Best Brake Pads?

[PsychoRallye]

I've said it before and I'll say it again, R4 pads are cheaper, and wear the rotors more. When they are "overnight cold" stopping with them is like trying to stick your feet out the door to stop the car. After 2-3 normal stops they heat up enough to act like somewhat normal brake pads. I liken it to the car having a "normal braking system" much like my Volvo.

Once they heat up, they eat less rotor, make less noise, and you'll never heat them up that much on the street, no matter how aggressive you think you are.

I use Pagids now, exclusively, rotors are more expensive.

EDIT: That is, I use the RS14 for track. I've not decided what to do for the times I am on the road.. dunno.

[xtn]

Originally Posted by Rennstore

......and a higher friction pad will indeed generate more heat than a comparable pad of lesser friction - when used to it's capabilities

Craig
www.Rennstore.com

No it won't. Not stopping the same mass of car over the same distance.

Now you may mean to say that you're generating the SAME heat over a shorter stopping distance, but that's only true if the control set of pads wouldn't lock up the tires anyway, and it's still the same amount of heat, just concentrated into a shorter timeframe if you're really stopping faster.

Now you might make the arguement that you're getting up to higher speeds since you can brake later, and therefore generating more heat because you're stopping from a higher speed, and that's an okay point, but still only true if the control pads wouldn't stop the car up to the tires' ability as I said above.

A higher friction pad just means you don't have to build as much cylinder pressure to achieve the same stopping power. It may FEEL better through the pedal, and durability may be improved, but it's not stopping you any faster or generating any more heat for a given decrease in speed.

Let's say you can stop from 0-60 in 100ft with your best high-friction pads. Now let me have enough cylinder pressure to squeeze a teflon pad coated in WD-40 against the disk surface hard enough to stop from the same speed over the same distance. The same amount of heat will be generated. Now that's silly, and it would probably take a 10-ton press to squeeze hard enough to do it, but it WOULD generate the exact same amount of heat energy.

[edaddy]

Originally Posted by Surferjer

+1

If what you care about is braking, ignoring dust, squeal, and rotor life, Hawk HP+ pads are awesome. So much better than OEM at stopping power it's unbelievable. I hear they may fade on the track, but for autoX and street they are amazing.

Can anyone else comment on Hawk HP+ pads? Surferjer is an AutoX champ, if I recall correctly, so these may be the very best pads for the street. Tire Rack has them. Any other (less expensive) sources?

[Rennstore]

Apply a temp strip to the front caliper of two Eli.

Run a pagid Orange pad at full boogie, then take another Elise, & run it the same way with a set of RS14/Black Pagids.

You will end up with higher temp readings with the RS14 Pagids, every time.

Additional information - Pagid friction graph:

[xtn]

That graph is telling us the friction level AT those temps. In other words, when they get to X temp, they have Y (unknown scale on the Y axis for that chart) friction level.

In other words the RS14 pads, for example, at any given temp they are bothed graphed at on the X axis, have a higher friction than the RS42 pads.

How on Earth can you read that graph to mean that the higher friction level creates more heat?

I can imagine you might get a higher temp reading on a rotor surface by using higher friction pads to stop faster (if the older pads couldn't achieve max braking allowed by the tires already) but that's only because the heat is generated in a shorter timeframe but the dissipation characteristics are still the same. Put simply, it's being heated faster but it can't spread the heat out faster too, so yeah you might read a higher peak surface temps. But that's ONLY if you've got the rubber to actually stop in a shorter distance because the old pads couldn't, and that STILL doesn't mean the pad is generating MORE heat energy.

So if it had been asked, "Are you seeing a bit higher peak rotor surface temps?" It's possible for the answer to be yes.

But since it was asked if they're generating any more heat, the answer is no.

I appreciate the debate, but whatever it is you think you know isn't jiving with basic physics, unless you agree with my analysis but we have only a symantics problem.

[Joetz]

Originally Posted by sl8anic

Did you notice any additional heat being generated by the new pads or any visible changes to rotors? EBC's site seems to promise a very good alternative to OEM's.

Do you have p/n's and $ info?

I am not sure what you mean by heat but I did notice that that holes in the rotors no longer get plugged up with dust. I would say there is probably about 50% less dust now than before. I don't have part numbers unfortunately since I did it quite some time ago.

[lotusmark]

Not to divert this thread, but when running rs14's on the track, since I've uprated my calipers andrunning slicks, my rotors are turning blue. Too much heat for the rotors?

