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PostPosted:Sun Feb 28, 2021 5:22 pm 
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1N again, the clamp would be the same 1N combined, and the body of the clamp would be pulled apart with a force of 1N


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PostPosted:Sun Feb 28, 2021 8:04 pm 
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This whole debate is going over my head (and I suspect a lot of other peoples too)

Would someone like to explain (in words of one syllable preferably) what "force" is in this context and what (if any) are the differences between "force" and "energy"?

What little I remember of O level physics, says energy may not be created or destroyed, it just gets moved around or stored. What you get out is what you put in (minus parasitic losses from gravity, friction, inefficiency or whatever, the energy still goes SOMEWHERE)

Apply this idea to a braking system and you are putting energy into a caliper by applying pressure to a hydraulic system. Assume for the sake of argument you are putting 1 unit of energy into your caliper, ignoring for the moment, any parasitic losses, this is then divided equally between 2 pistons whether by separate feeds to each piston WITHIN the caliper (fixed caliper) or by pressurizing the piston and caliper body (floating caliper) Either way there is only 1 unit of energy to go round so each piston (or pseudo piston) gets by my reckoning, 0.5 units of energy each.

Now take the single seized piston in a 2 piston fixed caliper. both pistons still get the 0.5 units each, but, because the piston is seized, the energy is contained and doesn't reach the pad. Most of this energy, I suspect will feed back to the other piston which WILL move and transfer itself through bending the disc and the bearing till it meets the other, now fixed, pad. Still 1 unit of energy used, just a load of parasitic losses from bending the disc and bearing. Then the whole lot gets dissipated through the disc and pad as heat energy to atmosphere, not lost or destroyed energy, just given away! Just a lot less efficiently than if both pistons were free.

So explain to the idiot (me) how you can put 1n in at the hose connection and get 2n out, 1n at each piston? When you only had 1n to start with?

Steve

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'73 2 door Toledo with Vauxhall Carlton 2.0 8v engine (The Carledo)
'78 Sprint Auto with Vauxhall Omega 2.2 16v engine (The Dolomega)
'72 Triumph 1500FWD in Slate Grey

Maverick Triumph, Servicing, Repairs, Electrical, Recomissioning, MOT prep, Trackerjack brake fitting service.
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Last edited by Carledo on Sun Feb 28, 2021 8:15 pm, edited 1 time in total.

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PostPosted:Sun Feb 28, 2021 8:12 pm 
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Well starting from the obvious fact that two single piston fixed calipers will give twice the brake effort of one, given Amontons' first law that area does not matter between two solids, the force has to be twice. So, if one piston in a fixed caliper gives 1N on the one pad, the two, either in two fixed calipers or one two piston fixed caliper give 2N on the pair.

Understanding that that is 2N of compression may actually be easier from the two screw g cramp analogy. If one screw when tightened by n turns takes the compression from 0 to 1N , then tightening the other screw by the same n turns (assuming all the components respond linearly) adds a second force of 1N to give a compression of 2N. It may be obvious also, that leaving the second screw alone, but tightening the first by an additional n turns will also give an additional 1N and a total of 2N compression (in a fully linear system).

The possibly interesting thing is that if you replace the screw in a g cramp with a hydraulic piston and a cylinder in the frame, then applying a pressure that gives 1N on the area of the piston will give 2N of compression: 1N from the piston pushing on one end and 1N from the frame pushing on the other. This is why the one piston sliding caliper gives the same brake effort as the two piston fixed caliper. But I continue to admit I got that wrong first time.

Graham

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The 16v Slant 4 engine is more fun than the 3.5 V8, because you mostly drive it on the upslope of the torque curve.

Factory 1977 TR7 Sprint FHC VVC 697S (Now all of, but still needs putting together)
B&Y 73 Dolomite Sprint UVB 274M (kids!)
1970 Maroon 13/60 Herald Convertable (wife's fun car).


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PostPosted:Sun Feb 28, 2021 8:30 pm 
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If you tighten one screw to 1n, then tighten the second screw from a start point of 1N as it won't be 0N given its got 1n bearing on it from the other screw( ,,,and add a further 1N of tightening ontop then yes there will be then 2N of force) so I can see where your angling from.


