A New GRG Formulation: Circle and Area from Binary Radial Modules

A New GRG Formulation: Circle and Area from Binary Radial Modules

Jun 16, 2026

A GRG Formulation: Circle and Area from a Binary Radial Framework

Generative Radial Geometry describes circular measures through a finite binary radial framework. Its working quantities are a radius, a full Turn, a binary resolution, and local chord–tangent constructions.

About the Current GRG Formulation

This page presents the documented binary radial formulation, with geometric definitions and an explicit verification of its measurement kernels. Earlier approaches, including Original Projection and previous Turns formulations, remain part of the research history.

The geometric construction, kernel accuracy, and computational efficiency are separate subjects. The analysis below establishes accuracy for regular symmetric sectors. It does not establish a performance advantage or prescribe an implementation.

The weighted area proposal is documented in A Symmetric GRG Kernel for Area. The geometric ratio notation and proofs below are an analytical clarification of the documented triangles and kernels.

Binary Turns and Resolution

A full rotation is one Turn. Resolve it into equal sectors. The nondegenerate chord–tangent formulas on this page require at least four sectors.

$$
N=2^b,\qquad b\ge2,\qquad R>0.
$$
$$
t_b=\frac1N,\qquad \text{half-sector width}=\frac1{2N}.
$$

One Symmetric Radial Sector

Let a be the distance from the centre to the internal chord, h half that chord, and e half the external tangent side. The tangent is perpendicular to the central radius. The two boundary rays intersect both the circle and this tangent.

$$
a^2+h^2=R^2,\qquad e=R\frac ha.
$$
Symmetric binary radial sector with chord and tangent side A circle with a symmetric radial sector. The internal chord lies between two points on the circle. The external tangent side lies on the tangent line perpendicular to the central radius. O R tᵦ tangent line external tangent side upper sector boundary lower sector boundary internal chord
Figure 1 — Symmetric sector: internal chord, central radius, and external tangent side.

Define dimensionless ratios from these constructed lengths:

$$
c_b=\frac aR,\qquad s_b=\frac hR,\qquad q_b=\frac ha.
$$
$$
c_b^2+s_b^2=1,\qquad q_b=\frac{s_b}{c_b},\qquad c_b=\frac1{\sqrt{1+q_b^2}}.
$$

These symbols denote geometric ratios. They do not specify a trigonometric software routine.

Circumference Kernels

$$
L_i=2h,\qquad L_t=2e.
$$

The arithmetic balance and the classical Snell–Huygens weighted balance give:

$$
\begin{aligned}C_m&=\frac N2(L_i+L_t)=NR(s_b+q_b),\\C_{SH}&=\frac N3(2L_i+L_t)=\frac{2NR}{3}(2s_b+q_b).\end{aligned}
$$

The weighting is a known local arc approximation. This formulation uses it as a kernel within the binary radial framework; the weighting itself is not claimed as an original GRG result.

Area Elements and Kernels

The internal triangle has base 2h and height a. The external triangle has base 2e and height R.

$$
\begin{aligned}I_b&=ah=R^2c_bs_b=R^2\frac{q_b}{1+q_b^2},\\E_b&=Re=R^2q_b.\end{aligned}
$$
$$
A_{\mathrm{in}}=NI_b,\qquad A_{\mathrm{out}}=NE_b.
$$

The arithmetic area estimate and the proposed internal–external weighting are:

$$
\begin{aligned}A_m&=\frac N2(I_b+E_b),\\A_w&=\frac N3(I_b+2E_b).\end{aligned}
$$
$$
A_w=\frac{NR^2}{3}\left(\frac{q_b}{1+q_b^2}+2q_b\right).
$$

Geometric Refinement Bound

The following identity is used to verify regular equal-sector bisection. It is a geometric consequence, not an attribution of a particular numerical algorithm to GRG.

At four sectors the half-sector triangle has equal legs, so q₂ = 1. Bisecting its direction gives the following ratio, obtained by normalising the boundary direction and adding it to the central unit direction:

$$
q_{b+1}=\frac{q_b}{1+\sqrt{1+q_b^2}}.
$$

The denominator is at least two. Induction therefore gives:

$$
0\lt q_b\le 2^{2-b}=\frac4N.
$$

This proves the scale bound needed for the global error estimates. It uses the regular bisection geometry; it does not require a precomputed full-Turn length or area.

