Sorry stw, a bit of a dumb question, but when it comes to hydraulic disc brakes, what gives a brake its power?

If I have a set of xt levers and match them with saint calipers, would these me more powerful than let’s say lx calipers on the same xt levers?

I’m assuming that the stopping power is dependant on the caliper, am I right or is there more to it?

Google “master and slave cylinders”, then bask in the glory of self directed learning…
🙂

DrP

BTW – the power comes from your fingers, the lever and cylinders transfer the power..

Combination of mechanical and hydraulic advantage. Lever length for example. Also, a bigger slave will have a similar effect to a smaller master.

Too powerful and you’ll sacrific feel and pad clearance.

In the simplest form, if you want more power just increase the diameter of the rotor.

That will make the single biggest difference.

More rotor = more area for the pads to work on = more friction.

In the simplest form, if you want more power just increase the diameter of the rotor.

That will make the single biggest difference.

More rotor = more area for the pads to work on = more friction.
Not really. Bigger rotors are more powerful because they generate more braking torque due to the larger radius, just like using longer spanner makes it easier to undo tight bolts. The surface area has almost no effect.

In the simplest form, if you want more power just increase the diameter of the rotor.

That will make the single biggest difference.

More rotor = more area for the pads to work on = more friction.

Swing and a-miss.

Now I think about it smaller surface area is actually better for brakes as you’re applying the same force to a smaller contact patch, resulting in more pressure.

Think pushing a drawing pin into wood with your hand vs trying to push a chopstick in with your hand.

#EDIT: but obviously with disks you need ample heat dissipation etc which is why most disc brakes are the size they are rather than skinny 5mm wide or somthing.

BTW – the power comes from your fingers, the lever and cylinders transfer the power..

No your finger supplies a force, this creates a pressure in the fluid, which then exerts more force through the larger slave pistons. This force acts on the brake pads and rotor, force * coeficienct of friction = moment (force) on the rotor (it’s not a torque, a torque is the moment of a couple, i.e. includeds the reaction force of the axle/bolts). Force x distance = energy, energy/time = power, distance/time = speed, therefore force x distance / time = force x speed = power.

So technicaly the power comes from the energy you put in pedaling.

Now I think about it smaller surface area is actually better for brakes as you’re applying the same force to a smaller contact patch, resulting in more pressure.

It’s not though, the actual equation derives via

force (friction) = pressure x area x coeffieicent

which obviously cancels to force(normal) x coefficient.

I’ve some silly lightweight rotors with huge cut outs, the power is notably down compared to conventional designs.

Now I think about it smaller surface area is actually better for brakes as you’re applying the same force to a smaller contact patch, resulting in more pressure.

Ah, so THAT’S why skinny tyres are grippier than fat ones 😉

“Power” is open to interpretation too, loads of people think the M810 Saint is more powerful than the new one- the actual maximum braking force of the M820 is higher but because it has a less immediate impact when you start to brake, it doesn’t feel as immediately strong. I suppose it’s a bit like a torque curve, you feel the area under the line rather than the line itself.

Trimix – Member

In the simplest form, if you want more power just increase the diameter of the rotor.

That will make the single biggest difference.

More rotor = more area for the pads to work on = more friction.

Nope… Increasing the diameter alone doesn’t change the contact patch at all, but does increase leverage and cooling surface.

Changing the land/sea ratio of the rotor (by using one with less holes, frinstance) has a more complicated effect, because yes you can increase the contact patch and you might expect that to increase friction, but the pressure is spread more widely. The relationship’s not going to be simple enough to just say “X is better” and have it work for all brakes.

I love STW forum….

Now I think about it smaller surface area is actually better for brakes as you’re applying the same force to a smaller contact patch, resulting in more pressure.

Swing and a-miss.
It’s not about the pressure applied, it is the friction between the rotor and brake pad surfaces.
Fr = ?R where ? is the coefficient of friction, and R is the reaction force of the rotor on the pads.
Area has no effect.
Decreasing the area applies more pressure but less contract area. Fr remains the same.
Increasing the area reduces pressure, but increases contact area. Fr remains the same.

BTW – the power comes from your fingers, the lever and cylinders transfer the power..

No your finger supplies a force, this creates a pressure in the fluid, which then exerts more force through the larger slave pistons.

You knew what I meant, pendant!

DrP

Just to clear this up…

The combination of slave and master cylinders plus pads and disks multiplies and transfers the power of your hands to apply it effectively at the disc.

