Kinematics for Wheeled Mobile Robots

Overview

A practical approach to deriving kinematics for wheeled robots.

Wheeled Mobile Robots

The kinematics of wheeled mobile robots describes the relationship between how the wheels move and how the robot moves, under some simplifying assumptions. Here, we assume:

The robot consists of a single rigid-body (although we could extend this to handle, for example, trailers)
- The location of the robot corresponds to the position and orientation of a frame attached to its body, called the body frame ($\{b\}$).
The robot instantaneously achieves its commanded velocity (there are no dynamics/inertia to worry about)
Friction is ignored. In each direction, the wheels either roll perfectly or slip completely.

When computing the robot kinematics, we can divide the problem into two parts:

Find the relationship between movement of the robot at its body frame and frames at each wheel location
Relate the movement of the wheels to the movement of the body at the wheel locations.

By dividing the problem into these two steps, we can create a procedure for deriving the kinematics for most wheeled robots regardless of the types or configurations of the wheels.

Kinematics Derivation Summary

Overall, the procedure has the following steps:

Label the body frame and each wheel frame
Find the transformations and adjoints between the body frame and each wheel frame
Write the body frame twist in each of the wheel frames
Substitute the wheel constraints into the equations relating the body-frame twist to the wheel-frame twist
Solve for the variables in the wheel constraints in terms of the body-frame twist: this is the inverse kinematics.
Invert the inverse-kinematics equation to find the forward kinematics (given the wheel state, how does the body move)?
Solve for the relationship between the robot's velocity $\dot{q} = (\dot{\theta}, \dot{x}, \dot{y})$ and its body twist $V_b$.

Notation

A transformation matrix converting vectors in frame $\{j\}$ to vectors in frame $\{i\}$ is written as $T_{ij}(\theta, x, y)$
- $T_{ij}$ can be used when the parameters of the transformation do not matter
An adjoint matrix converting twists in frame $\{j\}$ to twists in frame $\{i\}$ is $A_{ij}(\theta, x, y)$
- $A_{ij}$ can be used when the parameters of the transformation do not matter

Review of Transforms and Adjoints in 2D

A 2D transformation matrix, which can perform rotations and translations

\begin{equation} T(\theta, x, y) = \begin{bmatrix} \cos\theta & -\sin\theta & x \\ \sin\theta & \cos\theta & y \\ 0 & 0 & 1 \end{bmatrix}. \end{equation}

The inverse transformation matrix is

\begin{align} T^{-1}(\theta,x,y)&= \begin{bmatrix} \cos\theta & \sin\theta & -x\cos\theta - y\sin\theta \\ -\sin\theta & \cos\theta & -y\cos\theta+x\sin\theta \\ 0 & 0 & 1 \end{bmatrix}\\ &= T(-\theta, -x\cos\theta - y\sin\theta, -y\cos\theta + x\sin\theta) \end{align}

We write $T_{ij} = T(\theta, x, y)$ to indicate the transform from frame $\{j\}$ coordinates to frame $\{i\}$ coordinates.

The 2D Adjoint, which converts twists in one frame to another frame is

\begin{equation} A(\theta, x, y) = \begin{bmatrix} 1 & 0 & 0 \\ y & \cos\theta & -\sin\theta \\ -x & \sin\theta & \cos\theta \end{bmatrix}. \end{equation}

The inverse of the 2D adjoint is found by directly applying the Adjoint equation above to the inverse transformation

\begin{align} A^{-1}(\theta,x,y) &= A(-\theta, -x\cos\theta - y\sin\theta, -y\cos\theta + x\sin\theta) \\ &=\begin{bmatrix} 1 & 0 & 0 \\ -y\cos\theta + x\sin\theta & \cos\theta &\sin\theta \\ x\cos\theta+y\sin\theta & -\sin\theta & \cos\theta \end{bmatrix} \end{align}

Transforms Between Body and Wheels

The first step in the process is to find all the transforms and adjoints between the body frame and frames located at each wheel.

