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Old 09 November 2000 | 01:53 AM
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I have been thinking about uprated brakes and I am a little confused.

I am trying to figure out why larger discs give better stopiing power. My little mind keeps telling me that standard brakes will apply enough braking force to lock the wheels and just before this point will be maximum braking power. Therefore I am assuming larger discs would give more power without locking up. How can this be achieved? I am guessing something to do with the pressure=force/area equation (is that correct - A-Level physics was a long time ago now)?

Also, what exactly do 4 pot or 6 pot brakes do differently? Do they apply more pressure than standard brakes? But if this is so and the standard brakes supply enough pressure to lock the wheels, how is this better?

Hope some of you can shed some light on this for me
Rikki
Old 09 November 2000 | 01:56 AM
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He's right you know - I don't understand either. Standard brakes can lock the wheels, the anti-lock decides how much power is applied.

Do bigger disks just mean less fade as the surface area is larger and so has less friction/heat build up?

S
Old 09 November 2000 | 02:01 PM
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what if you haven't got ABS?!

I have just had a set of P1 calipers and
disks fitted to my MY94 WRX and there is
a huge difference, much firmer, it feels
'solid' (the best word I can think of).
Old 09 November 2000 | 02:25 PM
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I guess it is a case of over engineering. That is to say, the aftermarket brakes that are often in discussion on this board are designed to stop time and again from track speeds without wilting.

From a scientific viewpoint, if the standard brakes can apply enough restitution to lock the wheels at any point during the deacceleration to the stop, there is nothing to gain in uprated brakes. However as your tyres and brakes warm up, there is a point at which the brakes will not be able to cross the threshold of grip, and the car would be out braked by a car with bigger / better disks. This is by virtue of uprated brakes increased surface area and the directly related ability to dissipate more heat more quickly.

K "Ready to be corrected" F
Old 09 November 2000 | 02:53 PM
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mattski,
oooh! nice! I've been asking about better brakes for my MY94 WRX Wagon.

fancy emailing me (see profile) with a few details? eg; price (with and without fitting) where you got it done, plus pads n discs being used.

did you do anything to the rear brakes?

rude aint I
Old 09 November 2000 | 03:03 PM
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Markus, you have mail.

Old 09 November 2000 | 03:05 PM
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Well..... I think that there are two separate effects that occur when you put in larger discs:

1) Bigger discs heat up less when they absorb a given amount of kinetic energy (from the moving car). Consequently they are less prone to fade because their temperatures rise less and they can dissipate heat more easily (just like a larger radiator).

2) Bigger discs act as longer levers. It's like opening a door by pushing on the edge near the hinge (small discs) and then on the edge furthest from the hinge (large discs). Bigger discs allow more force to be applied more quickly, which allows the brakes to reach the optimum amount of braking sooner, with less effort. This results in the brake pedal feeling "firmer" and "more responsive".


That's why bigger discs feel firmer before they've had to deal with heavy braking.

I think a good test would be to race a standard car and a car with bigger discs down a straight and get them to both brake at the same time. Even with cold pads and discs, I bet the car with the bigger brakes would stop shorter because it will reach full braking (before the ABS cuts in) quicker.


IMHO and willing to be proved wrong as always.
Old 09 November 2000 | 03:10 PM
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bigger disks = more metal, so heat build up takes longer and as there's more surface area, they cool down quicker.

As to why 4 (or 6) small cylinders as opposed to 2 big cylinders allow more force to be transmitted given the amount of pressure you can apply with your foot, I have not a clue.

Something I've always wondered in addition, is why we don't just pit a bigger master brake cylinder in ? surely that would have a similar effect to increasing the number of braing cylinders at the wheel ?

Gary
Old 09 November 2000 | 03:13 PM
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Yep

My understanding is that kinetic energy loss means more heat to dissipate. Build up of heat equals brake fade.

braking at 30 mph is one thing. braking at 100 mph another - even without ABS locking the wheels not so easy then - Heat build up is very rapid.

Bigger brakes with bigger pad area dissipate heat more quickly

I guess

Isn't there another thread about this somewhere on this bbs?

JD
Old 09 November 2000 | 03:26 PM
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good stuff boys. why can the pads/discs not work as well when hot?

still no clearer on the 2/4/6 pot thing. anybody?
Old 09 November 2000 | 03:30 PM
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guys,

think the point behind more pots is a more even braking force over the area of the pad, so same force, but applied to larger area
I think this also helps to reduce the effect of getting uneven pad wear, but Im sure I will soon be corrected!

robski
Old 09 November 2000 | 03:50 PM
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I can understand what you are saying Robski but I am sure in physics I learnt that the (braking) force is spread evenly over the surface?
Old 09 November 2000 | 04:28 PM
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May be wrong, but..

