Rotation Matrix

Rotation Vector, A

A is a coordinate transformation (rotation) matrix.

Looking at the above example, we rotate the mobile link 90 degrees from the original position.

There are 2 coordinate frames.

Base Frame (F)
Mobile Frame (M)

M = {m_{1}, m_{2}, m_{3}}, F = {f_{1}, f_{2}, f_{3}}

where:

$m_{1}, m_{2}, m_{3}$ are unit vectors unique to the Mobile Frame ( $M$ )
$f_{1}, f_{2}, f_{3}$ are unit vectors unique to the Base Frame ( $F$ )
$1, 2, 3$ subscript is similar to how we use $x, y, z$ convention in a 3-Dimensional Space

Next, we know that point $p$ moved because we moved the mobile link. Given the coordinates of $p$ with respect to Mobile Frame (M), we wanna find out the coordinate of point $p$ with respect to Base Frame (F).

We can express a point, $p$ , in a coordinate system that is governed by the Mobile Frame ( $M$ ) as such:

[p]^{M} = p \cdot m_{1} p \cdot m_{2} p \cdot m_{3} = [p \cdot m_{1}, p \cdot m_{2}, p \cdot m_{3}]^{T}

Note that $p^{M}$ and $[p]^{M}$ are the same thing. $p$ is the physical vector, and $m_{1}$ , $m_{2}$ , $m_{3}$ are the unit axes of the frame. We are dotting the physical vector $p$ with each unit axis vector $m_{j}$ to get the coordinates. We also learn previously that dotting is similar to $cos (θ)$ , therefore it has a semantic meaning of finding the signed mangitude of that $j$ component of the point $p$ .

The formula that links point $p$ wrt F and $p$ wrt M is:

[p]^{F} = A [p]^{M}

$A$ here represents a coordinate transformation (rotation) matrix that maps transformation from M to F. $A$ can also be expressed in terms of it's own unit vectors, i.e. $A = [a_{1}, a_{2}, a_{3}]$ .

a_{j} = [m_{j}]^{F}

This formula can be seen as $[p h y s i c a l v ec t or]^{p er s p ec t i v e (o b ser v er)}$ . It tells us "how can vector $m$ be represented wrt base frame $F$ ?". This matrix $A$ allows us to transform a point with coordinates that are wrt $M$ , to a point with new coordinates that are wrt $F$ .

A = [m_{1}^{F}, m_{2}^{F}, m_{3}^{F}]

This formula is a result of combination of the earlier 2 formulas. What it means is that this rotation vector $A$ can be represented as a vector of $m_{j}^{F}$ in 3 components, $1, 2, 3$ , in this 3-Dimensional space.

Similar to how we connect the idea of point $p$ in frame $M$ , $[p]^{M}$ , we can do the same for $[m_{j}]^{F}$ . Let's work on $[m_{1}]^{F}$ first. What this means is we wanna find how can $m_{1}$ be expressed in terms of frame $F$ . Since $m_{1}$ may contain components from $f_{1}$ , $f_{2}$ , and $f_{3}$ , we can therefore express it as:

m_{1}^{F} = f_{1} \cdot m_{1} f_{2} \cdot m_{1} f_{3} \cdot m_{1}

If we continue this on for $j = 2, 3$ , we get:

A = f^{1} ∙ m_{1} f^{2} ∙ m_{1} f^{3} ∙ m_{1} f_{1} ∙ m_{2} f_{2} ∙ m_{2} f_{3} ∙ m_{2} f_{1} ∙ m_{3} f_{2} ∙ m_{3} f_{3} ∙ m_{3}

Substitute this back into the equation we get:s

A = f^{1} ∙ m_{1} f^{2} ∙ m_{1} f^{3} ∙ m_{1} f_{1} ∙ m_{2} f_{2} ∙ m_{2} f_{3} ∙ m_{2} f_{1} ∙ m_{3} f_{2} ∙ m_{3} f_{3} ∙ m_{3} 0.6 0.5 1.4

Note that $A^{- 1}$ is the coordinate transformation matrix that maps $F$ coordinates into $M$ coordinates, where $A^{- 1} = A^{T}$ .

Formula for Rotation Matrices

Fundamental Rotation Matrix is defined when a mobile coordinate frame $M$ that is obtained from fixed frame $F$ , rotates about one of $F$ 's unit vectors., as shown in the figure.

R_{1} (ϕ) = 100 0 cos (ϕ) s in (ϕ) 0 - s in (ϕ) cos (ϕ)

where:

$R_{1} (ϕ)$ is the resulting coordinate transformation matrix which maps mobile 𝑀 coordinates into fixed 𝐹 coordinates.
$f_{1}$ = $m_{1}$
$f_{2} + ϕ = m_{2}$
$f_{3} + ϕ = m_{3}$
$ϕ$ is the angle that $M$ rotates about $f_{1}$ as illustrated above

Likewise we get:

R_{2} (ϕ) = cos (ϕ) 0 - s in (ϕ) 010 s in (ϕ) 0 cos (ϕ)

R_{3} (ϕ) = cos (ϕ) s in (ϕ) 0 - s in (ϕ) cos (ϕ) 0 001

JJ

Explorer

Rotation Matrix

Rotation Vector, A

Formula for Rotation Matrices

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