Brachistochrone Time Calculator
Introduction to the brachistochrone, the curve of quickest descent
A bead is released from rest at one point and must slide, without friction and driven by gravity alone, to a second point that is lower down and off to one side. Which shape of wire gets it there soonest? The answer is not the straight ramp, and it is not any of the circular or parabolic tracks that intuition suggests. It is an arc of a cycloid — the curve traced by a marked point on the rim of a wheel rolling along a straight line — turned upside down so that the cusp sits at the release point. That curve is the brachistochrone, from the Greek brakhistos (shortest) and khronos (time).
This calculator solves the problem completely rather than quoting a rule of thumb. You give it the horizontal run, the vertical drop and the local gravitational acceleration; it solves the transcendental endpoint equation numerically for the rolling angle, recovers the radius of the generating circle, and returns the exact minimum descent time together with the geometry of the winning track. It also computes the two paths people usually compare against — the straight ramp and the drop-then-slide corner — so you can see the size of the advantage instead of taking the theorem on trust.
The problem has an unusually well documented origin. Johann Bernoulli published it as a challenge to "the sharpest mathematicians in the whole world" in the Acta Eruditorum of June 1696; solutions from Johann Bernoulli, his brother Jakob, Isaac Newton, Gottfried Leibniz and Ehrenfried Walther von Tschirnhaus appeared in the May 1697 issue. The techniques invented to attack it became the calculus of variations, which is why the brachistochrone still opens the variational chapter of most graduate mechanics courses.
How to use this cycloid descent solver
Three numbers define the problem, and a fourth control fixes the units.
- Horizontal run x — the horizontal separation of the two points. It must be zero or positive; the geometry is mirror-symmetric, so a run to the left is identical to a run of the same size to the right.
- Vertical drop y — how far the finish lies below the release point, measured downward as a positive number. Zero is allowed and is a genuine, finite-time case (see the worked notes below). A negative value is rejected: a bead released from rest cannot climb.
- Gravitational acceleration g — strictly positive. The Earth default is standard gravity, 9.80665 m/s2, which is an exact defined constant rather than a measurement, so it carries no uncertainty. Local free-fall acceleration at the surface actually ranges from roughly 9.78 m/s2 at the equator to 9.83 m/s2 at the poles; use the local value if you have one.
- Length unit — metres or feet. Switching it converts the run, the drop and the gravity field together, so the answer stays consistent. Standard gravity in imperial units is 32.174049 ft/s2.
Press Compute descent time and the result panel reports the descent time, the rolling angle θf in radians and degrees, the generating radius a, the depth of the lowest point of the arc, the arc length against the straight chord, the terminal speed, the isochronous (tautochrone) time, and the head-to-head comparison with the straight ramp and the drop-then-slide path. The scale drawing underneath shows the solved cycloid against those alternatives, and the buttons let you copy a permalink that reproduces the exact inputs or download the full breakdown as a CSV file.
The time is in seconds whenever x, y and the length part of g share a unit, because the only dimensional combination in the answer is the square root of a length divided by an acceleration. Nothing else about the bead matters: mass, size and material never enter the equations.
The brachistochrone formula and its Euler-Lagrange derivation
Put the origin at the release point, let x run horizontally and let y increase downward. Conservation of mechanical energy for a frictionless bead released from rest fixes the speed at every depth, independently of the shape of the wire:
The mass cancels, which is why the answer never depends on it. Since v = ds/dt and the arc element is ds = √(1 + y′2) dx, the total descent time is the functional
Minimising this integral is the founding problem of the calculus of variations. Writing the integrand as f(y, y′), the stationary curve satisfies the Euler-Lagrange equation
Here f contains no explicit x, so the Beltrami identity — the first integral obtained by differentiating f along the curve and substituting the Euler-Lagrange equation — applies and reduces the second-order problem to a first-order one:
Naming the integration constant (2a)−1/2 is a convenience: a then turns out to be the radius of the rolling wheel. The parametric solution through the origin is the inverted cycloid
where θ is the angle the generating wheel has rolled through. Substituting the parametrisation into dt = ds/v gives ds = a√(2 − 2 cos θ) dθ and v = √(2ga(1 − cos θ)), and the two factors of two cancel exactly, leaving the remarkably clean result
This is worth emphasising because it is a common transcription error: the radicand is a/g, not a/(2g). The descent time is simply the rolling angle multiplied by the natural time scale √(a/g) of the generating wheel.
What is left is to find the particular cycloid that threads your endpoint. The two parametric equations evaluated at θ = θf must reproduce (x1, y1), and eliminating a between them leaves one transcendental equation in one unknown:
No closed form for θf exists — any page that offers one is not solving the brachistochrone problem. The left-hand side is, however, provably a strictly increasing function of θf that sweeps from 0 to +∞ across the interval (0, 2π), because its derivative has numerator 2 − 2 cos θ − θ sin θ, which is non-negative on that interval. A root therefore exists, is unique, and can be bracketed for certain. This calculator exploits that: it runs a guarded bisection on (0, 2π) to full double precision, which cannot diverge or overshoot the way a bare Newton iteration can. Series expansions replace θ − sin θ and 1 − cos θ near the ends of the interval so that catastrophic cancellation does not corrupt extreme aspect ratios.