Example: Four Wheel Car

Figure 1: A Car With Four Wheels

The transformation matrices are:

\begin{align*} \begin{matrix} T_{b1}(0, L, D) & T_{b2}(0,L, -D) \\ T_{b3}(0, -L, -D) & T_{b4}(0,-L, D) \end{matrix} \end{align*}

The adjoints are

\begin{align*} \begin{matrix} A_{b1} &= \begin{bmatrix} 1 & 0 & 0 \\ D & 1 & 0\\ -L & 0 & 1 \end{bmatrix} & A_{b2} &= \begin{bmatrix} 1 & 0 & 0 \\ -D & 1 & 0\\ -L & 0 & 1 \end{bmatrix} \\ A_{b3} &= \begin{bmatrix} 1 & 0 & 0 \\ -D & 1 & 0\\ L & 0 & 1 \end{bmatrix} & A_{b4} &= \begin{bmatrix} 1 & 0 & 0 \\ D & 1 & 0\\ L & 0 & 1 \end{bmatrix} \end{matrix} \end{align*}

The inverse adjoints are

\begin{align*} \begin{matrix} A_{1b} &= \begin{bmatrix} 1 & 0 & 0 \\ -D & 1 & 0 \\ L & 0 & 1 \end{bmatrix} & A_{2b} &= \begin{bmatrix} 1 & 0 & 0 \\ D & 1 & 0 \\ L & 0 & 1 \end{bmatrix} \\ A_{3b} &= \begin{bmatrix} 1 & 0 & 0 \\ D & 1 & 0 \\ -L & 0 & 1 \end{bmatrix} & A_{4b} &= \begin{bmatrix} 1 & 0 & 0 \\ -D & 1 & 0 \\ -L & 0 & 1 \end{bmatrix} \end{matrix} \end{align*}

Body Twist in the Wheel Frames

Let $V_b = \begin{bmatrix}\dot{\theta} \\ v_x \\ v_y \end{bmatrix}$ be the body frame twist. Let $V_i = \begin{bmatrix}\dot{\theta} \\ v_{xi} \\ v_{yi}\end{bmatrix}$ be the twist in the frame at wheel $i$.

Then $V_i = A_{ib}V_b$

These equations map the body twist to the velocities at each wheel, and thus tell you how the wheels must move in order to achieve a desired body twist.

Example: Four Wheel Car

Wheel 1

\begin{align} V_1 &= A_{1b}V_b \\ \begin{bmatrix} \dot{\theta} \\ v_{x1}\\ v_{y1} \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 \\ -D & 1 & 0 \\ L & 0 & 1 \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} = \begin{bmatrix} \dot{\theta} \\ -D\dot{\theta} + v_x\\ L\dot{\theta} + v_y \end{bmatrix} \end{align}

Wheel 2

\begin{align} V_2 &= A_{2b}V_b \\ \begin{bmatrix} \dot{\theta} \\ v_{x2}\\ v_{y2} \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 \\ D & 1 & 0 \\ L & 0 & 1 \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} = \begin{bmatrix} \dot{\theta} \\ D\dot{\theta}+v_x\\ L\dot{\theta}+v_y \end{bmatrix} \end{align}

Wheel 3

\begin{align} V_3 &= A_{3b}V_b \\ \begin{bmatrix} \dot{\theta} \\ v_{x3}\\ v_{y3} \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 \\ D & 1 & 0 \\ -L & 0 & 1 \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} = \begin{bmatrix} \dot{\theta} \\ D\dot{\theta}+v_x\\ -L\dot{\theta}+v_y \end{bmatrix} \end{align}

Wheel 4

\begin{align} V_4 &= A_{4b}V_b \\ \begin{bmatrix} \dot{\theta} \\ v_{x4}\\ v_{y4} \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 \\ -D & 1 & 0 \\ -L & 0 & 1 \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} = \begin{bmatrix} \dot{\theta} \\ -D\dot{\theta} + v_x \\ -L\dot{\theta} + v_y \end{bmatrix} \end{align}

Types of Wheels

Wheels determine how the robot interacts with the ground. Each wheel type has controls and constraints. The controls depend on what actuators are connected to the wheel. The constraints depend on the wheel construction and the assumptions made (such as no-slip).

