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PostPosted:Mon Mar 01, 2021 9:33 pm 
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That's fair enough , but the first point with newtons 3rd law as stated that one piston producing 1N , fed through the strut and then causing a reaction of 1N naturally has the effect of 1N strut compression, if you then add a further 1N ontop of the 1N reactive force from piston 2, then the strut is compressed with 2N of force which feeds back through to piston 1 and the 1st piston force grows by 1N to 2N. ( if its fixed at a maximum 1N then it will be pushed back by the greater force untill it hits the caliper, then the reactive force grows to 2N)
I think we agree but have been casual with my terms admittedly.


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PostPosted:Mon Mar 01, 2021 9:45 pm 
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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.
You can indeed lift a house with the same force as that which will lift a feather. You just need a lot more area of slave cylinder and have to move the master cylinder a heck of a lot further. This is Pascal's principle, and it's very, very well tested and proven.

You can lift a house (and Archimedes claimed he could move a planet) with the same force as will lift a feather using levers. The one that you lift the house (or move the planet) with will be a tad longer, and you have to move the input end a bit further. So levers can be a pretty good analogy of hydraulics.

So with (only) two pistons giving one 1N each, it's the same force on the pedal as one piston giveing 1N, but you have to move the pedal twice as far, so (in deference to Steve's comment) it takes twice the energy, to close two pistons onto the disc as it takes to close up one.

And you could, indeed, do anything with one tiny size of hydraulic pump and some humongous slave pistons. Just some things would then move at the low end of tectonic speeds.

Graham

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PostPosted:Mon Mar 01, 2021 10:33 pm 
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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 :D

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 :wink: ) ??

Roger
If you have a piston with an area of one square centimeter [how do I get superscripts in this place] supplied with hydraulic fluid at 100 kilo Pascals of pressure pressing down (with the cylinder in a frame fixed to the table), that would apply a force of 10N, through the pads and against the table, and the frame would also pull up on the table with 10N. And that would be like what you've drawn.

But if you put another piston with the same 1 sq cm area and supplied with fluid at the same 100kPa pressure in another cylinder fixed to the table and pushing up underneath the bottom pad, that pushes up with an extra 10N. So the force on the table under the new cylinder is now 20N and the force on the two pads is 20N and the force from the frame holding the cylinder of the top piston pulling up in the table is 20N. That's because the force is the hydraulic pressure of 100kPa times the sum of the areas of the two pistons 2 x 1 sq cm. And that is 20N, not the 10N you had when you had one piston of 1 sq cm area and 100kPa of hydraulic pressure.

If you have a master cylinder of the same 1 square cm area, then, in both cases, there has to be a force of 10N on that. But there's some compliance from the pad in pressing at 10N, which means the piston has to move a bit in applying that force. So the master cylinder also has to move, and it moves twice as far with two pistons as with one. So, while it takes the same force, you move that force twice as far and do twice as much work with the 2 pistons as with one.

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 10:40 pm 
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That's fair enough , but the first point with newtons 3rd law as stated that one piston producing 1N , fed through the strut and then causing a reaction of 1N naturally has the effect of 1N strut compression, if you then add a further 1N ontop of the 1N reactive force from piston 2, then the strut is compressed with 2N of force which feeds back through to piston 1 and the 1st piston force grows by 1N to 2N. ( if its fixed at a maximum 1N then it will be pushed back by the greater force untill it hits the caliper, then the reactive force grows to 2N)
I think we agree but have been casual with my terms admittedly.
If you had used the quote button I might have some idea what you say is a fair point.

But there is no feedback in that sense. The pressure on the piston is defined in both cases as being such that it gives 1N of force on the area of the piston. The difference is how far you move the master cylinder to achieve that pressure that gives 1N on the area of one or two pistons, i.e. twice as far for two. The same as the two screw g cramp giving 2N of compression taking twice as many turns of the screw as the case where the compression is 1N.

