# Soapbox racer - heavier is better?

Hello all,

The company that I work for was involved in a soapbox competition over the weekend. During the design of the car, the consensus was that adding more weight inside the car (without affecting the shape) was a good thing. With more mass, the racer would have more momentum, so when the force of air resistance acted against it, the relative change in momentum would be less than for a lighter vehicle.

Anyways, we didn't make it to the finals, so now people are saying we should have made the car lighter, since adding more weight increases the amount of friction (neglecting air resistance). The analogy made was that four wheels roll down a hill are faster than four wheels on a car in neutral rolling down a hill. Personally, I'm not too sure about this one, because the air resistance on a car and the air resistance on four wheels is different. What about four heavier wheels vs. four lighter wheels both of the same dimensions on a windy day?

So what do you guys think? Is heavier better? I'm inclined to think that there is a happy medium - some amount of weight that doesn't increase the rolling friction too much but gives the car enough momentum so that it won't be stopped by a strong gust of wind.

Thanks!

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I found this very interesting - I've never done any soapbox racing and haven't previously thought about it. Here's my best guess:

I assumed the soapbox was a block sitting on a slope inclined at some angle $$\theta$$ from horizontal. It starts from rest with only two forces acting on it - gravity and drag (and at the beginning when it's not moving, drag is zero). I assumed the drag force is proportional to the square of the velocity (with constant of proportionality k). If you do the force balance in the direction of the hill, taking downhill as positive, and solve for the acceleration, you get this:

$$a=g sin\left(\theta\right) - \frac{k v^2}{m}$$

So the only place that the mass appears is in the denominator of a negative term - this means that the heavier cars will accelerate down the hill faster than lighter cars.

If you want to find the top speed of the car down the hill, you can set the acceleration to zero (find the speed at which the drag and gravity balance out). You get this relationship:

$$v=\sqrt{\frac{m g}{k} sin\left(\theta\right)}$$

So the heavier cars actually have a higher top speed!

I think the wheels, however, should be as light as possible, or more accurately they should have the least amount of rotational inertia as possible. They will have be accelerated, too, and aren't affected by drag the same way.

Including things like rolling friction and losses in the wheel bearings might change these effects slightly (and might prove that there is actually and optimal weight), but I suspect these effects are very small compared to gravity and drag. Both losses in the bearings and rolling resistance could be modeled as a force that is proportional to the load on the wheel.

Any other thoughts?

-Kerry

This is a good question, and there are several parts to the answer.

The downward (gravitational) accelerating force on a hill of slope angle θ is

Fg = d(mv)/dt = mg sin(θ)

Here, g = 9.81 m/sec2, and m is the total mass, including soapbox, driver, bricks, wheels, etc.
The primary retarding force is air drag. There are several types of air drag. The major one in automobile fuel efficiency is the turbulent air drag, proportional to velocity squared. The other one, at low turbulence is Stokes Law drag, linear in velocity. Both are independent of vehicle mass m, so it represents a relatively larger retarding force for light vehicles. See http://en.wikipedia.org/wiki/Drag_(physics [Broken])

Fd = -(1/2)ρACdv2

Here, ρ is the air density (about 1.2 Kg/m3), Cd is the drag coefficient. A is the frontal area (m2), and v is the velocity (m/sec).
Tire rolling resistance in automobiles is linear in both velocity and vehicle mass, and has a CR (rolling resistance coefficient) of about 0.01. Soapbox Derby hard rubber wheels probably have a lower CR. It is proportional to the force (Newtons) normal to the track. See http://en.wikipedia.org/wiki/Rolling_resistance. For all four wheels, we get

FR = -CRmg cos(θ)

The last force, other than friction, is the effect of the moment of inertia of the wheels. A solid disk wheel of mass mw has a moment of inertia I = (1/2) mwr2, where r is the radius of the wheel. If the wheel is rolling without slipping, r2 becomes v2. If all of the mass is in the rim (rubber), it is mwr2. When all of the mass is in the rim, half the total kinetic energy is in the rotation, and half in the translation. The best way to minimize this is to use light wheels. This creates a slowing down of the soapbox acceleration by an effective force (for a wheel with all the mass in the rubber). For all four wheels we get (recall that mw is included in m).

Fw = -(1/2)mwg sin(θ)

Thus if all of the soapbox mass were in the wheels, the net downward effective translational accelerating force would be +(1/2)mg sin(θ). Putting all these together we get

Fsum = mg sin(θ) - (1/2)ρACdv2 - CRmg cos(θ) - (1/2)mwg sin(θ)

All of these, the accelerating forces and retarding forces, are proportional to the total mass m except the air drag, and the wheel mass. Other than reducing the drag coefficient, the best way to improve acceleration is to load the soapbox up with bricks.

