Rigor of GRG

Rigor of GRG

Aug 22, 2025

Earlier Formulation

This page preserves a historical GRG method. Its notation and formulas belong to that version. The current binary radial formulation and verification are available here. Historical computational claims require separate validation.

Generative Radial Geometry (GRG)

A Constructive Alternative to Classical Circle Measurement

Introduction

Generative Radial Geometry (GRG) is a digital, constructive approach: it does not assume π in the build step. Circles are generated from the radius with finite operators (Pythagoras, chords, tangents); classical values (e.g. π) then emerge as limits or comparisons.

Two complementary builds

  • Quarter / Cartesian (Δx): work on one quarter of the disk, sum rectangles/trapezoids with \(\Delta x=R/N\), then ×4. Certified right/midpoint corridor.
  • Full turn / Radial (Δu): work on the full loop in turns \(u\in[0,1)\), step \(Δu=1/U\); equal-angle chords/tangents give in/out polygons (inscribed/circumscribed).

Notation in turns: \(c(u)=\cos(2\pi u),\ s(u)=\sin(2\pi u)\). π appears only when comparing to classical formulas.

Quarter coordinates (Δx, no trig in the step)

On the quarter, sample \(x_k=k\,\Delta x\) with \(\Delta x=R/N\) and heights by Pythagoras:

$$ y_k=\sqrt{R^2-x_k^2},\qquad k=0,\dots,N. $$

Area — Cartesian quarter (GRG)

Δx Trapezoidal (lower)

$$ A_{\text{quarter}}^{\text{trap}}(R,N)=\sum_{k=1}^{N}\frac{y_{k-1}+y_k}{2}\,\Delta x,\quad A_{\text{GRG}}^{\text{trap}}=4\,A_{\text{quarter}}^{\text{trap}}. $$

Projection corridor (certified bounds)

$$ A_{\text{quarter}}^{\mathrm{right}}=\!\sum_{k=1}^{N} y_k\,\Delta x,\qquad A_{\text{quarter}}^{\mathrm{mid}}=\!\sum_{k=1}^{N} y\!\left(x_{k-\tfrac12}\right)\Delta x, $$
$$ A_{\text{quarter}}^{\mathrm{right}}\le A_{\text{quarter}}\le A_{\text{quarter}}^{\mathrm{mid}},\qquad 4A_{\text{quarter}}^{\mathrm{right}}\le A_{\text{GRG}}\le 4A_{\text{quarter}}^{\mathrm{mid}}. $$

Circumference — Radial full turn (GRG)

With \(U\) equal sectors per full turn (step \(Δu=1/U\)), all chords are equal. The inscribed polygon length is:

$$ C_U \;=\; 2\,U\,R\,s\!\Big(\tfrac{1}{2U}\Big) \;=\; 2\,U\,R\,\sin\!\Big(\tfrac{\pi}{U}\Big)\quad\text{(classical cross-check)}. $$

On a quarter with \(N\) steps set \(U=4N\) and get \(C=8NR\,s\!\big(\tfrac{1}{8N}\big)\).

Radial area — in/out corridor on the full turn

Inscribed polygon area (lower, shoelace)

Let \(P_j=(x_j,y_j)\) be border points in CCW order, \(j=0,\dots,U-1\) and \(P_U=P_0\) with \(x_j^2+y_j^2=R^2\).

$$ A_{\text{in}}(R,U)=\tfrac12\sum_{j=0}^{U-1}\bigl(x_jy_{j+1}-x_{j+1}y_j\bigr)\ \le\ A. $$

Circumscribed polygon area (upper, tangents)

The tangent at \(P_j\) is \(x\,x_j+y\,y_j=R^2\). Intersecting tangents at \(P_j,P_{j+1}\) yields \(T_j\):

$$ \det_j=x_jy_{j+1}-y_jx_{j+1},\qquad T_j=\Bigl(\frac{R^2\,(y_{j+1}-y_j)}{\det_j},\ \frac{R^2\,(x_j-x_{j+1})}{\det_j}\Bigr), $$
$$ A_{\text{out}}(R,U)=\tfrac12\sum_{j=0}^{U-1}\bigl(T_j^xT_{j+1}^y-T_{j+1}^xT_j^y\bigr)\ \ge\ A. $$

Hence \(A_{\text{in}}\le A \le A_{\text{out}}\) and the corridor collapses as \(U\) increases.

Answering common objections

“This is nothing new”

GRG constructs perimeter and area from the radius alone in two certified ways: quarter/Δx (right–mid corridor) and full-turn/Δu (inscribed–circumscribed corridor).

“It is only an approximation”

Accuracy is controlled by \(N\) or \(U\); both builds provide bounding pairs that converge.

