REVIEW 4 major objections 5 minor 20 references
On the perimeter length determination of the eight-centered oval
T0 review · 4 major / 5 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read The perimeter of the eight-centered oval, the classical architectural approximation of an ellipse, is exactly given by a closed-form arcsine formula in the two semi-axes.
desk verdict A sound, short derivation of the eight-centered oval perimeter with a genuine notational flaw in Eq. (17) that needs fixing, but the geometry and numerics hold up. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The load-bearing construction is Honey's system of three osculating circles per quadrant: the major circle with radius $R=a^{2}/b$ centered at $g(0,b-a^{2}/b)$, the minor circle with radius $r=b^{2}/a$ centered at $e(a-b^{2}/a,0)$, and the intermediate circle with radius $p=(a+b)/2$ whose center $k$ is the intersection of two auxiliary circles. The proof's engine is the triangle $(gek)$: applying the law of cosines twice yields the intermediate angle $\alpha'$, the law of sines yields the intermediate angle $\alpha$, and the right triangle $(geo)$ connects these to the two outer center angles $\gamma$ and $\delta$. The perimeter identity $O=4(R\gamma+p\beta+r\delta)$ then turns the sine expressions for the angles into a closed-form arcsine formula.
What would settle it
For an extreme ratio, say $a=10$ and $b=1$, evaluate the three arcsine arguments in Table I; if any lies outside $[0,1]$, or if the three resulting angles do not sum to $\pi/2$, the arcs cannot form a continuous tangent quarter-oval and the formula fails.
Extended reading notes
Core claim
The central discovery is that the perimeter of the eight-centered oval, as constructed by Honey, is fully determined by the two ellipse semi-axes and is given by $O(a,b)=4[\arcsin(\gamma)a^{2}/b + \arcsin(\beta)(a+b)/2 + \arcsin(\delta)b^{2}/a]$, where the sine of each center angle is a rational-algebraic function of $a$ and $b$ (Table I). The three angles $\gamma$, $\beta$, $\delta$ are obtained geometrically: from the right triangle formed by the small and large osculating circles' centers and the ellipse center, and from the triangle formed by the three circle centers, using the law of cosines and the law of sines. Because the three arcs meet tangentially, the oval's perimeter is exactly the sum of the three arc lengths, giving a closed-form expression in $a$ and $b$. The paper verifies that when $a=b$ the formula collapses to the circle circumference $2\pi\rho$, and that for the Colosseum it reproduces the elliptic-integral perimeter to $1.85\times 10^{-4}\%$ relative error.
Load-bearing premise
The perimeter formula assumes the three arcs of each quadrant meet tangentially, so their lengths add without correction; this tangency is taken from Honey's construction and is not proved in this paper.
Editorial extensions
If this is right
- The perimeter of any structure built on Honey's eight-centered oval can now be computed directly from its two measured axes, with no elliptic integrals.
- For the Colosseum's semi-axes $a=94$, $b=78$, the formula gives 541.523 m, within one millimeter of the elliptic-integral value 541.524 m.
- The numerical tests show the relative error between the oval and ellipse perimeters stays below 0.029% for semi-axes between 1 and 10.
- Setting $a=b$ recovers the circle's circumference $2\pi a$, confirming the formula's behavior at the circular limit.
Reading between the lines
- Because the arcsine expressions are algebraic, the formula could serve as a fast, closed-form approximation of the complete elliptic integral of the second kind, with the 0.029% bound providing a known worst-case error.
- Honey's construction could be iterated by inserting additional intermediate circles between adjacent osculating circles, yielding N-centered ovals whose perimeters follow from the same trigonometric method.
