Olsztyn, 27th May, 2013

University of Warmia and Mazury in Olsztyn

Faculty of Geodesy and Land Management

Department of Satellite Geodesy and Navigation

ESSAY

Elementary rotation matrices in 3D space.

Daria Bruniecka

Geodesy and Satellite Navigation

1st year M.Sc. studies

Elementary rotation matrices in 3D space

1. Generated by modulo function

2. Prove that and check relation

R_i⁻¹(α)=R_i^T(α)=R_i(−α);i = 1,  2,  3

The elementary 3D rotation matrices are constructed to perform rotations individually about the three coordinate axes.

Rotation about z-axis (x, y, z = 1,2,3)

Z = Z’

- the angle of elementary rotation about z-axis

From Oxyz to Ox’y’z’ we have R₃ (+)

From Ox’y’z’ to Oxyz we have R₃ (-)

From above figure we have:

We calculate:

Therefore:

And

From the point of view of matrix calculations we have:

Analogical considerations give us:

Rotation about x-axis (x, y, z = 1,2,3)

X = X’

- the angle of elementary rotation about x-axis

From Oxyz to Ox’y’z’ we have R₁ (+)

From Ox’y’z’ to Oxyz we have R₁ (-)

From above figure we have:

We calculate:

Therefore:

And

The elementary rotation about x:

Analogical considerations give us:

Rotation about y-axis (x, y, z = 1,2,3)

Y = Y’

- the angle of elementary rotation about y-axis

From Oxyz to Ox’y’z’ we have R₁ (+)

From Ox’y’z’ to Oxyz we have R₁ (-)

From above figure we have:

We calculate:

And:

The elementary rotation about y:

Elementary rotation matrices in 3D space

1. Generated by modulo function

Common looking on elementary rotations:

Subroutine: (i, α, A_3x3); i = 1, 2, 3 = x, y, z

MOD(A,P) computes the remainder of the division of A by P

j = mod(i, 3) + 1

k = mod(j, 3) + 1

$$\text{mod}\left( i,\ j \right) = i - \frac{i}{j}j\ \ \ \ \ \ \ \ i,j \in N$$

For i -th row

A (i,i)=1 A(i,j)=0 A(i,k)=0

For j -th row

A (j,i)=0 A(j,j)= A(j,k)=

For k-th row

A(k,i)=0 A(k,j)= -A(j,k) A(k,k)=A(j,j)

1. for x-azis:

i = 1

j = mod (1,3) + 1 = 2

k = mod (2,3) +1 = 3

For 1^st (i) row:

For 2^nd (j) row:

For 3^rd (k) row:

1. for y-azis:

i = 2

j = mod (2,3) + 1 = 3

k = mod (3,3) +1 = 1

For 1^st (k) row:

For 2^nd (i) row:

For 3^rd (j) row:

1. for z-azis:

i = 3

j = mod (3,3) + 1 = 1

k = mod (1,3) +1 = 2

For 1^st (j) row:

For 2^nd (k) row:

For 3^rd (i) row:

Elementary rotation matrices in 3D space

1. Prove that and check relation

R_i⁻¹(α)=R_i^T(α)=R_i(−α);i = 1,  2,  3

For i=1, x-axis:

R₁(+β)=$\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & \text{cosβ} & \text{sinβ} \\ 0 & - \text{sinβ} & \text{cosβ} \\ \end{bmatrix}$

The inverse matrix:

$$R_{1}^{- 1}(\beta) = \frac{1}{\det{R_{1}(\beta)}}M^{T}$$

det R₁(+β) - determinant of a matrix

M - matrix of cofactors

det R₁(+β)= $\left| \begin{matrix} 1 & 0 & 0 \\ 0 & \text{cosβ} & \text{sinβ} \\ 0 & - \text{sinβ} & \text{cosβ} \\ \end{matrix} \right|$= cos²β+ sin²β=1

