New Determination

Aug 24, 2025

Quarter-area as a sum of trapezoids on one radius

Divide the single radius into N parts: $$\Delta r=\frac{R}{N},\qquad x_k=k\,\Delta r\;(k=0,\dots,N).$$ Define the heights on the circle by Pythagoras: $$b_k=\sqrt{R^2-x_k^2},\qquad b_0=R,\; b_N=0.$$

Trapezoids

The k-th trapezoid (for k=1..N) uses the adjacent heights \(B=b_{k-1}\) and \(b=b_k\) and has height \(\Delta r\):

$$A_k=\frac{(b_{k-1}+b_k)}{2}\,\Delta r \;=\; \frac{(B+b)}{2}\,\frac{R}{N}.$$

  • First trapezoid: $$A_1=\frac{(R+b_1)}{2}\,\frac{R}{N}.$$
  • Last trapezoid (degenerates to a triangle since \(b_N=0\)): $$A_N=\frac{(b_{N-1}+0)}{2}\,\frac{R}{N}=\frac{b_{N-1}}{2}\,\frac{R}{N}.$$

Total quarter area

$$A^{(Q)}_{\text{trap}}(N)=\sum_{k=1}^{N}\frac{(b_{k-1}+b_k)}{2}\,\frac{R}{N} \;=\; \sum_{k=0}^{N-1}\frac{\Delta r}{2}\big(b_k+b_{k+1}\big).$$ By symmetry for the full disk: $$A_{\text{trap}}(N)=4\,A^{(Q)}_{\text{trap}}(N).$$

Optional recursive update (computational shortcut)

You can compute the next height without re-evaluating the square root from scratch:

$$b_{k+1}^2 \;=\; R^2-(x_k+\Delta r)^2 \;=\; b_k^2-2\,x_k\,\Delta r-(\Delta r)^2.$$

Summary. Yes: the area of the quarter is the sum of all these trapezoids. The last one is automatically a triangle because \(b_N=0\). This matches your picture and the loop a computer would run.

GRG — 3D extension: the sphere (surface area) + error behaviour

Work-in-progress. GRG keeps growing while we build the site. The “joined method” (angular sectors + one-radius trapezoids) lifts naturally from the circle to the sphere.

Setup (hemisphere)

  • Choose two integers: Nz (vertical splits) and Nθ (angular sectors for circumferences).
  • Vertical step: $$\Delta z=\frac{R}{N_z},\qquad z_k=k\,\Delta z\ (k=0,\dots,N_z).$$
  • Latitude radius by Pythagoras: $$r_k=\sqrt{R^2-z_k^2}.$$
  • GRG circumference at radius \(r\) with angular split \(N_\theta\): $$C_{GRG}(r;N_\theta)=8\,N_\theta\,r\,\sin\!\Big(\frac{90^\circ}{2N_\theta}\Big).$$

Surface area by “frusta” along a meridian

Between two adjacent latitudes \(k \to k+1\) we approximate the spherical band by a conical frustum:

  • Meridian slant step (no π): $$s_k=\sqrt{(\Delta z)^2+\big(r_{k+1}-r_k\big)^2}.$$
  • Band area (average circumference × slant step, π-free):

$$A_k \;=\; \frac{C_{GRG}(r_k;N_\theta)+C_{GRG}(r_{k+1};N_\theta)}{2}\;\cdot\; s_k.$$

Hemisphere and sphere:

$$A_{\text{hemi}}(N_z,N_\theta)=\sum_{k=0}^{N_z-1} A_k,\qquad A_{\text{sphere}}(N_z,N_\theta)=2\,A_{\text{hemi}}(N_z,N_\theta).$$

Circle area reminder (why accuracy improves)

For the quarter of a circle we already use trapezoids on the single radius:

$$A^{(Q)}_{\text{trap}}(N)=\sum_{k=0}^{N-1}\frac{\Delta r}{2}\big(b_k+b_{k+1}\big),\quad \Delta r=\frac{R}{N},\ b_k=\sqrt{R^2-(k\Delta r)^2}.$$

This has quadratic error in \(\Delta r\) (order \(N^{-2}\)), so the accuracy rises fast. The same idea—“average of neighbours” along one axis—drives the sphere construction above.

Error behaviour (what shrinks, and why)

  • Angular sectors (per circumference): chord vs arc gives a relative error that scales like \( \mathcal{O}(N_\theta^{-2})\).
  • Vertical frusta (along a meridian): the trapezoidal averaging in \(z\) (via the slant step \(s_k\)) yields error \( \mathcal{O}(N_z^{-2})\).

Combined effect: the sphere estimate satisfies $$A_{\text{sphere}}(N_z,N_\theta) \to 4\pi R^2 \quad\text{as}\quad N_z,N_\theta\to\infty,$$ with total error behaving like $$\mathcal{O}\!\big(N_z^{-2}\big)+\mathcal{O}\!\big(N_\theta^{-2}\big).$$ (We keep formulas π-free in the computation; π appears only when comparing to the classical continuous value.)

