, counterclockwise (looking along they-axis towards the origin). ThenP⁰=RyPwhere the rotation matrix, Ry,is given by:

R_y = 2 66 4

cos y 0 sin y 0

0 1 0 0

sin y 0 cos y 0

0 0 0 1

3 77 5

3. Rotation about thez-axis by an angle _z, counterclockwise (looking along thez-axis towards the origin). ThenP⁰=R_zP where the rotation matrix, Rz,is given by:

Rz= 2 66 4

cos z sin z 0 0 sin z cos z 0 0

0 0 1 0

0 0 0 1

3 77 5

117

Note that the above transformations also apply to vectors.

You will also recall that

R_x¹ = R^T_x R_y¹ = R^T_y R_z¹ = R^T_z

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,R_xyz, is

Rxyz=RxRyRz= 2

4 C_yC_z C_yS_z S_y

S_xS_yC_z+C_xS_z S_xS_yS_z+C_xC_z S_xC_y C_xS_yC_z+S_xS_z C_xS_yS_z+S_xC_z C_xC_y

3 5 (9.1)

where

C_i = cos _i andS_i = sin _i fori=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 vectorv= (x; y; z)about a general axis with direction vectorrb (assume br 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 decomposingvinto two components: one parallel tobrand one perpendicular tobr. Let us denote v_k the component parallel to brand v_? the component perpendicular tor. You will recall from our study of vectors thatb

v_k = comp_brv

= v br kbrk²rb

= (v br)rbsincebris a unit vector Similarly,

v_? = orth_brv

= v (v r)b br

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])

LetT denote the rotation we are studying. We need to compute T(v).

T(v) = T v_k+v_?

= T v_k +T(v_?) sinceT is a linear transformation. Also,

T v_k =v_k

Figure 9.2: Rotation about a general axis through the origin, showing vectors on the plane of rotation (see [VB1])

since v_k has the same direction as brand we are rotating around an axis with direction vectorr(see …gure 9.1). Therefore,b

T(v) =v_k+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 = br v_? (9.2)

= br v Looking at …gure 9.2, we see that

T(v_?) = cos v_?+ sin w

= cos v_?+ sin (br v) and therefore

T(v) = v_k+T(v_?)

= (v r)b br+ cos v_?+ sin (br v)

= (v r)b br+ cos [v (v br)br] + sin (br v)

= (v r)b br+ cos v cos (v br)rb+ sin (rb v)

= (1 cos ) (v br)br+ cos v+ sin (br 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) andv= (vx; vy; vz)then

u v= 2

4 0 uz uy

uz 0 ux

u_y u_x 0 3 5v Proof.

u v =

i j k ux uy uz

vx vy vz

= 2

4 u_yv_z u_zv_y u_zv_x u_xv_z u_xv_y u_yv_x

3 5

= 2

4 0 u_z u_y u_z 0 u_x

u_y u_x 0 3 5v

Lemma 98 Using the notation of the previous lemma, we have

(u v)u= 2

4 u²_x u_xu_y u_xu_z uxuy u²_y uyuz

u_xu_z u_yu_z u²_z 3 5v

Proof.

u v=u_xv_x+u_yv_y+u_zv_z So,

(u v)u = (u_xv_x+u_yv_y+u_zv_z)u

= (u_xv_x+u_yv_y+u_zv_z) 2 4 u_x

u_y u_z

3 5

= 2

4 u²_x uxuy uxuz

uxuy u²_y uyuz

u_xu_z u_yu_z u²_z 3 5v

We are now ready to write T(v)as a matrix transformation.

T(v) = (1 cos ) (v r)b br+ cos v+ sin (rb v)

= (1 cos ) 2

4 u²_x uxuy uxuz

uxuy u²_y uyuz

uxuz uyuz u²_z 3 5v+

2

4 1 0 0 0 1 0 0 0 1

3

5cos v+ sin 2

4 0 uz uy

uz 0 ux

uy ux 0 3 5v

= 8<

:(1 cos ) 2

4 u²_x uxuy uxuz

uxuy u²_y uyuz

u_xu_z u_yu_z u²_z 3 5+

2

4 1 0 0 0 1 0 0 0 1

3

5cos + sin 2

4 0 u_z u_y u_z 0 u_x

uy ux 0 3 5

9=

;v

= 2

4 tu²_x+C tu_xu_y Su_z tu_xu_z+Su_y tuxuy+Suz tu²_y+C tuyuz Sux

tu_xu_z Su_y tu_yu_z+Su_x tu²_z+C 3 5v

where

b

r = (u_x; u_y; u_z) C = cos

S = sin t = 1 cos .

9.3 Assignment

1. Prove that the product of two orthogonal matrices is also an orthogonal matrix.

2. Prove equation 9.1 3. Prove equation 9.2

4. How would you implement rotation about an axis not going through the origin? First, consider the cases in which the axis of rotation is parallel to one of the coordinate axes. Then, consider the general case. In each case, derive the corresponding transformation matrix.

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