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 10:51 pm 
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Ok, I now get, after lots of deep thought, that energy and force are not the same. If i've read it right, force is what moves energy around (in laymans terms)

I understand mechanical advantage and force multiplication (Archimedes moving the earth etc) I can also cope with hydraulic advantage. There's quite a lot of it going on in a braking system. First mechanical advantage from the differing distance from the fulcrum (pedal pivot) of the pushrod and pedal pad (circa 10:1) I'm also aware that it matters not a bit for practical purposes that pressure in and effort out are on the same side of the fulcrum. Then comes a servo (vacuum?) advantage of around 1.75:1 and finally a hydraulic advantage, numbers unknown to me. But it will have been calculated to achieve a fine balance between pedal length and effectiveness of the system.

I realize I made a mistake in my single seized piston theory by treating one caliper in isolation rather than as part of the 8 piston system it in fact is. YES EIGHT pistons, the rear cylinders, because they are "floating" (they slide in the backplate) work exactly the same as a floating caliper, so the piston dia counts twice (a fixed cylinder in different design brakes has 2 pistons in opposition)
So the energy destined for the seized piston (but thwarted by it) would be spread around the rest of the entire sytem by which time it would be negligible in effect, you wouldn't notice it, or the infinitesimally shorter pedal. This is, in any case a purely theoretical case as I doubt it possible anyone brakes in real life with exactly the same pedal loading each time. But I still maintain that IF the pressure (or force, or energy) put in at the pedal is constant each time the same amount of energy is put into the system, whether it has 7 pistons or 8, it has to go somewhere. If 1/8 of the system is effectively removed the energy MUST be equally distributed around the 7 functional pistons.

I'm STILL not sure what you are both debating. I know every action has an equal and opposite reaction. But you have such a baffling array of "opposites" to choose from, how do you work out which ones cancel out? It seems to me you are debating the age old conundrum of the irresistable force and the immovable object. And in any case, the answer only works on spherical chickens in a vacuum!

Steve

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PostPosted:Tue Mar 02, 2021 7:56 am 
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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 :D

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 :wink: ) ??

Roger
If you have a piston with an area of one square centimeter [how do I get superscripts in this place] supplied with hydraulic fluid at 100 kilo Pascals of pressure pressing down (with the cylinder in a frame fixed to the table), that would apply a force of 10N, through the pads and against the table, and the frame would also pull up on the table with 10N. And that would be like what you've drawn.

But if you put another piston with the same 1 sq cm area and supplied with fluid at the same 100kPa pressure in another cylinder fixed to the table and pushing up underneath the bottom pad, that pushes up with an extra 10N. So the force on the table under the new cylinder is now 20N and the force on the two pads is 20N and the force from the frame holding the cylinder of the top piston pulling up in the table is 20N. That's because the force is the hydraulic pressure of 100kPa times the sum of the areas of the two pistons 2 x 1 sq cm. And that is 20N, not the 10N you had when you had one piston of 1 sq cm area and 100kPa of hydraulic pressure.

If you have a master cylinder of the same 1 square cm area, then, in both cases, there has to be a force of 10N on that. But there's some compliance from the pad in pressing at 10N, which means the piston has to move a bit in applying that force. So the master cylinder also has to move, and it moves twice as far with two pistons as with one. So, while it takes the same force, you move that force twice as far and do twice as much work with the 2 pistons as with one.

Graham
You've just stated that a 2 pot caliper provides twice the compresive force than a single piston sliding caliper when fed the same hydraulic pressure, contradicting your realisation that they provide the same clamping force for the same hydraulic pressure.


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PostPosted:Tue Mar 02, 2021 9:05 am 
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Damn, and I tried so hard to get it right :D

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 :wink: ) ??

Roger
If you have a piston with an area of one square centimeter [how do I get superscripts in this place] supplied with hydraulic fluid at 100 kilo Pascals of pressure pressing down (with the cylinder in a frame fixed to the table), that would apply a force of 10N, through the pads and against the table, and the frame would also pull up on the table with 10N. And that would be like what you've drawn.

