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# Re: Al Rotors - The definitive answer (Looong)

```May I offer a couple of the minor corrections you mention.

>Here's a lesson on metallurgy/kinetics/heat transfer/fluid dynamics for
>anyone interested - subject to correction as needed.
>
>Glen, you're right on the money with one clarification (I'll get to that
>later).  Eric, you're full of it.  You must be using a solar calculator
>in moonlight to do your "math".  Alan, very correct.
>
>The key to stopping a car from forward motion (ie possessing kinetic
>energy) is to somehow dissipate that energy.  Remember, energy can
>neither be created or destroyed.  Therefore we have to get it to change
>forms.  One way would be to run the car into a wall or a tree or
>something, and share that kinetic energy with the wall and the little
>atoms in the metal of the car itself.  Not good.  So the way we do it
>without destroying everything is through FRICTION.  Friction is the most
>important thing in stopping a car using brakes.  There are two places
>this friction is important - one is between the rotor and pads, the other
>is between the tire contact patch and the road.  If you think back to
>your classes (physics), a frictional force (in our case - stopping force)
>is equal to the normal force (normal as in perpendicular, not everyday)
>times the coefficient of friction.  So, there are ONLY two ways to
>increase a car's stopping force.  One is to increase the coefficient of
>friction.  The other is to increase the normal force.  Before somebody
>(Glen) jumps all over me on that one with, "What about swept area?"  Easy
>- to come up with a total stopping force on one wheel at the rotor, you must
>integrate the normal force at each point over the whole pad area.  Thus
>the bigger the pad surface area is, the bigger the overall normal force
>will be.

Yuo missed on this one.  The size of the pad surface doesn't have any
obvious bearing on the total normal force.  A small pad will operate at a
higher pressure to produce the same normal force, assuming a bunch of stuff
to simplify the analysis (constant coefficient of friction across the pad
surface, no change due to heating etc.)

>
>The first thing to look at is the rotor/pad interface.  The rotor/pad's
>job is to overcome the rotation of the tire/wheel/rotor/hub/etc
>(everything that's rotating) period.  The rotor/pad does not directly
>stop linear momentum of the car!  So, Eric, you are correct in saying
>that a lighter rotor would reduce rotational inertia of the rotating
>mass.  But you're dead wrong in saying that the mass difference from an
>aluminum rotor to an iron one will make a significant difference in
>stopping force due to reduced rotational inertia.  This effect will be
>totally insignificant.  Glen is right on the money saying that the wheel
>and tire have a much greater effect on this phenomenon.

I'm disappointed, Jeremy.  Why not actually do some simple calculations
that say what the effect is.  I'm inclined to agree with your guess, but it
still feels like a guess.

>The one thing
>Glen left out is the fact that the tire also has a frictional force
>acting on it that wants to continue to rotate the wheel/tire/etc.  This
>force is much greater than the rotational inertia of the rotational mass,
>and much much greater than the rotational inertia of the wimpy little
>rotor!  So where does this frictional force on the tire come from.
>Simple - multiply the normal force on the tire (gee, would that be
>weight?!!! as in MASS times acceleration

I guess you mean gravity here?

>) times the coefficient of
>friction between the tire contact patch and road surface.  Again,
>integrate the weight of the car over the contact patch - thus if the
>contact patch is bigger, you get more overall normal force.

Not true.  See above.  The total normal force on the 4 tires is the weight
of the car, period.

>Also note
>here that the rubber compound of the tire

... and the tread, and the carcass, etc.

>plays a key role in coefficient
>of friction.  So, the heavier the car, the larger the normal force for a
>given Cf and tire patch.  Thus a larger frictional force trying to keep
>the tire/wheel/etc. spinning.
>
>Obviously, if you have a light car with really good brakes (high Cf), you
>can easily lock up the tires.  This is because the normal force on the
>tires isn't big enough (small weight) to create enough friction to keep
>the tire spinning.  OTOH with a heavy car and bad brakes, the tires
>probably won't lock up and the car'll take a while to stop.
>
>We all know that friction works through the generation of heat.
>Friction's job is to take kinetic energy and change it into heat energy.
>So now heat comes into play.  At the rotors, heat can change the compound
>of the brake pads and thus reduce the coefficient of friction between the
>pads and rotor (fade).  Here's where I think Aluminum makes its most
>important contribution.  First of all, Aluminum is a softer metal than
>iron.  Because it's softer, the coefficient of friction between an
>Aluminum rotor and a given set of pads would be significantly higher than
>the Cf between the same pads and an iron rotor.  MORE STOPPING FORCE!
>This is, IMO, the main difference.

a power assist mechanism or adjustment of mechanical advantage of the brake

>as significant) is Aluminum's higher specific heat.  It was pointed out
>that an Aluminum rotor would get hotter than an iron one.  Not so.  The
>higher specific heat of the aluminum allows it to conduct heat AWAY FROM
>THE PADS to other parts of the rotor.

Specific heat is a measure of heat capacity, not conductance.

>This keeps the temperature at the
>pad lower than it would be with an iron rotor.

Aluminum's better conductance would tend to help this way.

>BECAUSE Aluminum also
>dissipates heat to air a lot better than iron.

Really?

>So now the heat that is
>better conducted away from the pad is now better convected and radiated
>from the rotor into the air, which will carry this heat away.
>
>So why aren't all rotors aluminum?  For one thing, aluminum is less stiff
>than iron.  And the stronger alloys (7075, 7079) tend to be very
>brittle.  But the main drawback is that aluminum has about a third of the
>cycle life of iron.  It is very subject to fatigue.  Since braking is a
>very cycle-intensive process, aluminum is not the best choice for your
>dependable, everyday commuter.  Since the racing life of a part (on a
>race car) is very small compared to a passenger car, aluminum would be a
>viable alternative.  Also, racing applications are tremendously sensitive
>to unsprung weight,

>where aluminum would provide yet another benefit.  I
>think the answer lies with composite alloys, which would allow the
>benefits and help remedy the weaknesses.
>
>Sorry this was so long.  I hope someone made it to the end.
>
>Jeremy R. King
>Senior Mechanical Engineering Student
>Suspension Team Leader - War Eagle Motorsports Formula SAE
>
>'86 VW Quantum GL5
>Auburn University, Alabama, USA
>Hometown - Reidville, South Carolina, USA

After slogging through all this, you're still guessing at the relative
effect of the rotor weight.

Richard Funnell,
San Jose, California
'83 urQ
'87 560 SL

```