M₁₁=${( - 1)}^{2}\left| \begin{matrix} \text{co}\text{sβ} & \text{sinβ} \\ - sin\beta & \text{cosβ} \\ \end{matrix} \right|$= cos²β+ sin²β=1 M₁₂=${( - 1)}^{3}\left| \begin{matrix} 0 & \text{sinβ} \\ 0 & \text{cosβ} \\ \end{matrix} \right|$=0 M₁₃=${( - 1)}^{4}\left| \begin{matrix} 0 & \text{cosβ} \\ 0 & - sin\beta \\ \end{matrix} \right|$=0

M₂₁=( − 1)³$\left| \begin{matrix} 0 & 0 \\ - sin\beta & \text{cosβ} \\ \end{matrix} \right|$=0 M₂₂=(−1)⁴$\left| \begin{matrix} 1 & 0 \\ 0 & \text{cosβ} \\ \end{matrix} \right|$=cosβ M₂₃=( − 1)⁵$\left| \begin{matrix} 1 & 0 \\ 0 & - sin\beta \\ \end{matrix} \right|$=sinβ

M₃₁=${( - 1)}^{4}\left| \begin{matrix} 0 & 0 \\ \text{cosβ} & \text{sinβ} \\ \end{matrix} \right|$=0 M₃₂=( − 1)⁵$\left| \begin{matrix} 1 & 0 \\ 0 & \text{sinβ} \\ \end{matrix} \right| = -$sinβ M₃₃=${( - 1)}^{6}\left| \begin{matrix} 1 & 0 \\ 0 & \text{cosβ} \\ \end{matrix} \right|$=cosβ

M=$\ \begin{bmatrix} 1 & 0 & 0 \\ 0 & \text{cosβ} & \text{sinβ} \\ 0 & - \text{sinβ} & \text{cosβ} \\ \end{bmatrix}$

$$R_{1}^{- 1}\left( \beta \right) = \frac{1}{\det{R_{1}\left( \beta \right)}}M^{T} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \text{cosβ} & - \text{sinβ} \\ 0 & \text{sinβ} & \text{cosβ} \\ \end{bmatrix}$$

Transposed matrix is as follows:

$$R_{1}^{T}(\beta) = \ \begin{bmatrix} 1 & 0 & 0 \\ 0 & \text{cosβ} & - \text{sinβ} \\ 0 & \text{sinβ} & \text{cosβ} \\ \end{bmatrix}$$

It is known that:

sin(−β) = −sin(β) and cos(−β) = cos(β)

So:

$$R_{1}\left( - \beta \right) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \text{cosβ} & - \text{sinβ} \\ 0 & \text{sinβ} & \text{cosβ} \\ \end{bmatrix}$$

To sum up the proof:

R₁⁻¹(β)=R₁^T(β)=R₁(−β)

For i=2, y-axis:

R₂(γ)=$\begin{bmatrix} c\text{osγ} & 0 & - sin\gamma \\ 0 & 1 & 0 \\ \text{sinγ} & 0 & \text{cosγ} \\ \end{bmatrix}$

det R₂(γ)=$\left| \begin{matrix} \text{cosγ} & 0 & - sin\gamma \\ 0 & 1 & 0 \\ \text{sinγ} & 0 & \text{cosγ} \\ \end{matrix} \right|$= cos²γ+ sin²γ=1

M₁₁=$\left| \begin{matrix} 1 & 0 \\ 0 & \text{cosγ} \\ \end{matrix} \right|$= cosγ M₁₂=$\left| \begin{matrix} 0 & 0 \\ \text{sinγ} & \text{cosγ} \\ \end{matrix} \right|$=0 M₁₃=$\left| \begin{matrix} 0 & 1 \\ \text{sinγ} & 0 \\ \end{matrix} \right| = - sin\gamma$

M₂₁=$\left| \begin{matrix} 0 & - sin\gamma \\ 0 & \text{cosγ} \\ \end{matrix} \right|$=0 M₂₂=$\left| \begin{matrix} \text{cosγ} & - sin\gamma \\ \text{sinγ} & \text{cosγ} \\ \end{matrix} \right|$= 1 M₂₃=$\left| \begin{matrix} \text{cosγ} & 0 \\ \text{sinγ} & 0 \\ \end{matrix} \right|$=0

