RocketGuy3

iTrader: (1)

I was debating with a friend about how the weight of a car affects its ability to accelerate. He was contesting that on a RWD car, adding weight to the rear, all else being equal, could make the car accelerate faster (in particular, give it a lower 0-60 time). I told him that is 100% false, and adding weight is ALWAYS bad for performance in dry conditions, at least unless the coefficient of friction between your tires and the road is >1...

As the argument got deeper, though, I decided to whip out a pencil and paper and hammer some stuff out. Being an electrical engineer, I expected to just reinforce basic physics principles that I had taken for granted in the car world for years, but what I found kind of confused the living **** out of me. I have to be overlooking something...

OK, so the formula for frictional force is:

Ff = Fn * u, where Ff = frictional force, Fn = normal force, and u = coefficient of friction, and
Fn = m * g, so
Ff = m * g * u, where m = mass, g = gravitational acceleration, u = same.

And the formula for acceleration, as everyone knows is:

F = m * a

This means that, assuming that the maximum motivational force for a car is limited by the frictional force between the tires and the road (and it has to be, right?):

F = Ff, therefore:
m*a = m*g*u

And the mass (m) cancels out on both sides, leaving you with:

a = g*u

So that means that the rate at which a car can accelerate (or corner, for that matter) is INDEPENDENT of the mass of the car and only depends on gravity and the frictional constant between the tires and the road? This makes NO sense to me, intuitively. I have to be doing something wrong, and it's probably something incredibly stupid and obvious, but I can't for the life of me figure out what it is.

Now to be fair, I guess i can make sense of this for acceleration. Once the tires hook up, after the first bit, the limiting factor is no longer the frictional force, so the F = m*a formula "takes over". But it still doesn't make sense when you think about cornering or braking.

Can anyone help a brotha out here?

Thanks in advance.

Whitigir

The Force from acceleration .....it is not equal to Frictional force

Acceleration force is affected by mass and friction. The more mass, the more friction, the less acceleration if the same force is applied. This is under the same gravity and friction coefficient.

Therefore, the heavier the car, the slower it can accelerate.

Adding weight to the rear is just adding weight to the whole object. Ofcourse the friction increase, with the same torque the car moves slower.

The same as a stock FRS with 1 driver do 0-60 vs FRS with 2 passengers in the rear. The one without extra weight will accelerate faster.

Only under a circumstance where the friction is loss, adding weight will increase friction, now the car moves faster. So you are correct

RocketGuy3

iTrader: (1)

The maximum force that can be used to accelerate the car is the maximum amount of force that the tires are capable of providing, which is the frictional force between the tires and the road.

If that is not the case, please explain why not.

Whitigir

The maximum force the tire can provide is affected by the road condition and the mass of the car. You are leaving out friction coefficient by saying that

gengar

You said yourself that you are calculating the theoretical maximum horizontal force that a tire can achieve on the road. Your calculations are not incorrect in this regard. So long as everything else remains equal, acceleration of a traction-limited wheel will be the same regardless of mass. But you have to understand what you said originally, which is that you're calculating the theoretical maximum.

I think the disconnect you are having is the practical application. It's quite rare for a car to exceed maximum friction in acceleration (unlike braking), so analysis based on theoretical maximum friction is usually irrelevant. Even the McLaren F1 LM in doing 0-100-0 in 11.5 seconds does the braking in less time than the acceleration. There are also so many other factors in determining acceleration. To start, obviously there are four wheels and not one. Funny I should bring up the McLaren F1, since in cars not like it, the weight on every single one of the four wheels will actually be substantially different because of the driver. Are all four wheels even being driven, or only two? What's the contact area between the tire and the ground, as that will change μ? What about the tire compound and tread pattern, which will also change μ? We can go on and on with this.

Anyway, the bottom line is that your friend's question becomes a much more interesting one if you assume that the car is not traction limited. And in reality, it usually will not be, not even on a 0-60 run.

No, that's not right - the limiting factor is always friction. You can't get more force in this context than friction will allow, period. What you seem to mean by "hook up" is the difference between kinetic and static friction. You'll recall from physics class that you used the same equation, so it's no different (just that you used μk or μs depending on the situation).

Whitigir

Gengar got that bug for you !

I8ABMR

Weight is bad always. What can make some platforms better than other for 0-60 times is related to weight transfer and grip to get off the line. An AWD car takes off the fastest since all wheels are pulling, RR cars like the 911 are very quick since the weight of the engine is right on top of the wheels. In the case of the RR car the weight transfer will increase the amount of grip ( friction ). In a FR car the rearward weight transfer will aid in take off but will not equal the type of grip you get from an AWD of RR car

4TehNguyen

iTrader: (1)

its not always the weight, its grip. Thats how the GTR seemingly defies physics with its near 4k curb weight. It used that extra weight to gain grip with its AWD system whose performance gains were much greater than the performance losses of the increased weight.

There seems to be something about 1g of lateral grip. The only way to surpass 1g is to have enough downforce. Thats how F1 cars can do 3-4g in a turn all day. Again its grip and downforce aids that

Whitigir

All of those grip you mentioned above is simply friction coefficient.

Gtr has better acceleration because the computer was programed to monitor all 4 wheels and apply the highest force to achieve maximum friction on each wheel. As soon as one start to slip (friction loss) even just milli seconds, the applied force is reduced, or distributed to other wheels.

Again, everything effect friction coefficient. The only thing you can control is the applied force before friction is loss. Weight is just simply another factor in thousands factor that change your friction coefficient. This has been Mastered by the GTR.

