# Which car will slide first?

1. Jan 25, 2009

### max604

So i was in this suv with my buddy the other day and we came to a red light on a downward slope. He used his hand brake to stop sliding and somehow we got into a debate about something and i put 50 bucks into a bet. Here it is :

-Assume we have 2 identical cars
-The car weights 1 ton itself
-1 of the car is carrying 1 ton of stock so it weights 2 ton
-We assume the stock has the same centre of gravity as the car.
-We put the 2 cars side by side on a flat road and slowly raise a downward slope
-Which car will start sliding down first?

I'm betting on both car slide at the same time and same speed because the weight is proportional to the force of gravity and the extra force balances out with the extra friction on the wheel and road. My buddy thinks the heavier car requires a bigger initial force to start sliding so the 1 ton car will go first.

2. Jan 25, 2009

### confinement

You are correct!

3. Jan 25, 2009

### rcgldr

The heavier car should start sliding first if static friction follows the same tire load sensitivity that applies to tires when cornering, that the coefficient of friction decreases with normal force:

4. Jan 26, 2009

### Ranger Mike

see dec 30, 2008 post race car physics

Tire traction- realize the more load placed on a tire, the more traction ..but...the tires coefficient of friction decrease. However, up to the design limit of the tire, its traction capacity ( ability to actually transmit force to the road) as opposed to dimensionless coefficient of friction, increases with vertical load.
simplified - vertical load on a given tire increases, the area of the rolling contact patch remains virtually constant, and so the unit pressure of the footprint increases. As the unit loading rises, the rubber has less resistance to frictional shearing and so the coefficient decreases. However, the curve is so gentle , if you graphed this, that the increase in vertical load overpowers the decrease in coefficient of friction.

5. Jan 26, 2009

### rcgldr

But not linearly as explained in the wiki link. So the forward force from gravity on a heavier car on a slope could become greater than the opposing traction capicity, if the vehicle was heavy enough and the slope steep enough.

6. Jan 26, 2009

### max604

omg dude that **** is way too deep for me. let's keep it to simple answers

7. Jan 26, 2009

### Georgepowell

rofl

8. Jan 26, 2009

### Ranger Mike

Jeff
Tire coefficient of friction vs. slip angle is linear
Tire coefficient of friction vs. vertical load is linear
Tire force vs. vertical load graph is linear
if you read my above post...it confirms the heavier car moves first..

9. Jan 26, 2009

### rcgldr

Not according to the wiki article which states that

Fy = c x Fz^s

Where
Fy = maximum horizontal force
c = constant (not coefficient of friction)
s = load sensitivity exponent, in the range .7 to .9 (less than linear, 1.0 would be linear).

coefficient of friction = Fy / Fz = c x Fz^s / Fz = c x Fz^(s - 1)

Note that race cars suspensions can be tuned by making anti-roll bars and/or springs relatively stiffer at one end of the car, shifting the vertical load towards the outside tire of the stiffer end, reducing the maximum grip at the stiffer end, to adjust oversteer or understeer.

10. Jan 26, 2009

### Ranger Mike

Ok ..I got an open nuff mind to take another look ..tho in my business, knowing things are linear ( straight line in my feeble mind) to a point , then roll off is all i require.

My comments came from attached graphs we use to figure best handling combination. Again, things look pretty straight line to a point...but in my opinion

the best way to tune a race car is to design it with proper Roll Centers and CG located for max handling..all other tuning is band aid at track side to get best compromise when changing RC is not practical..

There is only one thing that wins races..tires , tires , tires...we tune the chassis to maintain MAXIMUM tire contact patch at ALL times on the track. The one that does this best, wins...

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11. Jan 26, 2009

### rcgldr

Look at the left graph. For a 2000 lb car, the average load per tire would be 500 lbs / tire. For the car with double the weight the average load per tire would be 1000lbs, and the coefficient of friction is signifcantly less according to the left graph.

I'm not sure why, but for high downforce cars like Formula 1, optimal performance occurs when the camber is set negative enough that the inside tire temps are bit hotter than the outside temps on the front tires. For non-downforce cars, the goal is even temps across the tire tread.

12. Jan 27, 2009

### Ranger Mike

Ok Jeff.. I did some reading up on linearity and empirical definitions and u r right..good show..F1 chassis are the only real race cars MHO with proper suspensions..they have the longest links and best roll center locations, equal and parallel links..they still run huge static negative camber which is why the inside tire temps are 10 to 25 degrees hotter than the outside, depending on the track and geographic location of the tire measurement point. excessively hot temps will cause too much tire wear and reduce braking capacity.
If I remember correctly Indy series cars and un equal length parallel link suspensions which are good but not optimum..

typical door slammers like NASCAR still mandate " stock " truck trailing arm lengths and " stock" A-Arm configurations which are unequal and non parallel link suspensions that have pretty bad camber curves ( relative to F1) and limited on any roll center changing. You just about have to live with what you have so the effort to maintain equal tire temp on sides and middle are an effort to control the stagger ( tire pressure) build.