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PostPosted:Sun Feb 28, 2021 8:42 pm 
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Well starting from the obvious fact that two single piston fixed calipers will give twice the brake effort of one, given Amontons' first law that area does not matter between two solids, the force has to be twice. So, if one piston in a fixed caliper gives 1N on the one pad, the two, either in two fixed calipers or one two piston fixed caliper give 2N on the pair.

Understanding that that is 2N of compression may actually be easier from the two screw g cramp analogy. If one screw when tightened by n turns takes the compression from 0 to 1N , then tightening the other screw by the same n turns (assuming all the components respond linearly) adds a second force of 1N to give a compression of 2N. It may be obvious also, that leaving the second screw alone, but tightening the first by an additional n turns will also give an additional 1N and a total of 2N compression (in a fully linear system).

The possibly interesting thing is that if you replace the screw in a g cramp with a hydraulic piston and a cylinder in the frame, then applying a pressure that gives 1N on the area of the piston will give 2N of compression: 1N from the piston pushing on one end and 1N from the frame pushing on the other. This is why the one piston sliding caliper gives the same brake effort as the two piston fixed caliper. But I continue to admit I got that wrong first time.

Graham
Sorry, but that did not a) tell me what "force" is and how it differs from "energy", b) was not in laymans language. So it still didn't make a lot of sense. I still can't see how you can get 2N out at the business end without putting at least 2N in!

I realize you now understand how a floating caliper works. And for Jikovron (so you know it's not just you i'm picking on) a G clamp and a Vise are NOT the same, a G clamp is a free floating hand held device, like a floating caliper. A vise (not VICE) has 1 jaw fixed to the bench or whatever, so it more nearly resembles a fixed caliper with 1 seized piston!

Steve

PS, Amonton's law? My understanding ends around Murphy's!

_________________
'73 2 door Toledo with Vauxhall Carlton 2.0 8v engine (The Carledo)
'78 Sprint Auto with Vauxhall Omega 2.2 16v engine (The Dolomega)
'72 Triumph 1500FWD in Slate Grey

Maverick Triumph, Servicing, Repairs, Electrical, Recomissioning, MOT prep, Trackerjack brake fitting service.
Apprentice served Triumph Specialist for 50 years. PM for more info or quotes.


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PostPosted:Sun Feb 28, 2021 8:46 pm 
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My vice is sat on the floor as a free to roam clamp lol, a G clamp welded to a bench could be a vice in principle, if an ultra crap one.


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PostPosted:Sun Feb 28, 2021 10:36 pm 
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This whole debate is going over my head (and I suspect a lot of other peoples too)
Oh dear.
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. That and to explain, why ever it is he thinks that, he's wrong. It is absolutely certain that you have two pistons each pressing on something with 1N, the total force on whatever they are pressing on is 2N. And in that specific issue, it don't matter if they press on the same or opposite sides.
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Would someone like to explain (in words of one syllable preferably) what "force" is in this context and what (if any) are the differences between "force" and "energy"?
Again, oh dear.
Force can be described simply as a push or a pull. The SI unit is the Newton, and its that amount of push that will accelerate (where there are no other forces like friction or drag) a mass of 1kg at a rate of one meter per second per second: one meter per second squared. That is, a force of 1N increases the speed of something with a mass of 1kg by 1m/s in every second. So, because Earths gravity causes an acceleration of 9.81 meters per second squared, the force downward from a mass of 1 kg on the surface of the Earth or anything fixed and parallel to it, is 9.81 Newtons. We should then talk about the weight of things being in Newtons, but being humans and easily confused, we still use kilograms.

The simplest definition of energy is "the ability to do work". Energy is how things change and move. The SI unit is the Joule, and its where a force of 1 Newton moves its point of application by a distance of one meter.

As a free extra, power is the rate at which work is done. The SI unit is the Watt, and its 1 Joule per second, so where a force of 1 Newton moves its point of application by 1 meter in one second.
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What little I remember of O level physics, says energy may not be created or destroyed, it just gets moved around or stored. What you get out is what you put in (minus parasitic losses from gravity, friction, inefficiency or whatever)
True enough, but not relevant to this. Which is really all about force and reactions.
Quote:
Apply this idea to a braking system and you are putting energy into a caliper by applying pressure to a hydraulic system. Assume for the sake of argument you are putting 1 unit of energy into your caliper, this is then divided equally between 2 pistons whether by separate feeds to each piston WITHIN the caliper (fixed caliper) or by pressurizing the piston and caliper body (floating caliper) Either way there is only 1 unit of energy to go round so each piston (or pseudo piston) gets 0.5 units of energy each.

