09 ralliart vs 09 wrx
Jun 30, 2009, 10:33 PM
#106
Ok AMBY.
http://www.thestudentroom.co.uk/showthread.php?t=198733
Thats for you since you think I am not also in school for w/e reasons. Clearly outlines the role of brake pad area. Have a great day this argument is over.
http://www.thestudentroom.co.uk/showthread.php?t=198733
Thats for you since you think I am not also in school for w/e reasons. Clearly outlines the role of brake pad area. Have a great day this argument is over.
Jun 30, 2009, 10:49 PM
#107
That very thread proves you wrong. You're arguing that coefficient of friction is dependent on area, which it's not. This post shows that
Brake force is based on the size of the pistons, not the size of the pad as shown here
Perhaps you should read your own articles before you try to use them as proof to your unwavering knowledge and wisdom.
Edit: I have no idea what you're trying to show with the Integra post since it covers the basics of brake pads and says nothing about area or whatever.
Frictional braking force = Reaction force * Coefficient of friction
But Reaction force = Pressure * Area
So Braking force = Pressure * Area * Coefficient of friction
Assuming a hydraulic braking system, the 'pressure' in the equation will be the pressure of the brake fluid. Note that if it isn't a hydraulic braking system, the reaction force will not depend on the area of a brake pad, since it will be directly transmitted and so is constant - so the second equation will not apply. E.g. a brick will have the same reaction force whether it is on its side or its end. If it is a hydraulic system, however, it will be pressure that is constant, so the force will depend on area.
[Edit: just to make clear, by 'constant' I mean that it will not change if the area of the brake pads changes, not that it's not a variable].
But Reaction force = Pressure * Area
So Braking force = Pressure * Area * Coefficient of friction
Assuming a hydraulic braking system, the 'pressure' in the equation will be the pressure of the brake fluid. Note that if it isn't a hydraulic braking system, the reaction force will not depend on the area of a brake pad, since it will be directly transmitted and so is constant - so the second equation will not apply. E.g. a brick will have the same reaction force whether it is on its side or its end. If it is a hydraulic system, however, it will be pressure that is constant, so the force will depend on area.
[Edit: just to make clear, by 'constant' I mean that it will not change if the area of the brake pads changes, not that it's not a variable].
In order to calculate the amount of clamping force generated in the caliper, the incoming pressure is multiplied by the area of the caliper piston. In our example, the 558 PSI that had been generated at the master cylinder has traveled through the brake pipes and lines and is pushing against two 1.5" pistons per caliper. Therefore, the 'effective area' of the caliper will be equal to two times the area of a single 1.5" piston. Working the numbers reveals that 558 PSI will generate 2,068 pounds of clamp load {558 PSI x 1.84 in2 x 2}.
Edit: I have no idea what you're trying to show with the Integra post since it covers the basics of brake pads and says nothing about area or whatever.
Last edited by ambystom01; Jun 30, 2009 at 11:00 PM.
Jun 30, 2009, 11:02 PM
#108
Uhh that artical clearly states that the area is important if your using a hydraulic system.
I guess we've been talking about tractors this entire time. And when you leave area out of the equation I assumed they factored it into the u. So shoot me. I am still right either way.
I guess we've been talking about tractors this entire time. And when you leave area out of the equation I assumed they factored it into the u. So shoot me. I am still right either way.
Jun 30, 2009, 11:15 PM
#110
Frictional braking force = Reaction force * Coefficient of friction
But Reaction force = Pressure * Area
So Braking force = Pressure * Area * Coefficient of friction
Assuming a hydraulic braking system, the 'pressure' in the equation will be the pressure of the brake fluid. Note that if it isn't a hydraulic braking system, the reaction force will not depend on the area of a brake pad, since it will be directly transmitted and so is constant - so the second equation will not apply. E.g. a brick will have the same reaction force whether it is on its side or its end. If it is a hydraulic system, however, it will be pressure that is constant, so the force will depend on area.
But Reaction force = Pressure * Area
So Braking force = Pressure * Area * Coefficient of friction
Assuming a hydraulic braking system, the 'pressure' in the equation will be the pressure of the brake fluid. Note that if it isn't a hydraulic braking system, the reaction force will not depend on the area of a brake pad, since it will be directly transmitted and so is constant - so the second equation will not apply. E.g. a brick will have the same reaction force whether it is on its side or its end. If it is a hydraulic system, however, it will be pressure that is constant, so the force will depend on area.
I will put it this way:
Wider tires have better traction all things being equal. If wider tires over power your brakes, GET WIDER BRAKES.....
Your making this over complicated and losing yourself.
Last edited by 4g64fiero; Jun 30, 2009 at 11:20 PM.
