Just curious, wouldn't the tire be more likely to slip on braking?

archetype - 04 October 2009 09:49 PM

Just curious, wouldn't the tire be more likely to slip on braking?

It depends on the situation and the friction/traction available. In a light car like a Formula One, the tyres would be more likely to turn on their rims under braking. That's when the most weight is put on the tread, giving it the most bite on the road. Under accelleration, the rear wheels may gain more load as the nose lifts, but the wheelspin of tyres on pavement limits accelleration and the load on the bead.

Heavy vehicles like trucks (and possibly NASCAR stockers) are different. There, braking is, as always the limit of forward bite. Once a wheel locks, traction decreases by about half. Under accelleration, the truck's weight adds a great deal of traction to the treads, allowing them to stick hard to the pavement, while the much narrower bead slides on the steel rim. In this case, the tread has a lot more traction than the bead, so when traction is lost, that's where it happens.

archetype - 04 October 2009 09:49 PM

Just curious, wouldn't the tire be more likely to slip on braking?

I say yes.

Most people I read about say the first thing they notice when they get into an F1 car is the braking. I guess CART and other racers acclimate relatively quickly to the acceleration and lateral control of an F1 car, but apparently nothing can prepare them for the braking. They said 5.9G max this weekend for the Suzuka circuit. And I guess the left leg (and neck muscles) are what drivers recommend newcomer work on. I guess the amount of pressure needed for full braking is significantly more in F1 than in other race cars. I’m not sure why that would be.

Now that I think about it, perhaps because modulation control is of utmost importance (“feel”), so maybe they design the brakes in F1 to have a wider “field” of operation to allow for more modulation control by the driver. That’s my guess.

So relative to bead/spin, it’s the down force. I forget what the estimate max down force on an F1 car is, but it can be more than the weight of the car itself. (2K pounds?) So when you’re going 200mph and stomp on the brakes, you’re getting WAY more torque on the rim beads than you would under acceleration where the wheel can just spin. I’m not sure what the estimated max G force is on acceleration of an F1 car.

Oh – wait a minute. If you have more down force, you probably have more traction between the rim and tire – so maybe it’s the opposite, and rim spin is harder to achieve while braking.

Now I’m confusing myself.

At any rate, it would be interesting to compare max G acceleration/braking between CART and F1 to gain perspective. Like a Corvette vs. 911 on the skidpad. Where’s the suggestion box for TopGear?

According to Road & Track:
the base level Corvette generates 0.93G,
the Z06, 0.99G
and the ZR1, 1.10G on the skidpad.

Under braking the Corvettes do:

base 06 - 0 in 119 feet and 80 - 0 in 212 feet
Z06 Corvette 06 - 0 in 111 feet and 80 - 0 in 189 feet
ZR1 Corvette 06 - 0 in 106 feet and 80 - 0 in 183 feet (best of all cars tested)

For the Porsches,

Carerra GT, 0.99G, 60 - 0 in 124 feet and 80 - 0 in 199 feet
911 CarerraS, 1.00G, 60 - 0 in 111 feet and 80 - 0 in 193 feet
911GT2, 1.03, 60 - 0 in 108 feet and 80 - 0 in 187 feet
911GT3, 1.04, 60 - 0 in 106 feet and 80 - 0 in 186 feet
911 Turbo, 0.99, 60 - 0 in 110 feet and 80 - 0 in 194 feet

Under accelleration, the Corvetts do:

Base 0-60 in 4.3sec, 0-100 in 9.5 sec and ¼ mile in 12.6 sec. at 115.7 mph
Z06 0-60 in 3.4sec, 0-100 in 8.0sec and ¼ mile in 11.7 sec at 123.7 mph
ZR1 0-60 in 3.3 sec, 0-100 in 7.4 sec and ¼ miin 11.4 sec at 125.5 mph

The Porsches:

Carerra GT 0-60 in 3.6 sec, 0-100 in 7.0 sec and ¼ mile in 11.6 sec at 131.6 mph
911 CarerraS 0-60 in 4.3 sec, 0-100 in 910.3 sec and ¼ mile in 12.6 sec at 111.9 mph
911 GT2 0-60 in 3.4 sec, 0-100 in 7.7 sec and ¼ mile in 11.5 sec at 123.9 mph
911 GT3 0-60 in 3.7 sec, 0-100 in 8.3 sec and ¼ mile in 11.8 sec at 119.5 mph
911 Turbo 0-60 in 3.4 sec, 0-100 in 7.9 sec and ¼ mile in 11.7 sec at 121.2 mph


Hope that clears up some of the confusion.

archetype - 04 October 2009 09:49 PM

Just curious, wouldn't the tire be more likely to slip on braking?

Yup... your right. But they still don't.

archetype - 04 October 2009 09:49 PM

Just curious, wouldn't the tire be more likely to slip on braking?

In a simple answer, no.. most tire "spinning" incidents occur at the driven wheels.
The high G numbers in F1 and other high downforce cars is because of the aero drag combined with high brake forces. The 5.9 G at Suzuka occurs at a high speed braking down to a low speed, on a very high grip track. The 5.9 number is the initial strike of the brakes and declines from there.
The reason a tire is most likely to spin on a driven wheel is due to the torque being applied to the wheel which in turn is applying the said torque to the tire. Applying torque too sharply to a tire that has a lot of grip at a slower corner exit is the most likely cause of spinning a wheel on a tire. High speed corners tend not have tire on the wheel spin problems, it is the low speed high acceleration run that usually are the culprit.
Unlike braking where a too abrupt use of the brakes, merely will lock the wheel before it will spin the wheel on the tire and the speed of the rotation of the tire/wheel are at a very high level.

speedsense - 02 October 2009 11:57 AM

headbanginrockstar - 02 October 2009 10:16 AM

GreyWolf74 - 02 October 2009 08:32 AM

If there is rubber build-up on a wheel, it is probably cast-off rubber from tyres, which is very sticky when hot.

Is not the rubber compoud different from thread to sidewall to bead as far as "sticky" goes???

The construction is different and the amount of rubber used is greater in the bead. Though the compound it's self is most likely the same across the contact surface. Compound for contact area tends to be different that the rubber for the rest of the tire. Kinda like a donut laid over a bigger donut.

A lubricate isn't usually applied until the mounting process takes place. Typically they have to inflate the pressure to over 60 lbs to get the tire to bead.

This is the answer (highlighted in large text) to the original question. The bead has an interference fit with the rim (it has a smaller radius). It requires extreme pressure to seat the bead on the rim, and once seated, it stays there because it is being stretched over the rim like a rubber band. It then takes extreme force from a tire changing machine to break it loose again (someone mentioned this already). If you visit a tire shop and observe the work, you will hear a loud pop when the bead seats.

CHEMENG - 17 October 2009 12:13 PM

[The bead has an interference fit with the rim (it has a smaller radius). It requires extreme pressure to seat the bead on the rim, and once seated, it stays there because it is being stretched over the rim like a rubber band. It then takes extreme force from a tire changing machine to break it loose again (someone mentioned this already). If you visit a tire shop and observe the work, you will hear a loud pop when the bead seats.

Some of that stretch is to overcome the safety bead 'hump' which is slightly raised from the actual bead surface. It is what takes most of the excessive pressure to overcome.

An illustration of this 'hump' can be found here - http://www.actiontireco.com/scripts/basicwheel.asp