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I know 180mm rotors have more leverage than 160mm, but how can that make a difference on a bike?
It's to do with the fact they have more leverage!
For a given brake, force at calliper is constant, therefore the further from the hub you apply that force, the more effective the brake. They also disipate heat better as there's a larger mass/surface area. The trade off is less modulation and more weight.
I think you have answered your own question
Someone failed their Physics exam 😆
Dumbass
For the same reason that it is easier to undo a nut with a 200mm long spanner than a 50mm long spanner.
I'm sure the extra surface area for heat dissipation plays a small role.
The rotors are not 'more powerful'
Power comes from the caliper.
Larger rotors have more 'leverage' if you like, and turn the same pressure at the caliper into more braking force at the rim*
* I think! 🙂
Or to explain the same thing in a different way, force x distance = work done. You apply a fixed force to the brake. If the rotor is bigger then it's travelling further in a fixed time so you're converting more energy to heat so you slow faster for the same brake pressure.
"I know 180mm rotors have more leverage than 160mm, but how can that make a difference on a bike?"
Um... Why would it not make a difference on a bike? Do bikes occupy some parallel universe where normal physics doesn't apply?
Anyway, words 3-9 of your question constitute the answer.
Dumbass
Talk about saying how it is! lol
it does have more leverage, but there is less friction on the disk, as 2 surfaces travelling over each other have more friction when the speed differential is less.
larger rotors heat up less though
Why are 180mm rotors more powerful than 160mm?
Well.... the rotors are not more powerful.
Neither is the calliper.
So to answer your question they arn't.
So if you were clever enough to invent a system that used the rim as the braking surface then you'd have a brilliant brake.
Yeah, until it either wore out, got covered in crap, etc 😉
but there is less friction on the disk, as 2 surfaces travelling over each other have more friction when the speed differential is less
Not true - friction interface is an incredibly complicated subject and simple statements like this are mis-leading 😆
invent a system that used the rim as the braking surface
Now dont quote me on this but wasn't this 'invention' used before discs on bikes. I'm sure i used these things called 'v-brakes' and also some 'magura's' and before all of that i used 'cantilevers' but i'm quite positive that my disc brakes worked better than all of them 🙂
There is no extra leverage from the disc size nor is there any extra leverage from the calliper. These will always remain the same whatever size disc you run.
Picture this two bikes side by side. One with a 180mm disc, the other with 165mm. If both bikes started braking at the same time the reality is the larger disc would offer more surface area to the pad over the same distance covered under braking. Simple!
I'm sure i used these things called 'v-brakes' and also some 'magura's' and before all of that i used 'cantilevers'
No you couldn't have done - you wouldn't have stopped by now.
Birch1983 you are a dumbass too
Birch - FAIL!
Picture this two bikes side by side. One with a 180mm disc, the other with 165mm. If both bikes started braking at the same time the reality is the larger disc would offer more surface area to the pad over the same distance covered under braking. Simple!
Wrong 🙂 Have a think about it.
it's called mechanical advantage
Birch, harry tonight’s reading:
[url= http://en.wikipedia.org/wiki/Mechanical_advantage ]http://en.wikipedia.org/wiki/Mechanical_advantage[/url]
So if you were clever enough to invent a system that used the rim as the braking surface then you'd have a brilliant brake.
Yeah, until it either wore out, got covered in crap, etc
The best solution to this issue is to combine the two. You have a secondary rim near the actual rim, which you use as a braking surface. Erik Buell does this and calls it perimeter braking.
You have a think about your comment. Same speed initially and same total stopping distance. The larger disc will expose more surface area! Study physics.
can't believe some of the nonsense in here!! worse than the logo thread!
Or simply read this..
http://www.bikeradar.com/gear/article/buyers-guide-to-disc-brakes-45/
And pay particular attention to this....
Power varies with each calliper and its pad surface/leverage, but 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.
NOW GO STUDY PHYSICS DUMBASS!
What we really need is a disk rotor the same size as the wheel.
[i]You have a think about your comment. Same speed initially and same total stopping distance. The larger disc will expose more surface area! Study physics. [/i]
hi dylan
flow - Member
Dumbass
lol.
+1
So all you guys are saying that you use brakes?
😉
Yep, and Harry333, Birch1983, Peterpoddy and Clubber are thick as shite
Birch - do the pads grow in size? Does their surface area increase? LOL
Okee day, basic stuff, between a 180 and 160 braking system look at what stays constant and what changes.
[b]Constants[/b]
Force (the calliper applies the same amount of force regardless of rotor size)
Friction Surface (brake pads and the area the contact area doesn't change)
[b]Variables[/b]
Distance of the contact points (pads to disc) to the axle (moment of force)
(We are going to ignore thermal properties)
And then if you get some cheese and add that to the rotor you can fill your head with it.
So if you were clever enough to invent a system that used the rim as the braking surface then you'd have a brilliant brake
and if it were hydraulic it would be even better!
oh, they did that already - Magura HS33s 🙁
So if you were clever enough to invent a system that used the rim as the braking surface then you'd have a brilliant brake.
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.
All brakes are the same. That's why I have 203s on my rigid singlespeed and 160s on my Alpine 160 (the 160 is Orange telling you what brake to use).
I may be lying.
PJay- the coefficient of friction between pad material and disc material is different to those between brake block material and Al rims. The additional weight of a steel wheel needs to be taken into account as well, although it's relative longevity compared to an Al braking surface also needs to be taken into account.
Are ceramic rims still available?
Can one not have a grown-up discussion without some kiddy calling someone else a dumbass? 🙄
Yep, and Harry333, Birch1983, Peterpoddy, Clubber, [i][b]IGM and PJay[/b][/i] are thick as shite
Had to re write the thick as shite list
Buzz-lightyear
This is primary school stuff, hardly a grown up question.
If you don't like it you know what to do.
somebody mention torque
HS33s are pretty damn powerful - powerful enough to bend the seatstays on some bikes - are discs really more powerful?
Tiger6791 - MemberOkee day, basic stuff, between a 180 and 160 braking system look at what stays constant and what changes.
Constants
Force (the calliper applies the same amount of force regardless of rotor size)
Friction Surface (brake pads and the area the contact area doesn't change)Variables
Distance of the contact points (pads to disc) to the axle (moment of force)
(We are going to ignore thermal properties)And then if you get some cheese and add that to the rotor you can fill your head with it.
You forgot the thing that actually has the most effect, the increase in the swept length of disc per revolution.
Flow - read what is written before you pass judgement. You just added yourself to the idiots list.
Err nope, you are definitely a dumbass.
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.