Algorithm (summary)

  • For k = 0..Nz: compute \(z_k, r_k, C_k=C_{GRG}(r_k;N_\theta)\).
  • For k = 0..Nz−1: compute \(s_k=\sqrt{(\Delta z)^2+(r_{k+1}-r_k)^2}\) and $$A_k=\tfrac{C_k+C_{k+1}}{2}\,s_k.$$
  • Sum \(A_k\) for the hemisphere; double for the whole sphere.

Notes

  • The polar band automatically degenerates (like the “last trapezoid → triangle” in 2D), so the loop stays uniform.
  • To balance errors, pick \(N_z\approx N_\theta\) (e.g., 24–48 each gives already small relative errors).
  • Volume of the sphere can be done analogously using cylindrical shells with \(C_{GRG}(r;N_\theta)\) and a height term; we can add that next.
[Optional future figure: hemisphere bands as conical frusta with GRG circumferences]

GRG — Joined method for the circle: perimeter by angular sectors, area by single-radius trapezoids

Work-in-progress. Generative Radial Geometry (GRG) is evolving as we build the website. The idea below emerged today (23 August 2025) and we document it in real time. GRG is a young, living framework: definitions, figures, and formulas grow together with the site.

Setup (quarter of a circle)

Let the radius be R and consider the first-quadrant quarter of the circle. Fix an integer N ≥ 1. We couple two distinct discretisations:

  • Perimeter (angular division): split the right angle into N equal parts with step
    • $$ \Delta\theta=\frac{\pi}{2N} $$
  • Area (single-radius heights): split one radius into N equal parts with
    • $$ \Delta r=\frac{R}{N},\qquad x_k=k\,\Delta r,\qquad y_k=\sqrt{R^2-x_k^2} $$
  • Adjacent heights \(y_k,y_{k+1}\) form a trapezoid that tiles the quarter.

Formulas (π-free perimeter and trapezoidal area)

Quarter arc length (GRG, from angular sectors):

$$ C_N^{(Q)} \;=\; 2\,N\,R\,\sin\!\Big(\frac{\pi}{4N}\Big) $$

(full circle: \(C_N=8\,N\,R\,\sin(\tfrac{\pi}{4N})\)).

Quarter area (trapezoids on one radius):

$$ A_{\text{trap}}^{(Q)}(N)\;=\;\sum_{k=0}^{N-1}\frac{\Delta r}{2}\,\big(y_k+y_{k+1}\big),\qquad \Delta r=\frac{R}{N},\quad y_k=\sqrt{R^2-x_k^2},\; x_k=k\,\Delta r $$

Equivalently, with \(B_k:=y_k\) and \(b_k:=y_{k+1}\), each trapezoid is

$$ A_k=\frac{(B_k+b_k)}{2}\,\frac{R}{N},\qquad A_{N-1}=\frac{(B_{N-1}+0)}{2}\,\frac{R}{N} $$

Full disk by symmetry:

$$ A_{\text{trap}}(N)=4\,A_{\text{trap}}^{(Q)}(N) $$

Figure (joined method)

Blue = angular sectors for the perimeter. Red = heights on one radius and the adjacent trapezoids for the area. Dashed red rays start from the intersection points and do not, in general, coincide with the blue rays.

  • angular sectors (perimeter)
  • trapezoids on single-radius heights (area)

Formulas (π-free perimeter and trapezoidal area)

Quarter arc length (GRG, from angular sectors):

$$ C_N^{(Q)} \;=\; 2\,N\,R\,\sin\!\Big(\tfrac{\pi}{4N}\Big) $$

Full circle:

$$ C_N \;=\; 8\,N\,R\,\sin\!\Big(\tfrac{\pi}{4N}\Big) $$

Quarter area (trapezoids on one radius)

Using heights at \(x_k=k\,\Delta r\) with \(y_k=\sqrt{R^2-x_k^2}\), the quarter-area is:

$$ A_{\text{trap}}^{(Q)}(N)\;=\;\sum_{k=0}^{N-1}\frac{\Delta r}{2}\,\big(y_k+y_{k+1}\big) $$ $$ \Delta r\;=\;\tfrac{R}{N} $$ $$ x_k\;=\;k\,\Delta r $$ $$ y_k\;=\;\sqrt{R^2-x_k^2} $$

Equivalently, with \(B_k:=y_k\) and \(b_k:=y_{k+1}\), each trapezoid is:

$$ A_k\;=\;\frac{(B_k+b_k)}{2}\,\frac{R}{N} $$ $$ A_{N-1}\;=\;\frac{(B_{N-1}+0)}{2}\,\frac{R}{N} $$

Full disk by symmetry:

$$ A_{\text{trap}}(N)\;=\;4\,A_{\text{trap}}^{(Q)}(N) $$
Acknowledgments

Acknowledgments

Aug 23, 2025

Thanks to SANUSLIFE, the company where I now work. It’s a wonderful environment—a health-focused company that has cultivated a beautiful internal atmosphere. In this serene setting, I found the calm to finally shape an idea I had been contemplating for many years.

Lord, thank You for making me as I am. You give me enthusiasm for everything, and everything looks beautiful to me. I long to discover the world; I long to make it more beautiful, yet I am limited. Thank You for everything.

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.