Fellow Listers:
May I take a shot at clarifying the effects of varying piston sizes on the
braking forces. Being somewhat of a dimbulb, it is neccessary for me break
things down into simple terms before I can understand them. Hopefully, the
simplification below will be helpfull.
As you know, a brake system consists of, among other things, the master
cylinder, fluid lines, slave cylinders, and brake fluid. When the brake pedal
is depressed, the fluid is pressurized. Since a fluid is incompressable, the
pressure is the same throughout the system. The pressure is stated as "pounds
per square inch"
Pounds per square inch means just that - pounds PER square inch! To
illustrate, suppose you have a MC with a cross sectional area of 1/2 square
inch (approximately 0.8" diameter), and you apply a force of 100 pounds.
Dividing 100 by 1/2 gives 200 pounds per square inch. Now, suppose the slave
cylinder (or piston, for disc brakes) has a cross sectional area of 2 square
inches (approximately 1.6" diameter). 200 pounds per square inch times 2
square inches produces a force of 400 pounds. As someone pointed out, the
system has an effect much like a lever. And it doesn't matter if you have
one, two, four, or more, slave cylinders, the force relationship is the same
for all of them. Remember, fluid is incompressable, and the pressure is the
same throughout the system.
Therefore, a smaller MC, or a larger slave cylinder, will produce a greater
braking force. Dividing 100 pounds by 1/4 square inches, for example produces
400 PSI. 200 PSI applied to a SC with a 4 square inch area gives a force of
800 pounds. This sounds like you are getting something for nothing, and we
all know that doesn't happen. What are the tradeoffs? Very simply, the
tradeoff is movement. Just like with a lever, where we have to move the big
end a long way to get a small movement on the little end, the MC has to move
a larger distance than we get from the slave cylinders. It might help to
think of jacking up your car with a bottle jack. As you know, it takes a lot
of cranking to move the piston a little bit. We could do the analysis just as
well using movements, and fluid volumes, but it is so much easier to
understand, and to calculate, using pressure.
Of course, the hydraulic portion of the brake system is not the only place
where leverage comes into play. Typically, the brake pedal has a 5 to 1
ratio, from the pad to the MC pushrod. A force applied by the foot of 100
pounds is leveraged to the MC as 500 pounds. Also, within the brake
assemblies themselves there is leverage. Brake pads pressing against a 12
inch rotor will apply more braking force than pads pressing against a 10 inch
rotor. Likewise, 10 inch drums at the rear will provide more braking than 9
inch drums.
In a real world system such as we have on our LBCs, the front disc brake
pistons are much larger then the rear brake cylinders. A much larger force is
applied to the front pistons than is applied to the rear cylinders. The front
pads are of smaller area than the rear, and the rear shoes are leveraged
against the drums, so the difference is not as much as it might at first
appear. There is more braking power available in the front, though, to
compensate for the forward weight transfer under heavy breaking. If the
braking power at each end were the same, and the weight is lifted from the
rear wheels under braking, the effort required to lock them up would be
greatly reduced, increasing the tendency to lock up and slide. When a wheel
is sliding, the braking ability is severely reduced, and the car has an
annoying tendency to swap ends. Besides making it difficult to look cool, and
impress bystanders, it is also very hard to control a car when you are
looking where you've been instead of where you are going. This is something
to consider when contemplating swapping rear drum brakes for disc brakes. If
you don't make a corresponding improvement to the front brakes, you may be
setting up a dangerous situation.
Dan Masters,
Alcoa, TN
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