[Closed] Why are 180mm rotors more powerful than 160mm?

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 [b]Area[/b] 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 [b]because[/b] 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)

[img] [/img]

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,

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

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 ****ING 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?

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?

edit. I did agree but I've gone fuzzy headed. Bollocks, I need to think about this more..

pdw whats your answer to this?

its angular delta will remain the same as the wheel

composite pro, I had hoped that was implicit in what I said before..

Bigger distance with a smaller rotor, or shorter distance with a bigger rotor, I think for a given force at the calipers, the pads will have covered the same distance over the disc, as you've lost the same amount of kinetic energy either way.

this is the same conclusion that I came to above and the way you have written it makes it seem sensible. If its shorter distance (or less revolutions) to come to a stop this means you come to a stop faster, this is exercising me slightly..

If it takes one rev to stop and its doing 1 rev a second then the equation I listed above works fine, the time t which divides through the 2 pi radians is equal to the time t for the work done.

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

So for a 180 disc the term r goes up and hence I had assumed the 2 pi is multiplied the ratio of the radii (80/90) = 0.88, but actually I think both the number of radians goes down and the time goes down so its probably not the same distance travelled..

Am I making sense?
I think this is possibly only solved by integration.

the revolution (or rotation) is the same on either sized rotor (160 or 180), unless I am missing something drastic, which is quite possible.

it's summink to do with physics innit..?

Heechee. If a larger disc has more torque then it will stop your bike faster. Agreed? So if it stops faster the total number of revolutions from when you apply the lever till when it stops is less. Agreed?

I agree with the larger diameter increases torque. Not so sure the revolution or rotation is any different. I am no expert though, and would take no offense at being proven wrong.

YAWN!!!

Easy explanation is leverage applied to the disc verses the wheel size

Put massive 19" wheels on your chav corsa the brakes feel rubbish because the big wheel has more leverage against the disc go back to a smaller wheel 15" and the leverage on the disc is less so the brakes feel more powerful

Swap this around to fitting a bigger disc for the same size wheel the leverage is the same at the wheel but the bigger disc doesn't work as hard for the same braking effort at the caliper

Effectivly giving you more available power only to the point of how much grip you have before you lock up

Also bigger discs get rid of the heat better giving better braking as everthing heats up the disc can keep cooler as thereis more material to disapaite the heat

Didn't read all four pages so I may have missed the point

Toys - your equation shows that work done in a revolution is proportional to disc radius. Or to look at it another way, it's proportional to the swept distance i.e. The distance that the pad covers around the circumference of the disc. To stop a bike, you need to do the same amount of work however you do it, so whatever the disc size, the pads need to sweep the same distance. Which means more revolutions of a smaller disc i.e. A longer stopping distance.

pdw yeah I understand that, but look at the time taken to stop. In the work done equation there is a time term t, if you reduce the number of rotations by using a bigger disc, then it takes less time (or increase by using a smaller disc.)

All I am saying is that at first I thought I could equate the work done equations for a 160 and a 180 disc to find the reduced number of revolutions, (which is obviously the ratio between the disc radii) but actually you have two unknowns, the change in revolutions and the change in time, so my crap maths doesnt work. Needs integration, or possibly an approximation method. Its obviously too dull for anyone else but it has just interested me a bit..

Jebus! That's 5 min of my life I won't get back, and I was only skimming too!
They stop you quicker, cos they're bigger.
Night night.