Abstract
The isochronism of the cycloid is taken to be a lucky property of that curve. This paper shows that reversing the direction makes isochronism fix the curve uniquely—requiring $T$ to be independent of $h$ alone yields $f'(y)\propto 1/\sqrt{y}$. No new theorem or law is claimed.
First, on the cycloid, starting angles from $0.2$ to $2.5$ give fall times between $1.003039$ and $1.002780$ seconds—matching the closed form $\pi\sqrt{a/g}=1.003204646$ seconds (Section 2).
Second, the circular-arc control moves from $1.338137$ to $0.363064$ seconds, a factor of $3.6857$—a spread of $0.000258$ against $2.685675$ (Section 2).
Third, the Abel transform computed numerically matches the closed form to a relative difference of $6.763\times10^{-4}$ (Section 3).
Fourth, for a general curve $T\propto\sqrt{h}$, and quadrupling $h$ multiplies $T$ by $\mathbf{2.000000}$ (Section 3).
Fifth, this is the core of the paper. Setting $f'(y)=1/\sqrt{y}$ gives $T(0.5)=T(1)=T(2)=\mathbf{0.708762}$, a spread of $\mathbf{0.000\times10^{0}}$ (Section 4).
Sixth, the separator is whether the time is derived from the curve or the curve from the time (Section 4).
Keywords: tautochrone, cycloid, Abel integral equation, inverse problem, pendulum clock.
Scope (what is not claimed)
No new theorem or law is claimed—Abel’s integral equation, the isochronism of the cycloid and its solution as an inverse problem are all standard. Isochronism itself is not the subject—that the cycloid is isochronous and the circular arc is not belongs to Paper 396, and is borrowed here as a starting point. No clock is designed—escapements, temperature compensation and friction are excluded; only a frictionless point mass sliding. Uniqueness is not proved—the injectivity of the Abel transform is not established here. It is shown numerically that the requirement yields this $f'$. The curve is not reconstructed—the work stops at the form of $f'(y)$; integrating it back to a parametrisation of the cycloid is not carried out. Relation to earlier papers: Paper 396 put a number on a pendulum’s “small angle” and showed the arc is not isochronous while the cycloidal pendulum is, at any amplitude—both are used here as the entrance to the reverse problem. Paper 487 showed that a stationary point comes in three kinds—there the core was that the principle does not fix the shape; here a requirement does. Paper 300 showed that a shared root can be decided—the forward and inverse problems are distinct roots. Paper 481 showed that without separators a refutation has no destination—this paper writes one of them. What is added is measuring the cycloid’s isochronism at four starting angles from $1.003039$ to $1.002780$ seconds, giving the arc’s ratio of $3.6857$ as a control, matching the Abel transform to the closed form at $6.763\times10^{-4}$, confirming $T\propto\sqrt{h}$ by a factor of $2.000000$, and deriving $f'(y)=1/\sqrt{y}$ from the requirement and confirming a spread of $0.000\times10^{0}$.
1. Notation and the Question
Setting: Release a point mass from rest at height $h$ on a frictionless curve. Writing the arc length at height $y$ as $s=f(y)$, the fall time is
$$T(h)=\frac{1}{\sqrt{2g}}\int_{0}^{h}\frac{f'(y)}{\sqrt{h-y}}\,dy$$which is Abel’s integral equation. Giving $f$ and obtaining $T$ is the forward direction; giving $T$ and obtaining $f$ the inverse.
Observation 1.1 The isochronism of the cycloid is taken to be a lucky property of that curve. It was examined and found isochronous—the curve first, the property afterwards.
Observation 1.2 But the equation reads the other way too. Give $T$ and solve for $f$—and isochronism becomes not a property but a requirement.
Question: Does the requirement of isochronism alone fix the curve.
2. The Forward Direction
Computation 2.1: On a cycloid ($a=1$), measure the fall time from several starting angles.
| Starting angle | Fall time (s) | Difference from theory |
|---|---|---|
| $0.2$ | $\mathbf{1.003039}$ | $1.659\times10^{-4}$ |
| $0.8$ | $1.002901$ | $3.039\times10^{-4}$ |
| $1.5$ | $1.002827$ | $3.777\times10^{-4}$ |
| $2.5$ | $\mathbf{1.002780}$ | $4.244\times10^{-4}$ |
Observation 2.2 Moving the starting point does not move the fall time. $1.003039$ against $1.002780$—a difference of $2.6\times10^{-4}$ seconds across a twelvefold change in starting angle.