The wheel model describes, mathematically, the wheel's movement and is parameterized by the wheel variables (such as angular velocity). By substituting the wheel model into the wheel-body-twist equations and solving for the wheel variables we determine

What controls are needed to achieve a given twist
What twists are achievable and what are unobtainable

Conventional Wheel

A conventional wheel rotates about an axle, which causes it to move along the ground.

The velocity of the robot in the wheel frame is related to the wheel model by the following equation:

\begin{equation} \begin{bmatrix} v_{xi} \\ v_{yi} \end{bmatrix} = \begin{bmatrix} r \dot{\phi}_i \\ 0 \end{bmatrix} \end{equation}

Here we assume that the wheel does not slip. Integrating both sides of $v_{xi} = r \dot{\phi}_i$ over a period of one rotation reveals that the $x$ distance traveled over this period is equivalent to the wheel circumference.
We assume that the wheel does not slide (that is move in the direction of its axis). Hence $v_{yi} = 0$.
If the wheel's rotational velocity is actuated, we set $\dot{\phi}_i = u_i$ and can then control the wheel's angular velocity.

Mecanum and Omni Wheels

A mecanum wheel is driven via rotation along its $y$ axis and it slides along a vector offset from its $y$ axis by some angle $\gamma$ (see figure below)

Figure 2: Top-Down view of a Mecanum/Omni Wheel (it is omni when $\gamma = 0$).

The angle $\gamma$ determines which direction the wheel is free to slide in: when $\gamma = 0$ the wheel is called an omni wheel. Mecanum wheels typically have angles of $\gamma = 45^{\circ}$ (See Lynch, Park 2017).

Let $\dot{\phi}_i$ be the rotational speed of the wheel and $v_{si}$ be the speed in the sliding direction. The velocity of the wheel is related to its sliding and rotational velocities by

\begin{equation} \begin{bmatrix} v_{xi}\\ v_{yi} \end{bmatrix} = \begin{bmatrix} r\dot{\phi}_i - v_{si}\sin\gamma \\ v_{si}\cos\gamma \end{bmatrix}. \end{equation}

If the wheel's rotational velocity is actuated, we set $\dot{\phi}_i = u_i$
Typically, the wheel's velocity in the sliding direction is not actuated

Steered Wheel

A steered wheel is a conventional wheel whose angle relative to the chassis, $\psi_i$, can change. The wheel rolls without slipping along a vector in the $\psi_i$ direction.

\begin{equation} \begin{bmatrix} v_{xi}\\ v_{yi} \end{bmatrix} = \begin{bmatrix} r \dot{\phi}_i \cos\psi_i\\ r \dot{\phi}_i \sin\psi_i \end{bmatrix} \end{equation}

Often an additional constraint will be placed on $\psi$. For example, in a car with two front wheels being steered, the angle of the left and right wheels is related by a constraint (often implemented as a mechanical linkage).

Additional Constraints

Additional constraints can be added, for example we can necessitate that two wheels have some relationship to each other

Car with Four Mecanum Wheels

After solving for the body twist in the wheel frames, the next steps are to substitute the wheel equations into the equations relating the body twist to the wheel-frame twist and then solve for the inverse kinematics.

Continuing from the four-wheel car example, let's assume that the car has four actuated mecanum wheels.

