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"Recently I purchased and installed a remanufactured Dual Reservoir that was listed to replace a 73 Vette (1.00" bore for Non powered Disk Brakes)."
Is it possible your reman'd MC isn't actually the correct one? Some MC's had check valves in one or both outlet ports. IIRC, if split front disc and rear drum, the rear had a check valve, and if 4 drum, both had a check valve, on Pass cars and trucks I recall. 4 wheel disc used no check valves. I'm not sure what would result if that rear port had a check valve still installed.
You'd have to drill and tap the brass outlet port fitting and pull it out and check to be sure.
Is it? Or maybe something is just being misunderstood.
Let's take it to the extremes to demonstrate a point. Assume that we have a piston in a cylinder that, when pushed from one side to the other, displaces 10cc of fluid.
Now assume that we are going to pump that 10cc into chamber that measure 10cc in volume, and this 10cc of volume in the chamber is already filled with fluid. So when you pump the cylinder into the chamber you now have 20 ccs in a 10cc chamber, and the pressure this volume creates against the chamber wall.
Now assume that we are going to pump that 10cc into a chamber that measure 20cc in volume, and this 20cc of volume is already filled with fluid. So when you pump the cylinder into chamber you now have 30 ccs in a 20cc chamber, and the pressure this volume creates against the chamber walls.
Seems to me that 20cc in a 10cc chamber would exert more pressure than 30cc in a 20cc chamber.
Obviously the diameter of the brake line represents the difference in the size of the chambers.
apples and oranges...
the large line diameter only makes it flow faster (think orifice). if the volume changes in one cavity, but the diameter doesn't, then the pressure still remains the same. you are just transferring cc's from one cylinder to another because you are just transferring the force exerted on the face to an identical sized face in the other cylinder..
Bill
Thanks guys for the great discussion on braking & hydraulic theory.
So I exchanged the master cylinder for a new one and installed it today.
Problem gone!!
Apparently it was a defective "remanufactured" M/C.
the large line diameter only makes it flow faster (think orifice). if the volume changes in one cavity, but the diameter doesn't, then the pressure still remains the same. you are just transferring cc's from one cylinder to another because you are just transferring the force exerted on the face to an identical sized face in the other cylinder..
Bill
So if the 10 ccs were pumped into a 10 cc caliper though 100 feet of a 1/8" diameter line, to push a piston, there would be no difference in the pressure/travel at the piston as if that same 10 ccs were pumped into the 10 cc caliper to push the piston through 100 feet of an 8' diameter line?
Seems to me that to push the pistons in the calipers, you have to pressurize the entire line on the way to the calipers also and it would take just a bit more to pressure to push the pistons with the 8-foot line combination than the 1/8" line combination. But I could certainly be wrong.
So if the 10 ccs were pumped into a 10 cc caliper though 100 feet of a 1/8" diameter line, to push a piston, there would be no difference in the pressure/travel at the piston as if that same 10 ccs were pumped into the 10 cc caliper to push the piston through 100 feet of an 8' diameter line?
Seems to me that to push the pistons in the calipers, you have to pressurize the entire line on the way to the calipers also and it would take just a bit more to pressure to push the pistons with the 8-foot line combination than the 1/8" line combination. But I could certainly be wrong.
as long as the line/cylinders are full, could be a mile long; no difference. liquids are not compressible, therefore the 10cc is all that is going to be displaced
Bill
as long as the line/cylinders are full, could be a mile long; no difference. liquids are not compressible, therefore the 10cc is all that is going to be displaced
Bill
Yes, 10 cc is displaced if the piston is free to move to accomodate this volume.
But what happens when the piston pushes against the disk (like in a brake) and can no longer move forward to accomodate the volume, but only exert pressure against the disk?
At this point, isn't this pressure being exerted along the entire length (and diameter) of the line as well as in the caliper?
So wouldn't the PSI generated by the M/C piston be displaced over a larger area and exert less pressure on any given point, including the piston in the caliper?
At this point, isn't this pressure being exerted along the entire length (and diameter) of the line as well as in the caliper?
You are correct but I think you are misinterpreting or at least not realizing the significance of what you wrote.
In a hydraulic brake system the pressure everywhere is the same..... at the master cylinder, inside the brake line tubing (regardless of diameter or length), and at the caliper. No matter where you insert a pressure gauge, the reading will be the same.
So wouldn't the PSI generated by the M/C piston be displaced over a larger area and exert less pressure on any given point, including the piston in the caliper?
