Isn't the invariance of the zero angle what we have been talking about? What "measurement of orientation" is included in this debate?

By orientation, I mean the "angle of the microfacet" in the OP.

Good, ready to point at the link where you uploaded them?

I said I'm good to go.

**Notation:**
$$\vec{P}(t)$$ is the position vector of wheel element

**P** at time t in

**S**.

$$\vec{P'}(t')$$ is the position vector of wheel element

**P** at time t' in

**S'**.

$$\hat{P_t}(t)$$ is the unit displacement vector tangent to wheel element

**P** at time t in

**S**.

$$\hat{P_t}'(t')$$ is the unit displacement vector tangent to wheel element

**P** at time t' in

**S'**.

$$\vec{v_p}(t)$$ is the velocity vector of wheel element

**P** at time t in

**S**.

$$\vec{v_p}'(t')$$ is the velocity vector of wheel element

**P** at time t' in

**S'**.

$$\vec{T_1}(t,l)$$ is the position vectors of the elements of rod

**T1** at time t in

**S**. $$l$$ is distance along the rod.

Let $$v$$ denote the magnitude of the velocity of

**S'** relative to

**S** (this velocity is in the direction of the positive x-axis).

Let $$t'_0$$ denote the time in

**S'** when the rod

**T1** is touching wheel element

**P**.

**Pete's method recap:**
Determine whether $$\vec{v_p}(t=0)$$ is a scalar multiple of (ie is parallel to) $$\hat{P_t}(t=0)$$.

Determine whether $$\vec{v_p}'(t'=t'_0)$$ is a scalar multiple of (ie is parallel to) $$\hat{P_t}'(t'=t'_0)$$.

$$\vec{v_p}(t=0)$$ and $$\vec{v_p}'(t'=t'_0)$$ can be derived from $$\vec{P}(t)$$, the worldline of

**P**.

$$\hat{P_t}(t=0)$$ is obvious, but can be formally derived from the definition the wheel at t=0 if necessary.

Rod

**T1** is defined in terms of $$\hat{P_t}(t=0)$$.

$$\hat{P_t}'(t'=t'_0)$$ can be derived from the transformed

**T1**.

**Calculations:**
**Finding $$\vec{v_p}(t=0)$$**

This is trivial, as are much of the following calculations, but spelling it out points toward a general solution for times other than t=0.

$$\begin{align}

\vec{P}(t) &= \begin{pmatrix} r\cos(\omega t) \\ r\sin(\omega t)\end{pmatrix} \\

\vec{v_p}(t) &= \frac{d}{dt} \vec{P}(t) \\

&= \begin{pmatrix} -r\omega\sin(\omega t) \\ r\omega\cos(\omega t) \end{pmatrix} \\

\vec{v_p}(t=0) &= \begin{pmatrix} 0 \\ r\omega \end{pmatrix}

\end{align}$$
**Finding $$\vec{v_p}'(t'=t'_0)$$**

Transforming $$\vec{P}(t)$$ gives...

$$\vec{P'}(t) = \begin{pmatrix}\gamma(r\cos(\omega t) - vt) \\ r\sin(\omega t) \end{pmatrix}$$

Note that this is expressed as a function of t rather than t'. Given that

$$t = \frac{t'}{\gamma} - \frac{v}{c^2}r\cos(\omega t)$$

, I can't find a general closed form expression for $$\vec{P'}(t')$$. The same applies to the derivative:

$$\begin{align}

\vec{v_p}'(t) &= \frac{d}{dt'} \vec{P'}(t) \\

&= \begin{pmatrix}\frac{-r\omega\sin(\omega t) - v}{1 + \frac{vr\omega}{c^2}\sin(\omega t)} \\ \frac{r\omega\cos(\omega t)}{\gamma(1 + \frac{vr\omega}{c^2}\sin(\omega t))} \end{pmatrix}

\end{align}$$

But, we can still calculate the result, since we know that $$t=0$$ at $$t'=t'_0$$ on the worldine of **P**.

$$\begin{align}

\vec{v_p}'(t'=t'_0) &= \vec{v_p}'(t=0) \\

&= \begin{pmatrix}-v \\ r\omega / \gamma \end{pmatrix}

\end{align}$$
**Finding $$\hat{P_t}(t=0)$$**

Trivially, $$\hat{P_t}(t=0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$

If necessary, this can be formally derived by considering the direction of the displacement vector between **P** and a simultaneous wheel element a short distance away, and taking the limit as the distance from **P** approaches zero.
**Finding $$\hat{P_t}'(t'=t'_0)$$**

Rod **T1**:

$$\begin {align}

\vec{T_1}(t,l) &= \vec{P}(0) + l\hat{P_t}(0) + t\vec{v_p}(0) \\

&= \begin{pmatrix} r \\ l + tr\omega\end{pmatrix}

\end{align}$$

Transforming to **S'**:

$$\vec{T1'}(t',l) = \begin{pmatrix} \frac{r}{\gamma} - vt' \\ l + r\omega(\frac{t'}{\gamma} +\frac{vr}{c^2})\end{pmatrix}$$

$$\hat{P_t}'(t'=t'_0)$$ is parallel to the displacement vector between two points on the rod at $$t'=t'_0$$ with different values of $$l$$:

$$\begin{align}

\vec{T_1}'(t'=t'_0,l=l_0) &= \begin{pmatrix} \frac{r}{\gamma} - vt'_0 \\ l_0 + r\omega(\frac{t'_0}{\gamma} +\frac{vr}{c^2})\end{pmatrix} \\

\vec{T_1}'(t'=t'_0,l=l_1) &= \begin{pmatrix} \frac{r}{\gamma} - vt'_0 \\ l_1 + r\omega(\frac{t'_0}{\gamma} +\frac{vr}{c^2})\end{pmatrix} \\

\hat{P_t}'(t'=t'_0) &= \frac{\vec{T_1}'(t'=t'_0,l=l_0) - \vec{T_1}'(t'=t'_0,l=l_1)}{\left\|\vec{T_1}'(t'=t'_0,l=l_0) - \vec{T_1}'(t'=t'_0,l=l_1)\right\|

\end{align}$$

$$ \hat{P_t}'(t'=t'_0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$

**Results**
$$\vec{v_p}(t=0) = \begin{pmatrix} 0 \\ r\omega \end{pmatrix}$$

...is parallel to...

$$\hat{P_t}(t=0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$

i.e., in

**S** at t=0, the microfacet at

**P** is parallel to the velocity of

**P**.

$$\vec{v_p}'(t'=t'_0) = \begin{pmatrix}-v \\ \frac{r\omega}{\gamma} \end{pmatrix}$$

...is not parallel to...

$$\hat{P_t}'(t'=t'_0) = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$

i.e. in

**S'** at t'=t'_0, the microfacet at

**P** is

*not* parallel to the velocity of

**P**.