The next step is to substitute the mecanum wheel equations into the equations for the twist and solve for $\dot{\phi}_i$ and $v_{si}$ in terms of the body twist $\begin{bmatrix}\dot{\theta}\\ v_x \\ v_y\end{bmatrix}$,

Body Twist to Wheel Motion

Wheel 1

This wheel has a roller angle of $\gamma = -\frac{\pi}{4}$

\begin{align} \begin{bmatrix} \dot{\theta} \\ r \dot{\phi}_1 - v_{s1}\sin\frac{-\pi}{4}\\ v_{s1}\cos\frac{-\pi}{4} \end{bmatrix} &= \begin{bmatrix} \dot{\theta} \\ -D\dot{\theta} + v_x\\ L\dot{\theta} + v_y \end{bmatrix}\\ \begin{bmatrix} \dot{\phi}_1 \\ v_{s1} \end{bmatrix} &= \frac{1}{r} \begin{bmatrix} -L - D & 1 & -1 \\ r L\sqrt{2} & 0 & r \sqrt{2} \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{align}

Wheel 2

This wheel has a roller angle of $\gamma = \frac{\pi}{4}$

\begin{align} \begin{bmatrix} \dot{\theta} \\ r \dot{\phi}_2 - v_{s2}\sin\frac{\pi}{4}\\ v_{s2}\cos\frac{\pi}{4} \end{bmatrix} &= \begin{bmatrix} \dot{\theta} \\ D\dot{\theta}+v_x\\ L\dot{\theta}+v_y \end{bmatrix}\\ \begin{bmatrix} \dot{\phi}_2 \\ v_{s2} \end{bmatrix} &= \frac{1}{r} \begin{bmatrix} L+D & 1 & 1\\ r L \sqrt{2} & 0 & r\sqrt{2} \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{align}

Wheel 3

This wheel has a roller angle of $\gamma = \frac{-\pi}{4}$

\begin{align} \begin{bmatrix} \dot{\theta} \\ r \dot{\phi}_3 - v_{s3}\sin\frac{-\pi}{4}\\ v_{s3}\cos\frac{-\pi}{4} \end{bmatrix} &= \begin{bmatrix} \dot{\theta} \\ D\dot{\theta}+v_x\\ -L\dot{\theta}+v_y \end{bmatrix}\\ \begin{bmatrix} \dot{\phi}_3 \\ v_{s3} \end{bmatrix} &= \frac{1}{r} \begin{bmatrix} L+D & 1 & -1\\ -r L\sqrt{2} & 0 & r\sqrt{2} \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{align}

Wheel 4

This wheel has a roller angle of $\gamma = \frac{\pi}{4}$

\begin{align} \begin{bmatrix} \dot{\theta} \\ r \dot{\phi}_4 - v_{s4}\sin\frac{\pi}{4}\\ v_{s4}\cos\frac{\pi}{4} \end{bmatrix} &= \begin{bmatrix} \dot{\theta} \\ -D\dot{\theta} + v_x \\ -L\dot{\theta} + v_y \end{bmatrix}\\ \begin{bmatrix} \dot{\phi}_4 \\ v_{s4} \end{bmatrix} &= \frac{1}{r} \begin{bmatrix} -L-D & 1 & 1 \\ -r L \sqrt{2} & 0 & r\sqrt{2} \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{align}

Inverse Kinematics

Now that we have the relationship between the body twist and the velocity of the body in each wheel frame, we can now gather everything into a single equation:

\begin{align} \begin{bmatrix} \dot{\phi}_1\\\dot{\phi}_2\\\dot{\phi}_3\\\dot{\phi}_4 \\v_{s1}\\v_{s2}\\v_{s3}\\v_{s4} \end{bmatrix} &= \frac{1}{r} \begin{bmatrix} -L-D & 1 & -1 \\ L + D & 1 & 1 \\ L + D & 1 & -1 \\ -L - D & 1 & 1\\ r L\sqrt{2} & 0 & r \sqrt{2}\\ r L\sqrt{2} & 0 & r \sqrt{2}\\ - r L\sqrt{2} & 0 & r \sqrt{2}\\ - r L\sqrt{2} & 0 & r \sqrt{2}\\ \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{align}

The equation above tells us how the wheels will rotate and slide, given a twist applied to the body of the robot.

