# Heat Due to Friction In Brake Pads

1. Oct 30, 2004

### Jobistober

I am designing a bike frame and I want to know how much heat will be generated between the rubber brake pads and the alloy rims.

I have looked endlessly on the internet for formulas and I came here hoping someone could give me an answer. Im not looking for a definite answer at this point, just a general equation I can use.

2. Oct 30, 2004

### so-crates

A general equation hmmm, how about $$\Delta U = q + w$$ :-)

No really you can use it, but the devil is in the details. The amount heat must depend on the coefficient of kinetic friction between the brake pads and the frame. Also, assuming the chemical constitution of the brake pads/frame doesn't change(would be a fair assumption), then q must equation work input - work output. Only problem is you also have to consider how quickly the brake pads will conduct that heat into the air, as the system is not isolated. I don't have time to go much further than that.

3. Oct 30, 2004

### Jobistober

??

thanks 'crates

when you have some time, can you explain in more detail

Last edited: Oct 31, 2004
4. Oct 31, 2004

### Staff: Mentor

Its actually pretty simple: the brakes dissipate all of the kinetic energy of the car when it stops.

5. Oct 31, 2004

### Chronos

Friction is a real beast in the real world. It finds truly amazing ways to obey the laws of thermodynamics and energy conservancy. Heat is a major player, but other factors also come into play in a big way. In most mechanical systems, vibration [kinetic energy], not heat, is the preferred route to discharge energy. This is largely a curse for mechanical engineers. Heat is more easily managed than vibration.

6. Oct 31, 2004

### Jobistober

Lets not Get off subject here...

thank you all for your input, but let me remind you that I started this thread looking for general equations that I can use to find the heat generated due to friction between bicycle brake pads and its rims, just as I said in the first post.
Thanks

7. Oct 31, 2004

### Staff: Mentor

Its just as I said, the kinetic energy equation: e=1/2mv^2

8. Oct 31, 2004

### Jobistober

So

kinetic energy in joules is equal to heat in joules created by friction? Must I assume coefficients of friction, heat dissipation into the air, material makeup, etc have nothing to do with it?

9. Oct 31, 2004

### Chronos

You can do a straight up calculation, but the temperature you end up with will be higher than what occurs in reality. Energy is bled off through in a number of other ways, such as rolling resistance. The brake will also normally seize the rim at some point, transfering all the friction to the tires as they skid against the road surface. Having said all that, this is the formula

W = J x Q where W is the work [kinetic energy at initial velocity],
J is Joules constant, and Q is the total amount of heat generated.

What this does not tell you is how the heat is distributed between the brake pad and rim. To approximate that, you need to figure in the specific heat of the pad and rim material and their masses. The caloric energy is converted to temperature as follows

Q = mcT where Q is the caloric energy, m is the mass, c is the specific heat of the material and T is the temperature rise.

Calculate a temperature for both the rim and the pad assuming each will absorb the total heat energy. Of course neither will actually get that hot, but, it gives a starting point. You could split the difference, let the rim take half and the pad the other half, if you are curious what is more realistic. But, from a design perspective, I would be inclined to use the maximum possible temperature. I know for certain neither of them will get that hot, so my design has a nice fat safety margin. Hope that helps.

10. Oct 31, 2004

### Staff: Mentor

Since you already know the brakes are capable of stopping the car, you don't need to do any friction calculations.

But Chronos is right - going from how much heat is dissipated to how, precisely, it is dissipated is not an easy task.

11. Oct 31, 2004

### Jobistober

you input has helped Chronos, however could you specify what the value of joules constant is and what the units of T are in the equation Q=mcT please. I know I must sound like an idiot for asking such a dumb question and for that I apologize, but your help is greatly appreciated.

12. Nov 1, 2004

### Chronos

Joules constant is 4.184 Joules/calorie
T is in units of C [degrees centrigrade]
Suggestion: when you calculate the work done [removing the kinetic energy of the system] do it in Joules units.
Those are not dumb questions. The only dumb questions are the ones you never ask.

13. Nov 1, 2004

### Gokul43201

Staff Emeritus
Let me do a rough calculation with some assumed numbers. You can change them to those you prefer.

Mass of bike + rider = 220 lbs = 100 kg (I'm just using nice numbers...to make this easy for me )
Max. speed = 22 mph = 36 km/hr = 10 m/s

So, KE = 0.5*100*10^2 = 5000 J

Typically, most of this frictional loss comes from skidding. How tightly you squeeze the brakes, determines how much of the loss is at the brakes, and how much at the bottom of the tire.

