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I am making a 2D driving video game, and I would like to know the "simple" formula for calculating the friction force between the car and the road. I have read lots of friction diagrams involving balls rolling down inclined planes, but I'm having trouble applying it to this problem.

Consider a car driving on a horizontal road, with some coefficient of friction, $\mu$. Its velocity is $v$, mass is $m$, and a constant driving force is accelerating the car at $a\ \mathrm{m/sec}^2$ in the forwards direction. What is the friction force, $F$, that acts on the car in the backwards direction?

I guess that the friction force is proportional to the car's speed, but I can't explain this guess. The reason I think that is because of a handful of case studies. Say that the driver's foot is on the pedal such that the car will get up to $16\ \mathrm{m/sec}$:

  • If the car's velocity is $0\ \mathrm{m/sec}$, there is no force accelerating it backwards (it is stationary).
  • If the car's velocity is $8\ \mathrm{m/sec}$, the friction force accelerating it backwards must be less (in magnitude) than the driving force, and hence the car keeps accelerating.
  • If the car's velocity is $16\ \mathrm{m/sec}$, the friction force accelerating it backwards is equal (in magnitude) and opposite to the driving force, and hence the car maintains a constant velocity.
  • When the driver releases the pedal, the same friction force accelerates the car backwards, but now there is no driving force, so the car drifts slowly to a halt.

It would therefore make sense that $F \propto -\mu v$, such that in the first case, $F = 0$; in the second case $F = -\frac{1}{2}ma$; in the third and fourth cases, $F = -ma$, where $a$ is whatever acceleration is required to maintain a constant velocity of $16\ \mathrm{m/sec}$. But I can't figure out the exact relationship between $F$ and $v$, or explain why they are proportional.

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See my answer here, physics.stackexchange.com/questions/12874/… and to formulate the total retarding force $F$ you may want to combine aerodynamic, tire, and maybe or maybe not drive-train losses. That said, I'm not convinced this question is quite different enough from past questions so I'll wait before offering specific equation forms. –  AlanSE Aug 1 '11 at 14:34
    
Could you clarify your question a bit... the bullet 3 says that all engine power goes to friction between road and car? I would say that the biggest part of the car power goes to air drag and only a small portion goes to friction in the bearings etc. Do you want to know a formula for friction in the bearings or in the air or both? –  Juha Aug 2 '11 at 14:14
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I was using the word "friction" to refer to the force of the road on the car which retards its motion in the backwards direction, and assuming no other forces (such as air resistance). Remember, this is a simple 2D driving game. I am not trying to accurately model how a car works. I really just want to know if $F \propto -\mu v$ is on the right track. Having a look at rolling resistance on Wikipedia makes me think that it isn't; the friction force is (close to) constant for a given mass and surface. –  mgiuca Aug 3 '11 at 4:33
    
If it is solid-solid friction you can approximate it as a constant. And in this case it is. See the link that Zassounotsukushi posted above. It has a link teslamotors.com/blog/roadster-efficiency-and-range. See the figures, especially where is Wh/mile vs. mph with different losses (be careful both axis contain distance). Tires seem to have constant energy loss per mile... I am wondering how would F vs. mph figure look like if Wh/mile vs. mph looks like that. –  Juha Aug 3 '11 at 20:38
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Friction is irrelevant unless the car is actually sliding on the pavement, or burning rubber. You're talking about drag, not friction. There's aerodynamic drag plus rolling drag. When the driver presses the brakes, then there's friction. –  Mike Dunlavey Oct 13 '11 at 13:02
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7 Answers

Just to clarify, you mean the friction between air and the car (this is called drag, see e.g. stokes law http://en.wikipedia.org/wiki/Stokes%27_law). There is some friction in the bearings of the car but this is negligible to the resistance of the air.

The rest of the answer assumes that you mean drag.

You are right: the first degree approximation to drag force is that it is proportional to the velocity. You can also use higher order approximations (e.g. F = kv^3), but it makes the calculations quickly very difficult (depending of course how do you want to solve them).

How I would write this is (' means derivative respect to time, x is place, m and k are constants, a is acceleration)

f = ma + kv = mx'' + kx' => ax'' + bx' + c = 0

This is now a general differential equation (a,b,c are constants). This helps later if you have lots of forces in your system.

