This is why flyers will additionally use Safety Lines while they are learning to fly or practicing new and unfamiliar tricks.

### 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, and even be caught (as the pullies travel horizontally along the length of the trapeze). The lines leave the flyer's belt on each side, are looped over two pulleys at the top of the trapeze, and (usually) recombine as one thicker rope, which the operator holds. During normal operation, the operator pulls the rope to take up excess slack. When the flyer falls, the operator pulls down hard on the rope to exert a force upwards on the flyer, slowing their 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. In fact there is a mechanical disadvantage, since due to angle a, the operator actually excepts less force upwards on the flyer than they do downwards on the rope.

Since there is never much slack in the rope, as soon as the flyer falls, the lines exert an upward force. Therefore if the operator pulls on the ropes with a force such that the upward force (taking into account cos(a)) is equal to the flyer's weight, they will hang in mid-air. In other words, the flyer never builds up downward velocity during the fall, requiring no larger impulse to stop them. 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: No safety lines

Let us assume that the flyer falls from the extreme height of their swing (about 5m).Angle p is approximately 0.3 radians, so cos(p) = 0.95. Let us assume that in this case they weigh 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: $$ v^{2} = 2as\\ v^{2} = 2\times9.8\times5=98\\ v=9.9ms^{-1} $$ So the flyer hits the net at about 10ms

^{-1}(about 22 miles per hour).

*Notice that this did not depend on the flyer's mass.*

### Case 2: Light flyer

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 flyer can exert a maximum force on the safety lines of 500N.

$$ T\cos(p) = 500N\\ T = 526N $$ Since the operator weighs 700N this will not be enough to lift them from the ground, so they can support the child (and even provide enough force to lift them up… should he want to) and let them down at any speed they want to (ideally very slowly).

### Case 3: Heavy flyer

Again, using the same assumptions as in case 1, except the flyer now has a mass of 90kg, and the operator has a mass of 70kg.$$ F=ma $$

(taking downwards as positive)

$$
90g - 70g\cos(p) = 90\times a
$$
(all operator's weight hanging on lines)

$$
a = 2.56ms^{-2}\\
$$$$ v^{2} = u^{2} + 2as $$ We will assume starting velocity (u) = 0 so: $$ v^{2} = 2as\\ v^{2} = 2\times2.56\times5 = 25.6\\ v=5.1ms^{-1} $$ Although this is a reduction of about half, this still seems rather a large, especially given that people over 90kg will fall faster (up to max of 9.9ms

^{-1}- the answer to Case 1). What must be remembered however, is that the net is extremely deformable, and will absorb most of the impact, greatly reducing the risk of injury. This works since it stretches, and exerts the stopping 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 5ms

^{-1}, this would be a very acceptable speed for falling into the net.

In reality of course, the pulley is not completely efficient and there is a good deal of friction in the system. This means that the pulley itself resists the rope running through it and therefore slows the flyer’s fall. In other words it ‘helps’ the safety line operator. In most cases, the operator is able to slow the decent of the flyer much more that in our perfect world calculations with much less force.

If you are following carefully, you will have spotted that our calculation above for Case 3 involves not only the flyer falling into the net at 5ms

^{-1}, but the operator (putting their whole weight on the lines) being lifted up at the same speed. In reality, the operator would let the rope slip through their hands at the moment they feels themselves being lifted, so that they remains on the ground (and in a better position to control the flyer's descent). In this way they exert a force very slightly less than their weight on the lines: not enough to lift them up, but enough to slow the flyer as much as they can.