Once again,
In a stationary system, with a constant source of pressure, this is the case.
But the system does not have a constant pressure and the slave cylinder moves,
so it is not a stationary system.
The fluid will travel along the path of least resistance and seek the piston
when pressure is applied. Pressure on the input end (master cylinder)
shows up on the other end as movement of the slave cylinder system. The amount
of energy put into the system is going to be equal to the energy that
is gotten out of the system (see the law of conservation of matter and energy).
If the slave cylinder is smaller than the master cylinder, it stands
to reason that that amount of energy will be concentrated on a smaller area and
therefore will be expressed by a smaller piston as a greater force
applied to the brakes or clutch.
Now, that makes sense to me (May not be correct, but it certainly makes
sense)!!! :)
Joe (C)
"Mark J. Bradakis" wrote:
>
> If you take the same setup and apply the 10 psi to a 0.5 square inch
> piston you are now getting 10 lbs per 1/2 square inch which is 20 psi and
> you get 20 lbs. of force out of the system.
>
> No, you don't. This is nonsense. If you have a pressure of 10 psi (that's
> pounds per square inch), you have a pressure of 10 pounds per square inch,
> over every single bit of area in the entire system. The insides of the brake
> or clutch line, the bore of the fittings, against the base of the bleed screw,
> *any* area inside the system will have that same force.
>
> If that pressure is applied to 1 square inch, you get 10 pounds of force.
> If that pressure is applied to 2 square inches, you get 20 pounds of force.
> If that pressure is applied to 1/2 square inch, you get 5 pounds of force.
>
> The force in pounds is the product of the area times the pressure, simple,
> basic high school math. Write out the equation, and see how the units cancel:
>
> pounds pounds inch * inch
> Force ------- = Pressure ----------- x Area ------------
> 1 inch * inch 1
>
> This is classic Newtonian physics. The work that goes in is the same as
> the work that comes out. That is, negating the effects of mechanical friction
> and things like the force used to distort the clevis, the pins, etc. on every
> pedal application. Here I use work in the classic physics definition, a force
> applied over a distance.
>
> People are confusing volume, pressure, area and distance in the situation.
> They are all related by basic mathematical equations.
>
> mjb.
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