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Originally posted by RogerB:
Yes.
How many miles? What kind of driving?
Probably 3-4K, easy driving, mostly open roads, easy stops.
I have trouble understanding how surface area does not factor into the equation for stopping power. If that is so, why don't we just use a 1 inch rotor and pad? A few months ago, I noticed that the contour my daughter drives had very weak brakes. Upon examination, I noticed that only a small portion of the pad was contacting the rotor. I could tell by the shiny vs rusty area of the rotor. I put on new pads, and the problem was solved.
It's been a long long time since I had a physics class, but I thought the coefficient of friction multiplied by the surface area resulted in the total friction. Maybe I'm wrong, I had that class 34 years ago. 
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It's been a few years since i took my last physics class as well. But, Force = pressure x area. If you increase the area force increases as well, this is why our pistons in our master cylinder are much smaller than those in our calipers, we can apply less force at the pedal, and yet get more force upon the PADS. Originally posted by MFE:
Here's the formula for calculating torque at the rotor, from StopTech Systems, a leading aftermarket brake manufacturer:
Quote:
TW = PS x AP x �µ x 2 x RE
PS = Pressure of system; AP = Total Area of pistons in one half of caliper (one side of
opposed type or active (piston) side of sliding or floater type); �µ = Friction Coefficient; x 2, since
there are two sides of the rotor that the pads are exerting force against;
RE = Effective Radius of clamping force.
TW = PS x AP x �µ x 2 x RE
1. Torque created by the caliper on the rotor (at the wheel) = TW
Quote:
Braking torque: When we are talking about results in the braking department we are actually talking about braking torque - not line pressure, not clamping force and certainly not fluid displacement or fluid displacement ratio. Braking torque in pounds-feet on a single wheel is the effective disc radius in inches times clamping force times the coefficient of friction of the pad against the disc all divided by 12. The maximum braking torque on a single front wheel normally exceeds the entire torque output of a typical engine.
Notice pad surface area is not a factor. Coefficient of friction is, but the sizing of the pad does not change its coefficient of friction:
Quote:
Clamping force can only be increased either by increasing the line pressure or by increasing the diameter of the caliper piston(s). Increasing the size of the pads will not increase clamping force.
Ok, but i'm not talking about the part that you put in bold, i already know this. Of course the size of the pad isn't going to change the clamping force that is trasferred from the piston to the pad.
I know that: (Force of friction (static or kinetic))= (coef. of friction) X (Force-normal). And you can combine this with F=P x A and Tq= F x Radius, and you'll get the same formula that you've displayed.
Originally posted by alex_96GL:
by increasing the area you are decreasing the pressure accordingly. the force stays the same of course
How so? Pressure is applied by your foot ...through the master cylinder through the fluid to the piston in the caliper. The force stays the same? That's exactly what you're changing. If the force applied at a master cylinder piston is the same, for any two caliper pistons, the one with larger area exposed to the fluid will exert more force.
But i can't remeber, formula-wise, where the pad area plays into it.
I just can't think of all the formulas right now.
Stazi?
I really wish i hadn't sold my physics book, out of all the damn books i've had, that's the one i miss the most. 
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u r old...  
anyway:
work done by frictional forces (kinetic energy of the car converted into heat):
Work = F_friction x L distance traveled by the pad.
F_friction = �µ x F
�µ -- coefficient of friction,
F is the force applied to the pad.
Area of the pad is not in the picture, as you can see. Now, it's more complicated than this of course, because the pressure on the pad material plays its role too.
pressure x Area = F
Extremely high pressure (as in small Area) would cause glazing, overheating and other bad effects.
The above formulae helps to explain why area of the pad is immaterial, but it's not useful in practice. After all, you can transfer the same amount of kinetic energy into heat by increasing distance traveled. That's not what you want
That is why MFE quoted the equation for torque. Let's look at the other side of the pad first:
The other side of the pad is more interesting. Pascal's law states that the pressure in the liquid is equal in all directions, so the force applied to the pad is
F = PS x AP.
Two pads: PS x AP x 2
Friction: PS x AP x 2 x �µ
Tprque F_friction x R (effective radius which can be calculated as a distance from the axis of rotation to the center of mass of the pad's working surface)
TW = PS x AP x 2 x �µ x RE
amarv12:
we posted almost at the same time. your argument about pressure in the system applies to the other side of the pad i.e. caliper size. see above.
Last edited by alex_96GL; 10/08/04 01:40 AM.