Controls

Assume we control the rotation but not the sliding of each wheel, by letting $\dot{\phi}_i = u_i$. We can then partition the motion equation into two parts: one for control the other for sliding:

\begin{align} u &= H V_b\\ \begin{bmatrix} u_1\\u_2\\u_3\\u_4 \end{bmatrix} &= \frac{1}{r} \begin{bmatrix} -L-D & 1 & -1 \\ L + D & 1 & 1 \\ L + D & 1 & -1 \\ -L - D & 1 & 1\\ \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix}\\ v_s &= R V_b \\ \begin{bmatrix} v_{s1}\\v_{s2}\\v_{s3}\\v_{s4} \end{bmatrix} &= \begin{bmatrix} L\sqrt{2} & 0 & \sqrt{2}\\ L\sqrt{2} & 0 & \sqrt{2}\\ - L\sqrt{2} & 0 & \sqrt{2}\\ - L\sqrt{2} & 0 & \sqrt{2}\\ \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{align}

Examining the properties of these equations can provide insight into the robot's possible motions.

In this example, $H$ has full column rank (a rank of 3 in this case).
- Therefore the dimension of the Null Space of H is 0, which implies that only the 0 twist results in a 0 control vector
  - Therefore, every body twist maps to a unique control vector (and therefore wheel speed).
- Because every body twist maps to a unique control vector, there exists a control vector for every body twist.
  - Therefore the robot can controlled to move in any body twist direction, hence this robot is truly omnidirectional.
Suppose that the Null Space of $H$ were not empty (not the case here, just a thought experiment)
- There would exist multiple non-zero twists that would map to a control vector of zero.
- These twists (other than the zero twist) would involve pure slipping. We have no way of actuating the slipping, thus they would be unachievable by setting the wheel velocities.
Notice that $R$, in this case is rank(2), and it's null space is $\begin{bmatrix}0 \\ v_x \\ 0\end{bmatrix}$.
- Any body twist with velocity purely in the $x$ direction results in zero sliding of the wheels.
- If the body twist involves $y$ motion or rotational motion, the wheels necessarily slide.

Car with Four Conventional Wheels

Body Twist to Wheel Motion

Wheel 1

\begin{align} \begin{bmatrix} \dot{\theta} \\ r \dot{\phi}_1 \\ 0 \end{bmatrix} = \begin{bmatrix} \dot{\theta} \\ -D\dot{\theta} + v_x\\ L\dot{\theta} + v_y \end{bmatrix} \end{align}

Therefore:

\begin{equation} \begin{bmatrix} \dot{\phi}_1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{-D}{r} & \frac{1}{r} & 0\\ L & 0 & 1 \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{equation}

The other wheels are similar.

Inverse Kinematics

The kinematic equations are now

\begin{align} \dot{\phi} &= H V_b \\ \begin{bmatrix} \dot{\phi}_1\\ \dot{\phi}_2\\ \dot{\phi}_3\\ \dot{\phi}_4\\ \end{bmatrix} &= \frac{1}{r} \begin{bmatrix} -D & 1 & 0\\ D & 1 & 0\\ D & 1 & 0\\ -D & 1 & 0\\ \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{align}

and

\begin{align}\label{eq:slide} 0 &= QV_b\\ \begin{bmatrix} 0\\0\\0\\0 \end{bmatrix} &= \begin{bmatrix} L & 0 & 1 \\ L & 0 & 1 \\ -L & 0 & 1 \\ -L & 0 & 1 \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{align}

The H matrix has a rank of 2. Since it has 3 columns, this means it's null space has dimension 1.
The Null space of H is spanned by $\begin{bmatrix}0\\0\\1\end{bmatrix}$, which means that any $v_y$ component of the twist maps to zero wheel speed. For example, if the body of the robot were to be dragged along the $y$ axis, the wheels would not roll.
The sliding equation \eqref{eq:slide} adds an additional constraints to the motion.
To satisfy the constraints, $V_b$ must be in the Null space of $Q$ which is $\begin{bmatrix}0\\1\\0\end{bmatrix}$.
Thus only translating in the $x$ direction is permitted. In other words, it is not possible for this vehicle to rotate or move in the $y$ direction without violating this constraint.

