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.

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.

$$
a^2+h^2=R^2,\qquad e=R\frac ha.
$$

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.

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.

Circumference & Sphere

Aug 23, 2025

Why “Certainly on the Circumference” (and on the Sphere)

Category: Debate

Most ways to “draw a circle” are approximations: they split the arc into chords, use regular polygons, or compute points from arc lengths (where $\pi$ eventually appears). In GRG we do the opposite: we do not measure the arc, we enforce the distance. We fix a radius $R$ and choose a geometric direction; the generated point is the endpoint of a segment of length $R$ from a center $O$. By definition, that point is at distance $R$ from $O$: therefore it is certainly on the circumference. No approximation, no $\pi$ needed.

What we prove, in plain words

In 2D, every point built as $(x,y)=(R\cos\varphi,\,R\sin\varphi)$ identically satisfies

$$x^{2}+y^{2}=R^{2}.$$

It does not “tend” to the circle: it is on the circle.

In 3D, take a unit direction $\mathbf{u}$ and set $\mathbf{P}=R\,\mathbf{u}$. By construction $\|\mathbf{P}\|=R$: the point is certainly on the sphere

$$x^{2}+y^{2}+z^{2}=R^{2}.$$

The heart of it is: distance first. If the distance from the center is enforced to be $R$, every generated point belongs to the locus of points at distance $R$. That is the very definition of circle/sphere.

“Then what are sine and cosine for?”

Here they are not approximate numbers: they are ratios of sides in similar right triangles. They serve as a language to express the Cartesian projections of the chosen direction: \[ x=R\cos\varphi,\qquad y=R\sin\varphi. \] Even if, in a digital controller, you compute them with finite precision, the geometric law you implement remains exact: the commanded points are those of the ideal object defined by $x^{2}+y^{2}=R^{2}$ (or $x^{2}+y^{2}+z^{2}=R^{2}$).

If you wish to avoid direct computation of $\sin$ and $\cos$ altogether, there is an equivalent incremental form: rotate the vector $(x,y)$ by multiplying it with an orthogonal matrix. Orthogonal matrices preserve the norm: every step remains on the circle by construction.

“What about error accumulating at each step?”

Distinguish carefully:

Geometric law (GRG): exact. The norm is preserved at each step (rotations), therefore there is no geometric drift.
Physical implementation: it can introduce errors (backlash, compliance, friction, encoder resolution, numerical rounding). These are real-world errors, not flaws of the idea.

In digital control, the small loss/gain of norm due to rounding is removed with three simple techniques:

$\text{(i) adequate precision}\quad \text{(ii) occasional renormalization enforcing }x^{2}+y^{2}=R^{2}\quad \text{(iii) periodic recomputation from }\varphi.$

In all cases, the reference object remains the perfect circle/sphere.

“But the machine moves in straight lines, not curves!”

What you often see is the visualization (a polyline between samples). It is not the definition of the path. The GRG law commands instantaneous poses that lie on the circle/sphere. The drive’s interpolation between samples may be linear or spline-based: that is a control choice. If you want a “fuller” trajectory, increase the sampling rate or use an arc interpolator; in any case, the nodes remain exactly on the geometric locus.

Why $\pi$ is not needed

$\pi$ is needed when you measure the arc (lengths, areas, times along an arc). In GRG we do not measure the arc: we enforce the distance from the center and the direction. Therefore $\pi$ is irrelevant to guarantee that points lie on a circle/sphere. You may use $\pi$ later if you care about arc length, but not to construct the point.

Natural extension to the sphere (including the $Z$ axis)

What holds in $X\!-\!Y$ holds in $X\!-\!Y\!-\!Z$. A unit direction in 3D (obtained via Cartesian projections, without “walking on the surface”) is scaled by $R$. The result has norm $R$. That is all. Azimuth–elevation formulas are just a convenient way to obtain $(x,y,z)$. The key property remains the preservation of distance.

Practical consequences for robotics, CNC, and 3D printers

If your machine moves on Cartesian axes and you command $(x,y)$ (or $(x,y,z)$) according to the GRG law, the ideal path is exactly the circumference (or the sphere).

Any deviations you might measure on a real part come from mechanics and numerics: backlash, compliance, calibration, friction, quantization. Improve the machine, the controller, and the metrology, and you approach the ideal without changing the GRG law.

