Take RotD50 off the hull too, as a level set of the arc decomposition

[lispandfound]

Stacked on #221. Fifth in the stack.

The normal cones that gave RotD00 its minimum in #221 also cut the rotated peak into arcs — one per vertex of the origin-symmetric hull S, over each of which the peak is a single cosine radius · cos(x), x measured from that vertex's own direction. Those arcs tile the circle.

That makes the measure of directions whose peak is at or below a level a closed form, arc by arc: peak ≤ level holds exactly where |x| ≥ acos(level/radius), leaving at most two sub-arcs to clip against the arc's own bounds. RotD50 is the level where that measure is half the circle.

The sum of acos terms has no closed-form inverse — Σ acos(m/rₖ) = const with distinct rₖ — so the level comes from a safeguarded Newton root find rather than algebra. Exact in the sense any root is: to the last f64 the bracket resolves. The first guess is RotD100/√2, which the √2 bound puts at or below the answer, and a bisection takes over whenever a Newton step leaves the bracket or fails to halve it, capping the work at twice a plain bisection.

The √2 bound comes back exactly

This is the part I'd flagged as a wart in #221, and it's the main reason to do this.

f(θ) ≥ RotD100·|cos(θ−φ)| pointwise, and a pointwise inequality holds for quantiles too. The median of |cos| over the circle is cos(45°) exactly, because the directions where |cos| ≥ c measure 4·acos(c), which is half the circle at c = cos(45°). So RotD50 ≥ RotD100/√2.

	RotD100	RotD50	Bound
before #221	sampled	sampled	√2 exactly, but neither value exact
#221	exact	sampled	√2/cos(0.5°) ≈ 1.414267
here	exact	exact	√2 exactly

And linearly polarised records attain it at every angle now, not only at whole degrees — the β-dependent assertion I had to write in #221 goes away in both the Rust and the Python tests.

There's a second property worth naming: the statistics are now rotation invariant. Rotate a record by 0.3° and the grid RotD50 changes; the continuous one doesn't.

RotD50 has no orientation any more

As you asked. A continuous function reaches its median at an even number of directions, two at the fewest, so naming one would be arbitrary. The statistics row goes 6 → 5 columns, rotd50_orientation leaves ROTD_COMPONENTS, and psa_rotd180 returns 187 columns rather than 188.

What moves in the numbers

RotD50 shifts by 0.002% to 0.08% of itself across the seven benchmark records — measured, not estimated:

2024p950420_MWFS_HN_20   0.046%      3366146_DFHS_HN_20   0.014%
2024p220113_RLNS_HN_20   0.023%      3497857_HVSC_HN_20   0.002%
3497857_PARS_HN_20       0.077%      3528839_LPCC_HN_20   0.032%
nquist_freq              0.038%

Against a 1% benchmark tolerance, so the comparison needed no further slack beyond the RotD00/RotD100 allowance already in #221. Worth noting I can't give a clean a priori bound here the way I could for the extremes — a level set can have arbitrarily many components, so the sampled median's error has no half-degree argument behind it. The numbers above are measurement.

Verification

• arcs_tile_the_circle — the arc widths add to a full turn, on a square, a triangle, a noisy record and a 360-point circle. A mislabelled cone or an edge normal the wrong way round breaks the total.
• arc_integral_matches_the_perimeter — Cauchy's formula: integrating a convex body's support function over the circle gives its perimeter. The arcs put that integral in closed form, radius·(sin(end) − sin(start)) each, and a perimeter measured straight off the vertices has to match to 1e-12. This is the sharpest structural check of the decomposition.
• level_measure_matches_a_fine_sweep — closed-form measure against a count over 360,000 directions, at seven levels between the extremes.
• median_matches_a_fine_sweep — both directions: half of 360,000 sampled directions fall below RotD50, and the sweep's own median agrees.
• median_of_a_polarised_record_is_the_peak_over_root_two — the closed form the degenerate branch returns, at six polarisations on and off the grid.
• test_rotd_statistics now asserts the analytic answer. Two timesteps at (1,0) and (0,1) give a square symmetric hull; every normal cone spans 90°, so the median level solves acos(level) = 22.5° and RotD50 is cos(π/8) exactly. The 1 degree sweep could only reach 0.9238443540096138.
• The proptest suite checks on every generated record that RotD50 splits a 7200-direction grid in half.

Performance

No measurable change on either path. 200 stations × 20000 samples through RotD: 41–42 ms either side of this commit. pSA over 20 stations × 4000 samples × 4 periods: 25 ms either side.

rotd no longer sweeps at all — every statistic comes off the hull. That isn't visible in the timing because the cull over the raw timesteps dominates; the 180-angle sweep was about 3% of the work. rotd180_rows and psa_rotd180 still sweep, because the curve itself is what they return.

Verified

cargo test 49 passed, pytest 65 passed (97% coverage), plus ruff, ruff format --check, deptry, ty, numpydoc, cargo fmt --check, cargo clippy -D warnings, and vale clean across Python, Markdown and Rust.

🤖 Generated with Claude Code