Assume all of this loss is from skidding. The heat generated there is divided about evenly between the ground and the bottom of the tires.

So, about 2500J of heat goes into the tires, or about 1250 J per tire, if both wheels are skidding - if you lock both front and back brakes.

Since rubber is a lousy conductor of heat, I'm going to assume that all the heat absorbed by the rubber stays in that portion of the tire that makes contact with the ground. I have absolutely no idea what the thickness, width and typical softness of tires, so I'm making an arbitrary guess that the volume of rubber involved is about 5cc, so I guess it mass is about 25 gm (I'm guessing that the density of this rubber is about 5 g/cc).
Also, I've found the specific heat capacity of rubber to be between 1200 and 1700 J/K-kg. I'm going to use C = 1250 J/K-kg for niceness.

So we have, Heat =1250 J = m*C*dT = .025*1250*dT. That gives dT = 40 K or about 70 F. Clearly this is an overestimate...but it's like a worst case number. Now this is the increase in temperature of the tire due to skidding.

It is not the increase in temperature of the rim and brake padss due to the brakes. That can be done similarly, using the KE of the wheel (=mv^2, where m = mass of wheel). I'll get to it I find the time.

14. Nov 1, 2004

### Jobistober

Ok

Now considering everything everyone else has said, I am going to use these equations and you people can back me up. However this time, I am going to use as close to real world numbers as possible. Everyone is free to correct me if I am wrong!

Using actual numbers I have obtained on a run with my gravity bike:

total mass (Bike and rider) = approx. 300 lbs = 136 Kg
max speed = 47.7 mph = 21.32 m/s
KE = (1/2)136(21.32)^2 = 30908.88 J

Assuming the rims are made up of steel, mass and specific heat are:
c= 452 J/Kg/K
m= 4lbs = 1.81Kg

Also assuming braking forces are divided half and half between the two rims:
Q= 15454.44 J
15454.44= (1.81)(452)T
15454.44=818.12T

T = 18.89 K = 32 F

roughly 32 degrees ferenheit is the final temperature increase per rim, correct?

15. Nov 1, 2004

### Gokul43201

Staff Emeritus
This would be correct only if there's no skidding.

And even so, you'll have to make the following adjustments. The heat would be divided between the rim and the brake pads, roughly equally. And stainless steel, being a fairly poor conductor, the entire rim does not get heated immediately - only the portion of the rim that gets to be in contact with the pad (affected volume=rim thickness*pad width*wheel circumference)

This will give you an upper limit on the maximum instantaneous temperature reached by the rim. Radiation losses will be small and may be neglected.

You could also test this with some kind of strip thermometer, to confirm that it's not grossly off.

Last edited: Nov 1, 2004
16. Nov 2, 2004

### Chronos

Correct, Gokul. Which is what Job is looking for. He wants to know how hot the rims could get in the worst case. I should mention that the calculation does not take into account slope. You need both the slope and maximum initial velocity to get the right value for the total kinetic energy.

17. Nov 2, 2004

### Jobistober

how can slope be a factor in KE, would not slope only affect velocity, which is already a variable in the KE equation?

18. Nov 2, 2004

### Gokul43201

Staff Emeritus
Realistic slopes will have a small effect, because when you use energy conservation, you have to include the PE too (which, in the absense of a slope, remains constant)

19. Nov 2, 2004

### Chronos

It increases [downhill] or decreases [uphill] the potential energy of the system mass due to gravity. Picture the bicycle sitting on a table attached to a counterweight suspended off the end of the table across a pulley. How much counterweight do you need to add to keep the bicycle from moving when the table is perfectly level? None. How much counterweight do you need to keep it from moving when you tilt the table? It depends on the sine of the angle of inclination. If you tilt it 90 degrees, the counterweight must be equal in mass to that of the bicycle. So basically you just add the counterweight mass required to offset the inclination angle to the mass of the bicycle. So, you just substitute m = [m + m*sine(theta)], where theta is the angle of inclination, when calculating KE.

20. Nov 2, 2004

### wd40

I can presume you want the equations in order to determine the maximum heat generated in order to try and accomodate it in your design.
There seem to be an agreement that this is changing with many parameters. Unless you want to conduct a parametric study, I advise you to conduct an experiment at an extreme condition and extrapolate the data from there.
This might be a solution.

Luck.