Brake pad efficiency

I've been looking at the difference between disk and rim brakes in terms of efficiency (energy from hand to deceleration torque). It looks like it depends upon friction losses in the delivery system, the spring constant of the deliver system (how much the calipers are bending, etc.), the friction coefficient of the pad material, and the radius at which it is applied. Since both disk and rim brakes work by closing calipers the delivery system is irrelevant since you can always implement it for either disk or rim brakes.
The radius of application is a whopping disadvantage for disk brakes - 25% reduction in efficiency. But disk brakes do have the advantage of freely choosing pad and rotor material. However coefficient of friction (mu) appears to be somewhere between .4 and .6 for reasonable choices - not nearly enough to compensate.
But, but, but, mu is not a measure of efficiency. If you have a hard pad, you can get 100 newtons of force into it with less energy than a soft pad. You don't have to work squishing it first... However, I cannot find any easy reference on brake pad efficiency. It's all coefficient of friction. My suspicions is that it's pretty much the same for rim and disk brake pads, otherwise marketing would be all over it.
So for the better engineers/physicists among us, is my analysis correct? Anyone know of any study of brake pad efficiency?

On Fri, 29 Jan 2016 12:47:04 -0800 (PST),
wrote:
I've been looking at the difference between disk and rim brakes in
terms of efficiency (energy from hand to deceleration torque).

Ah, I miss Jobst.

On Friday, January 29, 2016 at 12:47:08 PM UTC-8, wrote:
I've been looking at the difference between disk and rim brakes in terms of efficiency (energy from hand to deceleration torque). It looks like it depends upon friction losses in the delivery system, the spring constant of the deliver system (how much the calipers are bending, etc.), the friction coefficient of the pad material, and the radius at which it is applied. Since both disk and rim brakes work by closing calipers the delivery system is irrelevant since you can always implement it for either disk or rim brakes.
The radius of application is a whopping disadvantage for disk brakes - 25% reduction in efficiency. But disk brakes do have the advantage of freely choosing pad and rotor material. However coefficient of friction (mu) appears to be somewhere between .4 and .6 for reasonable choices - not nearly enough to compensate.
But, but, but, mu is not a measure of efficiency. If you have a hard pad, you can get 100 newtons of force into it with less energy than a soft pad.. You don't have to work squishing it first... However, I cannot find any easy reference on brake pad efficiency. It's all coefficient of friction. My suspicions is that it's pretty much the same fo
r rim and disk brake pads, otherwise marketing would be all over it.
So for the better engineers/physicists among us, is my analysis correct? Anyone know of any study of brake pad efficiency?

Heat dissipation is another variable that needs to be considered. Also how much speed and over what amount of time will the brakes be applied for and amount of speed reduced.

On Fri, 29 Jan 2016 12:47:04 -0800 (PST), wrote:

I've been looking at the difference between disk and rim brakes in terms of efficiency (energy from hand to deceleration torque). It looks like it depends upon friction losses in the delivery system, the spring constant of the deliver system (how much the calipers are bending, etc.), the friction coefficient of the pad material, and the radius at which it is applied. Since both disk and rim brakes work by closing calipers the delivery system is irrelevant since you can always implement it for either disk or rim brakes.
The radius of application is a whopping disadvantage for disk brakes - 25% reduction in efficiency. But disk brakes do have the advantage of freely choosing pad and rotor material. However coefficient of friction (mu) appears to be somewhere between .4 and .6 for reasonable choices - not nearly enough to compensate.
But, but, but, mu is not a measure of efficiency. If you have a hard pad, you can get 100 newtons of force into it with less energy than a soft pad. You don't have to work squishing it first... However, I cannot find any easy reference on brake pad efficiency. It's all coefficient of friction. My suspicions is that it's pretty much the same for rim and disk brake pads, otherwise marketing would be all over it.
So for the better engineers/physicists among us, is my analysis correct? Anyone know of any study of brake pad efficiency?

Probably not formal measurement of brake pad effectiveness but
certainly there are a multitude of informal evaluations.

After all, how scientific does one need to be to discover that brake X
will skid the tires on dry pavement while brake Y won't stop when it
is wet?

From reading this site alone it appears that Disk Brakes with any sort
of pad will always stop when they are wet and muddy and while rim
brakes are far less effective in the wet that using "Coolstop" pads
improves their effectiveness immensely.
--

Cheers,

John B.

