Why a few turns of rope hold so much

Wrap a rope a few times round a post, and a light pull on the loose end can hold a heavy load on the other. Each extra turn does not add a fixed amount of grip. It multiplies the grip. That is why a few turns are enough, and why the size of the post makes no difference at all.

What the equation says

A capstan is the upright drum a ship's crew winds rope around, and the rule that describes it is called the capstan equation. Take a rope or flat belt wrapped over a fixed drum. One end carries the large force, the load. The other end carries the small force, the hold. The contact angle θ is how far round the drum the rope touches it, measured in radians. A full turn is 2π. The coefficient of friction μ is the ratio between the sideways force two surfaces can resist and the force pressing them together. At the moment the rope is about to slip, the load is the hold multiplied by e raised to the power μθ. Lubarda calls this “the Euler formula, or the belt friction (capstan) equation”. His reference list gives Euler's memoir on it as published in 1762, and a footnote adds that the German engineer Johann Albert Eytelwein included Euler's work in a book of 1808.

Why it matters

Attaway applies the equation to the devices rescuers use to lower people on ropes. His brake tube is a fixed tube the rope wraps around; he writes that the idea “was borrowed from the sailor's capstan”. He assumed, and says he did not measure, a coefficient of 0.25 for rope on aluminium. By his estimate one wrap of the tube holds the load at 10-to-1, two wraps at 50-to-1 and three at 250-to-1 (computed exactly: 10.6, 50.8 and 244; our arithmetic). For a 600-pound rescue load that means about 60 pounds of hold with one wrap, 12 with two, and “only two to three pounds” with three. The same law works against a rescuer hauling a load up. Over a rock edge, with a coefficient of 0.4, a turn of 135 degrees makes the rope carry “2.60 times more tension”, and a 600-pound load becomes “almost 1,560 pounds” in the rope (the formula gives 2.57 and about 1,540; our arithmetic).

Why it multiplies

Look at one tiny piece of rope on the drum, covering a small angle dφ. The tension pulls on both of its ends, and because the rope bends there, the two pulls do not quite line up. Together they press the piece into the drum with a force equal to the tension times dφ. Friction can resist at most μ times that press, so across the piece the tension can change by at most μ × T × dφ. The change is proportional to the tension already there. Where the rope is tight, it presses hard and friction can hold back a lot. Where it is slack, it presses lightly and friction holds back little.

A quantity whose growth is proportional to its own size grows exponentially, like money earning compound interest. Lubarda gives a second argument that needs no calculus. Split the wrap into two parts. The first part multiplies the hold by some factor, and the second part multiplies the result by another. So the factor for the whole wrap must be the product of the factors for its parts. The only smooth rule that turns adding angles into multiplying factors is an exponential. The same equilibrium on one small piece then shows that the number in the exponent is μ itself.

Why the drum's size drops out

Lubarda notes that students commonly expect a longer contact to give more grip. It does not. The tension at any point equals the drum's radius times the pressure there, so a drum twice as wide is pressed half as hard at every point. Its contact is twice as long, but each piece of it grips half as much, and the total comes out the same. He compares it to a block on a floor. The block needs the same push to slide whichever face it rests on, because the weight is spread over more area but pressed less hard. Attaway says the same of rope: the result is “independent of the radius of bend and the size of the rope”. He lists only three things the rope's friction depends on: the tension, the coefficient of friction and the total angle of contact.

What the equation does not say

The formula gives the most that one end can hold against the other, at the moment slip begins. Before that moment, Lubarda points out, friction alone does not fix the tension along the rope, so the formula does not tell you the tension actually there. It assumes a flat, weightless belt and one coefficient along the whole contact. If the rope is already sliding, the lower sliding coefficient applies. Real coefficients move too: Attaway warns that mud, water, ice and oil all change them. One wrap of his tube is 3π, not 2π, because the bends where the rope enters and leaves count as well (our reading of his Fig. 6). Computed exactly, e to the power 0.25 × 3π is 10.6 (our arithmetic); he prints 10.

In short

Each small piece of wrap can only change the tension in proportion to the tension it already carries. So equal amounts of extra angle multiply the load by equal factors. With μ = 0.25, each half turn multiplies it by about 2.19 (our arithmetic). The drum's radius cancels because a bigger drum is pressed less hard over a longer contact.