Please define which angle you're talking about. You appear to be referring to a different $$\phi_v$$ than is defined in my diagram.

I have defined the angle $$\phi_v$$ as an

**arbitrary** angle, call it $$\phi$$ if it confuses you. In the particular case I have explained to you, it is the angle between the velocity of the mirror and its plane, both measured in the frame S of the axle. As such, as

**already** explained (and agreed upon) $$\phi=0$$. In the frame of the ground S', $$\phi'=0$$, courtesy of the way angles transform between the two frames .

It seems you've given up thinking altogether. A simple application of common sense immediately shows that the microfacets do not move tangent to the ellipse in the ground frame.

Immediately next to the ground contact point, for example, the almost horizontal facets move almost vertically.

You will need to use math instead of insults in order to prove your point.

The equation of the ellipse, as given to you earlier on is

here. The parametric representation of the ellipse is:

$$x'=\gamma(V)(rsin(\omega t + \theta)+Vt)$$

$$y'=r(1+cos(\omega t + \theta))$$

I am quite sure that I have posted the above several times already.

The components of the velocity are given a few posts down the page, in post 39. It is easy to verify that the components of the velocity line up with the components of the vector tangent to the ellipse, i.e. the plane of the microfacet.

The vertical facets at the leading and trailing extremities of the ellipse have a horizontal compenent to their velocity.

Yes, you have repeated this true but irrelevant statement in order to further your point several times already. What matters is not the respective components, what matters is , when taken together, the velocity components line up with the tangent to the ellipse.

Rubbish. The orientation of the leading microfacet is vertical, otherwise it would not be leading.

Same nonsense repeated multiple times doesn't form a valid argument.