- The symmetry noted in the sine expressions (swapping $a$ and $b$ exchanges $\gamma$ and $\delta$) suggests the approximation error depends mainly on the axis ratio, so a one-variable error bound might be derivable.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript treats Honey's 1908 eight-centered oval as a piecewise-circular approximation to the ellipse with semi-axes a and b. For one quadrant it uses three circular arcs with radii R = a^2/b, p = (a+b)/2, r = b^2/a and center angles gamma, beta, delta. The paper derives trigonometric expressions for the sines of these angles (Eqs. (11), (14), (16); Table I) using the triangles (gek) and (geo), and then writes the perimeter of the oval as O = 4[arcsin(gamma) a^2/b + arcsin(beta)(a+b)/2 + arcsin(delta) b^2/a] in Eq. (17). It reports that for the Colosseum dimensions a = 94, b = 78 the oval perimeter is 541.523 m versus 541.524 m from the elliptic integral, a relative error of 1.85 x 10^-4%, and that for 1 <= a,b <= 10 the relative error stays below 0.029%. It also claims that the formula reduces to 2*pi*rho when a = b = rho.
Significance. If Eq. (17) is interpreted as intended, namely O = 4(R*gamma + p*beta + r*delta), this is a useful closed-form perimeter for a classical polycentric oval. The derivation is elementary, uses no fitted parameters, and is checked against an independent elliptic-integral benchmark for the Colosseum example. The paper also identifies a plausible historical gap, since Honey supplied radii but not center angles. However, the central formula is miswritten in a way that makes it ill-posed over part of the claimed domain, and the branch and tangency conditions needed for rigor are not explicitly established. These issues are local and fixable, but they affect the main result.
major comments (4)
- [III, Eq. (17)] The central formula (17) is not well-defined as printed. The quantities gamma, beta, delta are center angles, but only their sines are derived (Eqs. (11), (14), (16); Table I), and the notation arcsin(gamma) in Eq. (17) applies arcsin to an angle rather than to its sine. For a = 10, b = 1, which lies inside the stated test range, delta is about 1.09 rad > 1, so arcsin(delta) is not real; more generally arcsin(angle) is not equal to the angle. The intended perimeter is O = 4(R*gamma + p*beta + r*delta), or equivalently O = 4[R arcsin(sin gamma) + p arcsin(sin beta) + r arcsin(sin delta)] once the acute-angle branch is proved. This restatement is required before Eq. (17) can be evaluated.
- [II, Eqs. (11)-(16)] The identities arcsin(sin gamma) = gamma, arcsin(sin beta) = beta, and arcsin(sin delta) = delta used implicitly in Eq. (17) are valid only if gamma, beta, delta lie in (0, pi/2). The manuscript neither states nor proves this branch condition. It follows from the geometry of Fig. 2, for example from theta > alpha' and theta' > alpha for a,b > 0, but that verification is absent, so the domain of validity of Eq. (17) is not established.
- [II, Eqs. (3)-(4)] The perimeter calculation as the sum of three arc lengths assumes the three arcs meet tangentially at their junction points. This is not proved in the paper. It is an immediate consequence of the quoted Honey relations |ek| = p - r and |gk| = R - p, since internal tangency is equivalent to the distance between centers being the difference of the radii; the manuscript should state this explicitly, because it is essential to the exactness of O.
- [III, Remark] The circle-limit check contains a false displayed identity: '2 arcsin((4 - sqrt(2))/6) = 2 arcsin(7/9)' is not true. The correct identity is 2 arcsin((4 - sqrt(2))/6) = arcsin(7/9), obtained from sin(2 arcsin x) = 2x sqrt(1 - x^2). As printed, the subsequent cancellation that yields O = 2*pi*rho does not follow; this is probably a typographical slip but should be corrected.
minor comments (5)
- [III] The numerical error statements ('below 0.029%' for 1 <= a,b <= 10 and '1.85 x 10^-4%' for the Colosseum) are reported without a table or sample of computed values; a small table would make the claim verifiable.
- [II] In the sentence describing the large circle, 'Pour le cercle mineur (en rouge sur la Fig. 1), il emploie le centre g...' should presumably read 'cercle majeur'; as written it contradicts the preceding assignment of the blue circle to e.
- [I] The sentence 'il ne nous semble pas que ce calcul ... n'ait jamais encore été effectué ni publié' contains a double negative that makes the intended meaning ambiguous.
- [II] The derivation of sin(alpha') in Eq. (8) and sin(alpha) in Eq. (13) silently selects positive square roots; since the angles are meant to be acute, a one-line justification of the sign choice would improve clarity.