Now take the single seized piston in a 2 piston fixed caliper. both pistons still get the 0.5 units each, but, because the piston is seized, the energy is contained and doesn't reach the pad. Most of this energy, I suspect will feed back to the other piston which WILL move and transfer itself through bending the disc and the bearing till it meets the other, now fixed, pad. Still 1 unit of energy used, just a load of parasitic losses from bending the disc and bearing. Then the whole lot gets dissipated through the disc and pad as heat energy to atmosphere, not lost or destroyed energy, just given away! Just a lot less efficiently than if both pistons were free.

So explain to the idiot (me) how you can put 1n in at the hose connection and get 2n out, 1n at each piston?
Steve
The answer to all of that, and especially the last line, is hydraulics. It's all about the pressure in the fluid and the cross sectional area of the master and slave cylinders. Pressure is in Newtons per square meter (SI unit of pressure is the Pascal, which is 1N/sq.m). So you put in a hydraulic pressure of 1 Pascal and every meter squared that fluid touches is pushed by 1N of force. The important bits are the the pistons in the master and slaves. But Pascals are too small to be of any blumin use, and everybody uses the kilo Pascal (kPa)

So (using easy, but not necessarily representative numbers), if you apply 100N of force - equivalent to the weight of a 10.2kg mass - to a master cylinder of 1/100th of a square meter, you get 10kPa of pressure. If you and apply that hydraulic pressure to a slave cylinder with a piston of 1/10th of a square meter in area, you get 1000N or ten times as much force from the piston as you put on the pedal. Do remember that area goes with the square of diameter. However (ignoring how the hoses swell) the pedal has to move ten times as far as the piston. It's as well to go look at how a hydraulic lift works to grasp Pascals principle. It may be hard to see, but its exactly like a lever: If the distance from fulcrum to input (your arm) is ten times that from fulcrum to load, there's ten times the force you put in at the input applied to the load, but if you actually move the load, the input has to travel ten times as far. Technically, that's where the energy is, the pedal force being one tenth of the piston force but moving ten times as far give the same energy out as you put in.

So, at 10kPa of hydraulic pressure, every piston of 1/10 sq.m gives 1000 Newton or force. But, every time you multiply the number of pistons, if they all have to travel the same difference, the pedal has to travel the same number of times further.

So, if one piston is seized, and you press with the same force on the pedal then the one(s) that still work still give the same ten times as much force, i.e. 1000N, out. But, because one of the pistons doesn't move, the pedal has to move less, and energy out is still the same as the energy in. The important bit is that the ones that are not seized do not give more force out for the same force on the pedal because one has seized. And it's the force on the pad that determines how quick you stop (with the distance from pad to hub centre and the coefficient of friction between pad and disc, but according to Ammontons not, in any situation, the area of the pad).

I doubt that cleared much mud though.

Graham

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The 16v Slant 4 engine is more fun than the 3.5 V8, because you mostly drive it on the upslope of the torque curve.

Factory 1977 TR7 Sprint FHC VVC 697S (Now all of, but still needs putting together)
B&Y 73 Dolomite Sprint UVB 274M (kids!)
1970 Maroon 13/60 Herald Convertable (wife's fun car).


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PostPosted:Sun Feb 28, 2021 10:58 pm 
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If you tighten one screw to 1n, then tighten the second screw from a start point of 1N as it won't be 0N given its got 1n bearing on it from the other screw( ,,,and add a further 1N of tightening ontop then yes there will be then 2N of force) so I can see where your angling from.
It's not about the force on the second screw before you turn it. It's the same as with two pistons. If you have one piston on one side pushing with 1N, and the one on the other side not pushing, just reacting, then you have 1N of compression. But when the second piston also pushes with an additional 1N, the two 1N forces add together and the compression is 2N. So if you have one screw compressing enough to give 1N force, and you screw the other in the same distance again, the resulting force is 2N. It really is very, very simple.

