Thread: Handbrake conversion to disc : Defender

Originally Posted by uninformed

thanks John,

I knew when I posted my bit, that it would be open for correction and I was hoping it would.

You said "For a given normal force, the pressure is reduced if the area is increased, but the only reason we are interested in pressure for brakes and clutches is for the wear rate, which is a function of pressure x sliding velocity. If the brake doesn't slip there will be no wear to speak of."

"The disc pad doesn't need more pressure than a drum brake, it needs more force, because the effective radius is less and there is no self energisation like with leading shoes."

I think Im confused haha, What is the difference between pressure and force as we speak here?

This is a little like the chicken and egg. You can't have one without the other. Pressure in this or any other physical context, is the load (force) acting upon an area, i.e. (pressure = load / area)

This can be transformed into (load = pressure x area) Thus if we know the pressure and area, we could calculate the force. A hydraulic press creates force from pressure, but force is usually a result of some other action.

Load and weight are both forces.

Force = mass x acceleration
Weight = mass x acceleration due to gravity

Note laypeople use the term "weight" when they should use "mass".

We use the SI system of units, where the unit for mass is kg, and the unit for weight/force/load is Newton, a derived unit to simplify using [kg m / second squared], i.e. [mass x distance / (time x time)]

In the old imperial system, the unit for mass is "slug" and the unit for force is "lbf" (laypeople got that wrong as well).

In the old MKS (metre, kilogram, second) system (sometimes called metric system), which was replaced by the SI system, the unit for mass was kg and the unit for force was "pond".

Pressure in a fluid is a different context to what we were talking of but it is still load divided by area. Stress in a material is pressure acting within a solid, but unlike in a fluid it is not equal in all directions.

What I was trying to say, was that less area isnt more efficent as it needs more force to do the same job. (is that right)

I can't tell if you might still be misunderstanding something, or not. We need to totally forget the notion of area and pressure to understand friction and braking - they are "red herrings".

For an example take a box sitting on a level floor. The box has mass and weight. The normal force to solve the friction problem in this case is the weight of the box.

Friction is a reaction, and is the resistance to motion, or impending motion. If no force is applied to try and move the box, the friction force is zero. If we push against the box, but it fails to move (impending motion), the friction force is equal, but in the opposite direction, to the force pushing against the box. As we increase the force the box won't start to move until it is equal to the product of normal force (weight in this example) and the static coefficient of friction between the box and the floor.

It wouldn't matter if the box was sitting on its largest side or its smallest side, the results would be the same (assuming the same material and coefficient of friction on all sides). The pressure would be different between the box and the floor, and will be higher if it is sitting on a small side - as i have said, pressure doesn't change the friction force.

If we add some weight to the box, we increase the normal force, and have to push harder to move it along the floor.

Friction, or friction force is put to good use by brakes and clutches, to resist motion or impending motion. The motion my be linear or rotational. In the case of rotation, it is more convenient to use braking torque (friction force x radius).

Im thinking the stock drum brake requires less force than the x-eng disc brake to achieve the same braking torque?

Yes, if you mean the "normal force", and not the force applied at the brake lever, but only because of the smaller radius where the friction force is acting. It has nothing to do with the area of the friction material.

How the required normal force is produced is a mechanical problem. In both systems, the force applied to the brake lever is transferred and multiplied by a mechanism.

Many people tried and failed to produce a reliable disc handbrake and the reasons were that readily available and reasonably cheap calipers were either incapable of a higher enough normal force, or were not strong enough to withstand those forces.

Hope that helps.

The other thing I should have noted about disc brakes, is that the number of friction surfaces is two. So the total available braking torque = coefficient of friction x normal force x effective radius x number of friction surfaces.

With a drum brake, it is more complicated to explain because the force is not distributed uniformly over the angle subtended by the friction material on the shoes.

Drum brakes can work very well when they are in good condition, but a fair amount of maintenance is required to keep them working in that manner. Uneven wear, binding among the moving parts, contamination, etc. all contribute to reduce their effectiveness.

Disc brakes are well suited to hydraulic pressure or springs applying the needed force, while mechanical application is often less successful, which is why some vehicle have retained drum brakes for the handbrake.