Forward Kinematics

The forward kinematics relates the controls applied to the wheels to the body twist of the robot.

The control to twist relationship is found by solving $u = H V_b$ for $V_b$.

If $u$ is in the column space of $H$ (i.e., $u$ is either 0 or a control that actually causes movement) then $V_b = H^{\dagger}u$, where $H^{\dagger}$ is the Moore-Penrose Pseudoinverse.

Example: Four Omni-Directional Wheels

\begin{align} V_b = H^{\dagger}u &= \frac{r}{4} \begin{bmatrix} -\frac{1}{D+L} & \frac{1}{D+L} & \frac{1}{D+L} & -\frac{1}{D+L} \\ 1 & 1 & 1 & 1\\ -1 & 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} \end{align}

Analysis

The rank of $H^{\dagger}$ is 3, which means its its Null space dimension is 1.
$\mathrm{Null}(H^{\dagger})$ is spanned by $\begin{bmatrix}-1 \\ -1 \\ 1 \\1\end{bmatrix}$
The Null space corresponds to both front wheels spinning in the opposite direction from the rear wheels.
- This motion, intuitively, would put the chassis under compression or tension along the x axis and result in 0 net motion.
If $u$ is in the null space of $H^{\dagger}$ it is also in the null space of $H^T$ and is therefore orthogonal to every vector in the column space of $H$
- Therefore, there is no solution to $u = H V_b$ in this case.
- In other words, no body twist can be applied to the robot that makes the front wheels spin in the opposite direction of the rear wheels.

To see how the controls affect slipping, we write $v_s = RH^{\dagger}u$:

If you perform this operation, you will find that $Q=RH^{\dagger}$ is a $4\times4$ matrix with rank 2.
- Its nullspace is spanned by $\begin{bmatrix}0\\0\\1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\1\\0\\0\end{bmatrix}$.
- Thus there is no sideways sliding whenever the front wheels are moving at the same speed and the rear wheels are moving at the same speed.

World Frame Velocity

So far, we have solved for the body twist in terms of the wheel rotational velocities and vice versa. To be truly useful, we need to find the relationship between the wheel velocities and $\dot{q}$.

Since $\dot{q} = $(\dot{\theta}, \dot{x}, \dot{y})$ is the twist corresponding to a frame located at the body frame but aligned with the world frame, we can use the relationship $\dot{q} = A(\theta, 0, 0)V_b$ to relate the equations above to this velocity.

Example: Four wheel Robot

\begin{align} \begin{bmatrix} \dot{\theta}\\ \dot{x}\\ \dot{y} \end{bmatrix} &=\frac{r}{4} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} -\frac{1}{D+L} & \frac{1}{D+L} & \frac{1}{D+L} & -\frac{1}{D+L} \\ 1 & 1 & 1 & 1\\ -1 & 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} u_1\\u_2\\u_3\\u_4 \end{bmatrix}\\ &= \frac{r}{4} \begin{bmatrix} -\frac{1}{D+L} & \frac{1}{D+L} & \frac{1}{D+L} & -\frac{1}{D+L}\\ \cos\theta + \sin\theta & \cos\theta-\sin\theta & \cos\theta + \sin\theta & \cos\theta-\sin\theta \\ \sin\theta - \cos\theta & \cos\theta+\sin\theta & \sin\theta - \cos\theta & \cos\theta+\sin\theta \\ \end{bmatrix} \begin{bmatrix} u_1\\u_2\\u_3\\u_4 \end{bmatrix} \end{align}