If you want to improve disc brake power you can: –
1) increase the size of the disk to increase leverage and to reduce heat build up and speed cooling through greater surface area
2) reduce the transference of heat from discs and pads to the brake fluid. The cooler this is the better the brake will run
3) increase the friction between disc and pad by using a pad that increases friction especially at higher temperatures – replacing organic with sintered pads would be an example
4) increase the friction between disc and pad by increasing pad surface area by using a larger piston or more numerous pistons to push a larger pad
5) improve the mechanical efficiency of the interaction between lever, master and slave cylinders. Change the pivot point for actuation of the master piston etc…

An even dumber question maybe but how does the power of a brake beyond a certain point affect what happens in the real world?

What I mean is, presumably for each brake there’s a certain force needed for the pads to grip the disc so hard that the wheel locks up. At that point, depending on speed, conditions, etc, etc, the bike either stops or skids.

Let’s assume you brake hard enough to skid, what’s the effect of more ‘power’ if, beyond a certain amount of power, the tyres are going to lose grip first? If it only took 50% of the brake’s power to do that, what’s the rest for?

There was some connection to ABS on car brakes I was trying to get in here but I’ve totally lost my thread (yes, really) so can’t remember what it was.

^^^^^^^ I know that post is full of holes now that I’ve read properly it but I think there’s a gist and valid question somewhere in there 😀

What I mean is, presumably for each brake there’s a certain force needed for the pads to grip the disc so hard that the wheel locks up. At that point, depending on speed, conditions, etc, etc, the bike either stops or skids.

You want all the KE of your bike to turn into heat or ablation (pad wear), at the limit that your tyres can handle. Think ABS. Otherwise you may as well put a stick in your spokes!

boombip – once the wheel’s skidding, there’s no difference. Of course, the faster you are moving (more rotational energy in the wheel) the harder it is for your brake to lock the wheel into a skid.

If I remmeber correctly, the best braking point i.e. slowing the bike down in the shortest time possible, is just before you skid. The closer you are to the wheel ‘locking’, the closer you are to stopping in the shortest distance or time.

So how do rim brakes work ?

@06awjudd @thisisnotaspoon

A PRESSURE is the result of a FORCE being applied to a specific cross-sectional area, and is defined as FORCE per unit AREA, as in POUNDS per SQUARE INCH. For example, if a downward FORCE of 1000 pounds is applied evenly to a square plate of steel which measures 2″ by 2″ (4 square inches of area), then the PRESSURE applied to that block (Force per unit AREA) is determined by dividing the FORCE (1000 pounds) by the AREA (4 square inches), which is 250 pounds per square inch (“psi”).

If the same 1000 pound FORCE was applied to a plate which measured 2″ x 4″ (8 square inches), then the PRESSURE would be reduced to 125 psi because the area of the plate doubled. The same force is being applied over a greater area, resulting in a LOWER force per unit area.

The friction force can change because it relies on the normal force. The normal force can be increased by increasing mass or just by exerting a force so that the two surfaces are pushed together more.

This was my understanding of the physics at play (Written more clearly than I could at 5pm on Friday)

Can someone explain how this view is wrong?

You said yourself, you’re applying the same force. And it’s the force that matters.

So to sum up.

Jedimaster the force may be with you or it could be because of the pistons.

@scuzz @makecoldplayhistory

Cheers, guys – got the conversion of energy point and it all clicked. The fact that I had to ask probably reveals that I don’t ride fast enough to appreciate the benefits of stronger brakes 😆

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Can someone explain how this view is wrong?

Reducing the surface area of a pad would increase the pressure. The force exerted on the rotor remains constant.
While more pressure is exerted, it is exerted over a smaller area, so the net effect is zero.

The friction is calculated using Fr = ?R

? is the coefficient of friction between two surfaces (eg the steel of the rotor, and the brake pad). It is constant.
R is the reaction force exerted by the rotor on the pads (or vice versa)
Changing the surface area changes neither.
Therefore Fr remains the same, irrespective of pad size.

Hope this cleared it up.

The pressure of the fluid applies a force relative to the cross sectional area of the piston (f=pressure x area). But that’s only half the story. That force is then applied to the pads. The force creates friction between the pads and disc relative to the area of the pads. That friction then creates the braking torque which is also dependant on the diameter of the brake disc.

So the variables that determine braking torque are fluid pressure, piston area, pad area, disc diameter. And the thing that regulates all of this is the size of the riders balls!

The force creates friction between the pads and disc relative to the area of the pads.

No, that’s wrong, the area of the pads has no affect on the friction.

ahh I’m glad this “More rotor = more area for the pads to work on = more friction.” got corrected before I got here. It’s friday and I want to turn off my brain.

You all forgot the lever, altering pivot point alters leverage applied to master cylinder piston and where you squeeze on lever alters leverage as well.

Wow, just wow! Thank you all so much for your replies! I never thought that there was so much to it when it comes to hydraulic disc brakes lol!

I’m currently running a set of 8″ front 7″ rear rotors so I don’t think that I will be able to increase the power that way. I will have to experiment with different pad types I guess!
Thanks again!