{"id":"c33027d5-9d13-41fd-a741-28879316042e","arxiv_id":"1908.00783","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":5.0,"correctness_risk":"low","formal_verification":"none","parameter_count":0,"one_line_summary":"The perimeter of the eight-centered oval built from Honey's radii is expressed as four times the sum of three circular arc lengths, with the three center angles given explicitly from the ellipse's semiaxes.","lead":"This paper derives a perimeter formula for an eight-centered oval, a shape made of circular arcs that closely tracks an ellipse. The formula gives a simple approximation to the ellipse perimeter and could help archaeologists measure ancient oval monuments such as the Colosseum.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Eq. (17) as printed applies arcsin to the angles themselves rather than to their sines; for allowed parameters this expression is undefined (e.g., a=10,b=1 makes δ≈1.09 rad), so the central formula needs restatement plus an explicit proof of the acute-angle branch.","rationale":"The trigonometric derivation appears internally consistent: the distances (3)-(5) force the intermediate circle to be internally tangent to the two osculating circles, so the reader's tangency concern is not the weak point. The angle derivations via the law of cosines and sines check out, and the numerical agreement with the elliptic integral is plausible. The load-bearing gap is the statement of the central equation itself. Eq. (17) is written with arcsin applied to angles, although the construction supplies only their sines; this makes the formula ambiguous and, for parameter values the authors explicitly include in their tests, undefined. The missing branch proof is not merely formal, because without knowing γ, β, δ<π/2 one cannot replace arcsin(sin x) by x. I therefore recommend keeping the reader's conditional verdict, with the requirement to restate Eq. (17) using arcsin(sinγ), arcsin(sinβ), arcsin(sinδ) or simply γ, β, δ, and to add a one-line argument that these angles are acute. The a<b case also deserves a sentence, since Honey's Eqs. (3)-(4) as signed lengths become negative for b>a; symmetry restores them, but the domain should be stated explicitly.","tokens_in":7161,"tokens_out":20029,"duration_ms":194492,"concrete_test":"Evaluate Eq. (17) literally at a=10, b=1: compute δ from Eq. (16) (sinδ≈0.887, δ≈1.09 rad) and show that arcsin(δ) is undefined over the reals. Then evaluate the corrected expression O=4[arcsin(sinγ)a^2/b + arcsin(sinβ)(a+b)/2 + arcsin(sinδ)b^2/a] on a grid 1≤a,b≤10, comparing with the elliptic integral Eq. (2). If the corrected expression reproduces the reported maximum relative error below 0.029% and reduces to 2πρ when a=b=ρ, the intended formula is sound once restated; if not, the central claim fails.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The central perimeter formula (17) is O=4[arcsin(γ)a^2/b + arcsin(β)(a+b)/2 + arcsin(δ)b^2/a]. But γ, β, δ are defined as center angles, and the paper only derives their sines (Eqs. 11, 14, 16; Table I). The only way to evaluate (17) is to read arcsin(γ) as arcsin(sinγ), and similarly for β and δ. This is never stated. Taken literally, (17) is ill-posed: for a=10, b=1, inside the authors' own 1≤a,b≤10 test range, δ≈1.09 rad>1, so arcsin(δ) is not real; more generally, applying arcsin to an angle does not recover that angle. The corrected formula O=4[arcsin(sinγ)R + arcsin(sinβ)p + arcsin(sinδ)r] is valid only if γ, β, δ∈(0,π/2). The paper neither states nor proves this branch condition. It is true, since θ>α′ and θ′>α for all positive a,b, but that verification is absent. Because Eq. (17) is the central claim, the manuscript as written does not yet present a well-defined mathematical formula over the domain it claims to cover.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","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.","tokens_in":7443,"tokens_out":8331,"duration_ms":79628,"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":[{"comment":"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.","section":"III, Eq. (17)"},{"comment":"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.","section":"II, Eqs. (11)-(16)"},{"comment":"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.","section":"II, Eqs. (3)-(4)"},{"comment":"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.","section":"III, Remark"}],"minor_comments":[{"comment":"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.","section":"III"},{"comment":"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.","section":"II"},{"comment":"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.","section":"I"},{"comment":"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.","section":"II"},{"comment":"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.","section":"III"}],"recommendation":"major_revision","confidential_remarks":"The title page includes 'Pour citer cet article : J.-M. Ginoux, J.-C. Golvin, C. R. Acad. Sci. Paris, Ser. I 356 (2018) 1195-1202,' which suggests this arXiv posting reproduces or closely follows an already-published article. The editor may wish to verify duplicate-publication policy and whether the journal considers an arXiv reposting of a prior C. R. Acad. Sci. note as a new submission. This does not affect the technical assessment above."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Quick read: this note does what it says. It derives the three center angles for Honey's eight-centered oval and gives a perimeter formula. The trigonometric derivation checks out, and the numerical agreement with the ellipse perimeter is real — under 0.03% relative error for a,b in [1,10] and 1.85e-4% for the Colosseum. That's the useful part for archaeologists, architects, and CAD users who want a closed-form perimeter rather than an elliptic integral.