I've been thinking some more and realize I'm barking up the wrong tree a bit. The rider's hand grip energy is not translated into the deceleration torque. That's coming from the rotational energy in the wheel rubbing the pad. Instead, the rider is energizing the "spring" that constitutes the brake system. Once it's squeezed, you could, in theory, catch a ratchet and release your grip while the brake sat there scrubbing speed. In reality, you have to maintain a static force on the lever. I don't know biochemistry, but clearly maintaining more grip is more fatiguing than maintaining less, which is why higher MA levers feel so much better in actual use.
So it's back to a static force analysis. I'm pretty sure there's no way a harder brake pad in a disk brake could ever make up for the 25% reduction in torque.

It's all motivated by the oft made claim that disk brake provide more braking torque. It's even in the MTBR.com FAQ as a "generally agreed advantage".. Seems akin to declaring the earth is flat because lots of people say so....

On Fri, 29 Jan 2016 17:44:48 -0800 (PST), wrote:

I've been thinking some more and realize I'm barking up the wrong tree a bit.
The rider's hand grip energy is not translated into the deceleration torque.
That's coming from the rotational energy in the wheel rubbing the pad.

Nope. The kinetic coefficient of friction between the disk/rim and
the pads requires that a force normal to the direction of the
frictional drag be applied. This normal force comes from the rider
squeezing the brake levers. The frictional drag force is proportional
to this normal force as:
Frictional_drag_force = Normal_force * kinetic_coeff_of_friction
http://www.engineeringtoolbox.com/friction-coefficients-d_778.html

Ignoring hystersis, deceleration, inertia, heating, ablation, filth,
dirt, grease, water, grip strength, cable stretch, and all the other
variables that make such calculations a pain, if one knows the normal
force, disk diameter, rim diameter, and coefficient of friction, one
should be able to obtain a ratio of the rotational force to the lever
grip force. I hope.

The method is partly outlined he
https://www.sensorprod.com/news/white-papers/2010-03_ctb/wp_ctb-2010-03.pdf
Notice the hysteresis in the Fig 10 (bottom of last page) and the lack
of linearity, which should make comparisons rather awkward.

I tried to think of a way to easily measure the braking forces on a
functional bicycle instead of a test stand. Something like a force
sensor on the brake cables and a speed sensor. Generate a graph of
speed versus time at various brake lever grip pressures. That should
produce families of curves, where the fastest rate of deceleration at
a given lever grip pressure is the most efficient.

I'm feeling rather lousy due to a cold or flu, so I'll spare you my
bad math and deranged ideas.



--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

On Friday, January 29, 2016 at 2:47:08 PM UTC-6, wrote:
I've been looking at the difference between disk and rim brakes in terms of efficiency (energy from hand to deceleration torque). It looks like it depends upon friction losses in the delivery system, the spring constant of the deliver system (how much the calipers are bending, etc.), the friction coefficient of the pad material, and the radius at which it is applied. Since both disk and rim brakes work by closing calipers the delivery system is irrelevant since you can always implement it for either disk or rim brakes.
The radius of application is a whopping disadvantage for disk brakes - 25% reduction in efficiency. But disk brakes do have the advantage of freely choosing pad and rotor material. However coefficient of friction (mu) appears to be somewhere between .4 and .6 for reasonable choices - not nearly enough to compensate.
But, but, but, mu is not a measure of efficiency. If you have a hard pad, you can get 100 newtons of force into it with less energy than a soft pad.. You don't have to work squishing it first... However, I cannot find any easy reference on brake pad efficiency. It's all coefficient of friction. My suspicions is that it's pretty much the same for rim and disk brake pads, otherwise marketing would be all over it.
So for the better engineers/physicists among us, is my analysis correct? Anyone know of any study of brake pad efficiency?

If both can be squeezed hard enough to flip you on the head and kill you with head injuries, efficiency just doesn't matter. Controlling the hard braking is. The rest is technique. Or the amount of effort is comfort level.

