tautochrone curve

File:Huygens - Horologium oscillatorium, sive De motu pendulorum ad horologia aptato demonstrationes geometricae, 1673 - 869780.jpeg, Horologium oscillatorium sive de motu pendulorum, 1673]]

{{Quote box|width=30%|

quote=It was in the left hand try-pot of the Pequod, with the soapstone diligently circling round me, that I was first indirectly struck by the remarkable fact, that in geometry all bodies gliding along the cycloid, my soapstone for example, will descend from any point in precisely the same time.

|source=Moby Dick by Herman Melville, 1851}}

The tautochrone problem, the attempt to identify this curve, was solved by Christiaan Huygens in 1659. He proved geometrically in his Horologium Oscillatorium, originally published in 1673, that the curve is a cycloid.

{{Blockquote|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 ...{{cite book |last=Blackwell |first=Richard J. |title=Christiaan Huygens' The Pendulum Clock |publisher=Iowa State University Press |date=1986 |location=Ames, Iowa |isbn=0-8138-0933-9 |at= Part II, Proposition XXV, p. 69}}}}

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

$$\backslash begin\{align\}$$

x &= r(\theta - \sin \theta) \\

y &= r(1 - \cos \theta),

\end{align}

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 $\backslash pi\; /\; 2$. In modern terms, this means that the time of descent is $\backslash pi\; \backslash sqrt\{r/g\}$, where $r$ is the radius of the circle which generates the cycloid, and $g$ is the gravity of Earth, or more accurately, the earth's gravitational acceleration.

File:Isochronous cycloidal pendula.gif

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

File:CyloidPendulum.png]]

The tautochrone problem was studied by Huygens more closely when it was realized that a pendulum, which follows a circular path, was not isochronous and thus his pendulum 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 clock escapements could greatly reduce this source of inaccuracy.

Later, the mathematicians Joseph Louis Lagrange and Leonhard Euler provided an analytical solution to the problem.

For a simple 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 is isochronous.

In the tautochrone problem, if the particle's position is parametrized by the arclength {{math|s(t)}} from the lowest point, the kinetic energy is then proportional to $\backslash dot\{s\}^2$, and the potential energy is proportional to the height {{math|h(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:

{{block indent|1=$h(s)\; =\; s^2/(8r),$}}

where the constant of proportionality is $1/(8r)$. Compared to the simple harmonic oscillator's Lagrangian, the equivalent spring constant is $k=mg/(4r)$, and the time of descent is $T/4=\backslash frac\{\backslash pi\}\{2\}\; \backslash sqrt\{\backslash frac\{m\}\{k\}\}=\backslash pi\; \backslash sqrt\{\backslash frac\{r\}\{g\}\}.$ However, the physical meaning of the constant $r$ 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

{{block indent|1=$\backslash begin\{align\}$

dh &= s \,ds/(4r), \\

dh^2 &= s^2 \,ds^2/(16r^2) = h \left(dx^2 + dh^2\right)/(2r),\\

\left(\frac{dx}{dh}\right)^2&=\frac{2r}{h}-1

\end{align}}}

which eliminates {{math|s}}, and leaves a differential equation for {{math|dx}} and {{math|dh}}. This is the differential equation for a cycloid when the vertical coordinate {{math|h}} is counted from its vertex (the point with a horizontal tangent) instead of the cusp.

To find the solution, integrate for {{math|x}} in terms of {{math|h}}:

{{block indent|1=$\backslash begin\{align\}$

\frac{dx}{dh} &= -\frac{\sqrt{2r-h}}{\sqrt{h}}, \\

x &= -4r\int \sqrt{1-u^2} \, du,

\end{align}}}

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

{{block indent|1=$\backslash begin\{align\}$

x &=r(t - \sin t), \\

h &=r(1 + \cos t).

\end{align}}}

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

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 $g$, while a particle on a horizontal plane undergoes zero gravitational acceleration. At intermediate angles, the acceleration due to "virtual gravity" by the particle is $g\backslash sin\backslash theta$. Note that $\backslash theta$ is measured between the tangent to the curve and the horizontal, with angles above the horizontal being treated as positive angles. Thus, $\backslash theta$ varies from $-\backslash pi/2$ to $\backslash pi/2$.