It's to do with exerting even pressure over the area of the pad contact with the disc. 2 smaller pressure points ensure accurate contact better than 1 bigger 1.

Also has effect on pedal feel/pressure exerted as 2 smaller pots "use" less fluid than 1 bigger 1 (pythagoras & area of a circle x depth of pot). I say "use" should be displace I suppose.
Old 09 November 2000 | 04:49 PM
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Geek mode on

Bigger disks = more torque, or lever effect - hence more power. Simple as that.

As to fade, yes the bigger disks will dissapate heat more, but that's a secondary effect & would only be seen on the track.

Size of pad, surprisingly, has no effect - brake force = brake pressure * coefficient of friction. Size of pad, swept area, or no. of pistons does not, of itself, have an effect.

The number of pistons (4 or 6) has an advantage over 2 pots as follows

1) More pistons = more piston area = more pressure can be applied into a smaller movement - effectively making the caliper "higher geared" & thereore better modulating (but not actually more powerful for a given master cylinder lever ratio)

2) Differentially-sized pistons enable the pad to hit and stay square

3) (especially with 6-pots) enables the same piston area (relates to swept area) to be achieved with a narrower (therefore lighter therefore less unsprung weight and rotating inertia) disk

End Geek Mode
Old 09 November 2000 | 05:00 PM
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I disagree with getting brake fade only on the track, I regularly get brake fade on country roads (every day in fact). I don't like this, it scares me.

(And I can't get more than a couple of track laps before they stop working all together)

As to the 2 pot to 4/6 pot thing, it all sounds very nice but the reality is - if my brakes are cold, I can lock up the wheels at a quite respectable speed. I don't need more breaking just less fade.

Gary
Old 09 November 2000 | 05:04 PM
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ok

considering everything that has been written here, could somebody tell me if I am correct in saying:

larger discs would give no significant stopping advantage at a low speed, say 30 mph, where brake fade is not an issue, over standard discs. (going with the fact that they can both provide the maximum stopping force i.e. just before locking up - but is this the same force for each kind of disc?)
Old 09 November 2000 | 05:12 PM
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Rikki

Yep, agree with that

bigger pads = larger surface area to dissipate heat. All heat generated in pad area.

Dunno why hot brakes fade - ordinary road / trackday ones obviously do. But real racers have brakes that don't work properly till glowing red (F1 etc) - But then they won't work when cold !!!

Can't have everything

JD
Old 09 November 2000 | 05:42 PM
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Ahh

That makes sense (more pad out outer edge of disk therefore a bigger 'lever')

Read somewhere that the brake fade happens as at V high temperature's a chemical reaction between the steel disk and the pad cause a load of gas to be given off which is then trapped between the disk and the pad - therefore no pad touching disk.

Not sure whether to believe that or not though

Aren't F1 brakes carbon on carbon ?

Gary
Old 09 November 2000 | 05:42 PM
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I think Airmiles has put it best.

Bigger disks = greater leverage/torque.

Brake fade and heat dissipation are secondary.

BTW F1 (etc) set-ups use carbon fibre disks and pads which don't work too well below a certain temperature, but grip like a ba$tard when hot.


[This message has been edited by RichS (edited 09 November 2000).]
Old 09 November 2000 | 05:43 PM
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For all your fade questions...
Old 09 November 2000 | 06:25 PM
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RichS

Agree with your comment "large discs=more leverage". This should therefore mean that you would get a larger braking force with a shorter press of the brake pedal therefore
saving the person braking effort but in my understanding now, that doesn't explain why larger discs could stop you better as either size disc can ahieve maximum stopping force (i.e. just before lock up) . From what has been said here, I gather this is due to the heat dissipation thing??
Old 09 November 2000 | 06:40 PM
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Just like to add that surface area is NOT related to friction.(more disk and pad contact=faster stopping)

Its strange but true.

IE a wider tyre has more surface area in contact with the ground it will make the car handle better ,but if area is not in the formula for friction how come cars with wider tyres have more grip.
Any one care to explain this?
Old 09 November 2000 | 07:06 PM
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guy's

heat and it's related effects are the bane of any braking system.ie gas,warppeg,glazing boiling.
on the track 4/6 pod is whats needed. but on the road (for some of us) are generaly wasted.
to combat fade (heat)there are a number of things we can do for our road cars
1 vented discs (standard on scoobs)
2 grooved or cross drilled discs:- helps gas build up to escape and deglazes pads.
3 higher rated brake fluid dot 5 ,5.1:- higher boiling point 500 deg C,water reppelent,non-corrosive,silicon based.

my 2 cents john
Old 09 November 2000 | 07:50 PM
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Bumcrack.

