The Physics of Trapeze

Investigation 9: The Safety lines

Safety is extremely important on the trapeze. Even though the net is remarkably stretchy, and you could fall from right at the very top of your swing and not hurt yourself if you land on your back, it is most likely that (unless the fall is planned) you will fall awkwardly. Even a fall into the net will do considerable damage if you land on your head.

Professional trapeze artists do not use safety lines. This is not because they do not fall, but rather that they are practised at falling cleanly on their back. For example, in Cirque Du Soleil, I saw a performer just miss his double somersault with a half twist catch. He shot down into the net, but he tucked his head in at the last minute and landed on his back, so he was not hurt. The other reason is insurance. Professional trapeze artists are either flying at their own risk, or are covered by the company they work for. In trapeze schools, this freedom cannot be extended to beginners, since the premiums would be much too high. Insuring the trapeze is very expensive as it is! The flying trapeze school at the YMCA in Denver has had to close recently after 70 years of operation, after a man fell and broke his hip. There was some dispute regarding liability, and it was forced to close.

There is another non-safety reason why professionals prefer not to wear lines. The belt, which the lines attach to, needs to be tight, and can be restrictive in some tricks. Many people feel it is possible to achieve more height in the advanced swing without lines.

How do they work?

The operation of the safety lines is best demonstrated with a diagram. Basically, an operator stands on the ground next to the trapeze. The lines are clipped to the belt at the board, and allow the flyer to swing with them, and even be caught (they travel along the trapeze with the flyer). The lines leave the flyer's belt on each side, are looped over two pulleys at the top of the trapeze, and recombine as one, thicker rope, which the operator holds. During normal operation, the operator pulls the rope to take up excess slack. In the event of a mistake, during which the flyer falls, the operator pulls hard on the rope to exert a force upwards on the flyer, slowing his descent. It should be stressed that there is no mechanical advantage in the lines, that is to say that any force exerted on one end of the lines will be transmitted to the other end.

Since there is never much slack in the rope, as soon as the flyer falls, the lines exert an upward force on him. Therefore if the operator pulls on the ropes with a force equal to the flyer's weight (taking into account the cos a), he will hang in mid-air. In other words, the flyer never builds up speed during the fall, requiring no larger impulse to stop him. There are three cases to consider; firstly when there are no safety lines, secondly when the flyer is lighter than the operator, and thirdly when the flyer is heavier than the operator.

Case 1: Let us assume that the flyer falls from the extreme height of his swing (about 5m). a is approximately 0.3c, so cos a = 0.95. Let us assume that in this case he weighs 70kg. Ignoring air resistance:

No upward force, so acceleration is due to gravity (g) and is 9.8ms-1

v^2 = u^2 + 2as

We will assume starting velocity (u) = 0 so v^2 = 2as

v^2 = 2 x 9.8 x 5 = 98

v = 9.9ms-1

Notice this was independent of the man's mass.

Case 2: Using same assumptions from case 1, except flyer has a mass of only 50kg (child) and the operator has a mass of 70kg.

In this case, the child can exert a maximum force on the safety lines of 500N.

T (cos a) = 500N

T = 526N

Since the operator weighs 700N this will not be enough to lift his from the ground, so he can support the child (and even provide enough force to lift her up…should he want to) and let her down at any speed he wants to (ideally very slowly).

Case 3: Again, using the same assumptions as in case 1, with the flyer having a mass of 90kg, and the operator with a mass of 70kg.

F=ma (taking downwards as positive)

90g - 70g(cos a) (all operator's weight hanging on lines) = 90 x a

a = 2.56ms-2

v^2 = u^2 + 2as

We will assume starting velocity (u) = 0 so v^2 = 2as

v^2 = 2 x 2.56 x 5 = 21.8

v = 5.1ms-1

Although this is a reduction of about half, this still seems rather a large value to me, especially given that people over 90kg will fall faster (up to max of 9.9ms-1). What must be remembered however, is that the net is extremely bouncy, and will absorb most of the impact, greatly reducing the risk of injury. This works since it stretches, and exerts the force over a long period. Since the impulse needed to stop the descent is the same, and is given by I = Ft, you can either have a large force over a short time, or a smaller force over a long period. The net adopts the second option, reducing the force on the falling flyer to a minimum. Even if you would not want to fall to the ground at 5m/s, this would be a very acceptable speed for falling into the net.

If we briefly look at the forces on the operator, we see that the force upwards is greater than their weight, therefore there must be an upwards acceleration. In fact it is quite normal to see the operator pulled up off the ground by the force of the other person falling on the other end. The operator then lets the rope slip through his hands occasionally so he is not pulled too high.

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