Sprint rear brake bias valve.

[jikovron]

Its like Carledo pointed out, a G clamp/Vice will only compress something with force equal to the force that the screw inputs to the 'system', and a 2 pot caliper is like a g clamp with 2 screws same principle, the disc as I agree has negligible forces acting on it when a caliper is free to float only the tiny difference in piston stiction and slider pin resistance which would amount to fractions of a Newton.

[GrahamFountain]

Its like Carledo pointed out, a G clamp/Vice will only compress something with force equal to the force that the screw inputs to the 'system', and a 2 pot caliper is like a g clamp with 2 screws same principle, the disc as I agree has negligible forces acting on it when a caliper is free to float only the tiny difference in piston stiction and slider pin resistance which would amount to fractions of a Newton.

But you still evade the issue of how you can possibly think that two separate 1N forces can be applied one on each side of a disc, and not have a resulting force of 2N between them. The important point being it is two forces not one force and the reaction to it.

Graham

[jikovron]

It is 2 forces but each exists only because of the opposed force, 1N opposed to 1N tries to tear the caliper apart with a force of 1N ,in an odd 2N V 1N situation would push the weak piston back against its stop with a force of 1N, and total compressive force would be 2N.

[GrahamFountain]

It is 2 forces but each exists only because of the opposed force, 1N opposed to 1N tries to tear the caliper apart with a force of 1N ,in an odd 2N V 1N situation would push the weak piston back against its stop with a force of 1N, and total compressive force would be 2N.

But either force, with its own reaction of 1N, exists on its own.

And that is so even if the other piston is seized, and the bearing is so uncompliant that the piston that works cannot make the disc touch the other pad. Even in that case, the one remaining force, with the reaction from the bearing, still causes (half as much) drag on the disc. How long the bearing would last may be another question though.

So it so clearly is two separate forces, each with its own separate reaction.

I think you're confusing 2 forces in opposition with action and reaction. And they are not the same at all.

Graham

[jikovron]

Yes your on the right lines, in the case of the single pad, the brake force would drop to half as the reaction force is not via a the other pads friction but instead as you rightly point out via the near zero friction bearing , we only disagree on the point about force multiplication only,,,eventually it will click and you will kick yourself but for now the thread is derailed by this and agreeing to differ will suffice as a solution.

[GrahamFountain]

Yes your on the right lines, in the case of the single pad, the brake force would drop to half as the reaction force is not via a the other pads friction but instead as you rightly point out via the near zero friction bearing , we only disagree on the point about force multiplication only,,,eventually it will click and you will kick yourself but for now the thread is derailed by this and agreeing to differ will suffice as a solution.

1+1 = 2. So two forces each of 1N in opposition at the disc give 2N force on the two pads. And 1-1 = 0. So the two bending moments on the hub bearing cancel and there's no residual force there.

No "force multiplication", just simple addition and subtraction: 1+1 = 2 and 1-1=0. It's as simple as that, and if you can't get that much maths, there's just no helping you.

[jikovron]

Your still incorrect but I'm not going down a route of being disrespectful about it, a force is never a single force it always has an equal reaction built into it, they aren't 2 additive seperate forces there is only 1 in effect. it's just that's the easiest way to mentally visualise it as 2 forces in compound it takes abit of an adjustment to the thought process but honestly it will suddenly click.

[marshman]

I have been following this thread - with some difficulty !! I am not even sure I understand what Graham and jikovron are disagreeing about.

BUT I think the crux of the matter is this as I have always understood it force is a vector, i.e. it has magnitude AND direction. So with the example of a calliper, single piston (floating) or 2 piston (one either side) the result is the same. The hypothetical 1N force in the single piston will push on both ends. The piston face pushing onto the pad, pressing it onto the face of the disc, so lets call this +1N. The other end (actually the cylinder wall) pushes opposite direction so -1N. This is coupled by the "hook" shape of the sliding part of the calliper round to the other side of the disk and presses on that pad with a force of -1N. So regarding the overall system the net result outside of the calliper i.e. on the bearing etc is 0. BUT each pad sees a force of 1N and from their point of reference it is 1N pressing onto the disc - the pads are being forced in opposite directions.

Exactly the same is true of a two piston calliper. One piston exerts a force of +1N onto (or towards the disk) the other a force of -1N still onto or towards the disk because it is on the otherside facing the opposite direction. So again each individual pad only "sees" 1N and the total resultant force outside of the "system" is 0. To complete the circle (almost literally the force pushing on the backwall of the cylinder of each piston is transferred by the calliper casting to the "otherside" where again +1N cancels the -1N keeping Mr Newton and his laws happy.

So as far as I can see when you take account of vector forces then +1 + -1 does = 0, but each pad sees a force of 1N (just in the opposite direction which is OK as they face opposite ways) and the disc in the middle sees little overall effect other than getting a little hot and bothered!

Does that makes sense are am I talking rubbish

Roger

[jikovron]

The disagreement is simply that i believe 2 opposed piston forces (actually one force and its own reaction) of 1N will cause a compression to the disc/pads of 1N , however Graham believes they would add together to form a compressive force of 2N.

[marshman]

I would say that it is 1N on each pad but in opposing directions so following that to its logical conclusion it must be zero in the centre (thickness) of the disc - [difficult concept to visualise as it is very counterintuitive but it must be true because Mr Newton says so!]. I don't think 2N exists anywhere in the system as you must treat them as vector forces, i.e. observe not only the magnitude (size) but also the direction.

[jikovron]

The zero comes about I think due to force balance, ie there is no acceleration to anything however what I might add is that I'm pushing with a force one way and Graham is pushing with an equal force in the opposite direction which means we are crushing the thread with a force equal to what we're putting in

[marshman]

[GrahamFountain]

The disagreement is simply that i believe 2 opposed piston forces (actually one force and its own reaction) of 1N will cause a compression to the disc/pads of 1N , however Graham believes they would add together to form a compressive force of 2N.

This is absolutely and totally untrue. I know that one force and its reaction are not additive.

I do admit, I don't get what jiovron is disagreeing with properly.

It seems to be ether that he says the two pistons do not apply two separate forces, which is errant nonsense, or that, when the two pistons are each separately applying a force, the result is not the sum of their individual contributions, which is equally nonsensical.

Whereas, I say that there are two pistons each applying a separate force, in opposition, and the total is their sum. So 1N + 1N = 2N. Where's the possibility of error in that?

Graham

[GrahamFountain]

Oh, and that the result of those two forces, being in opposition, at the mounting of the disc, i.e. the bearing, is zero - when they are exactly equal.

Graham

[jikovron]

Ok using your thinking process; Each piston applies force , and the area of the closed end of the cylinder is the opposite reactive force, so in a sense a 2 pot caliper there are 4 opposed forces , but just 2 in a single piston sliding caliper hence why initially using your logic you postulated that a 2 pot would add each '1N' force to 4N and a single piston caliper to 2N , you squared that up but that leaves the issue of 4 forces V 2 equalling the same compressive force.