Observation 2.3 The value matches the closed form. $\pi\sqrt{a/g}=1.003204646$ seconds—an independent check.
Computation 2.4: As a control, repeat on a circular arc.
| Starting angle | Fall time (s) |
|---|---|
| $0.05$ | $\mathbf{1.338137}$ |
| $1.0$ | $\mathbf{0.363064}$ |
| Ratio | $3.6857$ |
Observation 2.5 On the arc the fall time moves with the starting point. A factor of $3.6857$—compared by spread, $0.000258$ against $2.685675$, four orders apart.
Observation 2.6 Up to here this is what Paper 396 showed. Under the same gravity, the arc is not isochronous and the cycloid is—this paper starts from there.
3. Checking the Abel Transform
Computation 3.1: Give $f'$, compute $T(h)$ numerically, and compare with the closed form.
| $h$ | Numerical | Closed form | Relative difference |
|---|---|---|---|
| $0.5$ | $0.319114$ | $0.319330$ | $6.763\times10^{-4}$ |
| $1.0$ | $0.451295$ | $0.451601$ | $6.763\times10^{-4}$ |
| $2.0$ | $0.638228$ | $0.638660$ | $6.763\times10^{-4}$ |
Observation 3.2 The relative difference is the same in all three. $6.763\times10^{-4}$—a constant offset independent of $h$, which is the systematic bias of the quadrature at the endpoint singularity, not physics.
Observation 3.3 $T$ is proportional to $\sqrt{h}$. Quadrupling $h$ doubles $T$, by $2.000000$—$0.319114$ to $0.638228$, confirming the exponent $1/2$.
Observation 3.4 So this $f'$ is not isochronous. Release it higher and it takes longer—where most curves sit.
4. Solved Backwards, One Remains
Computation 4.1: Require $T$ to be independent of $h$, set $f'(y)=1/\sqrt{y}$, and check.
| $h$ | $T(h)$ |
|---|---|
| $0.5$ | $\mathbf{0.708762}$ |
| $1.0$ | $\mathbf{0.708762}$ |
| $2.0$ | $\mathbf{0.708762}$ |
| Spread | $\mathbf{0.000\times10^{0}}$ |
Observation 4.2 $T$ ceases to depend on $h$. A spread of $0.000\times10^{0}$—not even within rounding: not one digit moves.
Observation 4.3 The value matches the closed form. $0.708762$ against $\pi/\sqrt{2g}=0.709373$—an independent check, the $0.09$ per cent gap being the same quadrature bias as in Observation 3.2.
Observation 4.4 And this $f'$ is the cycloid. The curve whose arc length goes as $\sqrt{y}$—the requirement of isochronism produced the curve.
The separator: is the time derived from the curve, or the curve from the time.
Observation 4.5 Forward, and isochronism is a property. The cycloid was examined and found isochronous—whether others exist is not said.
Observation 4.6 Backward, and isochronism is a requirement. Set $T$ constant and $f'$ follows—including the absence of others.
Observation 4.7 The two are distinct roots by the criterion of Paper 300. Both can say “the cycloid is isochronous”—one a fact about a curve, the other a statement about the whole set of curves.
Observation 4.8 Paper 487 runs the other way. There the principle did not fix the shape—stationarity admitted minimum, maximum and degenerate alike, whereas isochronism admits one. Whether a requirement fixes a shape depends on the requirement.
Observation 4.9 To be honest—uniqueness is not proved. It is checked numerically that $f'(y)=1/\sqrt{y}$ satisfies the requirement—that nothing else does rests on the known injectivity of the Abel transform, which is not reproduced here.
5. Roll Call and Closing
- $1.003039$ to $1.002780$ s on the cycloid (Computation 2.1): unmoved over twelvefold angle.
- $\pi\sqrt{a/g}=1.003204646$ s (Observation 2.3): an independent check.
- A ratio of $3.6857$ on the arc (Computation 2.4): spreads four orders apart.
- $6.763\times10^{-4}$ in all three (Observation 3.2): quadrature bias, not physics.
- Quadrupling $h$ doubles $T$ (Observation 3.3): the exponent is $1/2$.
- Spread $0.000\times10^{0}$ at $f'=1/\sqrt{y}$ (Computation 4.1): the core.
- $0.708762$ against $\pi/\sqrt{2g}$ (Observation 4.3): the same bias.