- [III] The paper does not specify the numerical method or software used to evaluate the elliptic integral (2); adding this information would make the reported comparisons reproducible.
Circularity Check
No significant circularity: the perimeter formula is a direct trigonometric consequence of Honey's 1908 construction and is checked against an independent elliptic-integral benchmark.
full rationale
The derivation chain is self-contained once Honey's construction is accepted. Sections II and III compute the three center angles from the fixed triangle (geo) and (gek) via the law of cosines and the law of sines (Eqs. 7-16), then form the perimeter as four times the sum of the three arc lengths (Eq. 17). Nothing is fitted to the elliptic-integral value: the comparison L(94,78)=541.524 versus O(94,78)=541.523 is an external numerical check, not an input to the formula. The only external ingredient is Honey's 1908 construction of the radii and centers (Eqs. 3-4), which is prior published work and not a self-citation. The citation to Golvin [5] supplies archaeological dimensions and historical context, but it is not load-bearing for the trigonometric derivation. The apparent arcsin-argument issue in Eq. (17) — where arcsin should be applied to the sines of the angles rather than to the angle variables themselves, with an unstated acute-angle branch assumption — is a correctness or typographical concern, not evidence that the result reduces to its own inputs. Accordingly, no genuine circular step is present.
Assumptions & free parameters
assumptions (5)
- standard math Euclidean law of cosines and law of sines apply to triangle (gek).
- domain assumption The osculating circle at (a,0) has center e=(a-b^2/a,0) and radius b^2/a; at (0,b), center g=(0,b-a^2/b) and radius a^2/b.
- domain assumption Honey's construction defines k as an intersection of the circles centered at e and g with radii |ek| and |gk|, and the intermediate arc has radius p=(a+b)/2 with tangency at the junctions.
- domain assumption The angles γ, β, δ lie in (0, π/2), so that arcsin(sinγ)=γ and similarly for β and δ.
- standard math The perimeter of the reference ellipse is given by the elliptic integral in Eq. (2).
Cite this review
Pith. "Pith review of On the perimeter length determination of the eight-centered oval." pith.science (2026). https://pith.science/paper/JIFCTUP3
@misc{pith2026190800783,
author = {Pith},
title = {Pith review of: On the perimeter length determination of the eight-centered oval},
year = {2026},
howpublished = {\url{https://pith.science/paper/JIFCTUP3}},
note = {Machine review of arXiv:1908.00783}
}
read the original abstract
On the perimeter length determination of the eight-centered oval. Several studies have shown that an eight-centered oval coincides almost perfectly with the ellipse constructed on the same axes and can be considered as a representation of the latter provided that the radii of the arcs of circles that compose it had been suitably chosen. Its perimeter's computation is then reduced to the simple sum of arc lengths of circles. However, it doesnot seem to us that this calculation, which could prove to be useful, has never been performed nor published. This note aims thus to present a geometric demonstration of the perimeter length determination of the eight-centered oval.
Reference graph
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Calcul de l’angle δ = θ′ − α avec θ′ = ˆe dans le triangle rectangle ( geo). o k e g i Γ Α ' Β Α ∆ /Minus2 2 4 6 /Minus6 /Minus4 /Minus2 2 Figure 2: Triangles ( geo) et ( gek) pour la détermination des trois angles aux centres. Au préalable, en appliquant le théorème de Pythagore au tria ngle rectangle ( geo), il est facile de déduire l’expression de la l...
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029%. Remarque. En posant a = b = ρ, l’expression (17) permet de retrouver le périmètre du cerc le de rayon ρ. L’expression (17) s’écrit alors : O (a, b ) = 4 ρ [arcsin (γ) + arcsin (β ) + arcsin (δ)] (18) Dans ce cas, les trois sinus (11, 14 and 16) deviennent égaux à sin (γ) = 4 − √ 2 6 , sin (β ) = 4 √ 2 9 et sin (δ) = 4 − √ 2 6 . En remplaçant ces sin...
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Reviewed August 14, 2026 · model on record in the stance chip above.
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