Typical objections, short answers

“Without $\pi$ I cannot make a circle.” False: a circle is the locus of points at distance $R$ from a center. No $\pi$ is needed to guarantee that distance.
“Sine and cosine are approximate numeric functions.” In geometry they are exact ratios; in control you can use incremental orthogonal rotations that preserve the norm by construction.
“Steps accumulate error.” Rotations do not change the norm. In digital control, rounding errors are contained with renormalization or periodic recomputation.
“Points on the sphere are not uniformly distributed.” Different topic: sampling uniformity does not affect the fact that each point lies on the sphere.

The key takeaway

GRG does not approximate the circle or the sphere: it imposes them. With a radius $R$ fixed and a direction chosen, the distance from the center is $R$ by construction: therefore the point is certainly on the circumference (or on the sphere). In a Cartesian system, the geometric precision of the trajectory does not depend on $\pi$ nor on arc approximations: it depends only on your machine and numerics.

Certainly on the Circumference — and the Sphere

Category: Rigor of GRG

GRG conventions. Angles in turns $u\in[0,1)$ with step $\Delta u=1/N$. Radial computations (perimeter & radial area) on the full turn; Cartesian area on the quarter (×4). Trig appears only as classical cross-check.

2D · Circle 3D · Sphere

Certainly on the Circumference

Circle $R$: $x^2+y^2=R^2$.
Direction (turns): $u\in[0,1)$ → radial pose (no arc lengths).
GRG border: certificate $x(u)^2+y(u)^2=R^2$ (Pythagoras on quarter or tangent intersections on turn).

Classical (optional): with $\varphi=2\pi u$: $x=R\cos\varphi,\ y=R\sin\varphi$.

Theorem

Every GRG point $P(u)$ satisfies $x(u)^2+y(u)^2=R^2$ ⇒ it lies on the circle.

Fig. — Radial construction / orthogonal update preserves norm.

Certainly on the Sphere

Sphere $R$: $x^2+y^2+z^2=R^2$.
Unit direction: build $\mathbf{u}$ constructively; $\mathbf{P}=R\,\mathbf{u}$ certifies $\|\mathbf{P}\|=R$.

Classical (optional): $(\varphi,\theta)$ → $x=R\cos\varphi\,\sin\theta,\ y=R\sin\varphi\,\sin\theta,\ z=R\cos\theta$.

Theorem

$\|R\,\mathbf{u}\|^2=R^2\|\mathbf{u}\|^2=R^2$ ⇒ point on the sphere.

Fig. — 3D orthogonal rotations preserve the norm.

GRG · Operative summary (turns & quarter)

Params. Pick $N$. Full turn: $u_j=j/N$ with $\Delta u=1/N$. Generate $P_j=(x_j,y_j)$ with $x_j^2+y_j^2=R^2$. Quarter: $\Delta x=R/N$.

Perimeter — full turn (radial)

Inscribed (lower)

\[ L_{\mathrm{in}}(N)=\sum_{j=0}^{N-1}\sqrt{(x_{j+1}-x_j)^2+(y_{j+1}-y_j)^2}. \]

Circumscribed (upper) — tangents at $P_j$: $x\,x_j+y\,y_j=R^2$; intersections $T_j$ with $\det_j=x_jy_{j+1}-y_jx_{j+1}$, $T_j^x=\frac{R^2(y_{j+1}-y_j)}{\det_j}$, $T_j^y=\frac{R^2(x_j-x_{j+1})}{\det_j}$.

\[ L_{\mathrm{out}}(N)=\sum_{j=0}^{N-1}\lVert T_{j+1}-T_j\rVert. \]

Radial area — full turn

\[ A_{\mathrm{in}}(N)=\tfrac12\sum_{j=0}^{N-1}(x_jy_{j+1}-x_{j+1}y_j),\qquad A_{\mathrm{out}}(N)=\tfrac12\sum_{j=0}^{N-1}(T_j^xT_{j+1}^y-T_{j+1}^xT_j^y). \]

Cartesian area — quarter (Δx) → ×4

Abscissas $x_i=i\Delta x$, $\Delta x=R/N$; heights $y_i=\sqrt{R^2-(i\Delta x)^2}$.

\[ A_{\mathrm{right,q}}=\sum_{i=1}^{N}y_i\Delta x,\quad A_{\mathrm{mid,q}}=\sum_{i=1}^{N}y\!\left(x_{i-\tfrac12}\right)\Delta x,\quad A_{\mathrm{trap,q}}=\sum_{i=1}^{N}\tfrac{y_{i-1}+y_i}{2}\Delta x. \]

Bounds: $A_{\mathrm{right,q}}\le A_{\mathrm{quarter}}\le A_{\mathrm{mid,q}}$; disk via ×4. π remains outside the computation; trig only as cross-check.