Three further quantities follow from a and θf and are all reported. The arc length of the track, the isochronous time to the low point of the cycloid, and the speed at the finish are
The middle expression is the tautochrone time: a bead released from rest anywhere on a cycloidal arc reaches the bottom after exactly that interval, no matter how high it started. Because it is computed from a alone it doubles as an independent sanity check on the solved radius, and the calculator prints it for that reason.
Finally, the two reference paths. A straight ramp of length √(x12 + y12) has constant along-track acceleration gy1/L, and a free vertical drop followed by a horizontal glide at the acquired speed gives a second benchmark:
Both must always exceed the cycloid time; that inequality is the theorem, and it is a good self-test. The calculator checks it on every run and flags a warning if it were ever violated.
Worked example: a 5 m run with a 2 m drop
Take a demonstration track whose finish is 5 m along and 2 m down from the release point, under standard gravity. Enter 5, 2 and 9.80665 with the length unit set to metres.
The endpoint ratio is x1/y1 = 2.5. Bisection on (0, 2π) converges to θf = 3.819665 rad, which is 218.85° — comfortably past π, so the track dips below the finish and climbs back up to it. The radius follows as a = 2 / (1 − cos 3.819665) = 1.124364 m, and the descent time is
The geometry that goes with it: the lowest point of the arc sits 2a = 2.2487 m below the release point, that is 0.2487 m below the finish. The track is 5.9932 m long against a straight chord of 5.3852 m, so the bead travels 11 % further. It arrives at √(2 × 9.80665 × 2) = 6.2631 m/s, exactly as it would on any other frictionless track between the same two heights — the cycloid does not create speed, it just acquires it earlier.
The straight ramp over the same points takes √(2 × 29 / (9.80665 × 2)) = 1.7196 s and the drop-then-corner path takes 1.4370 s. The cycloid is therefore 24.8 % quicker than the ramp and 10.0 % quicker than the corner. Note the ordering: at this aspect ratio the “obviously bad” corner path actually beats the straight ramp. Push the run out to 100 m over a 1 m drop and the cycloid needs 7.5552 s against 45.1623 s for the ramp, almost six times quicker, while a 1 m run over a 2 m drop reverses the ranking of the two reference paths again (0.7140 s for the ramp, 0.7983 s for the corner). Only the cycloid is guaranteed to come first every time.
Reading the result and comparing the three classic paths
The single most transferable number in the panel is the ratio Tramp/T, the factor by which the optimal track beats the naive one. It depends only on the aspect ratio x1/y1, not on the absolute size of the layout and not on g, so it carries straight over from a tabletop demonstration to a full-scale installation. The rolling angle tells you the shape: below π the track is a monotone descent; at exactly π the finish is the lowest point of the arc and the bead arrives moving horizontally; above π the track undershoots the finish and the bead climbs the last stretch.
| Horizontal run | Rolling angle θf | Cycloid time | Straight ramp | Drop then corner | Ramp / cycloid |
|---|---|---|---|---|---|
| 0.5 m | 0.7364 rad | 0.6533 s | 0.6583 s | 0.7185 s | 1.008 |
| 1 m | 1.4014 rad | 0.6941 s | 0.7140 s | 0.7983 s | 1.029 |
| 2 m | 2.4120 rad | 0.8245 s | 0.9032 s | 0.9580 s | 1.095 |
| 5 m | 3.8197 rad | 1.2934 s | 1.7196 s | 1.4370 s | 1.330 |
| 20 m | 5.1198 rad | 2.9755 s | 6.4185 s | 3.8320 s | 2.157 |
Two patterns are worth internalising. First, for a nearly vertical layout all three paths converge on the free-fall time √(2y1/g) and the advantage of the cycloid vanishes, because there is nothing left to optimise once the straight line is already the steepest possible descent. Second, the advantage grows without bound as the layout flattens: the ramp time diverges as the drop goes to zero, while the cycloid time stays finite all the way down to a horizontal finish.
Limitations of the frictionless bead idealisation
The output is an exact answer to an idealised question, and the gap between that question and a physical track is where essentially all of the error lives.
- Frictionless sliding, not rolling. This is the largest correction in practice. A solid sphere rolling without slipping puts 2/7 of the released potential energy into rotation, so its speed at any depth is √(5/7) ≈ 0.845 of the sliding value and every time is about 18 % longer. A rolling hoop is worse still, and the time-optimal shape for a rolling body is not exactly the cycloid.
- No friction and no drag. Sliding friction and air resistance remove energy that never comes back, and unlike the rolling correction they scale with path length — which is precisely where the cycloid is at a disadvantage, since its arc is longer than the chord.