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I have a couple of issues with what you wrote. Firstly, saying that any friction force is proportional to velocity is misleading. My short answer to this question would be $F=static+drag=\mu m g + k v^3$, which is what you wrote... sort of. The acceleration is then the remainder of the engine force minus the friction forces. You have all the right components, but I think you need to organize it better and reevaluate your answer. –  AlanSE Aug 1 '11 at 15:58
    
Ok, I'll clarify the friction/drag issue, but I still claim that the simplest approximation for drag forces is linear. The third power makes solving of the equations unnecessary difficult. I now assume that there is a slight contradiction in the question (friction force between the road and tire points forward and at bearings the friction is negligible compared to air drag) –  Juha Aug 2 '11 at 13:57
    
Can you give an example of any friction force at all that is proportional to velocity? The static and dynamic coefficient of friction are constant, not linear, so if you're looking to keep things simple, use the static/dynamic coefficient times $mg$ and the movement will be of a strictly $a t^2 + b t + c$ form in 1D. –  AlanSE Aug 2 '11 at 14:53
    
Example of a drag force that is linear in velocity is stokes law. I agree that friction constants are constants (solid-solid interface), I now mean solid-fluid "friction" that is called drag. The reason I am talking about drag is that the question was about simple car simulation where all the work produced by the engine is going to friction (which in real world is drag from the air). –  Juha Aug 2 '11 at 15:11
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You will have to track the motion of the wheels. If the wheels roll, then a small constant coefficient of friction applies. When the wheels slip, a larger coefficient of friction applies. Both do not really vary with speed.

Assuming you have hills (up/down) then you have to track how much weight % is on the front or the back to get your total friction/traction available. The rotational motion of the wheels get affected by the engine torque and the friction available.

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The confusing part in calculating the friction of a car tire on the road is that where the tire contacts the pavement it’s not moving relative to the pavement. For some reason it’s difficult for people to understand. If a car is going 60 m/h where the rubber contacts the pavement the speed is 0 and the top of the tire is going 120 m/h. So the coefficient of friction for the tire pavement contact is static. The coefficient of friction of a dry tire on dray pavement is approximately 1. Conditions: 1) Say you’re going 60 m/h and you slam on the brakes the static contact is broken and you start slipping so the mass of the car comes in to play and you apply kinetic friction formulas. 2) How hard do you step on the brakes before static friction gives in and that is momentum over kinetic friction. 3) If the car is turning on a curve the mass of the car and the radios of the curve will establish the angular momentum and this will give you the speed at which static friction gives and the car starts skidding.

Now your coefficient of static friction is almost arbitrary as far as the game goes. Consider the extremes a 5” wide car tire cold and smooth pavement Vs a formula One 18” wide sticky rubber and hot on a rough track. That gives you a pretty wide rage to play whit.

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This is utterly wrong! The "friction" of the tire (better named "sesistance to roll") is 99 % by deformation of the tire and mechanical hysteresis of that deformatin. –  Georg Aug 2 '11 at 10:01
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This is the figure I mean above in the comments for question. The units in the figure are a bit weird, but If I do this correctly...

W*h/mile = J/h*h/mile = N*mile/h*h/mile = N

So, y-axis has force and x-axis has velocity (you may want to check the above). This shows that tires have nearly constant friction force respect to velocity.

The figure is from www.teslamotors.com/blog/roadster-efficiency-and-range and the points go to Zassounotsukushi for revealing this web page (@Z you may want to make the correct answer, I am just posting this because of the figure).

F vs. v

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Good shot at it! I'll suggest a modified form for your equation here: $\frac{W \cdot h}{mile} = \frac{J}{h} \times \frac{h}{mile} = \frac{power}{speed}$. I don't know if this helps you very much (I admit I don't know what $N$ was supposed to mean), but it is an analog to power = speed x force. It also kind of shows force being the inverse of gas mileage.. which is a funny thought but all this is interpretation dependent anyway, but anyway, I digress... –  AlanSE Aug 5 '11 at 3:20
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You want a friction acceleration of the following form:

$$ A_f = - A - Bv - Cv^2$$

where you fit $A$, $B$, $C$ to three flat-road tests of how long a car will coast to a given reference speed (say 5mph) from 10mph, 45mph, and 65mph (1m/s \approx 2 mph).

$A$ is the slow-speed reduction in speed per second, whose origin is solid friction in interior components, $B$ is the exponential decay rate of moderate speeds, while $C$ gives the air drag for ultra-fast speeds. I estimate

$$A \approx {1\over 2} mph/s$$ $$B \approx {1\over 20 s}$$ $$C \approx {1\over 800} {1\over \mathrm{mph}s}$$

These estimates are based on the observations that it takes about 10 seconds to go from 10mph to 5mph, that if you are driving 40, it takes about 15 second to coast to driving 40/e, and the crossover for solid friction to air-friction is around 55mph, and correcting for mutual dependency. If you want better values, you need a car and a stopwatch.

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Your case study is erronous if you are only considering friction between the tyre and the ground. For a body that is rolling with uniform velocity, there is no need for friction to act. In rolling, there is no relative displacement at the point of contact, so we use static friction. Static friction is NOT $\mu_smg$, that is only its maximum value. For a body rolling uniformly, friction is zero in the absence of other forces.