But if you put another piston with the same 1 sq cm area and supplied with fluid at the same 100kPa pressure in another cylinder fixed to the table and pushing up underneath the bottom pad, that pushes up with an extra 10N. So the force on the table under the new cylinder is now 20N and the force on the two pads is 20N and the force from the frame holding the cylinder of the top piston pulling up in the table is 20N. That's because the force is the hydraulic pressure of 100kPa times the sum of the areas of the two pistons 2 x 1 sq cm. And that is 20N, not the 10N you had when you had one piston of 1 sq cm area and 100kPa of hydraulic pressure.

If you have a master cylinder of the same 1 square cm area, then, in both cases, there has to be a force of 10N on that. But there's some compliance from the pad in pressing at 10N, which means the piston has to move a bit in applying that force. So the master cylinder also has to move, and it moves twice as far with two pistons as with one. So, while it takes the same force, you move that force twice as far and do twice as much work with the 2 pistons as with one.

Graham
You've just stated that a 2 pot caliper provides twice the compresive force than a single piston sliding caliper when fed the same hydraulic pressure, contradicting your realisation that they provide the same clamping force for the same hydraulic pressure.
No, a two pot fixed caliper provides twice the compressive force of a single piston fixed caliper when fed the same hydraulic pressure. In the above example, the cylinder of the single piston pushing down is in a frame fixed to the table.

A single piston sliding caliper provides the same compressive force as a a two pot fixed caliper through the force on the cylinder applied though the slide to the opposite pad.

Graham

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Factory 1977 TR7 Sprint FHC VVC 697S (Now all of, but still needs putting together)
B&Y 73 Dolomite Sprint UVB 274M (kids!)
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PostPosted:Tue Mar 02, 2021 9:18 am 
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I'm STILL not sure what you are both debating. I know every action has an equal and opposite reaction. But you have such a baffling array of "opposites" to choose from, how do you work out which ones cancel out? It seems to me you are debating the age old conundrum of the irresistable force and the immovable object. And in any case, the answer only works on spherical chickens in a vacuum!

Steve
It's also not exactly clear to me what jikovron point is, other than whenever I say that two pistons in opposition give twice the compressive force of one, he says I'm wrong and comes up with some spurious argument or analogy that doesn't make any sense at all. Any time I explain why that's wrong or inapplicable he slides off into some other spurious, desperate argument, misrepresents what was said, or resorts to simple gainsaying. It's like Monty Python's being hit on the head lessons.

I just don't see why its not obvious to him that if you have the same pressure on twice the area you have twice the force (and do twice the work in applying that pressure). That to me, and I think to any sensible person, is obvious.

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:Tue Mar 02, 2021 9:25 am 
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YES EIGHT pistons, the rear cylinders, because they are "floating" (they slide in the backplate) work exactly the same as a floating caliper, so the piston dia counts twice (a fixed cylinder in different design brakes has 2 pistons in opposition)

Steve
Ah, I forgot that. But it's not 1/8th less work, because not all pistons are equal in area. Eric Blair's perspective of "some cylinders are more equal than others" may also apply.

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:Tue Mar 02, 2021 9:36 am 
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If you have a piston with an area of one square centimeter [how do I get superscripts in this place] supplied with hydraulic fluid at 100 kilo Pascals of pressure pressing down (with the cylinder in a frame fixed to the table), that would apply a force of 10N, through the pads and against the table, and the frame would also pull up on the table with 10N. And that would be like what you've drawn.

But if you put another piston with the same 1 sq cm area and supplied with fluid at the same 100kPa pressure in another cylinder fixed to the table and pushing up underneath the bottom pad, that pushes up with an extra 10N. So the force on the table under the new cylinder is now 20N and the force on the two pads is 20N and the force from the frame holding the cylinder of the top piston pulling up in the table is 20N. That's because the force is the hydraulic pressure of 100kPa times the sum of the areas of the two pistons 2 x 1 sq cm. And that is 20N, not the 10N you had when you had one piston of 1 sq cm area and 100kPa of hydraulic pressure.