Kernel Accuracy: Analytical Verification

Define the local integral below solely as an analytical reference for the exact sector measures. It is not an input to the measurement kernels.

$$
F(q)=\int_0^q\frac{dt}{1+t^2}.
$$

Parameterising the circular boundary by its slope gives exact sector arc length 2RF(q) and exact sector area R²F(q). Thus the fixed full-Turn reference measures satisfy:

$$
C_T(R)=2NRF(q_b),\qquad A_T(R)=NR^2F(q_b).
$$
$$
F(q)=q-\frac{q^3}{3}+\frac{q^5}{5}+O(q^7).
$$

Substitution into the circumference kernels yields:

$$
\begin{aligned}C_m-C_T&=NR\left(\frac{q_b^3}{6}-\frac{q_b^5}{40}+O(q_b^7)\right),\\C_{SH}-C_T&=NR\left(\frac{q_b^5}{10}+O(q_b^7)\right).\end{aligned}
$$

For area:

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

The area weighting cancels the cubic local residual. Combining these expansions with the geometric scale bound proves, for fixed radius and exact arithmetic:

$$
\begin{aligned}C_m-C_T&=O(N^{-2}),& C_{SH}-C_T&=O(N^{-4}),\\A_m-A_T&=O(N^{-2}),& A_w-A_T&=O(N^{-4}).\end{aligned}
$$

These are asymptotic error statements, not claims of exact equality at finite resolution or guarantees about floating-point performance.

Area Bounds

The geometric inner and outer triangles bound each exact sector. The two balanced area estimates provide a tighter bracket as well:

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

For a direct verification, write the normalised kernels as M(q) and W(q). Both differences vanish at zero, and:

$$
\begin{aligned}\frac{d}{dq}\bigl(F(q)-M(q)\bigr)&=\frac{q^2(1-q^2)}{2(1+q^2)^2},\\\frac{d}{dq}\bigl(W(q)-F(q)\bigr)&=\frac{2q^4}{3(1+q^2)^2}.\end{aligned}
$$

Both derivatives are nonnegative for 0 ≤ q ≤ 1, which covers the sectors considered here. Consequently the bracket follows for every b ≥ 2, rather than only for tested resolutions.

Refinement and Quarter-Step Sampling

$$
N_{b+1}=2N_b,\qquad t_{b+1}=\frac{t_b}{2},\qquad \Delta u=\frac{t_b}{4}=\frac1{4N}.
$$

The related area article specifies quarter-step sampling within a module. The verification above concerns the complete sector triangles. An implementation using internal samples must specify how those samples evaluate the same elements; the sampling parameter alone does not define an additional area kernel.

Binary refinement of a radial circle Three circles divided into increasing binary sector counts, showing refinement from coarse radial modules to a smoother circular construction. b = 2, N = 4 b = 3, N = 8 b = 4, N = 16
Figure 2 — Binary sector refinement at N = 4, 8, and 16.

Computational Efficiency and Research Scope

Binary resolution is compatible with binary angular indices and power-of-two scaling. A performance advantage for the complete calculation must nevertheless be measured.

A reproducible comparison must use equal accuracy targets, include generation or retrieval of the local measures, account for precomputation and memory, and specify arithmetic precision, rounding, hardware, and execution time. Stability in fixed-point arithmetic requires a separate error analysis.

The results established here concern the documented regular chord–tangent geometry and its kernels. Historical originality and implementation efficiency are separate investigations.

ERG – Radial (Turns)

ERG – Radial (Turns)

Sep 6, 2025

ERG — Radial (Turns)

Method block (turns-based). Angles are measured in turns (fractions of a full rotation, u∈[0,1)). We write c(u), s(u) for period-1 cosine/sine in turns (so c(0)=1, s(0)=0; c(1/4)=0, s(1/4)=1). No sexagesimal units and no external constants appear in the primary formulas.

