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For suggesting that that different sized rotors do give you different amounts of power? Because that has been my experience.
Compositepro what you torquing about.
That's the coat, Taxi!
Like these?
http://upload.wikimedia.org/wikipedia/commons/5/59/Buell_disk_brake.jp g" target="_blank">
The technical term you ar all looking for is moment.
Otherwise known as torque, or leverage to others.
force x distance. More distance = more moment = better braking.
Heat is a small factor
friction between the ground and tyre is all important.
Lesson over.
What a lot of posts considerning the answer was in post 1!!!!!!! lols
Having a think about it actually as the effect of leverage and swept area are both dependant on the increase in disc radius, and seeing as friction and moment are both linear calcs (for A level physics anyway) then the increase in dia will have the same proportional effect on both...
Frictional force at the pad/rotor interface tangential to the rotor rotation (Ff) is equal to the normal force (i.e. the force of the caliper pushing the pads onto the rotor, Fn) multiplied by the coefficient of friction (U). The coefficient of friction is a property of the interface materials, temperature etc. etc. I seem to remember 0.3 was the generic number to use in exams, I digress...
so Ff = Fn * U
Fn obviously depends how hard you pull on the brake lever and how good your hydraulics are. This is where heat comes into play. It plays havoc with your coeff of friction (think glazed pads) and messes up your hydraulics.
Ff * Rr = Fg * Rw
Where Fg = friction force at the ground/tyre interface tangential to the wheel rotation, Rw = wheel radius, Rr = rotor radius.
Rearranging the above equation to get Fg = (Ff * Rr)/Rw we can see that an increase in rotor radius (Rr) will increase the force at the tyre to ground interface for a given brake caliper force. This results in being able to stop quicker assuming the wheel doesn't lock resulting in a skid.
When Fg = Fn * U (Fn is the downwards gravitational force due to your mass and U is the coeff of friction between tyre and ground) the tyre will skid and you lose your advantage.
Simples
Yes we know this and your point is?
This analysis you have done is for statics. Seeing as 90% or more of the work you brake does is whilst the rotor is still rotating methinks the swept length of the disc will have an important effect. So larger dia disc, will pass more contact area between the pads.
My 160s on my 456 stop much better than the 203s on my Meta
...I was playing with oil levels on the meta and spilt fork oil all over the pads
If I am following this correctly, surely that [swept length] would only help dissipate more heat.
Forgot to add that coefficient of dynamic friction, i.e. contact surfaces moving relative to each other can also depend on velocity (as someone said on page one). Bigger disk gives a higher relative velocity although I can't say whether the coeff of friction will increase or decrease. Same principles apply though. For me, swept area of disk is a bit of an odd way of describing it.
surely that [swept length] would only help dissipate more heat.
Yeah maybe - the heat which is a product of the friction.
It cannot be considered in terms of forces alone, its an energy balance.
100% surely
And what about all those holes
Compositepro - Member100% surely
Yes. wooly thinking on my part.
Are we all OK? It has been over an hour since somebody has been insulted.
Bigger disk gives a higher relative velocity although I can't say whether the coeff of friction will increase or decrease. Same principles apply though. For me, swept area of disk is a bit of an odd way of describing it.
I don't see that it is odd, for one revolution the pads pass more length of steel surface if the disc is larger radius.
So if you were to consider friction with swept length, then you burn off more energy if you pass more disc through the pads.
Is it something akin to:-
The circumference of the disc increases by around 60mm for each step increase in size so if you brake for a fixed period with the same amount of force, more area will be swept, more heat will be generated (altough it will dissipate better) so you're put more frictions onto the disc.
That appears to be a more succint way of putting it what I've been waffling on about.
I just put your posts through a Richard Hammond translator.
I don't buy the swept area thing, I'll keep thinking, but I think it's a red herring. If we think of the infinitessimal change/moment in time, the maths is the same as a static model. When does it start changing?
That's probably clear as mud. Never did mechanics, and I'm not great at explaining things at the best of times, even when I do understand them!
Having a think about it actually as the effect of leverage and swept area are both dependant on the increase in disc radius, and seeing as friction and moment are both linear calcs (for A level physics anyway) then the increase in dia will have the same proportional effect on both...
Can anyone think of a system (linear/gearing/pulleys/whatever) that would let us separate the two?