[elptex]

xtn...physics don't agree with you. All materials have a different coefficient of friction and convert energy to heat in differing amounts. The formula is...



Where N is normal force and Uk the coefficient of friction (x being the point at which the object travels--this is just simple integration). The higher the coefficient of friction, the more energy that is being converted to heat, thus resulting in a greater amount of heat.

If all materials generated the exact same amount of heat from friction, such things like bringing the space shuttle back down to earth wouldn't be a problem, nor would there be any benefit from tires utilizing different percentages of compounds.

So yes, they (the brake pads) very well could be generating more heat if their coefficient of friction is higher than the stock pads.

[neckstrap]

I need to report a little miss-information. On a previous post someone said:

"First of all, if one does a cross-reference: Rear Pads for Standard 92-2002 Dodge Viper (not SRT-10) will fit Elise rear."

This is wrong. I bought some and they are too thick. The friction material on the new pads are about 10 mm thick. I will have to grind off at least a milimeter (maybe two) of each one to get them in. With the uneven surface, it's going to take much longer for the pads to seat now. What a pain!

Sands Museum got this right. I should have read his article first.

Paul

[sl8anic]

No problem man, I was just curious. I think I found some numbers:
* front: EBC Yellow DP4197/2R;
* rear: EBC Yellow DP4885/2R;

Less dust is always good. I suspect that more/less apparent dust is partially due to how "sticky" those brake particles are. So all pads may dust about the same but the dust just may not stick to the wheels as much.

As far as temps go - there are DEFINITE differences in temps for various pads under similar use. Whatever underlying reasons are (more friction or whatever) the temperatures do vary enough to be noticeable. The reason for asking is potential damage to rotors and brake lines.

Originally Posted by Joetz

I am not sure what you mean by heat but I did notice that that holes in the rotors no longer get plugged up with dust. I would say there is probably about 50% less dust now than before. I don't have part numbers unfortunately since I did it quite some time ago.

[tesprit]

Originally Posted by neckstrap

I need to report a little miss-information. On a previous post someone said:

"First of all, if one does a cross-reference: Rear Pads for Standard 92-2002 Dodge Viper (not SRT-10) will fit Elise rear."

This is wrong. I bought some and they are too thick. The friction material on the new pads are about 10 mm thick. I will have to grind off at least a milimeter (maybe two) of each one to get them in. With the uneven surface, it's going to take much longer for the pads to seat now. What a pain!

Sands Museum got this right. I should have read his article first.

Paul

Take the pads to a local machine shop and have them milled down. It shouldn't cost much (almost no set up and about a 5min job) and they will be perfectly flat when you get them back. All you need to do is tell them the thickness you want.

[Rennstore]

I added the graph as an aside, not an addendum to the discussion.

The RS14/Blacks do have more friction than almost any pad made, and do create more heat - luckily Elise owners have very light cars and are easy(ier) on the brakes.
In some non-Lotus cases, the Blacks have so much friction - and thus heat - that heat can be an issue and additional cooling is needed.

If there is a 'kit' for cooling Elise brakes - use it.
You will get longer pad and rotor life as well as more lap to lap consistency with regards to braking ---

[xtn]

Originally Posted by elptex

xtn...physics don't agree with you. All materials have a different coefficient of friction and convert energy to heat in differing amounts. The formula is...



Where N is normal force and Uk the coefficient of friction (x being the point at which the object travels--this is just simple integration). The higher the coefficient of friction, the more energy that is being converted to heat, thus resulting in a greater amount of heat.

If all materials generated the exact same amount of heat from friction, such things like bringing the space shuttle back down to earth wouldn't be a problem, nor would there be any benefit from tires utilizing different percentages of compounds.

So yes, they (the brake pads) very well could be generating more heat if their coefficient of friction is higher than the stock pads.

This is ridiculous.

Materials of different friction do NOT convert kenetic energy into heat energy in different amounts. They convert it at different RATES. The same amount of work must be done to accellerate (decellerate) a mass from one speed to another.

Yes, if you just apply two different pads with the same pressure to a rotor driven at a constant speed, the higher friction pad will convert more energy to heat. Let's say you're driving a testbed rotor on a workbench with a 5HP electric motor. The higher friction pad will be doing more of the dragging. It will get hotter. The division of the motor's power will be more heavily weighted to pulling against the higher friction material. This is what you're talking about, and it's correct. This is in a system with constant energy being applied into the system and asking two different pads to drain different amounts of energy out of it. Of course if you're asking the higher friction pad to drag off more Joules then it's going to make more heat.