The Unicycle Robot

A simple model for a mobile robot is a single wheel, rolling on the plane
Its steering angle and rotational velocity can be controlled.
It rolls without slipping

The method above is likely more complicated than needed to derive the kinematics for this system, but here goes:

There is only one wheel, its reference frame is coincident with the body frame, so all transforms/adjoints are the identity matrix.
We will let $\psi_1 = 0$, since the whole chassis of the robot is the wheel, hence the wheel cannot have an angle independent of the chassis angle.
The body frame twist in the wheel frame
\begin{equation} \begin{bmatrix} \dot{\theta}\\ v_{x1}\\ v_{y1} \end{bmatrix} = \begin{bmatrix} \dot{\theta}\\ v_x\\ v_y \end{bmatrix} \end{equation}
Substituting in the wheel constraints, with $\psi_1 = 0$ yields
\begin{equation} \begin{bmatrix} \dot{\theta}\\ \dot{\phi}_1\\ 0 \end{bmatrix}= \begin{bmatrix} \dot{\theta}\\ \frac{1}{r}v_x\\ v_y \end{bmatrix} \end{equation}
If our controls are $u_1 = \dot{\theta}$ and $u_2 = \dot{\phi_1}$ we then have the control equation
\begin{equation} \begin{bmatrix} u_1\\ u_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{1}{r} & 0\\ \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} \end{equation}
and the constraint equation $0 = v_y$.
The constraints indicate that the $v_y$ velocity is always 0, hence we can never accomplish a $y$ velocity in a body twist and therefore the robot is not omni-directional.
Taking the pseudo-inverse of the control matrix yields
\begin{equation} \begin{bmatrix} \dot{\theta}\\ v_x \\ v_y \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & r \\ 0 & 0 \end{bmatrix} \begin{bmatrix} u_1\\ u_2 \end{bmatrix} \end{equation}
from which we can see that no control can every give us a $v_y$ that is non-zero.
Finally, we should convert the body twist into $\dot{q}$:
\begin{align}\label{eq:gencar} \begin{bmatrix} \dot{q}\\ \dot{x}\\ \dot{y} \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \dot{\theta}\\ v_x\\ v_y \end{bmatrix}\\ &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & r \\ 0 & 0 \end{bmatrix} \begin{bmatrix} u_1\\ u_2 \end{bmatrix}\\ &=\begin{bmatrix} 1 & 0 \\ 0 & r \cos\theta \\ 0 & r \sin\theta \end{bmatrix} \begin{bmatrix} u_1\\ u_2 \end{bmatrix} \end{align}

Note that in the derivation above, we derived $\dot{q}$ in two steps: first (Step 7) we mapped control inputs to body twists. Then (Step 8) we mapped body twists to $\dot{q}$.

When we map the control inputs to the body twists, we see that we actually only can specify $\dot{\theta}$ and $v_x$, whereas $v_y = 0$.
There are many robots whose mappings to the body twist also have this property, even though the exact mapping from control inputs to body twist is different
Thus, we can create a generalized car model simply by using equation \eqref{eq:gencar}, (also note, we will drop $v_y$ and its associated column, since it is 0
\begin{equation} \begin{bmatrix} \dot{q}\\ \dot{x}\\ \dot{y} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & \cos\theta \\ 0 & \sin\theta \end{bmatrix} \begin{bmatrix} w \\ v \end{bmatrix}\\ \end{equation}
where the canonical control inputs $w = \dot{\theta}$ and $v = v_x$ are used
Although many types of wheel configurations lead to this model, wheel configuration matters when the original inputs have constraints (such as a maximum velocity constraint). Then the different mappings from original control inputs to canonical inputs leads to different minimum and maximum values for $w$ and $v$.

References:

Modern Robotics: Mechanics, Planning and Control, Kevin Lynch and Frank Park, Cambridge University Press, 2017