[Closed] Why do bigger discs brake better?

Leverage. A bigger rotor means that the caliper and braking surface are further from the centre of rotation (axle).

turning moments.

try to push a door open with one finger near the hinges, then try it again with the same force, but your finger near the handle.

Er, FOG, did you skip physics classes at school?

omin, our physics teacher was nicknamed Beaky,never understood a word he said!

my (teacher) dad's nickname at skool was also Beaky, or Captain Beaky from his more reverent pupils. He didn't teach physic though.

Back on topic, as above it's about leverage. And the other thing is the larger rotor conducts more heat away from the pads and disspiates/radiates it out to the air more efficiently, meaning it is a bit further down your enormo alpine descent before your brakes overheat and stop working.

Similar principal to Torque=Force*Radius

Don't forget you'll lose some modulation too.

Turn your bike upside down and spin the front wheel slowly.
Stop it by poking your finger through the spoke near the rim.
Spin it again, this time stop it by poking your finger through the spokes near the hub.

Now do you get it ?

Don't forget you'll lose some modulation too.

Err no you won't.

If anything it'll be better as you won't have to pull as hard to get the same braking force, making it easier to control.

205mm discs front and back for me always. Miles better!

Haze - Member
Similar principal to Torque=Force*Radius

Thought it might be, I was relying on memory from years ago and thought it may be a little less straightforward...

Because they look moto

So what we really need then is for someone to invent a brake that works on the rim!

So what we really need then is for someone to invent a brake that works on the rim!

Buell m/c's do or did this. Good idea, but never really that good when I tried them.

superfli - Member
> So what we really need then is for someone to invent a brake that
> works on the rim!
Buell m/c's do or did this. Good idea, but never really that good when I tried them.

Whoosh!

Even better would be one that works on the surface of the tyre.

[img] [/img]

avdave2 - top idea.

i am i right in thinking that with larger discs they just need less force (lever pulling) to apply the same stopping power as smaller ones?

I'm just beginning to get the whole 29er thing now...

So what we really need then is for someone to invent a brake that works on the rim!

Buell m/c's do or did this. Good idea, but never really that good when I tried them.

you mean like this

[img] [/img]

I always thought it was because a larger disc had a greater circumference so therefore increased the breaking area per wheel revoloution. Does this not play arole in the process?

isn't braking area constant ie the size of the pads? Altho I'd guess the longer circumference helps with heat dissipation

L_P 8pot and giganto rims? that just looks silly.

yes pad area [and force applied ???] is constant. The area that the pads cover, for one turn, will be greater with a bigger rotor than a smaller one due to larger circumference. Not sure whether leverage or greater surface area is the main factor tbh

leverage is the main factor. Simply put - at a greater radius, less force is required to decelerate the wheel by the same amount.

So - if it took 10 Newtons to cause a deceleration of 10 metres per second per second at a radius of 10cm, then it would only take 5 Newtons at a radius of 20cm to cause the same deceleration.

modulation does decreace with a bigger rotor,

Say you can apply a force of 1-10N wit your finger

Well with a 6" rotor you might need 10N to lock the wheel.

With a 12" rotor you only need 5N, so you've halved the 'modulation'.

I'm not disputing any of the physics, but on a bike surely isn't it modulation and heat dissipation that are the benefits of bigger rotors.

I always thought it was because a larger disc had a greater circumference so therefore increased the breaking area per wheel revoloution. Does this not play arole in the process?

Only with respect to heat dissipation, it's the moment of the force that has an effect (the distance from the axis of rotation).

isn't braking area constant ie the size of the pads?

Doesn't matter - it's a misconception that the surface area in contact has an effect on the force of friction between two solid surfaces.

cheers all

Doesn't matter - it's a misconception that the surface area in contact has an effect on the force of friction between two solid surfaces.

Or am I missing something blindingly obvious?

"it's a misconception that the surface area in contact has an effect on the force of friction between two solid surfaces."

Is that why skinny tyres are just grippy as fat tyres?

At 20 mph, the linear velocity of the disc where it contacts the pad is greater on a 8" disc than a 6" disc. It travels 33% faster, and the circumference increases by 33%.

Ignoring leverage, I can never work out what additional effect this has. Does this increased speed mean more friction?

Perhaps it heats up quicker, but the increased surface area improves heat dissipation by a bigger factor?

now I am back to confused 😕

Bigger tyres = more grip, hold something between my hands = more grip than holding it between fingertips.

Bigger tyres [u]run at lower pressures[/u] give larger contact patches and more grip. Dunno re the finger one!

Ignoring leverage, I can never work out what additional effect this has. Does this increased speed mean more friction?

Does it make any difference? The pressure (=force applied/pad area) applied will be the same (so there won't be any additional friction assuming the same pad material); a bigger disc may have the happy side effect of dissipating heat faster due to having a bigger surface area though.

Leverage is what's important, I think. My mechanical physics knowledge is a bit rusty, unlike the gas laws/flow etc...

Andy

To add to the confusion......if you are being pedantic then it doesn't matter the area of the pad when thinking about contact area. A large pad with a given force will not have any more area physically in contact with the disc compared to a small pad with the same force applied. The smaller pad will have more intimate contact therefore meaning the surface area is identical.

A large pad with a given force will not have any more area physically in contact with the disc compared to a small pad with the same force applied.

Counterintuitively, this is correct (Amonton's 2nd Law) - in which case, why bother with 4 or 6-pot brakes? Why not use ever-bigger rotors with normal 2-pot calipers?

Andy

friction has nothing to do with surface area, only coefficient of friction and normal force. However as you know heat build up/contaminants do affect the friction coefficient, so a large area pad increases the likelihood of contact and is more efficient as dissipating heat.

*shakes head and walks away*

Oh Amonton's 2nd Law, why didn't you say so in the first place of course 🙄

Bigger tyres run at lower pressures give larger contact patches and more grip

Amonton's 2nd Law

God Bless Wikipedia 😛

Andy

You gotta love it. It also gives us this

Contrary to earlier explanations, kinetic friction is now understood not to be caused by surface roughness but by chemical bonding between the surfaces.

L_P 8pot and giganto rims? that just looks silly.

😯 Wash your mouth out young man (woman?).

Actually very clever mounting the disc on the rim, moves all the braking forces off the hub and onto the rim. Means the spokes can be thinner and lighter as they have much lower forces applied to them. Lighter wheels is good 😀 .

good idea, but it's on a buell, ugh!

tyre pressure/size wise - for a given pressure the contact area will be the same.

a 3" tyre at 50 psi, under 100 pounds weight will have a 2 inch square footprint.

a 1" tyre, at 50 psi, under 100 pounds weight will have a 2 inch square footprint.

the footprints will just have a different shape (same for thin/wide car tyres). the shape of a footprint [i]may[/i] affect the grip, depends on the tyre.