Ah well ned what you think is not relavent. I happen to [b]know[/b] its a swept length thing (slong with the leverage thing).
If we think of the infinitessimal change/moment in time, the maths is the same as a static model.
Einstein lives!
Now I've often wondered about this, and even though about asking, but you generally get called something like Dumbass for asking silly questions on here.Why is a rim brake, which effectively uses a maximum sized rotor, less powerful than a tiny disc at the hub? Fair enough the rubber brake block is going to be less efficient that disc pads as you don't won't to wear your rims away too quickly, and cables may be less effective than an hydraulic system, but surely that can't account for all the difference.
It's down to how much mechanical advantage you can put in the system.
Pulling the levers on a disc brake moves the calipers a tiny distance - probably less than a mm. Pulling the lever on any rim brake moves it several mm. This flip side of this is that the same force on the levers exerts a much greater force at the calipers of disc brakes than on rim calipers.
You can use the high mechanical advantage of a disc caliper on a disc but not a rim for two reasons:
1. A disc can be made sufficiently true and with sufficiently consistent thickness to work with calipers with tiny amounts of travel. It would be pretty much impossible to true a rim to the tolerances required for a disc-style caliper.
2. A disc can withstand the forces exerted by such a caliper - a rim would probably collapse under that kind of force.
The much higher mechanical advantage of a disc brake lever/caliper outweighs the lower mechanical advantage resulting from the smaller diameter of the disc compared to the rim.
In fact, a well set up rim brake isn't that different from a well set up disc brake... until you cover it in mud and water. The higher forces in a disc brake do a much better job of clearing the braking surface.
zzzzz.
Then god created Eric Buell.
He would love this thread.
What we need is magnetic braking to go on our electric bikes
.
Hi all,
Would a longer brake lever make more of a difference to power that going up from 160mm to 180mm disks? After market levers are available... 😕
I love all of the joshing on here!
Ooooh, longer brake levers, then maybe we could use two or three fingers and apply more braking force with less effort per finger.This might even give us more control too.
Would a longer brake lever make more of a difference to power
Short levers can already supply an almighty amount of force to the piston so there probably wouldn't be a massive gain.
More modulation and potentially more tired hands (side effect of so many topis here).
Yep, and Harry333, Birch1983, Peterpoddy, Clubber, IGM and PJay are thick as shite
Err what?
But you post this:
.....the biggest difference is in disc (or rotor in the US) size. The bigger the disc, the more leverage your brake has on the wheel and the faster it can stop it. Each 20mm increase in size roughly equates to a 20 percent increase in power.
Which is exactly what I was saying.
Am I missing something here? Or were you trying to be funny? 😕
i think that we have 2 benefits to bigger rotors. Firstly the increased torque for the same force at the lever. Secondly we have a greater mass to absorb heat and a greater area to dissipated heat. But the op asked why larger discs are more powerful. Power is rate of energy transfer so i'm going with greater area as the answer to this question. But i know that this is a sad and pathetic attempt at being literal
This "swept area" that people are mentioning, at first I thought it sounded like plop but thiking about it if Work = Force x Distance then presumably the larger circumference of the larger disc will mean that the friction force does more work (i.e. converts more kinetic energy to mainly heat) per revolution of the wheel. Does that sound right?
well I called it swept length, the original analysis that I read in a paper about this 10 years ago called it that. Your description isn't a bad way of putting it. It appears to be the forum trend to just redescribe exactly what other people say...
surely with a more "powerful" brake the swept area could be less because being more "powerful" the wheel is brought to a stop quicker and so the disc has not been swept as much.
It appears to be the forum trend to just redescribe exactly what other people say...
I see your point and agree that there's a lot of that in this thread. On the other hand there's a difference between just saying "it's because of swept length" with no explanation and explaining a point based on concrete laws of physics to back it up.
The "swept area" and additional levarage are just two different ways of looking at the same problem.
If you look at the forces, the bike's decceleration is proportional to the static frictional force between the ground and the tyre (Newton's second law). This force is equal to (Rd/Rr) * Fd where Rd is radius of the disc, Rr is the radius of the wheel and Fd is the frictional force on the disc. So bigger disc (bigger Rd) equals bigger force.