We are confusing static friction and kinetic friction. That is to say, the tyre is static wrt the ground it rests upon, (unless it is skidding) and a brake pad is moving wrt the rotor. The F = uR doesn't hold true for static friction.
KF.
Old 09 November 2000 | 07:54 PM
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Addendum,
Anecdotal evidence would suggest that a smaller contact patch would yield *better* handling. The problematic parts of steering are exacerbated by wider tyres, and the greater turning moments exerted by the increased width. Witness the worsening of tram lining and lessening of self centering, with increased tyre width.
I am sure SDB can illucidate...
KF.

Edited for spelling...

[This message has been edited by KF (edited 09 November 2000).]
Old 09 November 2000 | 09:17 PM
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Have any of you lot tried braking hard from 140/150+ mph with your "standard but will lock the wheels" discs? If youd tried it youd appritiate the advantage gained from the additional leverage gained from the larger discs combined with their better heat dissipation. You maybe able to lock your wheels at 30mph but can you do it at 120? I doubt it and this is where the bigger discs come into their own. The additional pots do exactly what Air"geek"miles says they do.

I dont suppose you scooby owners can get your jap rubbish upto high enough speeds to appritiate bigger brakes

JOKING, im joking..
bren
Old 10 November 2000 | 11:15 AM
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Some more similies.

When thinking of the difference between 2pot, 4pot and 6pot brake calipers, imagine squeezing a cream eclair (one of the long ones).

A 2pot setup is like squeezing the eclair in the middle - lots of force at the middle, but not much exerted at either end.

A 4pot setup is like squeezing it with both hands - the force is more evenly spread across the length of the eclair. A 6pot is more of the same.

Although brake pads are strong, they are not infinitely rigid and hence will bend very slightly when squeezed by the calipers. Hence, more pots spreads the force more evenly across the width of the pad which equals more stopping power.

6pots also have the advantage gained by being smaller equalling less pedal travel.

Friction is most <I>definitely</I> (well, maybe not! See below) affected by surface area. The coefficient of friction is simply a number, but when applied to any experiment, area should always involved (I don't have the formulae to hand! ). Put a car with narrow tyres on a tilting ramp, tip it up until the car slides off. Change to wider tyres and the car will resist sliding until a higher angle. Equally, bigger pads give you better stopping power, otherwise all brake pads would be one molecule wide.

Smaller tyres might give better "handling" in as much as they have less inertia and are less resistant to being turned by the steering wheel, but these advantages are often negated by the extra grip achieved by wider tyres. As with everything in motor racing, it's a compromise.

Brakes work less well as they heat up for two reasons. The coefficient of friction of a steel disc drops as it gets warmer and a pad loses efficiency because it gives off gases that separate the pad from the disc. Both effects don't change the feel of the brake pedal, in otherwords, the pedal remains firm, but you don't stop. Once the heat gets into the brake fluid, then the pedal goes to the floor and you don't stop. I am undecided on which one is the more alarming.

F1 use carbon-carbon brakes (carbon pads, carbon discs) because as they get hotter, the coefficient of friction goes up, just like racing tyres get stickier as they get warmer.

Remember that a force might be the same, but pressure is applied to an area and that work is done over time. Two braking systems might be able to apply the same force, but a bigger pad/disc system can apply it more quickly and more evenly, hence shorter cold stopping distances on the same tyres.

[This message has been edited by DavidRB (edited 10 November 2000).]
Old 10 November 2000 | 11:49 AM
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DavidRB,

I think you missed the point. We know that surface area features in the equation for friction.
F = uR, wherein Force is a product of the coefficient of friction and the Normal reaction. The normal reaction is a unit of pressure, and hence directly related to area.
So, your assertion that u is unit less is correct, however your "proof" of a car on a slope isn't mathematically rigorous, and is also contradictory to your assertion...
KF.

EDIT: Excuse the sloppy physics. The normal reaction is of course nothing to do with pressure, and is merely a force, hence the fact that the coefficient is dimensionless. The normal force, is however related to the area of the object if we are calculating its force by application of pressure over area.

[This message has been edited by KF (edited 10 November 2000).]
Old 10 November 2000 | 12:18 PM
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KF: Sloppy physics and non-rigorous proofs are the refuge of those whose physics textbooks are buried in the loft somewhere. Along with most of my physics knowledge.

From what I remember, the static friction equation F=uR is simply "force=number x another force" with the two sides of the equation balancing out. Can't remember sliding friction, but I'm sure it has area in it. Funnily enough, there is a coefficient of static friction and one of sliding friction and the difference between the two is the reason that discs make lousy handbrakes. But I suspect you know that already.

Not sure where I've been contradictory, I was just trying to point out that bigger brake pads provide more friction than small pads.



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