Radial node

For a circle of radius r and an angle u (turns):

$$ P(u) \;=\; \bigl(r\,c(u),\; r\,s(u)\bigr), \qquad u\in[0,1). $$

Uniform partition over the full turn: $u_k = k/N$, step $\Delta u = 1/N$.

Fig. A — Radial node with turns arc Axes, a circle, two radii separated by an angle u in turns, an arc marking u, and the point P(u). u P(u) O
Fig. A — Radial point $P(u)=(r\,c(u),\,r\,s(u))$ with angle $u$ expressed in turns.

Circumference (equal-angle chords)

Each chord between $P(u_k)$ and $P(u_{k+1})$ has length $2r\,s(\Delta u/2)$. Sum of $N$ identical chords → product:

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

Interpretation: inscribed regular $N$-gon; $C_N$ increases with $N$ and converges to the circle’s circumference.

Fig. B — Regular N-gon from equal-angle chords (N=12) Circle of radius 100 with twelve equal-angle chords forming an inscribed regular polygon. N = 12, Δu = 1/N
Fig. B — Equal-angle chords tile the full turn (regular $N$-gon).

Area (equal-angle sectors)

The $N$ isosceles sectors (sides $r,r$ and angle $\Delta u$) each have area $\tfrac12 r^2 s(\Delta u)$. Sum → product:

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

$A_N$ increases with $N$ and converges to the circle’s area.

Fig. C — Sectors at equal angular step Two illustrative sectors on a circle, approximately 1/8 turn each, showing area by equal-angle sectors. Δu = 1/N (turns)
Fig. C — Area by sectors at equal angular step (sum of identical pieces).

How to use

  • Pick a precision N (e.g., 128–1024). If you store angles as fixed-point turns k/2^B, choose N dividing 2^B.
  • Step-doubling check: compare results at N vs 2N to gauge the remaining gap.

Pseudocode (2D)

Input: r, N
C = 2 * N * r * s( 1 / (2*N) )
A = 0.5 * N * r^2 * s( 1 / N )

Notes

  • Turns keep formulas simple and hardware-friendly (fixed-point angle indices, LUTs, SIMD).
  • Use this block when you want a pure angular build; for Cartesian builds see Δx Trapezoidal and Original Projection pages.

Angle unit & precision (turns, binary fixed-point)

We measure angles in turns: a full rotation is 1 turn. Hence quarter-turn = 1/4, half-turn = 1/2, etc. Trigonometric functions are period-1 in turns: c(u), s(u) with c(0)=1, s(0)=0, c(1/4)=0, s(1/4)=1.

Equal-angle partition

Choose an integer N (number of sectors) and set the angular step Δu = 1/N. The sampling angles are uk = k/N, k=0,…,N−1.

Binary fixed-point turns (B bits)

For hardware/software precision, store angles as u = k / 2^B turns with k ∈ {0,…,2^B−1}. The angular resolution is ΔuB = 1 / 2^B. If you want exactly N equal sectors, pick N that divides 2^B (e.g., N = 2^B), and use indices k = m·j with m = 2^B / N, j = 0,…,N−1.

Why more bits ⇒ more accuracy

  • Circumference (chords): CN = 2 N r · s(1/(2N)) is the perimeter of the inscribed regular N-gon and increases to the circle’s circumference. With N=2^B, the step halves when B→B+1 and the error typically drops ~×4 (quadratic in the step).
  • Area (sectors): AN = ½ N r² · s(1/N) is the area of N equal sectors and increases to the circle’s area; same step-squared behaviour.
0 1/8 1/4 3/8 1/2 5/8 3/4 7/8
Example with B = 3 bits ⇒ 23 = 8 ticks. Increasing B doubles the ticks and the precision.

Pseudocode (B-bit angles)

Input: r, B
N = 1 << B            // N = 2^B sectors
C = 2 * N * r * s( 1.0 / (2 * N) )
A = 0.5 * N * r^2 * s( 1.0 / N )

Implementation tip: if you keep an angle register k modulo 2^B, additions are wrap-around on B bits. For LUTs of size 2^M with M ≤ B, you can index by k >> (B−M).