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# Force and Pressure

Fluid is pressurised by the FORCE of a piston. The piston moves in a cylinder which is sealed to prevent fluid leaking out or air entering into the system. The piston must move with little friction in the cylinder as this will have an adverse effect on the system.

There is air on one side of the piston and hydraulic brake fluid on the
other side. The area of each piston is determined using the PiR^{2} formula. To obtain a measurement in inches from millimetres
simply multiply the millimetres by 0.03937. You then divide the
Diameter (D) by 2 to obtain the Radius (R) of the Piston and then
calculate the piston area in square inches by using the PiR^{2} formula. For example:

Piston Diameter (D) | Piston Diameter (D) | Piston Radius (R) | Piston Surface Area (A) |

14mm | 0.551" | 0.275" | 0.238 in^{2} |

15mm | 0.590" | 0.295" | 0.273 in^{2} |

16mm | 0.629" | 0.314" | 0.311 in^{2} |

17mm | 0.669" | 0.334" | 0.351 in^{2} |

18mm | 0.708" | 0.354" | 0.394 in^{2} |

19mm | 0.748" | 0.374" | 0.439 in^{2} |

20mm | 0.787" | 0.393" | 0.486 in^{2} |

Now that we have this figure we can start calculating other measurements.

**F - Force on Piston (lbf) / A - Area of Piston (in ^{2}) = P - Hydraulic Pressure (psi)**

The amount of pressure depends on how much **FORCE** you put on the piston.

If the piston in the above system has an area of 2 in^{2} and
the **FORCE** is 400lbf then the pressure is calculated using
the formula as follows:

400 (lbf) / 2 (in^{2}) = **200 (psi)**

A smaller piston gives a higher pressure. If you replaced the piston
with only a 1 in^{2} area in the same system with 400lbf
of **FORCE** then the pressure exerted is 400 psi. Pistons can
be used to multiply the **FORCE **in a hydraulic system. By
choosing pistons with different surface areas any relationship
with **FORCE** is possible.

A simple hydraulic braking system on a motorcycle is shown below.
The pistons in this system have two different surface areas with
the master cylinder piston having a surface area of 2 in^{2}
and the piston in the caliper has a surface area of 4 in^{2}.
With the same 400lbf of **FORCE** being applied to the master
cylinder the piston produces 200 psi. Remembering the hydraulic laws
we discussed earlier:

- Fluid cannot be compressed to a lesser volume, no matter how high the pressure.
- Pressure is equal over all surfaces of the containing system.

When looking at the two hydraulic laws above we know that the 200 psi in our example system will act equally on all surfaces within the system. The shell of the master cylinder, the shell of the caliper and the hoses that connect them both will have 200 psi acting upon them but they cannot move. However the piston at the caliper is able to move and will have the same 200 psi acting upon it.

As this piston has a 4 in^{2} surface area the FORCE it produces
will be 800lbf. The area has doubled so the FORCE is doubled.
400lbf of FORCE is pushed down onto 2 in^{2} which then
acts upon the 4 in^{2} surface. We must remember that only
the FORCE changes in this system – the pressure remains the same
at 200 psi. This is 2 in^{2} so with the other piston
having 4 in^{2} the FORCE produced has twice the surface
area to work upon and so twice the FORCE in lbs.

To increase the FORCE on the caliper piston we can decrease the area of the master cylinder piston area or increase the surface area on the caliper piston. The reverse is true to decrease FORCE on the caliper piston. We can either increase the area of the master cylinder piston area or decrease the surface area on the caliper piston.

So that covers the FORCE at work in your braking system and we know that by decreasing or increasing parts of the system you can alter the FORCE produced at the caliper but that does not increase pressure. Pressure is the constant in the braking system and cannot be altered internally. If you can produce 200 psi at the master cylinder hydraulic law requires that 200 psi is produced at the caliper.