a quick problem in physics: estimate the temperature of a rotor ...

emi323i · #1 (**permalink**) 11-23-2003, 12:40 PM

3500 lbs car is moving at 80mph. estimate/calculate the temperature of the front rotors after comming to a complete stop.

assumptions:

ignore the amount of heat dissipated into air during braking
ignore the amount of heat absorbed by the brake pad

the rotor is a 12.6" brembo rotor. info i found on the net it weighs 15.9lbs.

By asuming that 80% of the braking is done by the front wheel (therefor 40%/front rotor) i came to an increase in temperature of 126 degrees K (or C).

Can someone double check my answer...

here it is broken down:

Ek = m * v^2 /2 = [1590 kg*(128 km/h)^2 ] /2 = ... =1005 KJ

1005Kj * 40% = 401KJ (Energy distributed on one front rotor)

Specific heat of steel is .44 J/g*K

mass of rotor: 15.9lbs = 7200 g

change in temperature = (401*10^3)/(.44 * 7200) = 126 Kelvin

thanks

Zor · #2 (**permalink**) 11-23-2003, 02:52 PM

..er...why not just drive and test to see how hot they get with one of those laser heat guns

Bretts02 · #3 (**permalink**) 11-23-2003, 03:06 PM

the amount of heat lost to the air is not neglidgeble at all. The e36 specificly has brake cooling ducts in order to cool the rotors, your math is correct but it will clearly be less than that becuase the air is moving across the rotors quite fast

emi323i · #4 (**permalink**) 11-23-2003, 07:56 PM

i can drive it, but i don't have the cool laser temp gauge. and yes, the heat lost in air is not neglidgeable, of course, but this was theoretical... and surely the actual change in temperature would be significantly reduced... the important thing of this was the upper bound... the temperature would increase by a maximum of 126 degrees. all other aspects i chose to ignore would simply decrease this amount...

tomorrow, i'm gonna call brembo and tell them to screw off and send me two new rotors, cause there's no way in hell I warped those rotors by doing ocassional highway, 70-80mph stops.

thanks a lot guys.

Nitrous115 · #5 (**permalink**) 11-23-2003, 08:03 PM

Maybe you warped them by having your foot on the brakes while at a stop sigh/light after heating them up. If you do that then the part of the rotor where the pad is touching would cool faster then the rest of the rotor. Which could warp them.

czm3 · #6 (**permalink**) 11-23-2003, 10:02 PM

I used my laser senser on the rotor of my e46 after a track session. 662 deg. f.

emi323i · #7 (**permalink**) 11-24-2003, 09:53 AM

looks like i'm gonna have to swallow the cost... i talked to both brembo and race tech, the distributor, and they don't have any warranty on warped rotors ... $209/piece. it sucks... but i guess that's the price we have to pay for having the best technology. Since i'm unemployed, and the bmw is gonna be garaged for the winter, i'm gonna delay this expense 'till spring. perhaps even later... this sux really bad.

Alex Mc · #8 (**permalink**) 11-24-2003, 10:04 AM

Its a buttload more than a 126 Celcius increase.. more like 500+

emi323i · #9 (**permalink**) 11-24-2003, 10:10 AM

are you sure? how did you estimate that?

LITEWAITM3 · #10 (**permalink**) 11-24-2003, 05:04 PM

This is NOT a physics problem, its a dynamics and thermodynamic problem. To properly calculate the temperature you have to determine WHERE you are taking the temperature at, since the gradient will be different across the rotor. Air is a big factor!!! To CORRECTLY calculate this, and have thw calculations relate to the real temps if you took a probe to it, it would take about 2 pages of work.

emi323i · #11 (**permalink**) 11-25-2003, 10:28 AM

of course it would take pages to accurately calculate it.. all i wanted was a simple estimate. kinetic energy turns into heat. very simple concept.... sure there's rotational energy in the wheels and all the rotating parts, and etc. i wasn't trying to overdo it... like you said, i could just simple get a laser temp sensor, and get it over with.

SirTony76 · #12 (**permalink**) 11-25-2003, 01:55 PM

Well your calculations are wrong b/c the dimensions of energy you used are wrong...you need an acceleration dimenstion...and another distance dimension. The dimension of the length it took to stop, and the time it took you to stop. Then you can calculate that.
I am not sure I understood your work either, I think the better way to do it would be to calculate the energy that was lost while braking, then set that to the specific heat of the brakes. Then use the factors you need to arrive at the appropriate dimension.