“It lacks rigor”

Only algebraic steps (Pythagoras, chords, tangents, polygon sums). π is used solely for external comparison, not as an input.

“It is less practical”

Finite sums with elementary operations are ideal for digital computation and come with a transparent error model.

Conclusion

GRG complements classical geometry with a generative language. On the quarter, Δx gives fast areas with bounds; on the full turn, equal-angle chords/tangents give circumference and radial area—letting π emerge as a limit.

Circle with inscribed (lower) and circumscribed (upper) polygons; arrows suggest convergence. A_in A_out U ↑ ⇒ A_in ↑, A_out ↓ inscribed (lower) circumscribed (upper)
Common Objections

Common Objections

Aug 21, 2025

GRG (2D/Circle): A Generative, π-Free Reconstruction with Error Bounds — Reply to Common Objections

Abstract. Generative Radial Geometry (GRG) reconstructs a circle’s circumference and area from the radius alone using finite steps and exact border points (x²+y²=R² by construction). Angles are measured in turns \(u\in[0,1)\) with \(\Delta u=1/N\). The radial track (full turn) yields perimeter and radial area via inscribed/circumscribed polygons; the Cartesian track (quarter) yields area via Δx-trapezoids with a certified corridor. π is not assumed; it emerges as a limit.

Illustration on a quarter: left, chord sampling; right, Δx-trapezoids strictly under the arc. In GRG the radial computations (perimeter and radial area) run on the full turn; quarter is reserved for the Cartesian area (then ×4).

GRG construction — three coordinated modes

  • Radial perimeter (full turn): sample turns \(u_j=j/N\); build border points \(P_j=(x_j,y_j)\) with \(x_j^2+y_j^2=R^2\) via a GRG recurrence (no trig); close the loop with \(P_N:=P_0\). Sum chords $$ L_{\text{in}}=\sum_{j=0}^{N-1}\big\|P_{j+1}-P_j\big\| \quad\text{(inscribed, lower).} $$ Tangent intersections give a circumscribed polygon with length \(L_{\text{out}}\) (upper). Hence \(L_{\text{in}}\le C\le L_{\text{out}}\).
  • Radial area (full turn): use the same \(P_j\) (counter-clockwise). Inscribed polygon area (shoelace): $$ A_{\text{in}}=\tfrac12\sum_{j=0}^{N-1}\!\big(x_j y_{j+1}-x_{j+1} y_j\big) \quad\text{(lower).} $$ Circumscribed polygon from tangent intersections \(T_j\) (see below) gives \(A_{\text{out}}\) (upper): $$ A_{\text{out}}=\tfrac12\sum_{j=0}^{N-1}\!\big(T_j^x T_{j+1}^y – T_{j+1}^x T_j^y\big). $$ Hence \(A_{\text{in}}\le A\le A_{\text{out}}\).
  • Cartesian area (quarter only): abscissas \(x_i=i\,\Delta x\), \(\Delta x=R/N\); heights \(y_i=\sqrt{R^2-x_i^2}\). Right/midpoint/trapezoid give a Projection Corridor on the quarter; multiply by 4 for the disk.
Tangent intersections (no trig). La tangente in \(P_j\) è \(x\,x_j+y\,y_j=R^2\). L’intersezione delle tangenti in \(P_j,P_{j+1}\) è: \[ \begin{aligned} \det_j &= x_j\,y_{j+1} – y_j\,x_{j+1},\\[2pt] T_j^x &= \frac{R^2\,(y_{j+1}-y_j)}{\det_j},\qquad T_j^y = \frac{R^2\,(x_j – x_{j+1})}{\det_j},\\[2pt] T_j &= \bigl(T_j^x,\,T_j^y\bigr). \end{aligned} \] Usa lo shoelace su \(T_j\) per \(A_{\text{out}}\) e le distanze \(\|T_{j+1}-T_j\|\) per \(L_{\text{out}}\).

Orientation. Assumi l’ordine antiorario così che \(\det_j>0\); se i punti sono orari, adegua i segni.