However in our case we are not asking both pads to do the same amount of work. On a car we're talking about a different kind of system. We're talking about a condition wherein the vehicle is carrying so many Joules, and then we're asking the brake pads to disperse those Joules. When we stop a car from 60-0mph it is the same amount of work required no matter what pads you are using. The same number of joules must be dissipated.

Let's imagine that same test rotor on the workbench. Spin it up to X number of rpms, flip off the motor, and let it freewheel. Now stop it with both test pads at the same application pressure. The higher friction pad will stop the disk faster than the low friction pad will of course. But they both do the SAME amount of work and they both convert the SAME amount of kenetic energy into the SAME amount of heat energy. Now the heat generated PER UNIT OF TIME is more with the higher friction, but the total heat generated from the test is the same. The lower friction pad generates less heat per unit of time, but is also working longer to result in the SAME job done.

Your space shuttle analogy helps me make my point. The materials on the exterior of the space shuttle are not designed to be low friction so they'll generate less heat. If you were right that would be the case and they would make them out of teflon so the shuttle would slip through the air without making as much heat. But the total amount of heat generated in slowing down the shuttle from a given reentry speed is the same whether it's surface is low friction or high friction. If it were teflon coated it would slip through the air generating less heat PER TIME UNIT but it would also take longer to slow down and the overall heat generated by the time it stopped would be the same.

The fact is that the primary design criteria of the tiles on the space shuttle has to do with SURVIVING the heat, just like brake pads.

Please God Baby Jesus show me how to make my point understood!!!

The ONLY thing I can think of that means one pad might generate less heat energy than another pad is that the ones that squeel are obviouly converting at least a tiny bit of that kenetic energy into sound energy, and thus converting a tiny bit less of the total into heat!

xtn

[sl8anic]

Calm down XTN

Another possibility is that different pads absorb and transfer heat at various rates, thus the perceived temp differentiation.

[elptex]

You seem to be saying that, using simple numbers, 1 unit of heat is generated from both material A and material B, but that material A requires 1 unit of work and 1 minute, but that material B requires 2 units of work and 2 minutes. This would be the case such that material A has a higher coefficient of friction than material B.

That is true, absolutely correct, you are completely right...when you can apply that for a limit at infinity. What you appear to be saying is that material B will create the same amount of heat as material A...if you give it long enough. Yea, at infinity...and under the assumption that your material hasn't completely disintegrated due to the heat that has been produced. So sure, the same amount of heat is being produced, but at differing time intervals.

Lets plug in some numbers in a "real world" situation. We've got brake pad material A with a coefficient of friction of 1 (Uk) and weighs 1 pound (Normal force), and brake pad material B with a coefficient of friction of 2 and weighs 2 pounds. I'm coming up to the first turn in my lovely auto cross run and I've got to floor the brake pedal for 3 seconds. Lets see how much heat was generated by each pad after 3 seconds...

Brake pad material A: heat energy = Uk int( N, x ) = 1 Int( 1, 3 ) = 4.5 units of heat

Brake pad material B: heat energy = Uk int( N, x ) = 2 Int( 2, 3 ) = 9 units of heat

I'm using dx here to integrate along the traverse of time which makes the assumption that the work done on the pads is constant (=1) for all 3 seconds. This is a fair assumption as work could be applied as a controlled (constant) variable in a lab setting.

I don't want to be running around with incorrect information in my head so please correct me where needed.

[xtn]

Originally Posted by elptex

You seem to be saying that, using simple numbers, 1 unit of heat is generated from both material A and material B, but that material A requires 1 unit of work and 1 minute, but that material B requires 2 units of work and 2 minutes. This would be the case such that material A has a higher coefficient of friction than material B.

No. They both require one unit of work to be done.

The high friction brake pad might do the work in one minute (arbitrary control point). The Low friction pad might do the work in two minutes. It only has half the friction of the first pad. But to complete the stop from a given speed requires the SAME amount of work be done; the SAME number of Joules is converted from kenetic to heat energy. The spinning rotor on the work bench or a moving car - either way a certain number of Joules must be converted to heat to bring it to a stop from a given speed, period.

