A Symmetric GRG Kernel for Area

Jun 17, 2026

A Symmetric GRG Kernel for Area

The binary radial formulation combines complementary internal and external sector triangles. The area kernel uses one internal contribution and two external contributions, unlike the circumference kernel’s two-to-one chord–tangent balance.

Geometric Elements

Resolve one Turn into N equal symmetric sectors. Let a be the internal triangle’s height, h half its chord, and e half the external tangent side. Define q = h/a.

$$
N=2^b,\quad b\ge2,\quad R>0,\qquad a^2+h^2=R^2,\quad e=R\frac ha.
$$

$$
I_b=ah=R^2\frac{q_b}{1+q_b^2},\qquad E_b=Re=R^2q_b.
$$

Two Area Kernels

$$
A_m=\frac N2(I_b+E_b),\qquad A_w=\frac N3(I_b+2E_b).
$$

Accuracy for the Defined Triangles

The current formulation gives an analytical verification using a local integral as an exact sector reference, not as a kernel input.

$$
F(q)=\int_0^q\frac{dt}{1+t^2},\qquad A_T(R)=NR^2F(q_b).
$$

$$
\begin{aligned}A_m-A_T&=NR^2\left(-\frac{q_b^3}{6}+\frac{3q_b^5}{10}+O(q_b^7)\right),\\A_w-A_T&=NR^2\left(\frac{2q_b^5}{15}+O(q_b^7)\right).\end{aligned}
$$

Regular geometric bisection yields a scale bound. Together with the expansion, it proves second-order total error for the arithmetic kernel and fourth-order total error for the weighted kernel in exact arithmetic.

$$
0\lt q_b\le\frac4N,\qquad A_m-A_T=O(N^{-2}),\qquad A_w-A_T=O(N^{-4}).
$$

The same analysis proves the following bracket for these sectors:

$$
A_{\mathrm{in}}\le A_m\le A_T\le A_w\le A_{\mathrm{out}}.
$$

Sampling and Scope

$$
\Delta u=\frac1{4N}.
$$

The earlier version introduced this quarter-step sampling parameter inside each module. The proof here concerns the complete sector triangles defined above. A different sample-based area construction requires its own explicit definition and proof.

Internal–external weights are reversed between the circumference and area kernels. This diagram is schematic; the triangle definitions determine the calculation.

Research Status

The cancellation is established for the regular triangular elements above. This does not establish original authorship of the weighting, numerical stability of an implementation, or a computational advantage. Those questions require separate literature and implementation comparisons.

Earlier revisions of this page described fourth-order behaviour as a candidate. The current analytical clarification identifies the geometric assumptions under which that behaviour follows.

The Circle

Aug 18, 2025

The Circle — Radius-First (Three Methods)

Principle

The circle is generated from the center. The radius is the only primitive; the perimeter is the path traced by the radius, and the area is the surface swept by its rotation. We work in turns (fractions of a full rotation), not in sexagesimal degrees.

Angles in turns & radial node

Let $u\in[0,1)$ be an angle in turns (period $1$). We use period-1 trigonometric functions $c(u), s(u)$ with $c(0)=1, s(0)=0$ and $(c(\tfrac14), s(\tfrac14))=(0,1)$. The exact radial node is:

$$ P(u)=\bigl(r\,c(u),\; r\,s(u)\bigr). $$

Bit-friendly turns

Use fixed-point angles $u = k/2^B$. Picking $N$ that divides $2^B$ makes equal-angle steps exact ($\Delta u=1/N$).

Optional: LUT + interpolation for $c(u), s(u)$.

Method scope & scaling (quick key)

Radial Sectors (equal-angle, full turn) → no ×4.
Δx Trapezoidal (quarter tiling) → ×4 to get the full circle.
Original Projection (midpoint/right on a quadrant) → ×4.

Perimeter (equal-angle chords) is naturally full-turn (no ×4) and shared across methods.

Perimeter (shared across methods)

Equal-angle chords (full turn)

$$ \boxed{\, C_N \;=\; 2\,N\,r\,s\!\Bigl(\tfrac{1}{2N}\Bigr) \,} $$

Inscribed regular $N$-gon in turns. Monotone ↑ in $N$. No quadrant scaling.

Perimeter by equal-angle chords: $C_N = 2N r\, s(1/(2N))$ (full turn).

Area — three interchangeable builds (pick one)

Δx Trapezoidal (quarter → ×4)

$$ \Delta x=\frac{r}{N},\quad x_k=k\,\Delta x,\quad y_k=\sqrt{r^2-x_k^2} $$

$$ A_{\text{quarter}}(N)=\frac{\Delta x}{2}\!\left(y_0+2\sum_{k=1}^{N-1}y_k+y_N\right),\qquad A(N)=4\,A_{\text{quarter}}(N). $$

Under-approx on a concave arc; smooth convergence $O(N^{-2})$.

Original Projection (midpoint/right corridor, quarter → ×4)

$$ \Delta x=\frac{r}{N},\ \ x_{k+\tfrac12}=(k+\tfrac12)\Delta x,\ \ y(x)=\sqrt{r^2-x^2} $$

$$ A^{\text{quad}}_{\text{mid}}=\sum_{k=0}^{N-1} y\!\bigl(x_{k+\tfrac12}\bigr)\,\Delta x,\qquad A^{\text{quad}}_{\text{right}}=\sum_{k=0}^{N-1} y\!\bigl(x_{k+1}\bigr)\,\Delta x, $$

$$ A_{\text{mid/right}}=4\,A^{\text{quad}}_{\text{mid/right}},\qquad A_{\text{right}}\le A_{\text{true}}\le A_{\text{mid}}. $$

Midpoint is second-order; corridor shrinks with $N$.

Radial Sectors (equal-angle sectors, full turn)

$$ \boxed{\, A_N = \tfrac{N}{2}\,r^2\,s\!\Bigl(\tfrac{1}{N}\Bigr) \,} $$

Sum of $N$ congruent isosceles triangles with included angle $1/N$ turns (inscribed lower; monotone ↑ in $N$). No ×4.

Quarter area by Δx trapezoids; multiply by 4 for the full circle.

Radial sectors: $A_N=\tfrac{N}{2}r^2\,s(1/N)$ (full turn, no ×4).

Practical notes

Pick $N$. If angles are $k/2^B$ turns, choose $N\mid 2^B$. Use step-doubling ($N\!\to\!2N$) to gauge the gap.
Monotonicity. $C_N$ ↑ with $N$. Δx trapezoids ↑ toward area (under-tiling). Original corridor shrinks. Radial sectors ↑ (inscribed).
All in turns. No degrees, no π in the constructive steps.