# Leverage and Hydraulics

In the figure below, a force F is being applied to the left end of the lever. The left end of the lever is twice as long (2X) as the right end (X). Therefore, on the right end of the lever a force of 2F is available, but it acts through half of the distance (Y) that the left end moves (2Y). Changing the relative lengths of the left and right ends of the lever changes the multipliers.

The basic idea behind any hydraulic system is very simple: Force applied at one point is transmitted to another point using an **incompressible fluid**, almost always an oil of some sort. Most brake systems also multiply the force in the process. Here you can see the simplest possible hydraulic system:

### Simple hydraulic system

In the figure above, two pistons (shown in red) are fit into two glass cylinders filled with oil (shown in light blue) and connected to one another with an oil-filled pipe. If you apply a downward force to one piston (the left one, in this drawing), then the force is transmitted to the second piston through the oil in the pipe. Since oil is incompressible, the efficiency is very good -- almost all of the applied force appears at the second piston. The great thing about hydraulic systems is that the pipe connecting the two cylinders can be any length and shape, allowing it to snake through all sorts of things separating the two pistons. The pipe can also fork, so that one master cylinder can drive more than one slave cylinder if desired, as shown in here:

### Master cylinder with two slaves

The other neat thing about a hydraulic system is that it makes force multiplication (or division) fairly easy. If you have read How a Block and Tackle Works or How Gear Ratios Work, then you know that trading force for distance is very common in mechanical systems. In a hydraulic system, all you have to do is change the size of one piston and cylinder relative to the other, as shown here:

### Hydraulic multiplication

To determine the multiplication factor in the figure above, start by looking at the size of the pistons. Assume that the piston on the left is 2 inches (5.08 cm) in diameter (1-inch / 2.54 cm radius), while the piston on the right is 6 inches (15.24 cm) in diameter (3-inch / 7.62 cm radius). The area of the two pistons is Pi * r^{2}. The area of the left piston is therefore 3.14, while the area of the piston on the right is 28.26. The piston on the right is nine times larger than the piston on the left. This means that any force applied to the left-hand piston will come out nine times greater on the right-hand piston. So, if you apply a 100-pound downward force to the left piston, a 900-pound upward force will appear on the right. The only catch is that you will have to depress the left piston 9 inches (22.86 cm) to raise the right piston 1 inch (2.54 cm).

Next, we'll look at the role that friction plays in brake systems.