The frictional force, Fd is u * Fn where u is the coefficient of friction, and Fn is the normal force (i.e. the force being applied by the calipers, which depends on how hard you pull the lever).
u may depend on the slip velocity (rate of disc moving past pad) but not necessarily in an obvious way - it often goes down with increasing velocity. Coulomb's Law of Friction says it's independent of velocity, although that's just an empirical approximation. For the sake of argument, let's assume that it holds.
So, looking at the forces, the decceleration is proportional to the disc diameter. That's it. If I knew the coefficient of friction, the disc size, and the mechanical advantage of the lever/caliper, I could tell you how long it'd take to stop for a given force on the lever.
But, you can also look at the energy of the system. The above says that for a given force on the lever, a bigger disc will slow the bike down faster. That is, the bike will lose kinetic energy faster, and this is mostly being lost to heat in the brakes. A larger disc has a larger circumference and so for a given rotational speed, has a higher slip velocity between pads and disc. The rate at which friction generates heat is proportional to slip velocity and frictional force, so a bigger disc does indeed generate more heat for a given force and bike speed.
So, bigger disc means higher slip velocity, meaning greater generation of heat for a given frictional force meaning faster loss of kinetic energy. i.e. a more powerful brake.
Just two different ways of looking at the same problem.
There are also benefits in terms of heat dissipation from bigger discs, but that only means you can use the brakes for longer, rather than making them more powerful.
torque/moment/leverage. Call it what you will.
For the same amount of force at the lever, the pistons will exert the same amount of force on the disc. this equates to the same amount of frictional force on the disc. Co-efficient of friction is constant, and is dependant on the materials involved (static/dynamic difference only occurs at point of initial movement, aka stiction). therfore frictional force is equal. If this firctional force is applied at at greater moment arm, it induces greater "negative" torque into the hub to oppose its rotation.
A 160 can be more powerful than a 180, but only if you pull the lever more.
Mods, can we stop this now please?
Materials involved indeed has a great influence
Mods can we carry on please. There is a perfectly decent discussion going on here, aside from the earlier slagging which none of the recent posters have been involved with.
So, bigger disc means higher slip velocity, meaning greater generation of heat for a given frictional force meaning faster loss of kinetic energy. i.e. a more powerful brake.Just two different ways of looking at the same problem.
I disagree, the moment will be greater also - so its a combination of greater moment, and more energy being burned off.
I'm sourcing the definitive maths as we speak.
I see your point and agree that there's a lot of that in this thread. On the other hand there's a difference between just saying "it's because of swept length" with no explanation and explaining a point based on concrete laws of physics to back it up.
What he said. Thats why I used some simple equations to basically describe what other people had already said.
I was halfway through typing an answer and then PDW finally hit the nail on the head. You can't really say its due to the extra leverage [i]and[/i] extra heat disipation. Its two different ways to solve a problem.
Why close the thread? Its actually got less abusive as its gone on 🙂
It may have got less abusive, but the answer is a matter of physics which was established hundreds of years ago.
Yes, there is more heat build up on the bigger disc, due to the greater rotational speed, and that has an effect on the coefficient of friction as it will normally reduce with increased temperature. But assuming we're talking about a modern disc brake with good materials and at realistic bike speeds, the single greatest influence on braking power, assuming identical caliper/lever seup, materials, linear speed, bike and rider combined weight, is rotor size.
End. Of.
Nobody is disputing that, the question is why.
The answer is a combination of:
a) mechanical advantage due to leverage
b) swept length.
The reason why people are discussing it is to get a handle on understanding why this is. I fail to see why you find this irritating.
You can't really say its due to the extra leverage and extra heat disipation. Its two different ways to solve a problem.
Well you can say its down to extra heat dissipation only - but you cannot say its down to extra leverage only.
What I mean is the combination of leverage and swept length combine to make extra heat dissipation.
I disagree, the moment will be greater also - so its a combination of greater moment, and more energy being burned off.
But isn't that like saying the increase in speed of an accelerating car is somehow a sum of energy transfer from potential (in the unburnt fuel) to kinetic (piston movement) [i]and[/i] the driving force the wheel applies to the ground?! This doesn't make sense, they are two different ways of solving the problem and, as far as I can see cannot be combined. I'm fully open to be proved wrong and learn something though...
Ah your talking cars now unfortunately having bigger rotors on a car isn't always beneficial....