Hydraulic force question



You forgot to put a :-) behind your nonsense^Wstatement.
Or do you really not know the difference between inner and outer forces?
Nick
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ACK
NAK
To put an end to this funny discussion (I have put oil into the fire): With one piston, you have a reacting force on the other side of the bracket of exactly the same force coming from the piston (actio reactio; or summ of vectorial addition of forces must be zero). Now if you introduce a piston on the other side, that piston can only have the same force as the first piston (same pressure, same diameter) and thus just will compensate the actio. One piston actio, the other reactio.
Vector addition! Force has direction!
Think of having a C-clamp with two opposed spindles. Close one spindle with all the force you can bear with your hands. Now close the other spindle with all the force. Will there be a difference? Will you be able to turn the other spindle? No.
So it makes no difference regarding the forces wether there is one piston or two opposed pistons (same diameter).
Nick
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Nick Mόller wrote:

LOl. What you're saying then, is that 1 x one ton jack will lift as much as 2 x 1 ton jacks? Interesting hydraulic philosophy, I wonder what Bramah would say.
Why is it then, that comparable capacity single piston brake calipers have a far greater piston diameter than the equivalent 2 piston calipers? Brake calipers work on one simple formula:
total piston area x working pressure χ pad area = pad pressure psi.
Tom
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I didn't say this. If the jacks work in parallel, they lift 2 tons. But if you stack two jacks, will they lift 2 tons?
hint: The answer is "no"
Nick
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Nick Mόller wrote:

Exactly, but that is not the case in a 2 piston caliper, the forces aren't acting in tandem. Still I notice, you tend to ignore what you don't understand..
Tom
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What do you mean by "in tandem"? Parallel or serial?

Ignorance is often a subjective attribute. :-)
Nick
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On Wed, 1 Feb 2006 10:26:24 +0100, snipped-for-privacy@gmx.de (Nick Mόller) wrote:

Yes, in the case of lifting a car two stacked one-ton jacks would be limited to one ton. When you start pumping the jack on the top and reach the one ton pressure limit, even if you jimmy the relief on the upper jack the relief valve on the bottom jack would open to avoid overloading.
But NOT when you're talking about a brake caliper, that's not the right way to look at it - The object of the pressure would be BETWEEN the two stacked jacks, and it WILL see two tons of pressure - one ton from each side. They are in parallel, but the forces are opposing.
RIGID SURFACE | Jack | Pad -------- Rotor - Measure Force Here -------- Pad | Jack | RIGID SURFACE
Know what I mean, Vern?
--<< Bruce >>--
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wrote:

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Huh? I _read_ the response, and disagree. But I didn't intentionally respond. Sorry. LS
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snipped-for-privacy@earthlink.invalid says...

In other words, if you stacked your two one ton car jacks nose to nose, rather than nose to base, you'd be able to lift 2 tons??
In your ASCII diagram, no matter where in the stack you insert a force sensor, it will register the same value. If this were not the case there'd have to be some additional reaction to ground between the rigid surfaces.
Ned Simmons
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[stacked jacks]

They are in series. Sorry, but if you don't know the difference between serial and parallel, further discussion seems quite useless.

Um, ah! So if you stack two jacks and pump one with 1ton you will have 1 ton pressure between the two jacks. Then when you pump the second jack to 1 ton (disregarding it has already a "pressure" of one ton) the force will not increase, but the pressure between will be 2 tons? Get real!
This is my last attempt to explain: First, I ask you to read back a bit and try to understand my posting with the "two opposed spindles C-clamp" example. Thanks.
Now, you have to accept a fact of physics. "actio = reactio". Or, in plain english, "every force has a counter-force (or reacting force)". More verbose: Every force has a reacting force of the same value, but opposed direction and colinear (in line) with that force. Or, the vectorial summ of the forces must be zero. Accept that? OK
Low lets look at the single piston bracket:
----->DDD<----- R F
DDD is the disk R is the reacting force (actio = reactio, remember?) F is the force [I have left out the piston and the bracket]
Now the reacting force "R" is exactly the force "F" from the piston. If it would not be that way, you will move the disk. That doesn't happen. The resulting force is "conducted" (don't know your word for that, but you know what I mean) through the bracket. If F gets smaller, R will get smaller an have the save value (but still the opposite direction). Accept that? OK
Now to the two pistons:
<-----|----->DDD<-----|-----> Rl Fl Fr Rr
Rl: reacting force left side Fl: force left side | is the piston (I needed to draw it now, to show where each vector starts) DDD: disk Fr: Force right side Rr: reacting force right side
Now if you look at the drawing, you have two reacting forces (Rl and Rr). They do have opposed direction, same value and are colinear. They eliminate themselves. So I leave them out in the next drawing:
----->DDD<----- Fl Fr
Uuuups! Damned, that looks exactly like the drawing with a single piston. Just the names are a bit different. But names don't matter. I could have used my cats' names. :-)
And, believe it or not: **** IT IS EXACTLY THE SAME! ****
Nick,
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I can't believe this thread is still going. In your sketch, the vertical _force_ on the rotor is ZERO. That's what you _want_, unless your rotor is on a mighty strong bearing. That's what you _get_, whether you use two "rigid surfaces" and two jacks, or whether you use two "floating surfaces" rigidly connected to each other, and replace one jack with a spacer.
The _horizontal_ force on your rotor is your braking force. It is your friction coefficient times _two_ tons, because there are _two_ sliding interfaces. But there would be two interfaces even if you replace one jack with a spacer and let the caliper float.
Replace one of your "pads" with a (frictionless) _bearing_, and you cut your braking force in half. But the net vertical force _on_ the rotor is still zero, and the net vertical force _through_ the rotor is still one ton.
-- TP
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wrote:
[...]