Graham

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The 16v Slant 4 engine is more fun than the 3.5 V8, because you mostly drive it on the upslope of the torque curve.

Factory 1977 TR7 Sprint FHC VVC 697S (Now all of, but still needs putting together)
B&Y 73 Dolomite Sprint UVB 274M (kids!)
1970 Maroon 13/60 Herald Convertable (wife's fun car).


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PostPosted:Mon Mar 01, 2021 12:08 am 
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Not pushing but reacting , they are the same term , reacting with 1N is the same as pushing with 1N , just different words are used to describe the same thing interchangeably.

But yes if one is at 1N and the other reacting with 1N then there is 1N of compression, but the reaction can be said to be also simultaneously the other way round the reactive piston, is also a source of the force and the one that generated the force is reactive against the 'inactive' piston.

Yes the clamp , that does work as you say , however if it's tightened to 1N and the other is then tightened further by the same distance to cause a compression of 2N, both screws are then pressing equally with 2N .


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PostPosted:Mon Mar 01, 2021 12:37 am 
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My vice is sat on the floor as a free to roam clamp lol.
Then you are wasting 90% (roughly) of it's useful potential! You might as well have a G clamp!

A free vise is like a lever without a fulcrum, totally useless! As you probably know, but just wanted a comeback, for which I can't blame you! :lol: We all need a laugh now and again! :lol:

Steve

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'73 2 door Toledo with Vauxhall Carlton 2.0 8v engine (The Carledo)
'78 Sprint Auto with Vauxhall Omega 2.2 16v engine (The Dolomega)
'72 Triumph 1500FWD in Slate Grey

Maverick Triumph, Servicing, Repairs, Electrical, Recomissioning, MOT prep, Trackerjack brake fitting service.
Apprentice served Triumph Specialist for 50 years. PM for more info or quotes.


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PostPosted:Mon Mar 01, 2021 8:05 am 
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I know what you mean and it's true, however it's so huge and heavy that it does a reasonable job of most things, and I have used it as a massive mobile g clamp before! , I must mount a smaller one to the bench given its 12mm steel plate haha.
Ultimately nothing bad has gone on in the thread, there is a disagreement but guess that's the beauty of life in all its imperfections, there would be nothing to ponder without differences of opinion.


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PostPosted:Mon Mar 01, 2021 9:02 am 
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But yes if one is at 1N and the other reacting with 1N then there is 1N of compression, but the reaction can be said to be also simultaneously the other way round the reactive piston, is also a source of the force and the one that generated the force is reactive against the 'inactive' piston.
Yes. But, in the case where there are two pistons in a fixed caliper, one on each side of the disc, and each pushing with 1N of force, there is now 1N from one piston (and the reaction to that from the other) and an additional 1N from that other piston (and the reaction to that from the first). So the two forces from the two pistons add and there's a total of 2N on the area of two pads. So, by Amontons law that pad area is irrelevant only the force and CofF, there's twice as much brake effort as there was with one piston in a fixed caliper.

I think that two pistons, both pushing, give twice the force of only one should be absolutely obvious and so simple as to be virtually axiomatic. If it's not, then I really don't understand how it could not be, and I don't think there's anywhere to go from here. If it is, then you've finally understood.

As I said, the situation with the single piston sliding calliper is not so obvious, but it ends up being exactly the same as the two piston fixed calliper, and gives 2N over the area of the two pads and twice the brake force of the single piston fixed caliper.

Graham

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The 16v Slant 4 engine is more fun than the 3.5 V8, because you mostly drive it on the upslope of the torque curve.

Factory 1977 TR7 Sprint FHC VVC 697S (Now all of, but still needs putting together)
B&Y 73 Dolomite Sprint UVB 274M (kids!)
1970 Maroon 13/60 Herald Convertable (wife's fun car).


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PostPosted:Mon Mar 01, 2021 9:25 am 
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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.


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PostPosted:Mon Mar 01, 2021 9:28 am 
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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.


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PostPosted:Mon Mar 01, 2021 11:36 am 
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and then you get to the nitty-gritty part,,,what pads do you put in? and also what tyres your using? these 2 factors always get over looked....and the story continues :lol:

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