GRG Original Projection

Sep 4, 2025

Scientific correction — 13 October 2026

For the complete quarter-circle profile sampled uniformly along x, midpoint and trapezoidal area errors have order N^(-3/2), not N^(-2), because the endpoint derivative is singular. The right-endpoint error and midpoint/right corridor width have order N^(-1). The geometric upper/lower corridor remains valid in exact arithmetic. These rates concern the historical Cartesian rules, not the current weighted binary radial kernels.

Current reference: Circle and Area from Binary Radial Modules.

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.

Original Projection — historical method This page presents the original GRG idea: the circle is generated from the inside. The circle area is computed on a quadrant by projecting heights onto the x-axis and using midpoint rectangles (discrete upper); a matching right-endpoint rule gives a discrete lower, forming an under/over corridor that shrinks as $N$ increases (the midpoint area error is of order N^(-3/2) for this endpoint-singular profile). The perimeter is obtained from the sum of equal-angle chords on the quadrant (then $\times 4$). This method remains alongside other GRG blocks and is kept intact for comparison and continued study. See also Core Turns and Δx Trapezoidal.

Original Projection (Quarter → ×4)

Work on the quarter $x\in[0,r]$ with the profile $y(x)=\sqrt{r^{2}-x^{2}}$. Use uniform $\Delta x=r/N$. Midpoint rectangles provide a discrete upper estimate; right-endpoint rectangles provide a discrete lower. Multiply by $4$ to obtain full-circle values.

Setup

$$ \Delta x \;=\; \frac{r}{N}, \qquad x_k \;=\; k\,\Delta x \ (k=0,\dots,N), \qquad x_{k+\tfrac12} \;=\; \Bigl(k+\tfrac12\Bigr)\Delta x, \qquad y(x) \;=\; \sqrt{\,r^{2}-x^{2}\,}. $$

Quarter area: discrete corridor (upper vs lower)

Midpoint (discrete upper; area error of order N^(-3/2))

$$ A_{\text{quarter}}^{\text{mid}}(N) \;=\; \sum_{k=0}^{N-1} \Delta x \; y\!\bigl(x_{k+\tfrac12}\bigr), \qquad A_{\text{circle}}^{\text{mid}}(N) \;=\; 4\,A_{\text{quarter}}^{\text{mid}}(N). $$

Right-endpoint (discrete lower)

$$ A_{\text{quarter}}^{\text{right}}(N) \;=\; \sum_{k=0}^{N-1} \Delta x \; y\!\bigl(x_{k+1}\bigr), \qquad A_{\text{circle}}^{\text{right}}(N) \;=\; 4\,A_{\text{quarter}}^{\text{right}}(N). $$

Corridor: $A_{\text{circle}}^{\text{right}}(N) \;\le\; \text{Area(circle)} \;\le\; A_{\text{circle}}^{\text{mid}}(N)$, with midpoint/right corridor width of order N^(-1) for this profile.

Perimeter via equal-angle chords (quadrant → ×4)

Partition the quarter by equal angular steps in turns. Let the quarter have $Q$ equal-angle chords of step $\Delta u=\tfrac{1}{4Q}$. The sum over the quarter is

$$ C_{\text{quarter}}(Q) \;=\; 2\,Q\,r\,s\!\Bigl(\tfrac{\Delta u}{2}\Bigr) \;=\; 2\,Q\,r\,s\!\Bigl(\tfrac{1}{8Q}\Bigr), \qquad C_{\text{circle}} \;\approx\; 4\,C_{\text{quarter}}(Q) \;=\; 8\,Q\,r\,s\!\Bigl(\tfrac{1}{8Q}\Bigr). $$

Equivalently, with $N=4Q$ steps on the full turn: $C_{N}=2\,N\,r\,s\!\bigl(\tfrac{1}{2N}\bigr)$ (see Core Turns).

Figure — Quarter with midpoint and right rectangles

Quarter: midpoint (upper) vs right-endpoint (lower) Quarter arc of a circle (r=100) with Δx strips. Green: midpoint rectangles; orange: right-endpoint rectangles. Both lie inside the quarter; midpoint sits above right. x y midpoint (upper) right (lower)
Midpoint vs right-endpoint corridor on the quarter.

Why use this block

  • Explicit under/over corridor with midpoint/right rules on the quarter.
  • Area error of order N^(-3/2) for the midpoint rule on this complete profile with simple operations.
  • No angle handling; complements chord-based perimeter via equal-angle steps.

How to use

  • Choose $N$ (e.g., $200$–$1000$) for the quarter; multiply results by $4$ for the circle.
  • Step-doubling check: compare $N$ vs $2N$ to estimate the residual gap between upper and lower.
  • Pair with equal-angle chords on the quarter to report both area and perimeter consistently.

See also

Core Turns · Δx Trapezoidal · Applications / Circle

Original Projection

The original GRG area method: sample the quarter along the x-axis with uniform step $\Delta x=r/N$. At each sample, project the height on the circle graph $y(x)=\sqrt{r^{2}-x^{2}}$ and build discrete rectangles. Two rules create a shrinking corridor: midpoint rectangles (upper) and right-endpoint rectangles (lower). Multiply by $4$ for the full circle. No angles are required in this block.

Setup

$$ \Delta x \;=\; \frac{r}{N}, \qquad x_k \;=\; k\,\Delta x \quad (k=0,\dots,N), \qquad x_{k+\tfrac12} \;=\; \Bigl(k+\tfrac12\Bigr)\Delta x, \qquad y(x) \;=\; \sqrt{\,r^{2}-x^{2}\,}. $$

Quarter-area sums (→ ×4 for the circle)

$$ \textbf{Midpoint (upper):}\quad A_{\mathrm{mid}}^{\mathrm{q}}(N) \;=\; \sum_{k=0}^{N-1} y\!\bigl(x_{k+\tfrac12}\bigr)\,\Delta x $$
$$ \textbf{Right-end (lower):}\quad A_{\mathrm{right}}^{\mathrm{q}}(N) \;=\; \sum_{k=0}^{N-1} y\!\bigl(x_{k+1}\bigr)\,\Delta x $$
$$ \textbf{Circle totals:}\quad A_{\mathrm{mid}} \;=\; 4\,A_{\mathrm{mid}}^{\mathrm{q}}, \qquad A_{\mathrm{right}} \;=\; 4\,A_{\mathrm{right}}^{\mathrm{q}} $$
$$ \textbf{Corridor:}\quad A_{\mathrm{right}} \;\le\; A_{\mathrm{true}} \;\le\; A_{\mathrm{mid}} $$
$$ \textbf{Estimate (optional):}\quad A_{\mathrm{avg}} \;=\; \frac{A_{\mathrm{mid}}+A_{\mathrm{right}}}{2} \quad\text{(use half-gap as an error bar).} $$

Notes

  • $y(x)=\sqrt{r^{2}-x^{2}}$ is concave on $[0,r]$; hence midpoint gives an upper estimate and right-endpoint a lower one.
  • Increase $N$ to tighten the corridor; publish $A_{\mathrm{avg}}$ with its half-gap if you need an error bar.

See also: Δx Trapezoidal · Core Turns · Applications / Circle