Capstan Equation (Belt Tension Ratio)

Also known as capstan equation · Euler-Eytelwein equation · belt friction equation · rope on a bollard · T1/T2 = e^(mu theta) · windlass equation

F1F2=eμθ\frac{F_1}{F_2} = e^{\mu \theta}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Wrap a rope once round a post and it holds a few times what your hand can. Wrap it three times and it holds hundreds of times more. The relation behind that — Euler's, published in 1775, and independently Eytelwein's — is one of the most surprising results in elementary mechanics:

\[ \frac{F_1}{F_2} = e^{\mu\theta} \]

Where the exponential comes from

Take a tiny arc of the wrap, subtending dθd\theta. The tension changes across it by dFdF. The normal force pressing that element against the drum is FdθF\,d\theta, so the friction it can supply is μFdθ\mu F\,d\theta, and equilibrium gives dF=μFdθdF = \mu F\,d\theta. Every element of rope can add friction in proportion to the tension it is already carrying — the rope helps hold itself down — and a quantity whose growth rate is proportional to itself grows exponentially. Separate and integrate: ln(F1/F2)=μθ\ln(F_1/F_2) = \mu\theta.

The mistake this equation is famous for

θ\theta is in radians. A half wrap is π\pi, not 180. Put degrees into the exponent and you inflate the answer by a factor of e56.3μθe^{56.3\mu\theta} — at μ=0.3\mu = 0.3 and a half wrap that turns a correct 2.57:1 into something around 102310^{23}:1. Nobody catches it, because the wrong answer is enormous and enormous answers look safe. The calculator on this page takes your degrees and converts them, but do the conversion by hand every time you do this on paper.

What is not in the equation

Look at what is missing: the diameter of the drum, the diameter of the rope, the total force squeezing the drum, the length of contact. None of them appear. Only the coefficient of friction and the angle. That is why turns are so effective — each additional turn multiplies the ratio by the same factor — and why a small bollard holds as well as a large one. At μ=0.3\mu = 0.3, one turn gives about 6.6:1, two turns 44:1, three turns 286:1. A deckhand holding 250 N on the tail restrains better than 70 kN on the standing part.

The independence from drum diameter is genuine but not unlimited: a rope bent round a very small drum loses strength and life to the bending itself, which is why every rigging standard specifies a minimum drum-to-rope diameter ratio.

V-belts cheat, legitimately

Wedge the belt into a V-groove and the normal force is no longer simply FdθF\,d\theta — the two flanks squeeze inward, and the friction available rises to μ=μ/sin(β/2)\mu' = \mu/\sin(\beta/2) with β\beta the groove angle. A 38° groove gives μ3.1μ\mu' \approx 3.1\mu. Fed through the exponential, that threefold gain becomes an enormous gain in holding ratio, and it is the entire reason the V-section was invented.

Where else it turns up

Capstans and winches, band brakes (whose self-energising bite is this equation with the drum turning into the wrap), belt drives, the friction of a bowline round a bollard, the grip of a rope on a rock in climbing protection, and — one worth knowing — the difference between the two ends of a long conveyor. The same exponential explains why a band brake grabs violently in one direction of rotation and behaves gently in the other: reverse the drum and F1F_1 and F2F_2 swap ends.

Capstan Equation (Belt Tension Ratio)
F1F2=eμθ\frac{F_1}{F_2} = e^{\mu \theta}
θF1F2μ
Where
  • F1F_1= Tight-side tension (N)
  • F2F_2= Slack-side tension (N)
  • μ\mu= Coefficient of friction
  • θ\theta= Angle of wrap (°)