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

{{block indent|1=$\backslash frac\{d^2s\}\{\{dt\}^2\}\; =\; -\; \backslash omega^2s$}}

which, along with the initial conditions $s(0)=s\_0$ and $s\text{'}(0)=0$, has solution:

{{block indent|1=$s(t)\; =\; s\_0\; \backslash cos\; \backslash omega\; t$}}

It can be easily verified both that this solution solves the differential equation and that a particle will reach $s=0$ at time $\backslash pi/2\backslash omega$ from any starting position $s\_0$. 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:

{{block indent|1=$$

\begin{align}

-g \sin \theta & = \frac{d^2s}{{dt}^2} \\

& = - \omega^2 s \,

\end{align}

}}

The explicit appearance of the distance, $s$, is troublesome, but we can differentiate to obtain a more manageable form:

{{block indent|1=$\backslash begin\{align\}$

g \cos \theta \,d\theta &= \omega^2 \,ds \\

\Longrightarrow ds &= \frac{g}{\omega^2} \cos \theta \,d\theta

\end{align}}}

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

{{block indent|1=$$

\begin{align}

ds = \frac{dx}{\cos \theta} \\

ds = \frac{dy}{\sin \theta}

\end{align}

}}

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

{{block indent|1=$$

\begin{align}

ds & = \frac{g}{\omega^2} \cos \theta \,d\theta \\

\frac{dx}{\cos\theta} & = \frac{g}{\omega^2} \cos \theta\, d\theta \\

dx & = \frac{g}{\omega^2} \cos^2 \theta \,d\theta \\

& = \frac{g}{2 \omega^2} \left ( \cos 2 \theta + 1 \right ) d\theta \\

x & = \frac{g}{4 \omega^2} \left ( \sin 2 \theta + 2 \theta \right ) + C_x

\end{align}

}}

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

{{block indent|1=$$

\begin{align}

ds & = \frac{g}{\omega^2} \cos \theta \,d\theta \\

\frac{dy}{\sin\theta} & = \frac{g}{\omega^2} \cos \theta\, d\theta \\

dy & = \frac{g}{\omega^2} \sin \theta \cos \theta \,d\theta \\

& = \frac{g}{2\omega^2} \sin 2 \theta \,d\theta \\

y & = -\frac{g}{4\omega^2} \cos 2 \theta + C_y

\end{align}

}}

Substituting $\backslash phi\; =\; 2\backslash theta$ and $r\; =\; \backslash frac\{g\}\{4\backslash omega^2\}\backslash ,$, we see that these parametric equations for $x$ and $y$ are those of a point on a circle of radius $r$ rolling along a horizontal line (a cycloid), with the circle center at the coordinates $(C\_x\; +\; r\backslash phi,\; C\_y)$:

{{block indent|1=$$

\begin{align}

x & = r \left( \sin \phi + \phi \right) + C_x \\

y & = -r \cos \phi + C_y

\end{align}

}}

Note that $\backslash phi$ ranges from $-\backslash pi\; \backslash le\; \backslash phi\; \backslash le\; \backslash pi$. It is typical to set $C\_x\; =\; 0$ and $C\_y\; =\; r$ so that the lowest point on the curve coincides with the origin. Therefore:

{{block indent|1=$$

\begin{align}

x & = r \left( \phi + \sin \phi\right)\\

y & = r \left( 1 - \cos \phi\right)\\

\end{align}

}}

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

{{block indent|1=$$

\begin{align}

r & = \frac{g}{4\omega^2} \\

\omega & = \frac{1}{2} \sqrt{\frac{g}{r}} \\

T & = \pi \sqrt{\frac{r}{g}}\\

\end{align}

}}

(Based loosely on Proctor, pp. 135–139)

Niels Henrik Abel attacked a generalized version of the tautochrone problem (Abel's mechanical problem), namely, given a function $T(y)$ 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 $T(y)$ is a constant.

