The Official 356Talk Forum

The lines are like artiries before a heart attack sometimes, almost completely blocked. Check to see if this is the case, it makes it very hard to push the petal when they are like this. But they look great from the outside!

Mechanical advantage. It seems much more noticeable on a track bike. People often run into this changing m/c's on motorcycles. Everything needs to be thought of as a system, not a component.

Vic et al,
Hydraulic advantage is not that complicated, Think in terms of pressure as the common term.
So a 19mm M/C piston has 0.4395 in2 now we apply 50lbs. of force to the pedal and we get 50/ 0.4395=113.77 psi in the line. This same pressure is exerted on the 48mm caliper piston. Since the caliper piston has a greater square area (2.805 in2) than the M/C piston we will get more force, i.e. 113.77psi*2.805 in2=319.11 lbs. If you notice the ratio of the forces, pedal/ caliper piston is the same as the square area ratio. i.e. .04395/2.805=6.382 ! Keeping this ratio in mind, then a smaller M/C gives more advantage, a lager dia, less, while at the caliper end it is the opposite.
This is why the rear caliper pistons are smaller than the fronts because of the weight transfer to the front during braking which if they were the same diameters the rears would lock up too soon.
BTW, this also why the rear brakes of drum brake cars are single leading shoes while the fronts are double leading shoes.

Thanks all - lots to keep me busy here. BTW pads are new Mintex.
John

Ernesto Cabrera wrote:Smaller pistons have less mechanical advantage at the master cylinder so, there is less travel at the pedal but more pressure is required: P1/V1=P2/V2 if temperature is a constant.
Ernie

Pascal's Law might be a better choice for an incompressible fluid than the Ideal Gas Law, which is OK for an incompressible fluid that's far from the triple point. Unfortunately, your algebra is also incorrect.

PV=nRT => P1V1=nRT and P2V2=nRT then

For the same gas at the same temperature:

P1/V2 = P2/V1

Let F1= the pedal force required to generate a pressure P to stop a car. Let D1 = 19 mm, the diameter of a MC piston of diam 19 mm. Let F2 be the pedal force necessary to generate P in a car with MC piston diam D2 = 21 mm.

F1=PxA1 and F2=PxA2

Rearranging terms:

P = F1/A1 = F2/A2

Rearranging terms to solve for F2,, the pedal force required with the 21 mm MC.

F2 = F1(A2/A1)

Since we don't know F1, we can solve for the ratio of F2/F1 and express as a % difference in force.

F2/F1 = A2/A1

A1 = pi (19 mm)^2/4 and A2 = pi(21 mm)^2/4 then

F2/F1 = (21 mm/19mm)^2 x 100

= 122%

Then to stop in the same distance under identical conditions, it will take about 100 lbs of pedal force on the 21 mm MC versus 80 lbs of pedal force on the 19 mm MC.

Vic, you don't need to dust off that old textbook.

The Motivated Student may use these principles and relationships to better understand Larry's Lecture.

Kudos to professor Neil. I stand corrected on the equation (Its been more than 45 yrs since I used these formulas). The point is the mechanical advantage increases as the fluid is pumped from a smaller chamber to a larger one and vise versa as Ashley and Larry illustrated.

Best

Ernie

Vic,
I keep my slide rule on my desk at all times. Always check calculator answer with slide rule!
Conrad

In case I got lost in the various maths, the larger caliper piston gives more leverage, correct?

I agree with Jacques, replace the flex lines. Use TWO wrenches on the hard to flex interface and buy some flare nut wrenches.

viewtopic.php?f=1&t=42405&hilit=brake+lines

I agree also. Even the stainless steel braided hoses do not live forever. Ask me how I know!!
Conrad

Force equals pressure per area.

500PSI behind a smaller piston creates a greater force.

Pressure is defined as force divided by area. So force equals pressure times area. In symbols:

P = F/A

So F = P x A

Mike, I think you posted F = P/A