Stock brakes too weak?

[IMOA]

Originally Posted by qoncept

It's still absolutely wrong. Play with the calculator below. You'll see that when you increase the maximum available force you also increase the maximum static (rolling) friction. In other words, if your brakes are stronger, you can apply more force before you lock them up. It's a chain that starts at the pad and ends at the road.

It's right here, clear as day. Unsubbed.

http://www.ajdesigner.com/phpfrictio...imum_force.php

Example:
We'll make up a coefficient for brake pads on rotors since I reall have no idea.. lets say 0.8.

- Stock brakes apply 20 newtons of force, which gives us a max of 16 newtons of friction on the rotor. Tires on road is a coefficient of .9, and the 16 newtons we're applying gives us stopping force of 14.4 newtons.

- Megabrake applies 25 newtons of force. Same coefficient of .8 means the rotor sees 20 newtons of friction, and then the tire on the road sees 18 newtons of maximum static friction.

Of course, the friction required to stop a car increases exponentially. Same deal with power and acceleration.

That would work great if we were trying to work out how hard we had to push a car with it's handbrake on to get it to slide. Unfortunately thats not what we're talking about.

[Taxi]

For someone asking about if its right to swap between pads for autox, track, street, etc, you sure do seem to know a lot about how brakes work.

[qoncept]

Originally Posted by IMOA

That would work great if we were trying to work out how hard we had to push a car with it's handbrake on to get it to slide. Unfortunately thats not what we're talking about.

Sorry, there's a different equation for that. http://www.school-for-champions.com/...ling_start.htm

[Racecomp Engineering]

Originally Posted by qoncept

Since the invention of the wheel I suppose. Static friction isn't just a good idea, it's the law.

The law also states that if you exceed Fmax, than you have begun sliding. Fmax does not increase just because.



This plot kind of sums it up nicely.

- Andy

[qoncept]

Originally Posted by Racecomp Engineering

The law also states that if you exceed Fmax, than you have begun sliding. Fmax does not increase just because.

Here is the major difficulty. Just because it flies in the fact of everything we think makes sense doesn't make it untrue.

Yes, Fmax does increase. Look at the equation. Seriously, LOOK at it. That's EXACTLY what it says.

[IMOA]

No, it says that if the object weighs more you have to push it harder to get it to slide. Now look at how you're trying to use it.

[Racecomp Engineering]

Originally Posted by qoncept

Here is the major difficulty. Just because it flies in the fact of everything we think makes sense doesn't make it untrue.

Yes, Fmax does increase. Look at the equation. Seriously, LOOK at it. That's EXACTLY what it says.

It flies in the face of my statics, dynamics, and I can't remember how many physics courses I've had.

Fmax isn't really in the equation. You just know that if the calculated force of friction exceeds Fmax, then you are sliding and must now use kinematic mu. I'm talking wood blocks sliding on a table here.

- Andy

[Admiral Ballsy]

Originally Posted by qoncept

Yes, Fmax does increase. Look at the equation. Seriously, LOOK at it. That's EXACTLY what it says.

You're mixing up the equation defining the relationship of the pads to the disc with that defining the relationship between the tires and the road.

More force on the pads changes the dynamics of the pad-disc rotation/friction. It has nothing to do with the force defining the friction parameters of tire/road.

[qoncept]

Originally Posted by Racecomp Engineering

Fmax isn't really in the equation.

[Dave-ROR]

Originally Posted by qoncept

No, you can't exceed the limits of friction and stop any faster. But you can change the limit by changing the force applied to it.

I never said exceeding the limits of adhesion would stop you any faster.

[SubieNate]

Originally Posted by qoncept

Here is the major difficulty. Just because it flies in the fact of everything we think makes sense doesn't make it untrue.

Yes, Fmax does increase. Look at the equation. Seriously, LOOK at it. That's EXACTLY what it says.

Dude, you're so wrong it hurts.

Fmaxstatic=mu*fnormal.

For the tires, Fnormal is NOT NOT NOT the braking force. Fnormal is the weight pushing down perpendicular to the sliding/rolling surface.

Trust me. Your understanding is flawed. I went to school for this stuff and CSG_Mike, IMOA, Dave_ROR and Racecomp Engineering area all correct. Without changing the mass on the tire or it's friction coefficient, it's fmax will not change due to the applied force.

Nathan

[Racecomp Engineering]

Originally Posted by qoncept

Yes, that's one way to apply the equation when you're trying to find Fmax. You're still doing it wrong.

I do enjoy the discussion though.

- Andy

[wparsons]

Originally Posted by qoncept

It's still absolutely wrong. Play with the calculator below. You'll see that when you increase the maximum available force you also increase the maximum static (rolling) friction. In other words, if your brakes are stronger, you can apply more force before you lock them up. It's a chain that starts at the pad and ends at the road.

It's right here, clear as day. Unsubbed.

http://www.ajdesigner.com/phpfrictio...imum_force.php

Example:
We'll make up a coefficient for brake pads on rotors since I reall have no idea.. lets say 0.8.

- Stock brakes apply 20 newtons of force, which gives us a max of 16 newtons of friction on the rotor. Tires on road is a coefficient of .9, and the 16 newtons we're applying gives us stopping force of 14.4 newtons.

- Megabrake applies 25 newtons of force. Same coefficient of .8 means the rotor sees 20 newtons of friction, and then the tire on the road sees 18 newtons of maximum static friction.

Of course, the friction required to stop a car increases exponentially. Same deal with power and acceleration.

No, no, no. You're totally mixing up the equations. You shouldn't be adding the stopping force at the rotor to the stopping force of the tire on the road. That's like saying if you're towing a trailer the amount of force the tow vehicle can apply to the trailer is equal to the amount of force the two vehicle can apply to the road plus the amount of force the attachment point (hitch, chain, whatever) between the tow vehicle and trailer can sustain without braking. The reality is that the total force that can be applied isn't the sum, but the minimum of the two forces.

In this case, force applied to slow down a wheel doesn't change the amount of frictional force between the tire and the road. They're two different frictional forces. The only correlation is that the maximum braking force is the minimum of the two, not the sum.

Originally Posted by qoncept

The examples I just posted (see the bold) say that the wheels lock up at 14.4 newtons in the stock example, 18 newtons in the upgraded brake example.

The number you're calculating is maximum static friction. Increase force, increase maximum static friction. Stronger brakes increase the threshold of friction between the road and the tire before locking up.

So no, you don't just lock up the brakes easier.

You're using the equations 100% wrong.

First, if you want to see how much total force is being applied to the wheels from the brake pads you need to look at the amount of force being applied to them, the mu, the rotor speed and the diameter of the brake rotor. Just the mu and pressure isn't nearly enough.

Second, you simply can't apply more force to the rotor than the tires can to the road. Once the wheel locks the amount of braking force between the rotor and brake pad goes way down since the rotor speed is now 0. You can do all the math you want, but I'll give you a hint that the maximum amount of force you can apply through the rotor in a given situation is exactly the same as the maximum amount of force the tires can sustain without locking up.

[CSG Mike]

Originally Posted by qoncept

The examples I just posted (see the bold) say that the wheels lock up at 14.4 newtons in the stock example, 18 newtons in the upgraded brake example.

The number you're calculating is maximum static friction. Increase force, increase maximum static friction. Stronger brakes increase the threshold of friction between the road and the tire before locking up.

So no, you don't just lock up the brakes easier.

Increase maximum static friction on the rotor vs brake pad, not the tire vs the ground.