Atautochrone curve orisochrone curve (from Ancient Greek ταὐτό (tauto-) 'same' ἴσος (isos-) 'equal' and χρόνος (chronos) 'time') is thecurve for which the time taken by an object sliding withoutfriction in uniformgravity to its lowest point is independent of its starting point on the curve. The curve is acycloid, and the time is equal toπ times thesquare root of the radius of the circle which generates the cycloid, over theacceleration of gravity. The tautochrone curve is related to thebrachistochrone curve, which is also a cycloid.

The tautochrone problem, the attempt to identify this curve, was solved byChristiaan Huygens in 1659. He proved geometrically in hisHorologium Oscillatorium, originally published in 1673, that the curve is acycloid.

On a cycloid whose axis is erected on the perpendicular and whose vertex is located at the bottom, the times of descent, in which a body arrives at the lowest point at the vertex after having departed from any point on the cycloid, are equal to each other ...^[1]

The cycloid is given by a point on a circle of radius${\displaystyle r}$ tracing a curve as the circle rolls along the${\displaystyle x}$ axis, as:

Huygens also proved that the time of descent is equal to the time a body takes to fall vertically the same distance as diameter of the circle that generates the cycloid, multiplied by${\displaystyle \pi /2}$. In modern terms, this means that the time of descent is$\pi \sqrt{r/g}$, where${\displaystyle r}$ is the radius of the circle which generates the cycloid, and${\displaystyle g}$ is thegravity of Earth, or more accurately, the earth's gravitational acceleration.

This solution was later used to solve the problem of thebrachistochrone curve.Johann Bernoulli solved the problem in a paper (Acta Eruditorum, 1697).

The tautochrone problem was studied by Huygens more closely when it was realized that a pendulum, which follows a circular path, was notisochronous and thus hispendulum clock would keep different time depending on how far the pendulum swung. After determining the correct path, Christiaan Huygens attempted to create pendulum clocks that used a string to suspend the bob and curb cheeks near the top of the string to change the path to the tautochrone curve. These attempts proved unhelpful for a number of reasons. First, the bending of the string causes friction, changing the timing. Second, there were much more significant sources of timing errors that overwhelmed any theoretical improvements that traveling on the tautochrone curve helps. Finally, the "circular error" of a pendulum decreases as length of the swing decreases, so better clockescapements could greatly reduce this source of inaccuracy.

Later, the mathematiciansJoseph Louis Lagrange andLeonhard Euler provided an analytical solution to the problem.

For asimple harmonic oscillator released from rest, regardless of its initial displacement, the time it takes to reach the lowest potential energy point is always a quarter of its period, which is independent of its amplitude. Therefore, the Lagrangian of a simple harmonic oscillator isisochronous.

In the tautochrone problem, if the particle's position is parametrized by thearclengths(t) from the lowest point, the kinetic energy is then proportional to${\displaystyle {\dot{s}}^2}$, and the potential energy is proportional to the heighth(s). One way the curve in the tautochrone problem can be an isochrone is if the Lagrangian is mathematically equivalent to a simple harmonic oscillator; that is, the height of the curve must be proportional to the arclength squared:

${\displaystyle h(s)=s^2/(8r),}$

where the constant of proportionality is${\displaystyle 1/(8r)}$. Compared to the simple harmonic oscillator'sLagrangian, the equivalent spring constant is${\displaystyle k=mg/(4r)}$, and the time of descent is${\displaystyle T/4=\frac{\pi }{2}\sqrt{\frac{m}{k}}=\pi \sqrt{\frac{r}{g}}.}$ However, the physical meaning of the constant${\displaystyle r}$ is not clear until we determine the exact analytical equation of the curve.

To solve for the analytical equation of the curve, note that the differential form of the above relation is

which eliminatess, and leaves a differential equation fordx anddh. This is the differential equation for acycloid when the vertical coordinateh is counted from its vertex (the point with a horizontal tangent) instead of thecusp.

