Sprint rear brake bias valve.

[jikovron]

The horse has already bolted, it applied a growing force to the stable door which initially resisted with an equal reaction up untill the latch failed at 4.3KN, adding brake pads and tyre compound at this stage would likely raze the stable to the ground.

[GrahamFountain]

I'm sorry, its me trying to understand why jikovron thinks that two pistons each supplying a force only give the same force as one.

I think this is exactly where people are getting confused. The input into the hydraulic system is still the same - therefore the sum of the outputs must be the same - unless you're affecting pedal travel distance as well.

Don't be daft. The force from the hydraulic pressure on two pistons is twice the force on one piston. Two have twice the area of one, and pressure is force divided by area. So if you have the same pressure and twice the area, you have twice the force. But this only applies to the two pot fixed caliper.

Since the braking force from the disc is proportional to the force on the pads (that's why you stop more quickly if you press harder), if you have twice the force you stop twice as fast.

It does affect travel if one piston is seized, but how much depends on how many pistons of what diameters you started with. If you had 6 all the same (which you never do - front brake pistons are bigger than rear slave cylinders) the travel would go down by 1/6th.

Graham

[GrahamFountain]

I'll expand it into another concept bear with me on this, similar but maybe easier to visualize.
OK so a rope is pulled in tension between a U bar with tension of 1KN , that means that each end of the U bar at rope connection is resisting being drawn together with an equal but opposite 1KN force from each side, note 2 forces there not adding together.

Similarly if I attach a rope to a wall and I pull it away from the wall with 0.5KN (since covid I wouldn't manage a full 1KN) then I'm stretching the rope with a force of 0.5KN , and the wall is resisting in the opposite direction with a force of 0.5KN , and additionally between my feet and the wall the ground is being compressed by 0.5KN, now the crucial thing is the rope has no means to know (partly because its a rope) which end is the active force, as its being 'pulled' from both ends.

Ultimately you right though, being partly a semi intangible abstract concept there exists an explanation gap such that its becoming like we are trying to describe a colour in words.

I just don't see the relevance of any of that.

Two pistons in a fixed calliper give twice the force of one, when they are given the same hydraulic pressure. If you think that's not true, then explain why. And consider the case where the two pistons are both on the same side of the disc. If it's not clear that they give twice the force of one, then I don't know what's wrong with you.

Graham

[GrahamFountain]

The horse has already bolted, it applied a growing force to the stable door which initially resisted with an equal reaction up untill the latch failed at 4.3KN, adding brake pads and tyre compound at this stage would likely raze the stable to the ground.

Ah. No. I'm an avionic systems engineer. It's mechanical engineers you wind up.

[jikovron]

I'll take mild exception to the idea that there is something wrong with me, that's pushing civility a smidge what is ideally happening is that we are throwing ideas about till they stick in some manner.

Anything between two pistons is just a compressed strut by which force is transmitted from one to the other. 1N fed in one end is transfered into the other piston as 1N, however 1N from one side does not exist untill it has a 1N reaction.

You don't see the relevance of that despite the fact it is the same concept just the force roles were reversed for potential clarity nothing more.

[marshman]

Explain the forces on the two brake pads in this example where a brake disc is lying on a workbench with a brake pad either side and a 1.019kg weight on top. NOTE: Assume for the sake of this exercise the pads and disc have zero mass - and the disc is not touching the table top!.

I think there is a 1N force on the top pad and also a 1N force on the bottom pad - is this correct?

[jikovron]

That's a hair under 10N , but yes the force bearing down has an equal but opposite force bearing up from the table and all parts are being compressed with a force of just under 10N even though there are '2' 10Nish forces at play

[marshman]

That's a hair under 10N , but yes the force bearing down has an equal but opposite force bearing up from the table and all parts are being compressed with a force of just under 10N even though there are '2' 10Nish forces at play

Damn, and I tried so hard to get it right

So that agrees with what I thought.

So what is the difference, in terms of force, between what I have drawn - weight on a pad/disk sandwich on a table, and two opposing pistons - effectively replacing the table and the weight both exerting a force of 1N (10N ) ??

Roger

[jikovron]

Zero difference between them in essence , your example is like a caliper except the caliper body is replaced with gravitational pull, the weight is attracted to the earth through the table with 10N of force, and the earth through the table is attracted to the weight with a force of 10N in the opposite direction completing the force and its reaction.