If you have a master cylinder of the same 1 square cm area, then, in both cases, there has to be a force of 10N on that. But there's some compliance from the pad in pressing at 10N, which means the piston has to move a bit in applying that force. So the master cylinder also has to move, and it moves twice as far with two pistons as with one. So, while it takes the same force, you move that force twice as far and do twice as much work with the 2 pistons as with one.

Graham
You've just stated that a 2 pot caliper provides twice the compresive force than a single piston sliding caliper when fed the same hydraulic pressure, contradicting your realisation that they provide the same clamping force for the same hydraulic pressure.
No, a two pot fixed caliper provides twice the compressive force of a single piston fixed caliper when fed the same hydraulic pressure. In the above example, the cylinder of the single piston pushing down is in a frame fixed to the table.

A single piston sliding caliper provides the same compressive force as a a two pot fixed caliper through the force on the cylinder applied though the slide to the opposite pad.

Graham
Yes the frame and table is just the sliding caliper body in isolation,

2 bottle jacks of 10KN stacked ontop of each other can only press a captive frame apart with 10KN of force , however put them side by side in the captive frame and they then add together to produce 20KN of force.
That is as simple as I can explain it, and it is relevant.


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PostPosted:Tue Mar 02, 2021 12:04 pm 
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You've just stated that a 2 pot caliper provides twice the compresive force than a single piston sliding caliper when fed the same hydraulic pressure, contradicting your realisation that they provide the same clamping force for the same hydraulic pressure.
No, a two pot fixed caliper provides twice the compressive force of a single piston fixed caliper when fed the same hydraulic pressure. In the above example, the cylinder of the single piston pushing down is in a frame fixed to the table.

A single piston sliding caliper provides the same compressive force as a a two pot fixed caliper through the force on the cylinder applied though the slide to the opposite pad.

Graham
Yes the frame and table is just the sliding caliper body in isolation,

2 bottle jacks of 10KN stacked ontop of each other can only press a captive frame apart with 10KN of force , however put them side by side in the captive frame and they then add together to produce 20KN of force.
That is as simple as I can explain it, and it is relevant.

If you put two jacks in series between a rigid frame, like the caliper, and measure the force at the junction between them, then when you pump one up to the internal pressure that gives 10kN on its piston, you will measure 10kN at the junction. If you then do the same work again and pump the second up to the internal pressure that gives 10kN on its piston, the force will be 20kN. Moreover, if you connect the two jacks to the same hydraulic supply, and pump them both up to the pressure where one gives 10kN, together, they will give 20kN between them. And that is very simply because you have the same pressure on twice the area, and the force is equal to the pressure times the area.

I must admit that I don't immediately see where the slider is in the frame fixed to the table. But I also admit I was trying to fit in with a far from ideal analogy that started with a free weight, not a hydraulic system. I would think about it, if I thought it actually mattered.

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:Tue Mar 02, 2021 12:13 pm 
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If you put two jacks in series between a rigid frame, like the caliper, and measure the force at the junction between them, then when you pump one up to the internal pressure that gives 10kN on its piston, you will measure 10kN at the junction. If you then do the same work again and pump the second up to the internal pressure that gives 10kN on its piston, the force will be 20kN. Moreover, if you connect the two jacks to the same hydraulic supply, and pump them both up to the pressure where one gives 10kN, together, they will give 20kN between them. And that is very simply because you have the same pressure on twice the area, and the force is equal to the pressure times the area.

I must admit that I don't immediately see where the slider is in the frame fixed to the table. But I also admit I was trying to fit in with a far from ideal analogy that started with a free weight, not a hydraulic system. I would think about it, if I thought it actually mattered.

Graham
But in that example the second bottle jack will already have an internal pressure that gives a force of 10N caused by the the first jack.

With two pistons/cylinders in series the force must be the same in each and is effectively limited by the "strength" of the weakest one - weakest link in the chain. (i.e. a 10N and a 20N jack/piston/ram in series will be limited to the 10N of the weakest/smallest one). For totally separate pistons/cylinders - i.e in each in an independent calliper frame then the force is indeed the sum of each one.