Umm, hydraulic pistons have two important characteristics: force and stroke. Two identical cylinders with equal pressures placed in parallel double force, the same two in series double stroke. Sort of the same way identical batteries in parallel double current, the same in series double voltage. Likewise adding links to a chain makes it longer, not stronger. The discussion was about stationary brake calipers with opposed pistons--effectively hydraulic cylinders in series. So you would get a potential increase in stroke, but no increase in force. But since a fixed brake caliper needs no more total stroke than a floating caliper it's more comparable to replacing one long cylinder with two short ones stacked on top of each other. Someone did mention earlier in the discussion a caliper with parallel cylinders. He also mentioned that the cylinders are reduced in size--that keeps down the force on the pads and reduces the amount of fluid the master cylinder would need to move to apply the brakes. But that's a different arrangement. You mentioned two bottle jacks, but did not say if they were in series or parallel. So, yeah, under the right circumstances two, ten, or a hundred one ton bottle jacks push no harder than a single one ton jack. The formula is more correctly written as Force = piston cross sectional area X pressure - friction. Pad psi = force / pad cross sectional area. You'll note that pads follow the same rule: two pads in series have the same psi as one pad with the same area. But you get more friction since friction force = total pad area X pad psi.
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Ivan Vegvary wrote:

You are right. There is no force advantage to opposing cylinders.
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again, stop and think about it for a second. we have 100 psi. that is 100 pounds per square inch.
if we double the square inches, we double the force.
another way to think about it. assume we have a single piston, and 1000 psi of force being applied to from teh piston, and further assume that the brake pads are 10 square inches. with a single caliper, there are 2 pads, so that 1000 psi gets spread over 20 square inches, for 50 PSI on the pads. with 2 pistons, you would have 100 psi, and double the friction.
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snipped-for-privacy@hotmail.com says...

That'd be true if the single piston caliper were not free to center itself. But it can, and as a result applies the same pressure to the pads on both sides of the disc.
Ned Simmons
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Doug wrote:

You are not doubling the area because the force is colinear and shares a common ground. Because of this the force cannot exceed that generated by one piston. If it did it would drive the other piston back into its bore, but it cant because it is in equalibrium.
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Jesus there's a lot of shite on this thread.
1/ imagine a motorcycle front wheel and brake, it makes things easier.
2/ in the old days iy would be cable and shoes, now it's hydraulic and pads
3/ hydraulics give a mechanical advantage, the smaller "operating" cylinder vs the larger "operated" cylinder, but the flipside is you get less travel on the larger diameter cylinder
4/ if the "operating" cylinder displaces 1 cc of fluid, this doesn't necessarily mean there is any pressure of note in the system
5/ After you take up the slack on "off" or "released" brakes, then, apart from compressibility of the pad and friction material, generally speaking each extra amount of fluid displaced by the "operating" cylinder will inclrease pressure (yes, fluids are supposed to be incompressible, but brake lines expand etc etc
REALLY IMPORTANT from some of the shit mentioned above.
psi or any other pressure measurement is NOT a force measurement and is nothing like a force measure ment
single piston hydraulic press type system versus opposing (identical) pistin brake caliber type system, with the same pistons, piston area and pressures YOU DO NOT get double the force, you do not even get the force evenly applied to both sides cos it already is, what you do get, and all that you get is a floating caliper. this means you get a disk that is being gripped instead of deflected, this means a hard cast disk can be MUCH thinner = much lower unsprung weight
to the OP, the television person was talking complete bollocks
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snipped-for-privacy@blueyonder.co.uk says...

I think if you were to read the responses carefully you'd find only a few glaring errors in reasoning. Most everyone reached the same conclusion, though some in roundabout and not always clearly expressed ways.
Perhaps you'd prefer a thread on WMDs or the circumstances of Vince Foster's death <g>.
Ned Simmons
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Ned Simmons wrote:

you're prolly right, I tend to skim rather than read in depth

dunno who vince foster is, as far as wmd's go I have never had any illusions, if fact the only people who seem to be disconnected from reality are those that rely on the mainstream media H^H^H^H propoganda for their "facts"
the next 20 years sure are going to be interesting times.
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