[marshman]

So in both case there is a compression force on the disc of 10N.

[geeksteve]

I'm sorry, its me trying to understand why jikovron thinks that two pistons each supplying a force only give the same force as one.

I think this is exactly where people are getting confused. The input into the hydraulic system is still the same - therefore the sum of the outputs must be the same - unless you're affecting pedal travel distance as well.

Don't be daft. The force from the hydraulic pressure on two pistons is twice the force on one piston. Two have twice the area of one, and pressure is force divided by area. So if you have the same pressure and twice the area, you have twice the force. But this only applies to the two pot fixed caliper.

But you'd need to push the pedal harder for that to be true. Are you suggesting that if I got 1000 hydraulic rams I could lift a house, but it'd feel no different than one ram lifting a feather? That's just not true. If that was the case then there would only be one size of hydraulic pump - tiny! You need larger input volume for more output volume - and in doing so you're requiring a larger force to act on it.

[jikovron]

So in both case there is a compression force on the disc of 10N.

My opinion is that is true , if the caliper pistons are giving a 10N force , they interchange in the same manner as your example

[GrahamFountain]

I'll take mild exception to the idea that there is something wrong with me, that's pushing civility a smidge what is ideally happening is that we are throwing ideas about till they stick in some manner.

Anything between two pistons is just a compressed strut by which force is transmitted from one to the other. 1N fed in one end is transfered into the other piston as 1N, however 1N from one side does not exist untill it has a 1N reaction.

You don't see the relevance of that despite the fact it is the same concept just the force roles were reversed for potential clarity nothing more.

I'm finding difficult to deal with the spurious analogies and what appears to me as refusal to properly address the issues and response.

But in response to a reasonable point: It is true that "Anything between two pistons is just a compressed strut by which force is transmitted from one to the other. 1N fed in one end is transferred into the other piston as 1N, however 1N from one side does not exist until it has a 1N reaction." No force exist without its reaction, that's Newtons 3rd law.

But that is only a complete description of a situation where there is only one force applied at one end of that strut. However, it is entirely possible to apply separate forces at both ends and have them combine. In which case, if there is an external force applied at end A and an external force also separately applied at end B, then the compression is the sum of the two forces. The resulting force at end B is the sum of the force applied there and the reaction to the force at end A. Just as the resulting force at end A is the sum of the force applied there and the reaction to the force applied at end B. At every point in the strut, and including at the two ends, there is then the sum of the force applied at A and force applied at B.

Graham

[marshman]

I think this is exactly where people are getting confused. The input into the hydraulic system is still the same - therefore the sum of the outputs must be the same - unless you're affecting pedal travel distance as well.

Don't be daft. The force from the hydraulic pressure on two pistons is twice the force on one piston. Two have twice the area of one, and pressure is force divided by area. So if you have the same pressure and twice the area, you have twice the force. But this only applies to the two pot fixed caliper.

But you'd need to push the pedal harder for that to be true. Are you suggesting that if I got 1000 hydraulic rams I could lift a house, but it'd feel no different than one ram lifting a feather? That's just not true. If that was the case then there would only be one size of hydraulic pump - tiny! You need larger input volume for more output volume - and in doing so you're requiring a larger force to act on it.

The force IS the same because the pressure is the same. BUT in order to actually move the piston/ram and do any WORK (FORCE x DISTANCE) you would have to move the MASTER Piston or pump further, that is the difference. So in your example of 1 ram lifting a feather versus 1000 rams lifting a house, you would have to move the MASTER cylinder/piston 1000 times further using the same force to lift the house the same distance as 1 ram lifting your feather. Not Press Harder, but move further.

[marshman]

But that is only a complete description of a situation where there is only one force applied at one end of that strut. However, it is entirely possible to apply separate forces at both ends and have them combine. In which case, if there is an external force applied at end A and an external force also separately applied at end B, then the compression is the sum of the two forces. The resulting force at end B is the sum of the force applied there and the reaction to the force at end A. Just as the resulting force at end A is the sum of the force applied there and the reaction to the force applied at end B. At every point in the strut, and including at the two ends, there is then the sum of the force applied at A and force applied at B.

Graham

I really cant agree with that statement.

"the resulting force at end A is the sum of the force applied there and the reaction to the force applied at end B."

The force applied and the reaction force are the same thing, in the discussion it is 1N. You are counting exactly the same force twice.

Roger