Jun 30, 2009, 11:19 PM
#111
i'm coming in on the tail end of this one, but the brake pad itself would play some role in braking force. yes, the initial braking force is applied by the piston, but the actual force is distributed by the pad over a larger area than just the surface of the piston. So are you talking about the force from the pistons, or the actual force that's being applied to the rotor? there's no getting around the fact that you have to consider the pad because that's the point of contact between the rotor and the braking system. the larger the brake pad area, the less the force is on any one point. the piston pushes on the entire pad, not just one spot.
Jun 30, 2009, 11:27 PM
#114
Sure but given that pad area is not part of the braking force equation, it is irrelevant. If you had the same size pistons and merely increased the brake pad area, nothing would happen. You have to either change the force being applied (by increasing the size of the pistons, the pressure in the system) or the coefficient of friction (by using a different compound).
Jun 30, 2009, 11:51 PM
#117
i'm with you on this one. it's just common sense, unless he's talking about the braking force applied to the pad. if you're talking about braking force applied to the rotor, you HAVE to include the brake pad because it's like you said, the piston becomes the pad. the pad is what is causing friction with the rotor.
how can you talk about coeficient of friction without talking about the pad? the piston isn't causing friction with anything. it's just like wearing a snow shoe. you're not exerting force on the snow. if you were, it would be on a much smaller area, there for be a higher force and you'd sink into the snow. you're applying force to the snow shoe, which is applying a force which is distributed over a larger area to the snow. the pressure the piston is putting on the back of the pad and the amount of pressure the pad is putting on the rotor isn't the same. you have to take the pad area into account.
how can you talk about coeficient of friction without talking about the pad? the piston isn't causing friction with anything. it's just like wearing a snow shoe. you're not exerting force on the snow. if you were, it would be on a much smaller area, there for be a higher force and you'd sink into the snow. you're applying force to the snow shoe, which is applying a force which is distributed over a larger area to the snow. the pressure the piston is putting on the back of the pad and the amount of pressure the pad is putting on the rotor isn't the same. you have to take the pad area into account.
Jul 1, 2009, 04:55 AM
#118
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I think Amby is forgetting that the force gets distributed throughout the pad, therefore pad size does matter. However it would reach a point where the pad size is too large and the force distributed wouldn't be enough. But the point is you can't just ignore the fact that the pad makes contact, causing friction and exerting mechanical energy with heat as a by-product.
Jul 1, 2009, 09:03 AM
#119
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^ I'm no gearhead, but I did graduate from engineering... and I have to say that the two posts above make absolute sense if you put math/physics into this. Amby, just admit you are wrong this time =D Think about it, if the brake pad area had the diameter of a quarter dollar coin & even if the piston was the biggest thing you could install on a 3500 lb car, I can guarantee it's not going to stop the car running @ 100 mph.
Last edited by tipoytm; Jul 1, 2009 at 09:21 AM.
Jul 1, 2009, 09:18 AM
#120
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The pad and rotor is completely relevant for the coefficient of friction (CoF for now) calculations. A brake pad with 1sq unit of area will obviously have a smaller CoF than a 2sq unit of area pad made out of the same material - assuming that the force applied to the rotor is the same between the two. If size wasn't one of the most important factors (and the calipers ability to apply enough pressure), then there would be no reason for race cars to have HUGE brakes.
Ok - to be fair, *theoretically* you could have a smaller caliper/pad exert as much friction as a larger set, but the requirement for more exotic materials is necessary, or you would have significantly shortened component life.
Ok - to be fair, *theoretically* you could have a smaller caliper/pad exert as much friction as a larger set, but the requirement for more exotic materials is necessary, or you would have significantly shortened component life.
i'm with you on this one. it's just common sense, unless he's talking about the braking force applied to the pad. if you're talking about braking force applied to the rotor, you HAVE to include the brake pad because it's like you said, the piston becomes the pad. the pad is what is causing friction with the rotor.
how can you talk about coeficient of friction without talking about the pad? the piston isn't causing friction with anything. it's just like wearing a snow shoe. you're not exerting force on the snow. if you were, it would be on a much smaller area, there for be a higher force and you'd sink into the snow. you're applying force to the snow shoe, which is applying a force which is distributed over a larger area to the snow. the pressure the piston is putting on the back of the pad and the amount of pressure the pad is putting on the rotor isn't the same. you have to take the pad area into account.
how can you talk about coeficient of friction without talking about the pad? the piston isn't causing friction with anything. it's just like wearing a snow shoe. you're not exerting force on the snow. if you were, it would be on a much smaller area, there for be a higher force and you'd sink into the snow. you're applying force to the snow shoe, which is applying a force which is distributed over a larger area to the snow. the pressure the piston is putting on the back of the pad and the amount of pressure the pad is putting on the rotor isn't the same. you have to take the pad area into account.