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Brachistochrone Hoodie

STRAIGHT IS NOT THE FASTEST | Pixelated Physics

\[x = R(\theta - \sin\theta),\qquad y = R(1 - \cos\theta)\]
$54.99 USD
$54.99 USD
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Is the shortest path also the fastest? Not for a bead sliding under gravity. Johann Bernoulli’s 1696 brachistochrone problem is solved by a cycloid, the curve traced by a point on a rolling wheel. It drops steeply first, so the bead gains speed early. From rest to the cycloid’s lowest point, the straight line takes about 18.5% longer. Black pixel-art hoodie; equation, symbols and sources in the physics panel below.

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Details

Size & fitS–5XL

Unisex Heavy Blend hoodie (Gildan 18500), classic fit.

SizeWidthLengthSleeve
S202733.5
M222834.5
L242935.5
XL263036.5
2XL283137.5
3XL303238.5
4XL323339.5
5XL343440.5

Measurements in inches. Width and length ±1 in, sleeve (from center back) ±0.75 in.

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Delivery times & where we ship

Returns & replacements21 days

Every item is printed to order, so we can’t accept returns or exchanges for change of mind, or if you ordered the wrong size or colour.

Print defect, misprint, damage or the wrong item? We’ll send a free replacement, or a refund if you prefer. Report it within 21 days of delivery with a photo; no need to send it back.

Full refund & replacement policy

The physics

The fastest slide between two points under gravity is not a straight line but a cycloid, the curve traced by a point on a rolling wheel. The steeper start buys speed that more than pays for the longer path.

Which track gets a bead from one point down to another in the least time, with only gravity pulling it and no friction?

Not the straight line. The fastest track is a cycloid, the curve traced by a point on a rolling wheel: x = R(θ − sin θ), y = R(1 − cos θ). It drops steeply first, so the bead picks up speed early, and that speed pays for the longer path.

For a bead released from rest the rolling angle grows evenly in time, θ = t√(g/R), so it reaches the bottom after T = π√(R/g). The straight line between the same two points takes √(π² + 4)·√(R/g), about 18.5 % longer.

On the front, the amber beads on the curve and the cream beads on the straight line are drawn at the same six instants. When the curve bead arrives, the straight-line bead has covered π²/(π² + 4), about 71 %, of its track.

Johann Bernoulli posed the problem in 1696. Newton, Leibniz, Jakob Bernoulli and others all solved it.

Equations

\[x = R(\theta - \sin\theta),\qquad y = R(1 - \cos\theta)\]
\[\theta = t\sqrt{g/R}\]
\[T_{\text{cycloid}} = \pi\sqrt{R/g},\qquad T_{\text{straight}} = \sqrt{\pi^2+4}\,\sqrt{R/g}\]
SymbolMeaningUnit
\(x\) horizontal distance from the start \(\mathrm{m}\)
\(y\) depth below the start \(\mathrm{m}\)
\(R\) radius of the rolling circle that draws the cycloid \(\mathrm{m}\)
\(\theta\) rolling angle of that circle \(\mathrm{rad}\)
\(t\) time since release from rest \(\mathrm{s}\)
\(g\) gravitational acceleration \(\mathrm{m/s^2}\)
\(T\) time to reach the bottom point \(\mathrm{s}\)

Sources

  1. Johann Bernoulli (1696) Problema novum ad cujus solutionem Mathematici invitantur, Acta Eruditorum, June 1696, 269
  2. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Sec. 2.2 (the brachistochrone problem)
  3. Weisstein, Brachistochrone Problem, MathWorld

Source Bernoulli, Johann (1696). Problema novum ad cujus solutionem Mathematici invitantur. Acta Eruditorum, June 1696, 269.

Field notes

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