Except for a few trifling practical details, we could easily get around without the consumption of hydrocarbons and the production of greenhouse gases, simply making use of the force of gravity that is constant and inexhaustible. For example, suppose we must journey to a location 30 miles away and at the same elevation. We keep our vehicle at the top of a ramp 100 ft. high. It only has to be hoisted there once. When we wish to travel, we shove off and when we reach the foot of the ramp we are going at about 55 mph. The speed attained is v = √(2gh), where g is the acceleration of gravity (32 fpsps or 980 cm/s2). Our 30-mile trip will require about 33 minutes, quite comparable to the capability of an ordinary motorcar, but now with effortless quietness and unsurpassable economy. At the destination, we slow the vehicle by letting it climb a ramp, and it stops when it reaches an elevation of 100 ft. To return, we simply turn the vehicle around and push off. The only problem is friction, but many proposals we hear today for energy and other purposes demand the solution of no less a problem for their realizations.
The theoretical basis of this proposal is the principle of Conservation of Energy. When we wish to travel, we convert some potential energy into kinetic energy, which is associated with a speed v. We move by the agency of the kinetic energy without using up any of it, since no force acts on the vehicle. Finally, we convert the kinetic energy we have enjoyed back into potential energy. Gravity is a reliable source of potential energy, E = mgh, where h is the distance the mass m has been raised. We could just as well use a spring, or compressed air in a cylinder, but gravity is a good choice.
A quicker journey can be obtained by continuing the ramp to a depth h halfway to the destination, and then climbing a similar ramp to the terminus. Speed increases steadily as the vehicle descends, again a √(2gh), and decreases similarly on the ascending half. However, the length of the journey increases as we go deeper. The journey time, I think, is given by T = 4(D2/4 + h2)1/2/√(2gh). As h→0, T increases to infinity, and as h→∞, T also increases without limit. There must be some minimum time for a certain h, which we can find by setting dT/dh = 0. Solving this equation, we find h = D/2√3 and T = 8√(D/2)/√(2g). For our 30-mile journey, we find T = 281 s or 4.7 minutes. The average speed is 383 mph, which is certainly rapid transit.
The Brachistochrone Transit Company claims to be able to beat this time, making the trip in 2.9 minutes, at an average speed of 614 mph. It does this by means of a cunningly shaped ramp, developed for it by Indian engineers. The shape of the ramp is a company secret, but let's see if we can find out what it is.
We take the x-axis as horizontal, the y-axis as vertically downward in the direction of gravity, and the origin as the starting point. The shape of the ramp is given by the curve y = y(x), where y(0) = 0, and y(X) = Y, where the ramp must pass through the point (X,Y). The speed is v = ds/dt = √(2gy). The arc length of the curve is s, and ds = √(1 + y'2) dx, where y' = dy/dx. The time required to cover ds is ds/v, or √(1 + y'2)/√(2gy). Therefore, the time required to move from x = 0 to x is T = (1/√(2g))∫(0,x)[√(1 + y'2)/√y]dx. We must find the y(x) that makes this time a minimum.
This celebrated problem was proposed by John Bernoulli (1667-1748) in 1696, at the very beginning of the rise of analysis and calculus. It was solved by Newton, Leibniz, and Bernoulli. The surprising result was that the curve was a cycloid, the curve traced out by a point on the circumference of a rolling wheel. J. L. de Lagrange (1736-1813)later (1788) showed a general method for attacking such problems, and we follow his analysis. Leonhard Euler (1707-1783), Bernoulli's student, also made valuable contributions.
Let us consider first the problem of finding extreme values (maxima and minima) of the integral I = ∫(0,1)F(x,y,y')dx. Here, y(x) is an unknown function. In minimizing a function f(x), we find a single value of x by setting df/dx = 0. Here we have a much more difficult problem, since we must find an infinity of values y(x). Nevertheless, we can reduce the problem to a minimization with respect to a single value by an ingenious artifice. Suppose y(x) is the function we are seeking. Let g(x) be another continuous function such that g(0) = g(1) = 0. Then, y(x) + εg(x)is a neighboring function that approaches y(x) when ε→0. If this is placed in the integral, then I is a function of ε, and for an extremum dI/dε = 0.
