Circumference & Sphere

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.

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.

GRG for ellipses

GRG for ellipses

Aug 22, 2025

GRG for Ellipses — Constructive Area, Perimeter, and Equal-Area Sectors

The “Variable Radius” in GRG

The effective radius depends on the chosen direction.

  • From the center (angle \(\phi\) from the major axis):

    $$ r_c(\phi)=\frac{ab}{\sqrt{(b\cos\phi)^2+(a\sin\phi)^2}}. $$

  • From the focus (true anomaly \(\theta\)):

    $$ r_f(\theta)=\frac{p}{1+e\cos\theta},\qquad e=\sqrt{1-\frac{b^2}{a^2}},\qquad p=\frac{b^2}{a}. $$

  • Without trigonometry (slope \(m=\tan\phi\)):

    $$ r_c(m)=\frac{ab\,\sqrt{1+m^2}}{\sqrt{\,b^2+a^2m^2\,}}. $$

Ellipse Area in GRG

A1) Cartesian strips in the first quadrant

Split the semi-axis \(a\) into \(N\) parts: \(x_i=i\,a/N\) with \(\Delta x=a/N\). Heights \(y_i=b\sqrt{1-x_i^2/a^2}\).

$$ A_{1/4}^{(N)}=\sum_{i=1}^{N} y_i\,\Delta x,\qquad A^{(N)}=4\,A_{1/4}^{(N)}. $$
A1 — Cartesian strips in the first quadrant x–y axes, ellipse centered at (400,250) with a=240, b=160; shaded vertical strip inside Q1. +y +x Cartesian strips in Q1; then ×4
Quarter-by-strips: sum \(y_i\,\Delta x\) in Q1 and multiply by 4 for the ellipse area.

A2) Radial wedges from the center (by quarters)

With \(\Delta\phi=90^\circ/N\) and \(\phi_i=i\,\Delta\phi\):

$$ A_{1/4}^{(N)}=\sum_{i=0}^{N-1}\tfrac12\,r_c(\phi_i)\,r_c(\phi_{i+1})\,\sin\Delta\phi, \qquad A^{(N)}=4\,A_{1/4}^{(N)}. $$
A2 — Center-based wedges (quarter symmetry) Rays from center to ellipse boundary; the wedge between two rays approximates an area piece. C +x +y Center-based wedges (quarter symmetry)
Sum center-based wedges over a quarter (then ×4). Summing the chords between successive boundary points gives the corresponding perimeter.

A3) Radial wedges from the focus (natural for orbits)

With \(\Delta\theta=360^\circ/M\) and \(\theta_i=i\,\Delta\theta\):

$$ A^{(M)}=\sum_{i=0}^{M-1}\tfrac12\,r_f(\theta_i)\,r_f(\theta_{i+1})\,\sin\Delta\theta. $$
A3 — Focus-based sectors (equal areas → equal times) Sectors with apex at a focus; each filled sector has the same area increment ΔA (Kepler-2). F F′ periapsis → Equal-area sectors from the focus (constant ΔA; angle widths vary)
Equal-area stepping from the focus: constant \(\Delta A\) gives uniform time steps (Kepler’s 2nd law). Near periapsis the angular gaps shrink; away they widen.

Perimeter (sum of chords)

P1) Chords from the center (by quarters)

With \(\Delta\phi=90^\circ/N\):

$$ P_{1/4}^{(N)}=\sum_{i=0}^{N-1}\sqrt{\,r_c(\phi_i)^2+r_c(\phi_{i+1})^2-2\,r_c(\phi_i)\,r_c(\phi_{i+1})\cos\Delta\phi\,}, \qquad P^{(N)}=4\,P_{1/4}^{(N)}. $$

P2) Chords from the focus (full ellipse)

$$ P^{(M)}=\sum_{i=0}^{M-1}\sqrt{\,r_f(\theta_i)^2+r_f(\theta_{i+1})^2-2\,r_f(\theta_i)\,r_f(\theta_{i+1})\cos\Delta\theta\,}. $$

GRG Error Estimate (after-the-fact)

With step doubling the error typically decays ~quadratically. Method-agnostic (π-free):

  • Compute \(S_N\) and \(S_{2N}\) with the same operator (A1–A3, P1–P2).

  • Accelerated approximation:

    $$ S_\star \approx S_{2N}+\frac{S_{2N}-S_N}{3}. $$

  • Error estimate:

    $$ \varepsilon \approx \frac{|S_{2N}-S_N|}{3}. $$

Practical Pipeline (geometry → orbits)

  • Geometry: use A1 or A2 for area and P1 for perimeter with quarter symmetry.

  • Orbits: use A3. For uniform time steps, enforce constant-area sectors:

    $$ \tfrac12\,r_f(\theta_k)\,r_f(\theta_{k+1})\,\sin(\theta_{k+1}-\theta_k)=\Delta A. $$

  • The arc length per step is the P2 chord between successive points.

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.
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.