The way I look at is it doesn't matter what combination of rotors/brake pads/meme lines you decide to use since there is a finite amount of pressure you can actually apply without locking the wheels. The only real difference is quality equipment will let you hit this point a lot sooner.
good systems have good feel, so you can know how much pressure to apply in a linear way. i have driven cars where it was not linear. makes it much harder. drum brakes are also not linear, because they are self actuating a little.
good brakes are predictable, reliable in that they don't fade or wear quickly, and hopefully not heavy because unsprung weight has a huge affect on performance
Luke Wilson
Smooth, consistent and effective performance.
Jayden Young
>What exactly makes a braking system good?
Resistance to heat, contact area, ability to apply force, more or less in that order.
8 piston calipers and 2" bore master cylinders are great and all, but if you rotors are 1/32" thick, it won't help you much.
Cooper Thompson
Yes brakes can only be so good before they just cause a car to skid. A good braking system will maximize downwards forces on the front axle. The best way to do this is by putting more weight on the front axle, which is why FWD is so popular in areas with snow.
Other than that there is not much than can improve brakes other than better tires. If braking is continuous, like when you drive down a big hill, improved cooling will help in braking.
Robert Hall
they work fine on a grand prix bike
Nolan Wilson
Contact area doesn't matter, larger brake area only helps to dissipate heat. The trade off is the extra size creates more angular momentum that will take more energy to stop. Fortunately better surface area will help lose more heat than the corresponding increase in energy required to stop.
Friction is N (normal which is the force you apply) time mu which is a coefficient determined by the material the brakes are made of.
Thermal convection is directly related to the surface area of an object, the more surface area the better the heat dissipation.
P.S. just about everything on a car is cooled by convection.
Carson Jenkins
Oh come on, those are at least 1/8" thick. World of difference.
So you're saying that if I have 8 square inches of contact area of pad on a rotor, as compared to 2 square inches of contact area on an identically sized rotor, I would have equal stopping power on both rotors, but one would just dissipate heat better? Assuming equal clamping force, diameter/thickness of rotor, all that exactly the same, just a difference in how much pad contacts the rotor itself.
James Wright
That is exactly what I am saying.
Same rules of friction apply to tires however, there are a few hang ups there. Unlike brakes which are nice and even surfaces that operate under relatively clean condition, tires are in contact with the dirty road, so a wider tire will give you a better chance of touching a nice clean grippy bit of road than a skinny tire, effectively giving better traction.
When you deflate tires, say for off roading, you quite literally start to grip the surface you are driving on. As the wheel turns it will push off objects below it. As you increase your tire pressure the car will rely less on this pushing force and more on friction alone, this is why cyclists have very high pressure bike tires.
Adam Williams
So essentially, if I took a tire on an axle, spun it at 20 RPM, and used a 1/4" wide piece of steel to slow it to a stop, contacting the tread area with a force of say 100 lbs, it would bring the tire to a stop in the same amount of time as if I used a 12" wide piece of steel with that same 100 lbs of force. The only difference would be that the 1/4" wide piece of steel would become significantly hotter, but both ways the tire would come to a stop in the exact same amount of time.
I guess it just takes an engineer to figure out how that all makes sense.
Julian Jackson
>I guess it just takes an engineer to figure out how that all makes sense. I guess it does. It doesn't make sense to me at all.
If you have the same 100 lbs force applied to a larger area, then the pressure (force per area) decreases. As a result, there's less friction on a given piece of the contact area but the sum of the friction over the whole area is the same. The change in pressure and the change in area offset each other exactly so that only the total force matters.
James Jones
>So you're saying that if I have 8 square inches of contact area of pad on a rotor, as compared to 2 square inches of contact area on an identically sized rotor, I would have equal stopping power on both rotors, but one would just dissipate heat better? Assuming equal clamping force, diameter/thickness of rotor, all that exactly the same, just a difference in how much pad contacts the rotor itself.
Op here, that is exactly the point of this thread. I am trying to understanding how the effectiveness of braking is measured since at the end of the day all braking systems are bound by that finite amount of pressure before the wheels lock.
Luke White
I think he was implying the the piece was still applying the 100lbs at any point it is touching, i.e. force was the same and pressure was greater.
Grayson Ward
Its pretty simple really. The psi(lbs per square inch) of the acting force decreases linearly as area of the friction surface increases. It helps to think of the braking system as a simple hydraulic system which is just a matter of increasing or decreasing mechanical advantage by using different sized cylinders.
Virtually any modern braking system is able to lock up all 4 tires with ease therefore tire size and compound is much more important on a cold stop distance, however where a good brake system really shines is on a racetrack where multiple aggressive high speed brake applications occur in a short amount of time which creates an extraordinary amount of heat.
Jack Anderson
Nah you want more (static) weight over the rear wheels for better braking. It will obviously move forwards as you brake. Old 911s stop like a motherfucker
Blake Collins
Is this sarcasm?
Landon Phillips
Pounds are a measure of force. If you apply that force over a larger area that's less pressure, not more. His example is right, I was just explaining why it's right.