Using this varied function causes I to vary by δI = ∫[(∂F/∂y)εg(x) + (∂F/∂y')εg'(x)]dx. We can now get rid of g'(x) by integrating the second term by parts. This gives [∂F/∂y')g(x)](1,0) - ∫[(d/dx)(∂F/∂y')]g(x)dx. Now, δI = ∫(1,0)[(∂F/∂y) - (d/dx)(∂F/∂y')]εg(x)dx. Since εg(x) can be chosen at will, the expression in square brackets must be zero. Therefore, the integral is minimized if ∂F/∂y - (d/dx)∂F/∂y' = 0. This is a second order differential equation for the unknown function y(x), called the Euler-Lgrange equation. It is, in general, difficult to solve.
A very useful special case is the one in which F(x,y,y') is not an explicit function of x, as in the brachistochrone problem, where F(x,y,y') = √(1 + y'2)/√y. In this case, we can show that an integral of the equation is the expression E = F(y,y') - y'(∂F/∂y') = c. In fact, forming dE/dx and using the Euler-Lagrange equation, we get dE/dx = 0, or E = c, a constant.
In Hamilton's Principle of dynamics, F(y,y') is the Lagrangian, T - V. The integral E we have just found then becomes E = T - V - 2T, since y'(∂L/∂y') = y'(∂T/∂y') = 2T, since T is a homogeneous function of y' of degree 2. then, E = -(T + V) = c, or T + V = constant. This constant is the total energy H.
For the brachistochrone problem, this gives √[(1 + y'2)/y] - y'2/√[y(1 + y'2)] = c, which can be solved for y' to get y' = √[(1/c2y) - 1]. From this, we obtain x - b = ∫dy/[(1/c2y) - 1], where b is a second constant of integration. If we let 1/c2 = k, and y = ku, this integral becomes x = b = k∫√[u/(1-u)]du. The integral is easily performed if we substitute u = sin2(θ/2), where θ is a new parameter. Then, we get x - b = (k/2)(θ - sin θ), and, of course, y = (k/2)(1 - cos θ) by using the half-angle formula. These are just the parametric equations for the cycloid, where θ is the angle of rotation of the wheel. Taking b = 0 makes the curve pass through the origin, the starting point, where there is a cusp. It is intuitively satisfying that the journey begins with a free fall.
The parameter k must be chosen so that the cycloid passes through point (X,Y). To find k, we first take the ratio y/x = (1 - cos θ)/(θ - sin θ). This ratio varies from 0 at θ = 2π to infinity at θ = 0. Therefore, for any (X,Y) there will be some value of θ at which the cycloid passes through (X,Y). When θ < π, the curve will fall monotonically. For greater θ, the curve will pass through a minimum and rise to the destination. If Y = 0, then θ = 2π, and we will have a complete loop of the cycloid. The diameter of the rolling wheel is k.
In another article (Curves), I show that the brachistochrone is also the tautochrone. That is, the time taken to reach the bottom from any point on the curve is the same. The time is calculated in that article by straightforward integration, with the result that the time from cusp to cusp is (2π)√(k/2g). The period of a cycloidal pendulum is the same as the period of a small-amplitude pendulum of length 2k.
In mechanics, the variable x is the time t. The function F(t,y,y') is the Lagrangian, L = T - V, where T is the kinetic energy my'2/2 and V is the potential energy V(y). The motion must be such that the action, ∫Ldt, is an extremum (usually a minimum). The Euler-Lagrange equations are then (d/dt)(∂L/∂v) - ∂L/∂y = 0, where the velocity v = y'. With our Langrangian, this becomes (d/dt)(mv) + dV/dy = 0, or dp/dt = -dV/dy, which is just Newton's second law, with the force derived from the potential energy V. The momentum is p = ∂L/∂v. These relations can be generalized to provide a firm foundation for mechanics.
The BTC is actively soliciting government support, so necessary for the progress of free enterprise, and states that if it is successful, construction can begin as soon as the problems of friction and tunnelling to about 9 miles depth are overcome, which it is confident can be done. Lawyers and accountants are currently working on the problems. Tickets are available for the first journey for $100 standard, $200 first class. First class passengers will receive a complimentary T-shirt. The trip is not recommended for those with dicky tickers. Small, unmarked bills, please.
The term brachistochrone comes directly from Greek: brachistos, shortest, and chronos, time.
R. Courant, Differential and Integral Calculus, Vol. II (London: Blackie and Son, 1936). Chapter VII.
C. Lanczos, The Variational Principles of Mechanics (Toronto: The University of Toronto Press, 1949). A more complete treatment of the calculus of variations, especially as applied to Hamiltonian mechanics. The brachistochrone is mentioned, but not solved.
Composed by J. B. Calvert
Created 14 December 2004
Last revised 18 December 2004