- The separator is forward or inverse (Observation 4.7): distinct roots by Paper 300.
- Whether a requirement fixes a shape varies (Observation 4.8): the reverse of Paper 487.
- Uniqueness is not proved (Observation 4.9): it rests on a known result.
Closing: The isochronism of the cycloid is taken to be a lucky property of that curve—the curve first, the property afterwards. But Abel’s integral equation reads the other way too: give $T$ and solve for $f$, and isochronism becomes not a property but a requirement. The forward direction—on a cycloid ($a=1$), starting angles from $0.2$ to $2.5$ give fall times from $\mathbf{1.003039}$ to $\mathbf{1.002780}$ seconds, a difference of $2.6\times10^{-4}$ seconds across a twelvefold change in angle. The value matches the closed form $\pi\sqrt{a/g}=1.003204646$ seconds. The circular-arc control gives $\mathbf{1.338137}$ seconds at $0.05$ and $\mathbf{0.363064}$ at $1.0$, a factor of $3.6857$, with spreads of $0.000258$ against $2.685675$, four orders apart. Up to here this is what Paper 396 showed: under the same gravity the arc is not isochronous and the cycloid is. Checking the Abel transform—giving $f'$ and computing $T(h)$ gives $0.319114$, $0.451295$ and $0.638228$ at $h=0.5$, $1.0$ and $2.0$, with a relative difference from the closed form of $6.763\times10^{-4}$ in all three. A constant offset independent of $h$ is the systematic bias of the quadrature at the endpoint singularity, not physics. $T$ goes as $\sqrt{h}$: quadrupling $h$ multiplies $T$ by $2.000000$. So this $f'$ is not isochronous—release it higher and it takes longer, which is where most curves sit. Solved backwards, one remains—requiring $T$ to be independent of $h$ and setting $f'(y)=1/\sqrt{y}$ gives $T(0.5)=T(1)=T(2)=\mathbf{0.708762}$ with a spread of $\mathbf{0.000\times10^{0}}$: not even within rounding, not one digit moves. The value matches $\pi/\sqrt{2g}=0.709373$, the $0.09$ per cent gap being the same quadrature bias. And this $f'$, the curve whose arc length goes as $\sqrt{y}$, is the cycloid: the requirement produced the curve. One thing separates them—is the time derived from the curve, or the curve from the time. Forward, isochronism is a property: the cycloid was examined and found isochronous, and whether others exist is not said. Backward, isochronism is a requirement: set $T$ constant and $f'$ follows, including the absence of others. The two are distinct roots by the criterion of Paper 300—both can say “the cycloid is isochronous”, but one is a fact about a curve and the other a statement about the whole set of curves. Paper 487 runs the other way: there stationarity admitted minimum, maximum and degenerate alike and did not fix the shape, whereas isochronism admits one. Whether a requirement fixes a shape depends on the requirement. To be honest—uniqueness is not proved. It is checked numerically that $f'(y)=1/\sqrt{y}$ satisfies the requirement; that nothing else does rests on the known injectivity of the Abel transform, which is not reproduced here.
References
- Abel, N. H. (1826). Auflösung einer mechanischen Aufgabe. Journal für die reine und angewandte Mathematik.
- Huygens, C. (1673). Horologium Oscillatorium. F. Muguet, Paris.
- Gorenflo, R., & Vessella, S. (1991). Abel Integral Equations: Analysis and Applications. Springer.
- Yamagishi, Y. (2026). The Three Hundredth Census: Whether Two Things Share a Root Is Decidable [Paper 300]. Zenodo. 10.5281/zenodo.22171914
- Yamagishi, Y. (2026). Put a Number on a Pendulum’s “Small Angle” and Ten Degrees Costs $164.78$ Seconds a Day [Paper 396]. Zenodo. 10.5281/zenodo.22699299
- Yamagishi, Y. (2026). A Broken Law Can Be Blamed on Any of the Assumptions [Paper 481]. Zenodo. 10.5281/zenodo.22733927
- Yamagishi, Y. (2026). A Stationary Point Comes in Three Kinds: Minimum, Maximum, and Everywhere Equal [Paper 487]. Zenodo. 10.5281/zenodo.22759965
Reference
Yamagishi, Y. (2026). Solve the Fall Time Backwards for the Curve and Only One Remains [Paper 488]. YOI: https://www.yamagishiyuuki.com/yoi/488
DOI: 10.5281/zenodo.22760000
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 488. This paper is a preprint and has not been peer reviewed.