Understanding SE(3) Twists I am just learning about twists to represent 3D velocities, and I have two questions:
1) Wikipedia defines a twist as "an angular velocity around an axis and a linear velocity along this axis". To represent a twist mathematically, this requires defining a point in 3D space (3 coordinates), the direction of the axis passing through this point (2 coordinates), and the ratio of the linear and angular velocities. This is 6 numbers in total. However, it seems that you could also represent this same 3D velocity just by defining linear velocity about the axes (3 coordinates), and a rotational velocity about the three axes (3 coordinates), so that the linear and rotational velocities are defined independently. This is also 6 numbers in total. So why do we define 3D velocities as twists, rather than using my version? Is it just that there are some nice mathematical properties of twists that my version does not have? Or is it that my version is actually fundamentally wrong in some way?
2) Let's say I have an object with positive x- and y- linear velocity, but zero z- linear velocity. The object is also spinning around its z-axis, but there is no rotation about the x- and y- axes. You can think of this as a ball sliding across the floor, whilst spinning about its vertical axis. I am struggling to understand how it is possible to represent this 3D velocity using "an angular velocity around an axis and a linear velocity along this axis". In this example, it seems clear to me that the linear velocity is acting orthogonally to the axis of rotation of the angular velocity. The ball is rotating about its z-axis, and so it doesn't seem possible to also define the linear velocity in terms of this z-axis, because the linear velocity only has components in the x- and y- axes. What am I missing?
Thank you!
 A: I read the first part of your question several times but it was not that clear to me. However, I believe that the problem is that you have mixed up the concept of vehicular velocity (which is a 3-dimensional vector) and twist (which can be seen as a 6-dimensional vector). If you had more specific questions I will edit the post accordingly.
1. Assume a one-parameter spatial motion where $\Sigma_0$ is the fixed system and $\Sigma$ is our moving system. Now, Apply this motion to a point $X$ with position vector $\mathbf{x}$ in the moving system $\Sigma$. The the position vector of the aforementioned point will be $\mathbf{x}_0$ in the fixed system $\Sigma_0$ and is obtained as follows:
$$\mathbf{x}_0 = \mathbf{A}\left(t\right)\,\mathbf{x} + \mathbf{b}\left(t\right)$$
where $A(t)$ is a path on $\mathrm{SO}(3)$ and $b(t)$ is a path in $\mathbb{R}^3$. Now, having in mind that $\mathbf{x} = \mathbf{A}(t)^T\left(\mathbf{x}_0 - \mathbf{b}(t)\right)$ holds, one can derive a velocity vector (called vehicular velocity) in the following way (we just omit the variable $t$ for the sake of simplicity)
\begin{align}
\begin{aligned}
\dot{\mathbf{x}}_0 &= \dot{\mathbf{A}}\mathbf{x} + \dot{\mathbf{b}}\\
                   &= \dot{\mathbf{A}}\mathbf{A}^T\left(\mathbf{x}_0 - \mathbf{b}\right) + \dot{\mathbf{b}}\\
                   &= \dot{\mathbf{A}}\mathbf{A}^T\mathbf{x}_0 + \mathbf{\dot{\mathbf{b}}} - \dot{\mathbf{A}}\mathbf{A}^T\mathbf{b}
\label{eq}
\end{aligned}
\end{align}
Since $\mathbf{A}\in\mathrm{SO}(3)$ we have $\dot{\mathbf{A}}\mathbf{A}^T = -(\dot{\mathbf{A}}\mathbf{A}^T)^T$ and hence it gives
$$\dot{\mathbf{A}}\mathbf{A}^T = \left(\begin{array}{ccc} 
0    & -q_3 & q_2\\
q_3  & 0    & -q_1\\
-q_2 & q_1  & 0
\end{array}\right).$$
Now, considering the concept of cross product matrix, the vehicular velocity $V_f$ will be
$$V_f := \dot{\mathbf{x}}_0 = \mathbf{q}\times\mathbf{x}_0 + \widehat{\mathbf{q}},$$
where $\mathbf{q} = \left(q_1,q_2,q_3\right)^T$ and $\widehat{\mathbf{q}} = \mathbf{\dot{\mathbf{b}}} - \dot{\mathbf{A}}\mathbf{A}^T\mathbf{b}$. Note that $V_f$ is indeed a 3-dimensional vector. On the other hand a twist is usually defined as $\left(\mathbf{q},\widehat{\mathbf{q}}\right)^T$ and hence a 6-dimensional vector. In an algebraic way the twist is defined as the following dual vector
$$\underline{\mathbf{q}} = \mathbf{q} + \epsilon\,\widehat{\mathbf{q}},$$
where $\epsilon$ is a nilpotent element with the property $\epsilon^2 = 0$. Writing the twist in this way allows it to fall within the language of screw theory (which gives some certain theoretical and computational advantages. As an example, one can easily do spatial transformations via dual quaternions, c.f. Dual Quaternions). Finally, to summarize this part, we know that every spatial motion is a screw motion (Chasles theorem) where the object rotates around an axis and moves along this axis. In this way the twist will describe the linear velocity and angular velocity of this screw motion while vehicular velocity is the tangent vector to the helical curve (that is obtained by applying the spatial motion to a sample point $X$).
2. The spinning ball, as defined in the problem, in fact performs a planar Euclidean motion (which can be thought of as a path in $\mathrm{SE}(2)$). In order to obtain its twist and vehicular velocity one only needs to do the following substitutions in the above computations:
$$\mathbf{A} = \left(\begin{array}{ccc} 
\cos(\theta(t))  & -\sin(\theta(t)) & 0\\
\sin(\theta(t))  & \cos(\theta(t))  & 0\\
0             & 0             & 1
\end{array}\right),\quad
\mathbf{b} = 
\left(
\begin{array}{c}
b_1(t)\\
b_2(t)\\
0
\end{array}
\right).
$$
Doing the calculations reveals the following twist (which describes the angular and linear velocities)
$$\underline{\mathbf{q}} = \left(
\begin{array}{c}
0\\
0\\
\dot{\theta}
\end{array}
\right) + 
\epsilon\,
\left(
\begin{array}{c}
\dot{\theta}\,b_2 + \dot{b}_1\\
-\dot{\theta}\,b_1 + \dot{b}_2\\
0
\end{array}
\right).$$
Applying the twist to an arbitrary sample point on the ball with coordinates $(x,y,z)^T$ gives the follwoing vehicular velocity
$$\left(
\begin{array}{c}
(b_2 - y)\dot{\theta} + b_1\\
(x - b_1)\dot{\theta} + b_2\\
0
\end{array}
\right).$$
Finally, as the last remark, note the involvement of $\dot{\theta}$ in both angular and linear velocity components of the twist (this must be an answer to the last part of your question in part 2).