M₃₁=$\left| \begin{matrix} 0 & - sin\gamma \\ 1 & 0 \\ \end{matrix} \right| =$sinγ M₃₂=$\left| \begin{matrix} \text{cosγ} & - sin\gamma \\ 0 & 0 \\ \end{matrix} \right|$=0 M₃₃=$\left| \begin{matrix} \text{cosγ} & 0 \\ 0 & 1 \\ \end{matrix} \right|$=cosγ

M=$\begin{bmatrix} \text{cosγ} & 0 & - sin\gamma \\ 0 & 1 & 0 \\ \text{sinγ} & 0 & \text{cosγ} \\ \end{bmatrix}$

$$R_{2}^{- 1}\left( \gamma \right) = \frac{1}{\det{R_{2}\left( \gamma \right)}}M^{T} = \begin{bmatrix} c\text{osγ} & 0 & \text{sinγ} \\ 0 & 1 & 0 \\ - sin\gamma & 0 & \text{cosγ} \\ \end{bmatrix}$$

$$R_{2}^{T}\left( \gamma \right) = \begin{bmatrix} \text{cosγ} & 0 & \text{sinγ} \\ 0 & 1 & 0 \\ - sin\gamma & 0 & \text{cosγ} \\ \end{bmatrix}$$

$$R_{2}^{}\left( - \gamma \right) = \ \begin{bmatrix} \text{cosγ} & 0 & \text{sinγ} \\ 0 & 1 & 0 \\ - sin\gamma & 0 & \text{cosγ} \\ \end{bmatrix}$$

To sum up the proof:

R₂⁻¹(γ)=R₂^T(γ)=R₂(−γ)

For i=3, z-axis:

R₃(α)=$\begin{bmatrix} \text{cosα} & \text{sinα} & 0 \\ - sin\alpha & \text{cosα} & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

Det R₃(α)= $\left| \begin{matrix} \text{cosα} & \text{sinα} & 0 \\ - sin\alpha & \text{cosα} & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right|$= cos²α+ sin²α=1

M₁₁=$\left| \begin{matrix} \text{cosα} & 0 \\ 0 & 1 \\ \end{matrix} \right|$= cosα M₁₂=(-1)³$\left| \begin{matrix} - sin\alpha & 0 \\ 0 & 1 \\ \end{matrix} \right| = sin\alpha$ M₁₃=$\left| \begin{matrix} - sin\alpha & \text{cosα} \\ 0 & 0 \\ \end{matrix} \right|$=0

M₂₁=(-1)³$\left| \begin{matrix} \text{sinα} & 0 \\ 0 & 1 \\ \end{matrix} \right| = - sin\alpha$ M₂₂=$\left| \begin{matrix} \text{cosα} & 0 \\ 0 & 1 \\ \end{matrix} \right|$=cosα M₂₃=$\left| \begin{matrix} \text{cosα} & \text{sinα} \\ 0 & 0 \\ \end{matrix} \right|$=0

M₃₁=$\left| \begin{matrix} \text{sinα} & 0 \\ \text{cosα} & 0 \\ \end{matrix} \right|$=0 M₃₂=$\left| \begin{matrix} \text{cosα} & 0 \\ - sin\alpha & 0 \\ \end{matrix} \right|$=0 M₃₃=$\left| \begin{matrix} \text{cosα} & \text{sinα} \\ - sin\alpha & \text{cosα} \\ \end{matrix} \right|$= 1

M=$\begin{bmatrix} \text{cosα} & \text{sinα} & 0 \\ - sin\alpha & \text{cosα} & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

$$R_{3}^{- 1}\left( \alpha \right) = \frac{1}{\det{R_{3}\left( \alpha \right)}}M^{T} = \begin{bmatrix} \text{cosα} & - sin\alpha & 0 \\ \text{sinα} & \text{cosα} & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

$$R_{3}^{T}\left( \alpha \right) = \begin{bmatrix} \text{cosα} & - sin\alpha & 0 \\ \text{sinα} & \text{cosα} & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

$$R_{3}^{}\left( - \alpha \right) = \ \begin{bmatrix} \text{cosα} & - sin\alpha & 0 \\ \text{sinα} & \text{cosα} & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

To sum up the proof:

R₃⁻¹(α)=R₃^T(α)=R₃(−α)