I have allowed that the SURFACE temp will be hotter on the higher friction pad, because it doesn't have two minutes to let the heat soak out evenly. It's doing the work in half the time so the surface of the pad will get hotter. But the surface temperature is NOT a standalone indicator of the total heat generated.

Let me try this analogy:

Let's set our car up using electric motors to do the braking. When we want to stop we unload the current through an electric heater element as the load. With me so far? We control the braking force (friction) with the load current we allow through the load.

Now we can stop the car in 1 minute and the heater element will be burning brightly for that 1 minute. Or we can stop the car in two minutes and the heater element will burn half as brightly but twice as long. It's the same amount of Joules converted from kenetic to heat.

Now lets go back to our normal car, and lets assume we could lock up the tires already before we put on our great new pads with twice the friction. So having twice the friction does NOT allow us to stop any faster or in any shorter distance. It only means we have to press half as hard on the pedal to achieve maximum braking. And of course the feel of initial bite and the durability characteristics may be different blah blah blah. But we can't stop any faster or any shorter than we could before.

So despite the fact that the car feels better, the pedal travel required is less, etc., still for a given reduction in speed for a given mass of vehicle (same joules present in the system) the SAME amount of heat is produced. X kilograms traveling at Y meters per second has Z joules of energy, period. To bring Y down to zero, or to any particular lower value, requires that all or some of those joules be converted to heat by the brakes. In a test where X and Y are constant (repeated braking test in same vehicle from same speed) then Z (amount of energy) is also constant.

There is only Z joules of energy in the system. When you brake to 0mph you've converted them all to heat via the brakes. This is what happened with your old pads. This is also what happens with your new high-friction pads. No amount of pad changing can change the number of Joules that are in the system waiting to be converted to heat when you mash the pedal. By the time you come to a stop, you've converted all the energy to heat. It's the SAME amount of heat wether you use your old brake pads, your new brake pads, or your feet on the asphalt.

[elptex]

I typo'ed 2 work units in the part you quoted. Show me some science and math as I have done and then I'll consider accepting your arguement.

Cheers

[DreamCar]

There's a bit more complicating the equations here.

Elptex - your 3 second braking example doesn't work because the high friction pads would have the car going much slower: If 3 seconds of low friction pad took you from 60-30mph, the high friction pad would get you to 0mph. This is what XTN is trying to say I believe.

XTN - the assumption that both pads can only brake up to the friction point of the tires - don't forget that front may reach the ABS limit, while the rear may not, therefore higher friction pads all round may actually stop you faster by getting both front & rear tires closer to the adhesion of the tires instead of just the front.

Temperature of the pad while running is a function of heat input by the friction minus heat output by the brake cooling air. Low or high friction pad both have the same heat energy created, but the high friction pad does it over a shorter time. Shorter braking time means less air has travelled over the brakes, and the temp will rise higher. You have the same heat gain, but less heat loss = more temperature rise.

That's how I see the physics, either way...

[xtn]

Okay lets say our roughly 900kg car is going 40mph or 18m/s.

E = 0.5 * m * v^2

145,800 joules = 0.5 * 900kg * 18^2m/s

So our car, traveling 40mph, has 145,800 Joules, or Newton-meters, of kenetic energy. At that mass and speed it will always have the same kenetic energy. To stop from 40mph to 0mph requires that 145,800 joules of energy be converted from kenetic to heat.

Exactly 145,800 Newton-meters of work must be done to stop the car, no matter what method is used. So you'll get out exactly 145,800 Joules (a unit of energy; can be kenetic OR heat and still call it Joules) or 3.482373e4 calories, or 138.19 BTUs, or 0.001381917 Therms. Doesn't matter what unit of heat energy you use. There is no math to do. One Joule of kenetic energy is equal to one joule of heat energy. 145,800 joules is the the maximum amount of heat you will get from decellerating 900kg from 40mph to 0mph.

When you brake from 40 to 0 in a 900kg vehicle, you generate 138.19 BTUs. I don't care what kind of brake pads you use. The frictional level of the stopping medium is NOT a factor in the heat energy calculations.

Does that help?

[sl8anic]

Again, same amount of energy needs to be removed from a system, but how quickly and efficiently that's done is a function of the pad (it's friction coefficient). Obviously, if the heat is not removed/transferred away properly - it'll build up causing possible damage to rotors/lines/pad material.

And THAT was my original question - any significant difference in temperatures observed for a particular set of pads.

Regardless guys, this turned into a total thread-jack.