\n\nThe paper's main defect is in Eq. (17). As printed, it applies arcsin to the angles γ, β, δ themselves, but those are angles in radians, not sines. For a=10, b=1, δ≈1.09, so arcsin(δ) is undefined. The intended meaning is clear from Table I and the circle-limit check: they mean arcsin(sinγ), etc. All three angles are in fact acute, so that recovers the angle, but the branch condition is never stated or proved, and the formula as written is not well-defined over the claimed domain. This is a notation/typo problem, not a geometric one, but it is load-bearing in the sense that Eq. (17) is the central claim. A referee should insist on restating it as O = 4[γ R + β p + δ r] or equivalently 4[arcsin(sinγ)R + ...] with an explicit acuteness proof.\n\nThere is also a minor arithmetic typo in the circle-limit remark: the identity should be 2 arcsin((4−√2)/6) = arcsin(7/9), not 2 arcsin(7/9). The final 2πρ is correct, so this is cosmetic.\n\nNovelty is modest. The calculation is a routine application of the law of cosines and sines once Honey's construction is given. The authors do not verify whether Herrera-Samper (2015) or Mazzotti (2017) already contains a perimeter formula; if either does, the novelty claim collapses. They also withhold the algebraic derivation, which is a small annoyance but not a flaw.\n\nWho is this for? People working with polycentric ovals in historical architecture or technical drawing. It is a short note, not a major contribution. But it is correct in substance, and the presentation issues are fixable in one round. I would send it to a specialized venue and let a referee require the Eq. (17) restatement and a citation check. A desk rejection would be too harsh.","headline":"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.","tokens_in":7938,"tokens_out":4831,"would_cite":false,"duration_ms":46926,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["51M25","51N20"],"pacs":[],"model":"deepseek-v4-flash","headline":"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.","keywords":["eight-centered oval","perimeter formula","ellipse approximation","Honey construction","arc length","central angles","Colosseum","trigonometry"],"falsifier":"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.","tokens_in":6973,"feed_emoji":"📐","tokens_out":11349,"duration_ms":90856,"temperature":0.7,"pith_summary":"The paper establishes a closed-form formula for the perimeter of the eight-centered oval, a classical architectural curve that approximates an ellipse using eight tangent circular arcs. Working from Honey's 1908 construction, the authors derive the three central angles of one quarter of the oval using the laws of sines and cosines on the triangle formed by the three circle centers. The perimeter is then exactly four times the sum of each arc's radius times its angle, yielding $O(a,b) = 4[\\arcsin(\\gamma) a^{2}/b + \\arcsin(\\beta) (a+b)/2 + \\arcsin(\\delta) b^{2}/a]$ in terms of arcsines of algebraic expressions in the semi-axes $a$ and $b$. The formula matches the true ellipse perimeter to within 0.029% for axes from 1 to 10, and for the Colosseum ($a=94$, $b=78$) it gives 541.523 m versus 541.524 m from the elliptic integral. This gives architects and archaeologists a simple, direct way to compute perimeters of polycentric oval buildings without elliptic integrals.","feed_headline":"Eight-centered oval perimeter formula matches ellipse","feed_subtitle":"Arc-length sum derived from Honey's construction stays within 0.029 percent of the true ellipse.","key_machinery":"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.","core_discovery":"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.","pith_inferences":["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."],"forward_implications":["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."],"supporting_citations":[{"why":"Supplies Honey's construction of the three tangent circles (centers and radii) that the perimeter formula is built on.","marker":"[7]"},{"why":"Supplies the Colosseum semi-axes $a=94$, $b=78$ used in the numerical comparison.","marker":"[5]"},{"why":"Provides the arcsine addition identities used to verify the circle limit $O=2\\pi\\rho$.","marker":"[2]"},{"why":"Supports the identification of the Colosseum as an eight-centered oval, motivating the approximation test.","marker":"[14]"}],"fun_headline_variants":["Closed-form perimeter for eight-centered oval","Eight-centered oval perimeter: geometric proof","Oval perimeter formula matches ellipse within 0.029%","Perimeter of eight-centered oval: exact arc sum","Eight-centered oval perimeter: closed-form in semi-axes"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"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.","fun_headline_variants_meta":{"raw":{"variants":["Closed-form perimeter for eight-centered oval","Eight-centered oval perimeter: geometric proof","Oval perimeter formula matches ellipse within 0.029%","Perimeter of eight-centered oval: exact arc sum","Eight-centered oval perimeter: closed-form in semi-axes"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000273,"raw_usage":{"total_tokens":1597,"prompt_tokens":869,"completion_tokens":728,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":485,"completion_tokens_details":{"reasoning_tokens":656}},"tokens_in":485,"tokens_out":728,"duration_ms":6794,"temperature":1.0,"reasoning_tokens":656,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T15:33:12.020999+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"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.","supporting_citations":[{"cited_title":"Berndt, Ramanujan’s Notebooks, vol","cited_arxiv_id":null,"evidence_quote":"Supplies Honey's construction of the three tangent circles (centers and radii) that the perimeter formula is built on."},{"cited_title":"loi des cosinus","cited_arxiv_id":null,"evidence_quote":"Supplies the Colosseum semi-axes $a=94$, $b=78$ used in the numerical comparison."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Provides the arcsine addition identities used to verify the circle limit $O=2\\pi\\rho$."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Supports the identification of the Colosseum as an eight-centered oval, motivating the approximation test."}],"review_version":1}