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Originally posted by alex_96GL:
u r old...
You got that right! I sometimes feel it too. Physically and mentally.
Anyway, I found this discussion at sportcompactcarweb.com that might be applicable.
http://www.sportcompactcarweb.com/editors/technobabble/0212scc_techno/
Their last statement is "Bigger rotors generate more brake torque, which makes the pedal effort feel lighter without affecting pedal travel." I would assume that bigger rotors would mean more contact area between the rotors and the pads, which would mean that surface area does matter. ???
95 Contour Zetec, ATX (wrecked, sitting in the barn)
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2000 Taurus SEL 3.0L Duratec
1994 Crown Victoria LX 4.6 V8
1993 Chevy Silverado 4.3 V6
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Originally posted by bentleywarren:
Their last statement is "Bigger rotors generate more brake torque, which makes the pedal effort feel lighter without affecting pedal travel." I would assume that bigger rotors would mean more contact area between the rotors and the pads, which would mean that surface area does matter. ???
I didn't look at the link, but larger diameter rotors have larger radius obvioulsy, and since Torque= radius x Force, the brake torque increases. You can now apply the same force farther away from the center, which means more torque.
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I cannot believe that all this esoteric discussion is really necessary. I mean, how much area does this taper really represent? It's effect is apparently minimal relative to the effects of pad material transfer, pad/rotor mating, and the increase in adherent friction (not abrasive friction) that comes with these two events. See THE NATURE OF BRAKING FRICTION As for the area discussion, look at it this way. If you have two brake pads of the same material, and all else is held equal, including the force at the brake pedal, the pressure applied to each pad is the same. In the larger pad, the lbs per square inch applied to the rotor is less than in the smaller pad, but the increase in area compensates for the lower psi. Friction is the same. Any clearer?
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Yes, thanks.
I get it now....also, i misunderstood what alex96GL was saying in his first post when i first read it.
I believe that i understand.
So then is it correct to say that the main reason to choose a pad of larger area is to lengthen the life of the pad.....since you could effectively get the same brake torque from a smaller area pad, but it would wear through quicker?
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Originally posted by amarv12:
Yes, thanks.
I get it now....also, i misunderstood what alex96GL was saying in his first post when i first read it.
I believe that i understand.
So then is it correct to say that the main reason to choose a pad of larger area is to lengthen the life of the pad.....since you could effectively get the same brake torque from a smaller area pad, but it would wear through quicker?
I'm not sure on this one, but it seems that a larger pad would handle heat better. The smaller pad would "heatsoak" more quickly and then pass more heat to the caliper piston and into the fluid, etc. And yeah, now that I think of it, concentrating all that pressure in a small area, it seems like you'd wear the rotor and pad faster, too.
Function before fashion.
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Originally posted by alex_96GL:
Area of the pad is not in the picture, as you can see. Now, it's more complicated than this of course, because the pressure on the pad material plays its role too.
pressure x Area = F
Extremely high pressure (as in small Area) would cause glazing, overheating and other bad effects.
I know I have to work on my communication skilzzz
I am sure a lot of research went into the size and shape of the pad. heat dissipation, gas removal, chip resistance... all those issues are essential.
while we are on the subject of size, notice how Todd recommends small piston diameter in this thread where as from the stopping power considerations we would like to have it as big as possible. this just goes on to show that simple mechanics 101 considerations have very little to do w/ actual product. nevertheless, understanding of mechanics 101 is an essential step.
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Originally posted by alex_96GL:
Originally posted by alex_96GL:
Area of the pad is not in the picture, as you can see. Now, it's more complicated than this of course, because the pressure on the pad material plays its role too.
pressure x Area = F
Extremely high pressure (as in small Area) would cause glazing, overheating and other bad effects.
I know I have to work on my communication skilzzz
I am sure a lot of research went into the size and shape of the pad. heat dissipation, gas removal, chip resistance... all those issues are essential.
while we are on the subject of size, notice how Todd recommends small piston diameter in this thread where as from the stopping power considerations we would like to have it as big as possible. this just goes on to show that simple mechanics 101 considerations have very little to do w/ actual product. nevertheless, understanding of mechanics 101 is an essential step.
A large piston would require lots of pedal deflection to achieve the same force applied to the rotor.
Function before fashion.
'96 Contour SE
"Toss the Contour into a corner, and it's as easy to catch as a softball thrown by a preschooler." -Edmunds, 1998
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