The force exerted by the caliper piston is the product of the pressure multiplied by the area of the piston. Increase either the line pressure or the piston area and the force goes up.
This is how it works. Really. And what you wrote above is the opposite of how brake systems really work. It's also independant of the tubing size for the reason I stated at the beginning of this posting.
Last edited by jim lockwood; 06-28-2015 at 07:25 AM.
as Jim said, the pressure is equal in all directions, but as long as nothing else moves (like the walls of the tube expand/distort) the only possible movement is the pistons, and it is this pressure and POTENTIAL movement that pushes the shoe/pad against the drum/rotor.
Bill
You are correct but I think you are misinterpreting or at least not realizing the significance of what you wrote.
In a hydraulic brake system the pressure everywhere is the same..... at the master cylinder, inside the brake line tubing (regardless of diameter or length), and at the caliper. No matter where you insert a pressure gauge, the reading will be the same.
The force exerted by the caliper piston is the product of the pressure multiplied by the area of the piston. Increase either the line pressure or the piston area and the force goes up.
This is how it works. Really. And what you wrote above is the opposite of how brake systems really work. It's also independant of the tubing size for the reason I stated at the beginning of this posting.
So, if by using pedal leverage, I push on a Master Cylinder with a bore of 1 inch and it generates a force of 3,500 psi at the end of a stroke of 1 inch into a sealed volume of 1 cubic inch that is already full of fluid, I can expect that that same bore and stroke would generate 3,500 psi into a 10,000 cubic inch volume that is already full?
If this is the case, why would anyone ever need to alter the bore or stroke of a brake system as "one size fits all," so long as your line is big enough to get the fluid to transfer at speed. And the line would not need to be all that big because you would never transfer more than 0.79 cu of fluid (1" x pi x 0.5^2).
I'm not a fluid engineer, just a scientist and to me this seems to fly in the face of reason.
So, if by using pedal leverage, I push on a Master Cylinder with a bore of 1 inch and it generates a force of 3,500 psi at the end of a stroke of 1 inch into a sealed volume of 1 cubic inch that is already full of fluid,
I think were you may be making a mistake in logic, if that 1 cubic inch is full of fluid and there is no air in the system, you WON'T get an inch of stroke. You won't get any stroke because there's nothing in the system that will compress.
The only way I see you'd have more pressure with a small line vs larger line is if when the cylinder is stroked, you have fluid rushing from the MC cylinder toward the wheel cylinders. This isn't the way brakes work though unless there is something wrong with the system.
It might be true that in the instantaneous moment from initial pedal pressure until the brakes are firmly planted against the rotors or drums, you would/could have an instant higher rise in pressure from the small line but it wouldn't maintain itself. Not even a split second.
I think that would be classified as "response" time.
I'll leave that one to the hydro experts and the rocket scientists as I barely have an 8th grade education. Otherwise, I agree with what they've posted.
So, if by using pedal leverage, I push on a Master Cylinder with a bore of 1 inch and it generates a force of 3,500 psi at the end of a stroke of 1 inch into a sealed volume of 1 cubic inch that is already full of fluid, I can expect that that same bore and stroke would generate 3,500 psi into a 10,000 cubic inch volume that is already full?
first of all, you have to have movement on the receiving end (a moving wheel cylinder piston to reflect the moving piston on the MC end), otherwise there is just an increase of pressure in the system. remember, fluids are not compressible like gases; so unless there is movement there is only an increase in pressure.
If this is the case, why would anyone ever need to alter the bore or stroke of a brake system as "one size fits all," so long as your line is big enough to get the fluid to transfer at speed. And the line would not need to be all that big because you would never transfer more than 0.79 cu of fluid (1" x pi x 0.5^2).
brake systems are 'balanced' by the ratio of the master cylinder piston diameter to the wheel cylinder diameters, and they are not equal sizes. the corvette (and others) use smaller wheel cylinders on the rear than on the front because the front brakes provide proportionally more braking than the rears and therefore they have a larger MC to wheel cylinder diameter ratio than the MC to the rear diameter ratio. this is to keep the rears from locking up before the fronts. otherwise, ALL the piston diameters in every cylinder would be the same diameter. the pressure in each cylinder is the result of the different diameters.
I'm not a fluid engineer, just a scientist and to me this seems to fly in the face of reason.
Once the pads start pushing on the rotors, they are no longer moving, but building pressure against the rotors, so yes, at this point there is just an increase in the pressure of the system. As long as the pads are moving forward, they are just displacing the incoming fluid and not applying any real braking force.