RocketGuy3

iTrader: (1)

You're right that I forgot to state my other assumption in the OP, which was that I'm assuming the hypothetical car we're talking about always has enough power to break traction (at least in first gear).

But if the car is not traction-limited how does the question become more interesting? In that case, we're back to F = m*a, where the F is fixed, and you are increasing the m, so the a has to decrease.


Yeah, I know the difference between kinetic and static, what I meant was that the car would no longer be able to produce enough torque to spin the tires after a certain point. I guess it sort of goes along with the point you were making above.

In any case, thanks for the explanation, I think I see where I was going wrong now as far as acceleration goes, and as I suspected, I was indeed missing something that should have been obvious.

BUT, in cornering conditions, the car IS traction limited, is it not? So how in that case, how does weight affect a car's ability to corner? Again, according to those equations, it seems like it should not. It would seem like a well designed chassis and grippy tires are ultimately all that matter in cornering -- not the weight.

RX_330

The heavier car will have greater inertia. When cornering, the car naturally wants to continue on a linear path, but the tires are resisting that. The heavier car is going to resist the circular path more because of its greater inertial mass, and so the tires will have to fight it more.

RocketGuy3

iTrader: (1)

Right, but the tires also have more grip as a result of having more weight. Both the amount of centripetal force and the amount of frictional force are directly proportional to the weight of the car, therefore the impact that the car's mass has is cancelled out by the formulas I showed in the OP.

I have (or at least *had*) the same intuitive understanding of how this works that you have, but the most elementary physics equations don't seem to back up our thinking here. That's why I was hoping someone who understands physics better than I do could help me see the light here.

gengar

Because we've moved from a situation in which mass is irrelevant to one in which mass is relevant. Your friend's question is of no consequence in the former, but at least we have some room to do analysis in the latter.

Remember, what we're saying now is that the car is not traction limited all the time. In reality, a car's driven wheels are probably going to be traction limited some of the time - just not necessarily the entire time during a 0-60 or 0-100 run. For example, one way you cold pose the new question could be: Is there some hypothetical situation in which increasing the mass sufficiently increases the available friction when the car is traction limited so that it makes up for the additional mass that slows down the acceleration when the car is not traction limited? Probably not a lot of realistic application, but if we massage the numbers enough, surely we can up with some crazy scenario. For one, you alluded to the possibility of crazy scenarios yourself in the OP when you brought up dry conditions.

You also have to take into account all the other factors, including some I mentioned before - weight distribution, changing μ due to varying factors, etc. Just using F=ma doesn't work here. A car is not a completely rigid block moving on a frictionless surface with no air resistance.

Again, theoretically, you're completely correct. When traction limited, max friction (μ*m*g) = centripetal force ( m*(v^2)/r ) , so from that top cornering speed is v = sqrt(μ*g*r), which is independent of mass.

But once again, the reality is different from this theoretical equation: A car is not some completely rigid block, and there are a lot of external factors/forces at play especially when we're talking about high-performance cars or even race cars. For example, once again, μ is not constant because tires deform. When load on a tire increases, μ generally decreases (although typically "less" than the increase in m, so overall friction still increases). This effect is particularly relevant during cornering, because the weight distribution typically puts more load on the outer tires. When cornering, differing slip angles of the tire against the road also cause deformation, and I have the (totally unconfirmed) feeling that this deformation is also greater if the load is greater. In any case, even when m is not explicitly in the equation, if m affects μ, then m matters.

Another big factor is downforce. When you have an additional normal force to increase max friction, that totally changes the equation. If we just represent downforce as N, now we have max friction μ*(m*g + N), so the max cornering speed suddenly becomes v = sqrt (μ*r*(g+N/m)) , so now we see that m is in fact relevant (and greater m decreases v).

As I said before, we could probably go on and on with regard to the differences between reality and theory, but these seem like the big ones as far as having a car rather than a completely rigid block.

gengar

Also, for the people who get obscenely bored by people scribbling junk on chalkboards in classrooms (totally understandable), here's what I'm talking about as far as tire deformation (slow mo of drag cars starting off):


Also here at 10min 58sec (click on the link directly to auto to that point):


Here's one for cornering - not in super slow mo and kinda hard to see, but hopefully it sorta illustrates the point.

Whitigir

Gengar gave a lot of effort for this. Thanks gengar. He said it very clear. μ Is not constant. It is not only the tires deform that varies it, it is also mass, aero drag, velocity, road conditions, etc....

Also in cornering, gengar said that μ kinetic is different than static μ. The both also affected by mass. A lighter car brake at a longer distant than a heavier one, assume everything else being the same

μ Will always change, and mass is just one of many factors that can affect it. You can only control the applied forces to maximize your friction as needed.

FE. ABS, the car is programed to pulstate the brake force to achieve maximum friction without losing it (controled applied force)

The same as slip differential. As long as a slip is occured, the mechanism will transfer the torque to achieve maximum friction before losing it, or losing too much of it.

It goes on forever wheather u add weight or decrease weight to gain friction here, or lose fiction there, or downforce, or anything else. You are just in the box of μ.

So under your perfect condition where a 1000 lbs block assumed with optimal forces applied to accelerate on the optimal friction (max friction allowed by the tire). You are correct that the block is not affected by mass. Because you just assumed optimal forces for optimal friction. Which realistically not existed. Because the optimal forces you assumed totally ignored μ as your maximum friction changes, you also changes your applied forces.