To find the solution, integrate forx in terms ofh:

where${\displaystyle u=\sqrt{h/(2r)}}$, and the height decreases as the particle moves forward. This integral is the area under a circle, which can be done with another substitution${\displaystyle u=\cos (t/2)}$ and yield:

This is the standard parameterization of acycloid with${\displaystyle h=2r-y}$. It's interesting to note that the arc length squared is equal to the height difference multiplied by the full arch length${\displaystyle 8r}$.

The simplest solution to the tautochrone problem is to note a direct relation between the angle of an incline and the gravity felt by a particle on the incline. A particle on a 90° vertical incline undergoes full gravitational acceleration${\displaystyle g}$, while a particle on a horizontal plane undergoes zero gravitational acceleration. At intermediate angles, the acceleration due to "virtual gravity" by the particle is${\displaystyle g\sin \theta }$. Note that${\displaystyle \theta }$ is measured between the tangent to the curve and the horizontal, with angles above the horizontal being treated as positive angles. Thus,${\displaystyle \theta }$ varies from${\displaystyle -\pi /2}$ to${\displaystyle \pi /2}$.

The position of a mass measured along a tautochrone curve,${\displaystyle s(t)}$, must obey the following differential equation:

${\displaystyle \frac{d^{2}s}{{dt}^2}=-{\omega }^{2}s}$

which, along with the initial conditions${\displaystyle s(0)=s_0}$ and${\displaystyle s^{\prime }(0)=0}$, has solution:

${\displaystyle s(t)=s_0\cos \omega t}$

It can be easily verified both that this solution solves the differential equation and that a particle will reach${\displaystyle s=0}$ at time${\displaystyle \pi /2\omega }$ from any starting position${\displaystyle s_0}$. The problem is now to construct a curve that will cause the mass to obey the above motion.Newton's second law shows that the force of gravity and the acceleration of the mass are related by:

The explicit appearance of the distance,${\displaystyle s}$, is troublesome, but we candifferentiate to obtain a more manageable form:

This equation relates the change in the curve's angle to the change in the distance along the curve. We now usetrigonometry to relate the angle${\displaystyle \theta }$ to the differential lengths${\displaystyle dx}$,${\displaystyle dy}$ and${\displaystyle ds}$:

${\displaystyle \begin{array}{r}ds=\frac{dx}{\cos \theta }\\ ds=\frac{dy}{\sin \theta }\end{array}}$

Replacing${\displaystyle ds}$ with${\displaystyle dx}$ in the above equation lets us solve for${\displaystyle x}$ in terms of${\displaystyle \theta }$:

Likewise, we can also express${\displaystyle ds}$ in terms of${\displaystyle dy}$ and solve for${\displaystyle y}$ in terms of${\displaystyle \theta }$:

Substituting${\displaystyle \varphi =2\theta }$ and$r=\frac{g}{4{\omega }^2}$, we see that theseparametric equations for${\displaystyle x}$ and${\displaystyle y}$ are those of a point on a circle of radius${\displaystyle r}$ rolling along a horizontal line (acycloid), with the circle center at the coordinates${\displaystyle (C_x+r\varphi ,C_y)}$:

Note that${\displaystyle \varphi }$ ranges from${\displaystyle -\pi \le \varphi \le \pi }$. It is typical to set${\displaystyle C_x=0}$ and${\displaystyle C_y=r}$ so that the lowest point on the curve coincides with the origin. Therefore:

Solving for${\displaystyle \omega }$ and remembering that${\displaystyle T=\frac{\pi }{2\omega }}$ is the time required for descent, being a quarter of a whole cycle, we find the descent time in terms of the radius${\displaystyle r}$:

(Based loosely onProctor, pp. 135–139)

Niels Henrik Abel attacked a generalized version of the tautochrone problem (Abel's mechanical problem), namely, given a function${\displaystyle T(y)}$ that specifies the total time of descent for a given starting height, find an equation of the curve that yields this result. The tautochrone problem is a special case of Abel's mechanical problem when${\displaystyle T(y)}$ is a constant.