On 1/29/2016 5:47 PM, Barry Beams wrote:
On Friday, January 29, 2016 at 12:47:08 PM UTC-8, wrote:
I've been looking at the difference between disk and rim brakes in terms of efficiency (energy from hand to deceleration torque). It looks like it depends upon friction losses in the delivery system, the spring constant of the deliver system (how much the calipers are bending, etc.), the friction coefficient of the pad material, and the radius at which it is applied. Since both disk and rim brakes work by closing calipers the delivery system is irrelevant since you can always implement it for either disk or rim brakes.
The radius of application is a whopping disadvantage for disk brakes - 25% reduction in efficiency. But disk brakes do have the advantage of freely choosing pad and rotor material. However coefficient of friction (mu) appears to be somewhere between .4 and .6 for reasonable choices - not nearly enough to compensate.
But, but, but, mu is not a measure of efficiency. If you have a hard pad, you can get 100 newtons of force into it with less energy than a soft pad. You don't have to work squishing it first... However, I cannot find any easy reference on brake pad efficiency. It's all coefficient of friction. My suspicions is that it's pretty much the same fo
r rim and disk brake pads, otherwise marketing would be all over it.
So for the better engineers/physicists among us, is my analysis correct? Anyone know of any study of brake pad efficiency?

Heat dissipation is another variable that needs to be considered. Also how much speed and over what amount of time will the brakes be applied for and amount of speed reduced.


You're right that the questions have layers of complexity.

For example while both rims and discs move through the air
to dissipate heat, at the extreme the maximum btu load and
maximum temperature on a rim before tire failure is lower
than for a disc. That is to say discs heat up faster and
hotter aebe but it doesn't really matter to braking effect.

I can't answer OP either.

--
Andrew Muzi
www.yellowjersey.org/
Open every day since 1 April, 1971

On 1/29/2016 7:44 PM, wrote:
I've been thinking some more and realize I'm barking up the wrong tree a bit. The rider's hand grip energy is not translated into the deceleration torque. That's coming from the rotational energy in the wheel rubbing the pad. Instead, the rider is energizing the "spring" that constitutes the brake system. Once it's squeezed, you could, in theory, catch a ratchet and release your grip while the brake sat there scrubbing speed. In reality, you have to maintain a static force on the lever. I don't know biochemistry, but clearly maintaining more grip is more fatiguing than maintaining less, which is why higher MA levers feel so much better in actual use.
So it's back to a static force analysis. I'm pretty sure there's no way a harder brake pad in a disk brake could ever make up for the 25% reduction in torque.

It's all motivated by the oft made claim that disk brake provide more braking torque. It's even in the MTBR.com FAQ as a "generally agreed advantage". Seems akin to declaring the earth is flat because lots of people say so...

"you could, in theory, catch a ratchet and release your grip"

I generally stay away from telling people how or what to
ride, but please don't try that on a front brake.


--
Andrew Muzi
www.yellowjersey.org/
Open every day since 1 April, 1971

On 1/29/2016 9:20 PM, Jeff Liebermann wrote:
On Fri, 29 Jan 2016 17:44:48 -0800 (PST), wrote:

I've been thinking some more and realize I'm barking up the wrong tree a bit.
The rider's hand grip energy is not translated into the deceleration torque.
That's coming from the rotational energy in the wheel rubbing the pad.

Nope. The kinetic coefficient of friction between the disk/rim and
the pads requires that a force normal to the direction of the
frictional drag be applied. This normal force comes from the rider
squeezing the brake levers. The frictional drag force is proportional
to this normal force as:
Frictional_drag_force = Normal_force * kinetic_coeff_of_friction
http://www.engineeringtoolbox.com/friction-coefficients-d_778.html

Ignoring hystersis, deceleration, inertia, heating, ablation, filth,
dirt, grease, water, grip strength, cable stretch, and all the other
variables that make such calculations a pain, if one knows the normal
force, disk diameter, rim diameter, and coefficient of friction, one
should be able to obtain a ratio of the rotational force to the lever
grip force. I hope.

The method is partly outlined he
https://www.sensorprod.com/news/white-papers/2010-03_ctb/wp_ctb-2010-03.pdf
Notice the hysteresis in the Fig 10 (bottom of last page) and the lack
of linearity, which should make comparisons rather awkward.

I tried to think of a way to easily measure the braking forces on a
functional bicycle instead of a test stand. Something like a force
sensor on the brake cables and a speed sensor. Generate a graph of
speed versus time at various brake lever grip pressures. That should
produce families of curves, where the fastest rate of deceleration at
a given lever grip pressure is the most efficient.

I'm feeling rather lousy due to a cold or flu, so I'll spare you my
bad math and deranged ideas.




I think you want deceleration rate and not speed per se
although you could, given enough speed samples in time,
chart deceleration probably.

--
Andrew Muzi
www.yellowjersey.org/
Open every day since 1 April, 1971