Error control (π-free statements)

  • Perimeter corridor: \(L_{\text{in}}(N)\nearrow C \le L_{\text{out}}(N)\searrow\). Il gap si riduce con \(N\).
  • Radial area corridor: \(A_{\text{in}}(N)\nearrow A \le A_{\text{out}}(N)\searrow\).
  • Cartesian corridor (quarter): \(A_{\text{right,q}}\le A_{\text{mid,q}}\le A_{\text{quarter}}\le A_{\text{trap,q}}\) (arco concavo); disco via \(4\times\).
Classical cross-check (optional, non usato da GRG per calcolare). Con trigonometria classica: \[ \begin{aligned} C &\ge 8\,N\,R\,\sin\!\Bigl(\tfrac{\pi}{4N}\Bigr),\\ C &\le 8\,N\,R\,\tan\!\Bigl(\tfrac{\pi}{4N}\Bigr), \end{aligned} \] da cui un gap \(=\Theta(N^{-2})\).
(Turns equivalence, funzioni a periodo 1) con \(s(u):=\sin(2\pi u)\), \(t(u):=\tan(2\pi u)\): \[ \begin{aligned} C &\ge 8\,N\,R\,s\!\Bigl(\tfrac{1}{8N}\Bigr),\\ C &\le 8\,N\,R\,t\!\Bigl(\tfrac{1}{8N}\Bigr). \end{aligned} \] Stesso ordine \(O(N^{-2})\) per il trapezoide cartesiano su un quarto concavo. Sono controlli esterni, non prerequisiti di GRG.
ERG Cartesian

ERG Cartesian

Aug 20, 2025

ERG Cartesian — Quarter Chords to Sphere (GRG)

From a quarter construction we derive circumference, sphere surface and volume using discrete, angle-free operators. All figures below are drawn in a strict Cartesian frame (linear x–z axes, equal scale).

Quarter circumference (equal-angle chords)

For a radius $\rho$, divide the $90^\circ$ arc into $N$ equal angles. Replace each arc step with a chord. The discrete GRG quarter circumference is:

$$ C_{1/4}^{\mathrm{GRG}}(\rho;N)=2\,N\,\rho\,\sin\!\Bigl(\tfrac{45^\circ}{N}\Bigr) \;=\;2\,N\,\rho\,s\!\Bigl(\tfrac{1}{8N}\Bigr). $$
Quarter arc with equal-angle chords — strict Cartesian Axes and a true quarter circle of radius 100; chords join consecutive points at 15° steps (N=6). Coordinates are (x=ρ cosθ, z=ρ sinθ). xz
Equal-angle chords on a true quarter circle (example $N=6$). Axes and lengths are Cartesian.

Quarter section → sphere surface (horizontal slices + quarter chords)

Slice the sphere horizontally into $N$ levels of thickness $\Delta z=\tfrac{r}{N}$. At level $k$:

$$ z_k=\frac{k}{N}\,r, \qquad \rho_k=\sqrt{\,r^2-z_k^2\,}, \qquad S_{1/8}(N)=\sum_{k=1}^{N} C_{1/4}^{\mathrm{GRG}}(\rho_k;N)\,\frac{r}{N},\quad S(N)=8\,S_{1/8}(N). $$
Unwrapped surface S1/8(N) — strict Cartesian Bars of constant height Δz with width proportional to ρ_k (and hence to C_{1/4}^{GRG}(ρ_k;N) for fixed N). Axes are linear. local width ∝ ρk z
Area under the bars equals $S_{1/8}(N)$ up to the constant factor $2N\,s(\tfrac{1}{8N})$ common to all strips for fixed $N$.

Quarter area and sphere volume (stack quarter areas)

Quarter area for radius $\rho$ (base split into $N$ parts; rectangles on circle heights), and octant volume via the same $\Delta z$:

$$ A_{1/4}^{\mathrm{GRG}}(\rho;N)=\sum_{i=1}^{N}\Bigl(\tfrac{\rho}{N}\Bigr)\sqrt{\,\rho^{2}-\Bigl(\tfrac{i\,\rho}{N}\Bigr)^{2}\,}, \qquad V_{1/8}(N)=\sum_{k=1}^{N} A_{1/4}^{\mathrm{GRG}}(\rho_k;N)\,\tfrac{r}{N},\quad V(N)=8\,V_{1/8}(N). $$
Octant volume from stacked quarter-disk slabs — strict Cartesian Quarter disks of radii ρ_k = sqrt(r^2 - z_k^2) stacked along z; each slab has thickness Δz. x z
Stacking quarter-disk slabs of radius $\rho_k$ (Cartesian x–z plane). The sum of their quarter areas times $\Delta z$ gives $V_{1/8}(N)$.

ERG — Original Projection (Cartesian)

$$ \Delta x=\frac{r}{N},\quad x_k=k\,\Delta x,\quad x_{k+\tfrac12}=\Bigl(k+\tfrac12\Bigr)\Delta x,\quad y(x)=\sqrt{\,r^{2}-x^{2}\,}. $$
$$ A_{\text{quarter, mid}}(N)=\sum_{k=0}^{N-1} y\!\bigl(x_{k+\tfrac12}\bigr)\,\Delta x,\qquad A_{\text{quarter, right}}(N)=\sum_{k=0}^{N-1} y\!\bigl(x_{k+1}\bigr)\,\Delta x. $$

Below: true Cartesian rectangles for $r=100$, $N=5$ (so $\Delta x=20$). Heights are exactly $y=\sqrt{r^2-x^2}$ at the mid/right samples.

