Why are 180mm rotors more powerful than 160mm?

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 and 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) and 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….

Also some brakes don’t need to dissipate heat to work better

Well you can say its down to extra heat dissipation only – but you cannot say its down to extra leverage only.

Yes you can. Where is the gap in my analysis of the forces involved?

Ah your talking cars now unfortunately having bigger rotors on a car isn’t always beneficial….

I just thought I’d throw a spanner in the works and confuse things further by turning the conversation to an accelerating car rather than a decelerating bike 😀

Phototim – Member
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) and 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…

I see what you saying, sorry that my fault in not defining it more strictly which I did in the next post.

What I mean is the combination of leverage and swept length combine to make extra heat dissipation.

The gap in your analysis is that it is not a static problem, its dynamic.

I’ve been corrected by my colleague in the office, swept Area is actually the correct term.

If you think swept area does not have an effect then can you explain why multi piston calipers – with bigger pads work better than single piston calipers with smaller pads?

They don’t necessarily work better bigger pads can have a greater effect more pistons doesn’t mean more power

yeah sorry the pistons are only there to load the pads evenly. Its all about pad area.

Looking at this another way. If you take your brake pad and place it on a flat piece of steel with a constant force, and push it along 500mm, it will take x amount of energy. Then you have to push it a further 60mm. This will take extra energy. No extra leverage or machanical advantage I can see, but requires more energy. I think this is what toys is getting at.
I’m no engineer so may be talking rubbish.

If you think swept area does not have an effect then can you explain why multi piston calipers – with bigger pads work better than single piston calipers with smaller pads?

They’re not inherently more powerful. What they do offer is better heat dissipation, as the heat is generated evenly across a bigger area of pad.

Frictional force is independent of area, and it’s the force that determines how fast you slow down.

Bigger rotor’s are more expensive, and expensive things work better.
It’s physics.

Looking at this another way. If you take your brake pad and place it on a flat peice of steel with a constant force, and push it along 500mm, it will take x amount of energy. Then you have to push it a further 60mm. This will take extra energy. No extra leverage or machanical advantage I can see, but requires more energy. I think this is what toys is getting at.

Yes. The pad on a bigger rotor covers more distance on each wheel rotation, so for a given force it produces more heat, meaning you must have slowed down more (you’ve lost more kinetic energy).

This because of (not in addition to) the additional leverage of a bigger disc.

If analysis of the forces didn’t yield an answer that was consistent with analysis based on conservation of energy, it really would be news worthy 😀

Tony, that is exactly it. Anyway I have gone back to first principles to solve this which is the only way to do it.

Work done in a rotational system is the integral of Torque through its rotational distance. (Equation culled from wikipedia)

Or Power is given by torque x angular velocity.

Therefore power is dependant on the torque (which is proportional to the disc radius) and angular velocity (which is proportional to the disc radius) .

Therefore the effect of disc radius is to increase the torque, and the angular velocity which multiplied together increase power.

Frictional force is independent of area,

true

and it’s the force that determines how fast you slow down.

not quite. its the (force x radius) x the angular velocity

If analysis of the forces didn’t yield an answer that was consistent with analysis based on conservation of energy, it really would be news worthy

Ahh but you have forgotten about the time aspect, your analysis is static. Go back and do it again.

Or Power is given by torque x angular velocity.

Therefore power is dependant on the torque (which is proportional to the disc radius) and angular velocity (which is proportional to the disc radius)

BZZZZT.

Angular velocity is rate of change of angle – it’s not proportional to radius.

So fixing that, power is dependent on torque (proportional to disc radius). Which gets you back to the same conclusion as mine above.

Ahh but you have forgotten about the time aspect, your analysis is static. Go back and do it again.

That’s why I said “consistent”. To fully confirm the equivalence of the two approaches you do indeed need to take into account time.

Arse, there is more to come..

AHAHAHAHHAHA, I CAN’T BELIEVE THIS THREAD IS STILL GOING, SORT IT OUT YOU BUNCH OF **** LOSERS!!!!!

Don’t forget that a bigger disc is heavier, completely negating any stopping advantage it may offer….

DrP

(you know I’m playing, right…?!)

Double arse, I think I’ve convinced myself of your theory now pdw.

Definitions:

1) r = disc radius
2) Ff = frictional force = Force applied by pads x Coeef Friction
3) SL = Swept length = circumf of disc (or effective circumf of disc which might be half way across the friction surface) for one rev is 2 x pi x r
4) T = torque = Ff x r
5) Wf = Work done by Friction = Ff x Sl
6) Av = angular velocity – for 1 rev is 2pi/t
7) P= Power = T x Av
8,) Wt = P x time = work done by torque.

Wf = Ff x 2 x pi x r
Wt = Ff x r x (2 x pi)/t x t

The t’s cancel and so work done by friction = work done by torque.

Funny how as soon as you sit down and write it out, it all makes sense..
Doh.

The friction on the disc x the dia is what makes the torque.The work is that torque (or friction) applied for a period of time ( which equates to a distance travelled over the disc)

That was a good bout, well played everybody, well played. Good to get the brain working once in a while.

(high five pdw)

In future I’m saying anything until I’ve proved the maths to myself..

bigger disks are more powerful because they cost more. if they weren’t more powerful nobody would bother paying the extra would they? 💡 😉

how do you guys find helmets big enough to fit all them brains in?

i use a modified dustbin

I used to brake my bmx with my trainers, would a bigger shoe-size have provided more torque i wonder?

I did the same sunnrider – did you use sintered or organic compound soles?

I think I need a smaller lid after that..

In fact I have realised that the swept length thing is a massive red herring. As assuming everything else is equal the increase in torque will quite obviously reduce the angular displacement to stop the bike by a factor which will make the swept length exactly equal for any size disc.

As assuming everything else is equal the increase in torque will quite obviously reduce the angular displacement to stop the bike by a factor which will make the swept length exactly equal for any size disc.

But to sweep a given area on a disc with a greater circumference means the wheel hasn’t rotated as much…

I don’t understand your statement? Can you explain it a bit more?

edit actually i think i do.

Thats exactly what I mean – “reduced angualr dispalcemnt” means “hasn’t rotated as much”

Yes I think I just reversed your statement, probably because in true STW style I hadn’t read it properly.

Yes. The pad on a bigger rotor covers more distance on each wheel rotation, so for a given force it produces more heat, meaning you must have slowed down more

But if you have stopped in a shorter distance with the bigger rotor then does that not negate the difference in distance travelled by the rotor?