- Uniform, constant gravity. The field is treated as a constant vector of magnitude g. That is excellent at laboratory and civil-engineering scales and wrong at planetary or orbital scales.
- The bead never leaves the wire. The track is assumed able to pull as well as push. On an open chute the bead would separate from the surface wherever the required centripetal force exceeds what gravity can supply.
- The cusp is unbuildable. The optimal curve is vertical at the release point. Every real track has to round that corner off, which gives back a little of the theoretical advantage.
- Released from rest. A non-zero launch speed changes the problem. The solution is still a cycloid, but with the origin shifted to a virtual release point above the actual start.
- Negative drops are rejected, not extrapolated. If the finish is above the release point there is no gravity-only solution from rest, and the calculator says so instead of returning a number.
Treat the output as a hard lower bound. No physical descent between the same two points under the same gravity can be quicker, which makes it the right yardstick for scoring a real design, and a poor predictor of the actual stopwatch reading.
Common questions about brachistochrone descent times
Why is the fastest descent curve a cycloid rather than a straight line?
A straight ramp is the shortest path between the two points, but it is not the quickest one. A track that dives steeply at the start trades height for speed early, so the bead covers the gentler remainder at a high speed. Applying the Euler-Lagrange equation to the descent-time integral and reducing it with the Beltrami identity gives y multiplied by one plus the square of dy/dx equal to a constant 2a, and the curve that satisfies it is the cycloid with x = a(theta - sin theta) and y = a(1 - cos theta). Johann Bernoulli posed the problem in the Acta Eruditorum of June 1696 and five solutions were published in May 1697.
What is the brachistochrone descent time formula?
Once the cycloid through your endpoint is known, the descent time is T = theta_f times the square root of a divided by g, where a is the radius of the generating circle, theta_f is the rolling angle at the endpoint and g is the gravitational acceleration. The quantity under the root is a divided by g, not a divided by 2g: the factor of two in the speed and the factor of two in the arc length cancel exactly. For the half cycloid theta_f is pi and the time is pi times the square root of a divided by g.
Can the end point sit at the same height as the start?
Yes, and the descent still takes a finite time. With zero vertical drop the solution is the complete cycloid arch, theta_f is 2 pi, and the time is the square root of 2 pi x divided by g. The bead leaves a cusp where the track is vertical, so it accelerates immediately, dives to a depth of x divided by pi, and climbs back to the starting height with exactly the speed it needs. A straight horizontal line takes infinite time, so the calculator reports no speed-up ratio for this case.
How much time does the cycloid actually save over a straight ramp?
For a 5 m horizontal run and a 2 m drop under standard gravity of 9.80665 metres per second squared, the cycloid takes 1.2934 s against 1.7196 s for the straight ramp, a saving of 24.8 percent, even though the cycloid arc is 5.9932 m long against a 5.3852 m chord. The saving grows with the aspect ratio: for a 100 m run with a 1 m drop the cycloid needs 7.5552 s against 45.1623 s, almost six times quicker.
What is the tautochrone property and why does it matter here?
The same cycloid solves a second classical problem. A bead released from rest at any point of a cycloidal arc reaches the lowest point after pi times the square root of a divided by g, whatever height it started from. Christiaan Huygens published this result in Horologium Oscillatorium in 1673 and used it to shape the cheeks of a pendulum clock so that its period would not depend on amplitude. This calculator reports that isochronous time next to the descent time as an independent check on the solved radius.
What does this calculator deliberately ignore?
Everything except uniform gravity. The bead is a point mass released from rest on a rigid frictionless wire in a vacuum, and it never leaves the track. A rolling ball is slower because part of the released potential energy goes into rotation, sliding friction and air drag add more, and no real track reproduces the vertical tangent at the cusp. Treat the answer as a lower bound that no physical descent between the same two points can beat.
Sources. Standard acceleration of gravity, 9.80665 m/s2 exactly, a defined constant with zero uncertainty: NIST, The NIST Reference on Constants, Units and Uncertainty — standard acceleration of gravity, physics.nist.gov. Derivation, cycloid parametrisation, the descent time T = θf√(a/g), the arc-length relation and the tautochrone result: C. E. Mungan, Review of the Brachistochrone Problem, Physics Department, United States Naval Academy, 2012, usna.edu. Textbook treatments of the Euler-Lagrange equation and the Beltrami first integral: H. Goldstein, C. Poole and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, chapter 2; C. Lanczos, The Variational Principles of Mechanics, 4th ed., Dover. Historical record: Johann Bernoulli’s challenge in the Acta Eruditorum, June 1696, with five solutions published in the May 1697 issue; C. Huygens, Horologium Oscillatorium, 1673, for the tautochrone. Every numerical value quoted on this page was recomputed from these equations rather than copied from a secondary source.
Curve Racer: can you pick the runner-up?
Each race puts your chosen track against the true brachistochrone over a randomly generated pair of endpoints. You cannot beat the cycloid — that is the theorem — so the game is to finish within 3 % of it. Which shape comes closest depends on how steep the layout is, and the answer changes every round.