Quarter — midpoint (upper) vs right-endpoint (lower), strict Cartesian Quarter circle r=100. Δx=20. Green bars at x_mid = (k+1/2)Δx with exact heights sqrt(r^2 - x_mid^2). Orange outlines at right endpoints x=(k+1)Δx. x y
Midpoint (filled, upper) vs right-endpoint (outlined, lower) — heights computed from $y=\sqrt{r^2-x^2}$ with $r=100$, $\Delta x=20$.
The Circle

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 (Δu = 1/N turns) Cartesian axes; circle r=100; sample radii and dashed chords (example N=12).
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 (under the arc) Cartesian x–y axes; quarter arc with Δx verticals and interior trapezoids. Δx trapezoids
Quarter area by Δx trapezoids; multiply by 4 for the full circle.
Radial sector tiling (equal-angle sectors, full turn) Two example sectors on a unit circle (illustrative N=8).
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.
Origin of GRG

Origin of GRG

Aug 18, 2025

Historical context. This page records an earlier stage of GRG. Personal motivations and earlier terminology are distinct from the current documented formulation.

From the very beginning, the problem of the circle left me dissatisfied: its area depends on π, an irrational number that makes the calculation incomplete and always “open.” I could not accept that the most perfect figure in geometry was based on an infinite and never-ending value.
It may sound strange, but I spent entire nights trying to find a possible solution, constantly reasoning about all of geometry and about the fateful circle.

The area of a rectangle is easy to understand: base times height. We simply repeat the base line along the height, like a raft made of logs placed side by side. Figuratively and geometrically, everything is simple.
With the triangle, things are already more complicated: in theory we should do the same, but with a point and a line—or alternatively, with two lines meeting at a point. In the first case we try to merge two incompatible entities (the point has no dimension, the line has one); in the second, we end up with an “accumulation of matter” at the vertex and a stretching on the opposite side. This may not make sense in an analogical way, but in my mind a digital vision of geometry was already emerging.

One night I asked myself: how did they really solve the triangle? They could not project a point onto a line, so they used a trick: they projected the base along the normal towards the opposite vertex and closed the open sides. In this way, they actually duplicated it. Try cutting a triangle out of paper and see. Now the solution was acceptable: we had returned to the rectangle, base × height, and since it was doubled, the formula became (base × height) ÷ 2. An artifice gave the solution.

But what about the circle? Why does it have that formula?
The circle should be conceived as a concentric extrusion of circumferences towards the center or vice versa… but then we face the original problem: from a point it becomes a line. At first I thought: maybe π arises precisely from this unnatural fusion—something impossible.

I also tried imagining a radius revolving around a pivot to form the circle. But again, there was too much “material” at the sides and compression at the center.

Then I realized I needed a trick, like with the triangle. I thought: let’s imagine a rubber circle, cut it along one radius, and stretch the circumference onto a straight base. At that point I get a triangle with base = circumference (2πr) and height = r. The area of this triangle is (2πr × r) ÷ 2, which simplifies to πr².
At first I was happy with this discovery, but then I told myself: if the solution is like this, it has no true geometric foundation! I can really duplicate a triangle, but not a circle.

And here came the turning point: I cannot duplicate it, but I can halve it.

I looked at the semicircle. What do I have? I know the radius, of course, but I also know the diameter!
And the area? Excluding the previous artifices, only one logical path remained: projecting the progressive heights of the arc onto the diameter. That was the only coherent development.
Do I have those positions? Looking at the circle… yes, I have them all! Because I can apply triangulation. By drawing several radii, I realized that I could determine each point with absolute precision, without using π.

I noticed that I could use Pythagoras or trigonometry to compute the values. But soon I realized that I wanted to solve not only the area but also the circumference. For the circumference, I could divide the angle into N equal parts and calculate the chords subtended by the radii: the more divisions, the closer the chords matched the arc, until for N → ∞ they perfectly overlapped.

For the area I needed equally spaced heights along the diameter; for the circumference I needed equally spaced chords along the angle. For this reason, I decided: Pythagoras for the area, trigonometry for the circumference.

As my reflection progressed, I realized I did not even need the semicircle, but only a quarter of a circle. And from there my mind started to wander further: to the sphere, and eventually to geometry as a whole.

Before, we solved the circle with simpler figures. Now, in my mind, the opposite was happening: the circle itself was generating all the others.
This is where Generative Radial Geometry begins.