In the case of a floating single piston calliper or a two piston calliper (one each side) you have a series connected system. I think that the two pistons should be regarded as one as they are physically coupled to each other via the pad/disc/pad "sandwich".

Roger

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PostPosted:Tue Mar 02, 2021 12:47 pm 
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If you put two jacks in series between a rigid frame, like the caliper, and measure the force at the junction between them, then when you pump one up to the internal pressure that gives 10kN on its piston, you will measure 10kN at the junction. If you then do the same work again and pump the second up to the internal pressure that gives 10kN on its piston, the force will be 20kN. Moreover, if you connect the two jacks to the same hydraulic supply, and pump them both up to the pressure where one gives 10kN, together, they will give 20kN between them. And that is very simply because you have the same pressure on twice the area, and the force is equal to the pressure times the area.

I must admit that I don't immediately see where the slider is in the frame fixed to the table. But I also admit I was trying to fit in with a far from ideal analogy that started with a free weight, not a hydraulic system. I would think about it, if I thought it actually mattered.

Graham
But in that example the second bottle jack will already have an internal pressure that gives a force of 10N caused by the the first jack.

With two pistons/cylinders in series the force must be the same in each and is effectively limited by the "strength" of the weakest one - weakest link in the chain. (i.e. a 10N and a 20N jack/piston/ram in series will be limited to the 10N of the weakest/smallest one). For totally separate pistons/cylinders - i.e in each in an independent calliper frame then the force is indeed the sum of each one.

In the case of a floating single piston calliper or a two piston calliper (one each side) you have a series connected system. I think that the two pistons should be regarded as one as they are physically coupled to each other via the pad/disc/pad "sandwich".

Roger
I agree with this, but just to further add if i could the force would be limited to 10N by the weaker piston, but if you carried on jacking the 20N side then it would overload the smaller piston and push it back against its stop and then the compressive strut force would grow to 20N total.


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PostPosted:Tue Mar 02, 2021 1:25 pm 
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I don't think these strained analogies are moving us forward, though that's clearly an irony in a discussion about brakes. And my value of Young's modulus of elasticity isn't up to this level of strain - too much stress.

This is the best I can manage as a drawing of what I see as going on between a single piston and two pistons both in fixed callipers - remembering that the caliper is, in both cases, fixed to the same structure as the disc.
It shows the main pressures (blue), forces (red), and moments (green).

Image

As I hope can be seen, with the single piston caliper, there's unit force on one pad and a moment on the hub as a result of that. There's also a different moment on the caliper mount - same force but different length and opposite rotation.

But with the two piston caliper, there's the same force on two pads, i.e. twice the total. It's spread over twice the area, but Amontons' law says that area of contact between the two surfaces doesn't matter. Hence in that situation of twice the force, there's twice the brake effort. This is all that really matters in the analysis of brake effect - its the hydraulic pressure times the sum of the piston areas.

If there was a second fixed pad in the case of the single piston caliper and the hub bearing had enough movement to let the disc push up against it there would still be the same force between the two pads as there was on the one pad, i.e. half of what there was with two pistons (hence half the compression in the thickness of the disc), but over twice the surface area. And, as said, area not mattering, the brake effort would be the same as before. It would not give the same effort as the two piston caliper, because there would be half the force, so half the effort.

What happens in the single piston sliding caliper is the same as the two piston fixed caliper because, as said, the force from the back of the cylinder, which is free relative to the disc and suspension, is hooked around and applied through the slider to the second pad.

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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:Tue Mar 02, 2021 1:46 pm 
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In the second drawing , the green moments equal zero force all of the force is contained by the caliper body and if you unbolt the caliper it wouldn't move.
In the first example the caliper body is forced away from the disc and the reaction force comes from the disc itself.
Compression forces through the pads are the same for both examples, but braking effort is halfed in the first example as there only half frictional effort (equivalent to replacing one pad with one of zero co-ef in example 2.


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