\subsection{The projected phase across zero shear}
\label{sec:angular-projected-phase}

The reciprocal core radius depends on both shear and the
active wall. Its comparison with the physical transverse
radius can therefore be made uniformly across zero shear.
The useful cancellation is the same one encountered in
Section~\ref{sec:small-section-projection}: after rescaling,
the difference vanishes identically when the small wall
amplitudes and the scale vanish. Here the active wall keeps
the limiting state away from zero.

For $X=(n,s,w_2,w_3)$, let
\[
 \mathcal E(X)=\mathcal P(X)
       -\frac{\sqrt{s^2+(w_2-w_3)^2}}{\sqrt3},
\]
where $\mathcal P$ is given by~\eqref{eq:projection-phase}.
We use maximum norms as above.

\begin{lemma}
\label{lem:angular-projection-error}
Fix $c>0$ and $M_0\geq c$. There are $\delta>0$ and
$B\geq1$ such that, whenever
\begin{equation}
 \begin{gathered}
 0<a\leq\delta,\qquad |s|\leq M_0a,\qquad
 ca\leq w_2\leq M_0a,\\
 |n|,|w_3|\leq\delta a,
 \end{gathered}
 \label{eq:angular-projection-error-domain}
\end{equation}
$\mathcal E$ is $C^2$ near $X$, and
\begin{equation}
 |\mathcal E(X)|\leq B(a^2+|n|+|w_3|).
 \label{eq:angular-projection-error-value}
\end{equation}
For every $D\geq0$ and direction $v=(v_n,v_s,v_2,v_3)$
satisfying
\[
 |v_s|,|v_2|\leq D,\qquad
 |v_n|\leq D|n|/a,\qquad |v_3|\leq D|w_3|/a,
\]
the ordinary differential obeys
\begin{equation}
 |D\mathcal E(X)v|
      \leq2BD\left(a+\frac{|n|}{a}+\frac{|w_3|}{a}\right).
 \label{eq:angular-projection-error-derivative}
\end{equation}
\end{lemma}

\begin{proof}
Use the scaled function $F(u,z)$ of
\eqref{eq:projection-rescaling}, now on the compact set
\[
 \mathcal K_{\mathrm a}
    =\{z:|z_1|\leq M_0,\ c\leq z_2\leq M_0\}.
\]
At $u=0$, its three radicands are $1$,
$3(z_1^2+z_2^2)$ and $z_1^2+z_2^2$.
They are strictly positive on this compact slice.
Moreover $F(0,z)=0$. Compact smoothness and the mean
value theorem therefore give $\delta>0$, $B\geq1$ with
\begin{equation*}
 \begin{split}
 |F(u,z)|&\leq B\|u\|,\qquad \|DF(u,z)\|\leq B,\\
 |DF(u,z)(0,\delta z)|&\leq B\|u\|\|\delta z\|
       \quad(z\in\mathcal K_{\mathrm a},\ \|u\|\leq\delta).
 \end{split}
\end{equation*}
For the last bound, the tangential differential vanishes
at $u=0$; integrate its change in the normal direction,
using a bound on $D^2F$ on a compact neighborhood.

For fixed $a>0$, put
\[
 u=(a,n/a,w_3/a),\qquad z=(s/a,w_2/a),\qquad
 b_a=a+|n|/a+|w_3|/a.
\]
The domain conditions give $z\in\mathcal K_{\mathrm a}$
and $\|u\|\leq\delta$, while $\|u\|\leq b_a$.
By~\eqref{eq:projection-scaled-error},
$\mathcal E(X)=aF(u,z)$. Its smoothness and
\eqref{eq:angular-projection-error-value} follow at once.

In direction $v$, the scaled directions are
\[
 \delta u=(0,v_n/a,v_3/a),\qquad
 \delta z=(v_s/a,v_2/a).
\]
Separate these directions before estimating the differential:
\[
 |D\mathcal E(X)v|
    \leq aB\|\delta u\|+aB\|u\|\|\delta z\|
    \leq B(|v_n|+|v_3|)+BDb_a
    \leq2BDb_a.
\]
This is~\eqref{eq:angular-projection-error-derivative}.
\end{proof}

For the state~\eqref{eq:angular-projective-state}, define
\[
 f_{\mathrm p}(\rho)=\frac2{r(Y(\rho))},\qquad
 r(Y)=\sqrt{(M-1)^2+V^2}.
\]
The estimate $V\geq2$ keeps this radius positive
throughout the radial interval.

\begin{theorem}
\label{thm:angular-projected-phase}
There is $h_*>0$ such that for every $0<h\leq h_*$
and $L\geq0$ one can choose $K>0$ before choosing
$q>2$, with the following property. For all sufficiently
small $a>0$, uniformly for $|\theta|\leq L$,
$0\leq\zeta\leq a^2$ and $a\leq\rho\leq2a$,
$f_{\mathrm p}$ has an ordinary derivative and
\begin{equation}
 \begin{split}
 |f_{\mathrm p}-F_{\theta,\zeta}|&\leq Ka^2,\qquad
 |f_{\mathrm p}'-F_{\theta,\zeta}'|\leq Ka,\\
 \frac1{4\sqrt3}\leq f_{\mathrm p}'
       &\leq\frac{2L+3}{\sqrt3}+1.
 \end{split}
 \label{eq:angular-projected-phase-comparison}
\end{equation}
\end{theorem}

\begin{proof}
Take the height threshold below those in
Propositions~\ref{prop:angular-fast-phase}
and~\ref{prop:angular-cutoff-state}, and below $1/16$.
In Lemma~\ref{lem:angular-projection-error}, choose
$c=1/2$ and $M_0=M_L$. Its constants $\delta,B$
depend only on $L$. Put $D=8C(L+1)$.

At time $T$, Proposition~\ref{prop:angular-cutoff-state}
gives $n\leq a^2$, $w_3\leq2a^2$ and all the
directional bounds of the lemma with $v=X'$.
Choose $a$ so that $2a\leq\delta$. Then
$e(\rho)=\mathcal E(X(\rho))$ has an ordinary derivative,
and the lemma yields
\[
 |e|\leq4Ba^2,\qquad |e'|\leq8BDa.
\]
Both are bounded by $K_{\mathrm e}a^2$ and
$K_{\mathrm e}a$, respectively, with
$K_{\mathrm e}=8BD$ chosen before $q$.

The negative constraint branch and positive height persist
on an ordinary neighborhood of each initial point.
Equation~\eqref{eq:projection-phase-identity} therefore gives
the local identity
\[
 f_{\mathrm p}(\rho)=e(\rho)
       +\frac{\varrho(H_{h,T}(Z_{\theta,\zeta}(\rho)))}{\sqrt3}.
\]
It holds for derivatives as well as values.
For $a\leq1/9$, the condition $\zeta\leq a^2$
implies $\zeta\leq a/9$, so
Proposition~\ref{prop:angular-fast-phase} applies to the
second term. Set $K=K_{\mathrm e}+K_0$ to obtain the
two comparison estimates.

Finally decrease the scale threshold so that
$Ka\leq1/(4\sqrt3)$. Comparing the total error directly
with~\eqref{eq:angular-reference-slope} proves the
derivative bounds in~\eqref{eq:angular-projected-phase-comparison}.
\end{proof}

The actual reciprocal core radius is thus strictly increasing
along every radial family, including $\theta=0$ and
$\zeta=0$. The active wall supplies both the positive
reference slope and the regularity of the projection.
These estimates provide the initial radius and direction
for the remaining core passage and the next entrance.