Abel's solution begins with the principle of conservation of energy – since the particle is frictionless, and thus loses no energy to heat, its kinetic energy at any point is exactly equal to the difference in gravitational potential energy from its starting point. The kinetic energy is $\backslash frac\{1\}\{2\}\; mv^2$, and since the particle is constrained to move along a curve, its velocity is simply $\{d\backslash ell\}/\{dt\}$, where $\backslash ell$ is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height $y\_0$ to a height $y$ is $mg(y\_0\; -\; y)$, thus:

{{block indent|1=$$

\begin{align}

\frac{1}{2} m \left ( \frac{d\ell}{dt} \right ) ^2 & = mg(y_0-y) \\

\frac{d\ell}{dt} & = \pm \sqrt{2g(y_0-y)} \\

dt & = \pm \frac{d\ell}{\sqrt{2g(y_0-y)}} \\

dt & = - \frac{1}{\sqrt{2g(y_0-y)}} \frac{d\ell}{dy} \,dy

\end{align}

}}

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

Now we integrate from $y\; =\; y\_0$ to $y\; =\; 0$ to get the total time required for the particle to fall:

{{block indent|1=$$

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 called Abel's integral equation and allows us to compute the total time required for a particle to fall along a given curve (for which $\{d\backslash ell\}/\{dy\}$ would be easy to calculate). But Abel's mechanical problem requires the converse – given $T(y\_0)\backslash ,$, we wish to find $f(y)\; =\; \{d\backslash ell\}/\{dy\}$, from which an equation for the curve would follow in a straightforward manner. To proceed, we note that the integral on the right is the convolution of $\{d\backslash ell\}/\{dy\}$ with $\{1\}/\{\backslash sqrt\{y\}\}$ and thus take the Laplace transform of both sides with respect to variable $y$:

{{block indent|1=$$

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

}}

where $F(s)\; =\; \backslash mathcal\{L\}\; \{\backslash left[\; \{d\backslash ell\}/\{dy\}\; \backslash right\; ]\}$. Since $\backslash mathcal\{L\}\; \{\backslash left[\; \{1\}/\{\backslash sqrt\{y\}\}\; \backslash right]\}\; =\; \backslash sqrt\{\{\backslash pi\}/\{s\}\}$, we now have an expression for the Laplace transform of $\{d\backslash ell\}/\{dy\}$ in terms of the Laplace transform of $T(y\_0)$:

{{block indent|1=$$

\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 $T(y\_0)$. Once $T(y\_0)$ is known, we can compute its Laplace transform, calculate the Laplace transform of $\{d\backslash ell\}/\{dy\}$ and then take the inverse transform (or try to) to find $\{d\backslash ell\}/\{dy\}$.

For the tautochrone problem, $T(y\_0)\; =\; T\_0\backslash ,$ is constant. Since the Laplace transform of 1 is $\{1\}/\{s\}$, i.e., $\backslash mathcal\{L\}[T(y\_0)]\; =\; \{T\_0\}/\{s\}$, we find the shape function $f(y)\; =\; \{d\backslash ell\}/\{dy\}$:

{{block indent|1=$$

\begin{align}

F(s) = \mathcal{L} {\left [ \frac{d\ell}{dy} \right ]}

& = \sqrt{\frac{2g}{\pi}} s^{\frac{1}{2}} \mathcal{L}[T_0] \\

& = \sqrt{\frac{2g}{\pi}} T_0 s^{-\frac{1}{2}}

\end{align}

}}

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

{{block indent|1=$\backslash frac\{d\backslash ell\}\{dy\}\; =\; T\_0\; \backslash frac\{\backslash sqrt\{2g\}\}\{\backslash pi\}\backslash frac\{1\}\{\backslash sqrt\{y\}\}$}}

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

(Simmons, Section 54).

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

{{Reflist}}

• {{cite book |last=Simmons |first=George |title=Differential Equations with Applications and Historical Notes |publisher=McGraw–Hill |year=1972 |isbn=0-07-057540-1}}
• {{cite book |url=https://books.google.com/books?id=6UIZBvs0diIC |title=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 |last1=Proctor |first1=Richard Anthony |year=1878}}

• [https://mathworld.wolfram.com/TautochroneProblem.html Mathworld]

Category:Plane curves

Category:Mechanics

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