It is this pressure that is squeezing the pads together that applys pressure to the friction material that squeezes on the rotor. Once the friction material "beds against the rotors," this pressure would be distributed along the entire system and if you were making 3,500 psi at the m/c and pumping this pressure into the remainder of the system that had 3,500 square inches of area, the pressure should be 1 pound over each of those square inches. Not 3,500 pounds over each of the 3,500 square inches.
But if the system only included 3.5 square inches the pressure should be 1,000 pounds over each of those square inches (i.e., 3,500 psi / 3.5 si = 1,000 pounds).
But again, I'm just a scientist, not a fluid engineer.
Once the pads start pushing on the rotors, they are no longer moving, but building pressure against the rotors, so yes, at this point there is just an increase in the pressure of the system. As long as the pads are moving forward, they are just displacing the incoming fluid and not applying any real braking force.
It is this pressure that is squeezing the pads together that applys pressure to the friction material that squeezes on the rotor. Once the friction material "beds against the rotors," this pressure would be distributed along the entire system and if you were making 3,500 psi at the m/c and pumping this pressure into the remainder of the system that had 3,500 square inches of area, the pressure should be 1 pound over each of those square inches. Not 3,500 pounds over each of the 3,500 square inches.
But if the system only included 3.5 square inches the pressure should be 1,000 pounds over each of those square inches (i.e., 3,500 psi / 3.5 si = 1,000 pounds).
But again, I'm just a scientist, not a fluid engineer.
I understand the concept of hydraulic multiplication. But you encouraged some research.
First of all, I could not find a single discussion on what happens to the pressure of a hydraulic system once the slave cylinder can no longer move, but additional pressure is applied to the master cylinder. All discussions show the volume/displacements matching with resultant pressure/distance on the pistons. So...
In relation to what I was getting at, there is also the concept of "volume displacement" and "line capacitance" and a larger line has more volume to be displaced. A larger line also results in more frictional losses. Also, I think that what they call "brake line expansion" would also include the initial movement of the wheel cylinders before applying pressure to the rotor.
I understand the concept of hydraulic multiplication. But you encouraged some research.
First of all, I could not find a single discussion on what happens to the pressure of a hydraulic system once the slave cylinder can no longer move, but additional pressure is applied to the master cylinder. All discussions show the volume/displacements matching with resultant pressure/distance on the pistons. So...
In relation to what I was getting at, there is also the concept of "volume displacement" and "line capacitance" and a larger line has more volume to be displaced. A larger line also results in more frictional losses. Also, I think that what they call "brake line expansion" would also include the initial movement of the wheel cylinders before applying pressure to the rotor.
it's as simple as my first attachment pic of a brake system; the more pressure applied to the pedal, the more pressure against the moving/unmoving piston. frictional line losses and hose/tube expansion factors are minimal-to-nonexistent in a car.
Bill
Your graph of master cylinder pressure and wheel cylinder pressure shows the transient response to a change in brake pedal force, which as Bill previously (and very correctly) pointed out, CAN be a function of the brake line diameter.
But notice that at t=0 and at t=.5, the two pressures are identical, AS THEY MUST BE in a hydraulic system where there is no fluid in motion. That is, after the transient dies out, the pressure everywhere in the hydraulic system is the same. (Have we heard that before? Yep, I think so.)
You are over-thinking how a brake system works.
At equilibrium, the pressure in the line is the force applied to the master cylinder piston divided by the area of the MC piston.
And, at equilibrium, the force applied by the caliper piston to the pad is the previously calculated line pressure multiplied by the area of the caliper piston multiplied by the number of caliper pistons pushing on the brake pad.
it's as simple as my first attachment pic of a brake system; the more pressure applied to the pedal, the more pressure against the moving/unmoving piston. frictional line losses and hose/tube expansion factors are minimal-to-nonexistent in a car.
Bill
And they are irrelevant in doing the calculations which will predict brake system performance.
If you want to design a brake system, pay attention to:
1. the target maximum brake pedal force
2. the leverage ratio of the brake pedal arm
3. the area of the master cylinder piston
4. the area and number of caliper pistons
5. the shape of the brake pad and its radial position out from the wheel rotation axis
6. the coefficient of friction of the brake pad against the rotor.
7. the ratio of areas of front caliper pistons to rear caliper pistons
8. the F/R weight bias of the vehicle at rest
and ignore:
1. brake line diameter
Last edited by jim lockwood; 06-29-2015 at 08:07 AM.
06-26-2015, 07:40 PM
response time might be lessened by a larger line if a large amount of fluid needs to be moved; but pressure, no...