When a car accelerates, the friction force is forwards. For a body to accelerate, there must be an external force. There are two external forces here, gravity (useless for forward acceleration), and friction. The car 'pushes' back on the ground, and the ground exerts an equal and opposite forward push. Thus push manifests itself as friction. A similar thing happens while walking; friction need not always oppose net motion(it opposes relative motion). In a car, the treads do not move with respect to the ground for the duration of their contact, so there is no relative motion to oppose. The friction is free to choose its direction based on the other forces.

So, in each of your cases, the frictional force will be (mass of car)x(acceleration) forwards.

If the car is skidding, then you need to apply kinetic friction, which will be constant and equal to $\mu_kmg$. It will oppose the spinning of the wheels.



What you actually want is the drag forces due to viscosity and the internal friction forces(at the axle).

The internal friction force will probably be a constant for the car, you could approximate its torque to $\tau=r_{axle}\times\mu_{axle}mg$. Now, due to this torque, there will be a frictional force $$\frac{\tau}{r_{wheel}+\frac{I_{wheel}}{m_{car}r_{wheel}}}$$ (I is moment of inertia). I calculated this with similar concepts as "balls rolling down inclines", so I doubt you want the derivation here. Now, this force will be backwards, and it will retard the car. Since you're programming for a game and need a formula with values plugged in, I'll give you typical values of any constant you need(ask in the comments=).

Due to drag, there will be a drag force $\approx Bv$, where B is some constant.

Really, the easiest thing to do would be to set the retarding force as $$A+Bv+Cv^2+Dv^3....$$. You may set higher order terms to zero if you wish. What I've given above lets you calculate $A$ (since the frictional force described above is constant), and tells you where $B$ comes from, but such a formula would be better for programming purposes (The $v2$ etc terms come from the fact that cars are not simple bodies). You can fiddle around with values till you get something that feels realistic in your simulation (then use a variety of nearby values for different cars). I suggest you keep A to be the calculated value, though. Right now, it would be good to create a method getFriction(Car c, Velocity v), and have some dummy code in it till you have finished with the rest of the program. Then you can plug in the formulae and fiddle with them to see how the end result looks.

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I stumbled on a reference which summarizes what I think is the most correct answer to this question. You have several options, but I think we should only use terms that we have a physical reason to write. Here's the reference:

http://usna.edu/Users/physics/schneide/Buick.htm

They use a lot of unnecessary details like time between stopping that we're not interested in. I agree with their graph but not their equation. So here is my equation to explain their graph. I also limited it to flat roads (no hills).

$$F = \frac{P_0}{v} + \mu mg + c \rho A \frac{v^2}{2} $$

From the reference, known values about their car are:

  • The weight, $m=1800 kg$
  • The front area $A = 3 m^2$

I report these because there is no direct measurement available. I would then use their data to evaluate the three coefficients that determine the relative friction from each thing.

my reference

They put a constant term in the transmission factor (the 1/v term). I don't like this, because I want clean mathematics, so I'm bunching that long tail of transmission with the friction coefficient.

$$ \mu (1800 kg) (9.8 m/s^2) = 200 N + 250 N = 450 N \rightarrow \mu = 0.026 $$

$$ \frac{P_0}{15 m/s} = 500 N - 250 N = 250 N \rightarrow P_0 = 3750 W$$

$$ c (3 m^2) (1.3 kg/m^3) \frac{(31 m/s)^2}{2} = 500 N \rightarrow c = 0.27 $$

These are all consistent with what the link claimed, aside from the cases where I willfully used a different kind of definition. Another good thing that these all have physical interpretations, which those units are suggesting. I will avoid getting into the exact interpretation because I feel like there's space for quibbling.

I plotted this on Wolfram alpha. This is my altered version of that link.

my image

This fits expectations fairly well. Take note, however, of the 1/v term. That represents fuel consumption due to constant loads (thus, units of power, of course). That might not be relevant if you're looking for a force, but it can be kind of interpreted as a force. It's a force the engine is exerting against itself (to some fraction of that number) due to idling. It is also the constant electronic loads on the battery... and the charging of the battery itself. It's not the friction of the wheels on the road of air on the car. If you're only interested in those then you might do well to just take the last two terms. If you do that, however, there is no concept of maximum gas mileage, nor should there be. What you're going to do with these terms depends on the application. I just believe this to be the best available option so I posted it.

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$P_0$ should actually be $4000 W$ - you used $15 m/s$ as the limiting speed, when the OP specified $16 m/s$. –  AJMansfield Jun 21 '13 at 19:12
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