Abel's solution begins with the principle ofconservation of energy – since the particle is frictionless, and thus loses no energy toheat, itskinetic energy at any point is exactly equal to the difference ingravitational potential energy from its starting point. The kinetic energy is$\frac{1}{2}m{v}^2$, and since the particle is constrained to move along a curve, its velocity is simply${\displaystyle d\ell /dt}$, where${\displaystyle \ell }$ is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height${\displaystyle y_0}$ to a height${\displaystyle y}$ is${\displaystyle mg(y_0-y)}$, thus:

In the last equation, we have anticipated writing the distance remaining along the curve as a function of height (${\displaystyle \ell (y))}$, recognized that the distance remaining must decrease as time increases (thus the minus sign), and used thechain rule in the form$d\ell =\frac{d\ell }{dy}dy$.

Now we integrate from${\displaystyle y=y_0}$ to${\displaystyle y=0}$ to get the total time required for the particle to fall:

${\displaystyle T(y_0)={\int }_{y=y_0}^{y=0}dt=\frac{1}{\sqrt{2g}}{\int }_0^{y_0}\frac{1}{\sqrt{y_0-y}}\frac{d\ell }{dy}dy}$

This is calledAbel's integral equation and allows us to compute the total time required for a particle to fall along a given curve (for which${\displaystyle d\ell /dy}$ would be easy to calculate). But Abel's mechanical problem requires the converse – given${\displaystyle T(y_0)}$, we wish to find${\displaystyle f(y)=d\ell /dy}$, from which an equation for the curve would follow in a straightforward manner. To proceed, we note that the integral on the right is theconvolution of${\displaystyle d\ell /dy}$ with${\displaystyle 1/\sqrt{y}}$ and thus take theLaplace transform of both sides with respect to variable${\displaystyle y}$:

${\displaystyle \mathcal{L}[T(y_0)]=\frac{1}{\sqrt{2g}}\mathcal{L}\left[\frac{1}{\sqrt{y}}\right]F(s)}$

where${\displaystyle F(s)=\mathcal{L}\left[d\ell /dy\right]}$. Since$\mathcal{L}\left[1/\sqrt{y}\right]=\sqrt{\pi /s}$, we now have an expression for the Laplace transform of${\displaystyle d\ell /dy}$ in terms of the Laplace transform of${\displaystyle T(y_0)}$:

${\displaystyle \mathcal{L}\left[\frac{d\ell }{dy}\right]=\sqrt{\frac{2g}{\pi }}{s}^{\frac{1}{2}}\mathcal{L}[T(y_0)]}$

This is as far as we can go without specifying${\displaystyle T(y_0)}$. Once${\displaystyle T(y_0)}$ is known, we can compute its Laplace transform, calculate the Laplace transform of${\displaystyle d\ell /dy}$ and then take the inverse transform (or try to) to find${\displaystyle d\ell /dy}$.

For the tautochrone problem,${\displaystyle T(y_0)=T_0}$ is constant. Since the Laplace transform of 1 is${\displaystyle 1/s}$, i.e.,$\mathcal{L}[T(y_0)]=T_0/s$, we find the shape function$f(y)=d\ell /dy$:

Making use again of the Laplace transform above, we invert the transform and conclude:

${\displaystyle \frac{d\ell }{dy}=T_0\frac{\sqrt{2g}}{\pi }\frac{1}{\sqrt{y}}}$

It can be shown that the cycloid obeys this equation. It needs one step further to do the integral with respect to${\displaystyle y}$ to obtain the expression of the path shape.

(Simmons, Section 54).

• Beltrami identity
• Brachistochrone curve
• Calculus of variations
• Catenary
• Cycloid
• Uniformly accelerated motion

• Simmons, George (1972).Differential Equations with Applications and Historical Notes. McGraw–Hill.ISBN 0-07-057540-1.
• Proctor, Richard Anthony (1878).A Treatise on the Cycloid and All Forms of Cycloidal Curves, and on the Use of Such Curves in Dealing with the Motions of Planets, Comets, etc., and of Matter Projected from the Sun.

• Mathworld

• Plane curves
• Mechanics

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