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω

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This is a good analysis Kerry, but I think we can't neglect rolling friction. If the wheels are frictionless, then the heavier car will always win, but since this doesn't happen we see that these effects are important. Since we're dealing with an actual vehicle in reality at a competition, we should try to think about this in the correct context.

You can calculate the rolling friction of your wheels by some simple experiments.
http://www.lastufka.net/lab/cars/why/xrollf.htm
But be aware that the friction will be different for different surfaces, so asphalt may produce more higher friction than a board.

Once you can calculate your coefficient of rolling friction, you can find the negative acceleration due to friction on your vehicle as a function of the mass from:

$$F=\mu_{k}\times N = \mu_{k} \times mgCos\theta$$

$$a=\frac{F}{m}=\frac{\mu_{k} \times mgCos\theta}{m}=\mu_{k} \times gCos\theta$$

where $$\mu_{k}$$ is the coefficient of rolling friction, g is gravity, and theta is the incline of the road.

This is a force of friction, so it opposes your downhill acceleration, and we see that the heavier the car is, the more friction it has.

You should be able to combine these to get an expression for total acceleration. I would suggest using calculus to get a look at that function and see how the mass influences the acceleration and the top speed for different time intervals.

cepheid
Staff Emeritus
Gold Member
This is a force of friction, so it opposes your downhill acceleration, and we see that the heavier the car is, the more friction it has.
I'm a bit confused by how your final expression shows this. It looks like the mass cancels out.

The mass cancels out of the acceleration, but not the friction. Hey wait, that does seem strange.

I think Bob S has the best reply so far. If you want to understand the physics, try to work through his equations.

Bob, is my response correct?

cepheid
Staff Emeritus
Gold Member
All of these, the accelerating forces and retarding forces, are proportional to the total mass m except the air drag, and the wheel mass. Other than reducing the drag coefficient, the best way to improve acceleration is to load the soapbox up with bricks.
Yeah, this paragraph of Bob's seems to clear things up. Even if you take friction into account, the the heavier car is not a detriment.

The mass cancels out of the acceleration, but not the friction. Hey wait, that does seem strange.

I think Bob S has the best reply so far. If you want to understand the physics, try to work through his equations.

Bob, is my response correct?
Yup. Your expression for the rolling friction force is the same as my tire rolling friction force. They both increase linearly with the vehicle mass, and are independent of velocity.

RonL
Gold Member
There should be a few bells ringing in the minds of anyone interested in electric cars. A little suction to the frontal area and a positive displacment in the draft area.  Dale
Mentor
I don't know about soapbox cars, but pinewood derby cars have an advantage if they have their weight in the rear. (More potential energy).

DaveC426913
Gold Member
Does the length of the track not play in?

Intuitively, I'm thinking a heavier car starts off with a lower acceleration, but ends up with a higher velocity.

(No that can't be right. That implies the acceleration is not constant - which can't be right. If one car ends up farther ahead at the end, it would have show this higher acceleration right from the starting line. And that means the length of the track does not change the outcome of the race.)

Has anyone questioned if the coefficient of friction (both rolling and static) depends upon weight?

clustro
This is a very interesting piece of physics. I learned about this a few weeks ago when I was watching that old movie Cool Runnings on tv, about the Jamaican bobsled team. In the movie, it is revealed that John Candy's character was banned from bobsledding because he cheated by putting weights in the front of the sled to make it go faster.

This absolutely puzzled me; the force of gravity (so I erroneously thought) was the only significant force at play, and thus adding more weight seemed immaterial, even detrimental. But lo and behold, when drag terms are added in, it does indeed prove that greater acceleration is produce with increasing mass.

Although there were other benefits of greater mass I read. A heavier sled can get greater total momentum behind it during the running start, and a heavier sled is easier to maneuver/handle. Whether that carries over at all into this problem, I don't know.

Has anyone questioned if the coefficient of friction (both rolling and static) depends upon weight?
The coefficient doesn't (its a constant), but the total frictional force does depend on mass.
However, so does gravity. When you divide by m to get the total acceleration, the masses would cancel. The kv^2 damping term would still get divided by m.

Don't try it.Remember, that there is friction between the kart and the wheels, putting a lot of pressure on it would make it go really really slow(I remember from my mistake with my 8th grade project).Of course, there is air resistance but at low-ish speeds, its not going to matter as much as keeping your wheels free.

Also, you will increase the rolling friction.

The coefficient doesn't (its a constant), but the total frictional force does depend on mass.
The coefficients of friction are not constant with load. Only nominally and ideally are they constant. When a slim margin exists between competitors, small variation could make a difference.

On top of this, no one has brought up unsuspended load as effecting friction, have they?

I agree it looks like the OP's basic intuition is correct -- more weight in the car means more momentum, so proportionally the decelerating effect of air resistance is reduced. Then again, if you can make the car very streamlined, air resistance might not matter that much anyways for the speeds that you will be considering.

Just wanted to add something else that I didn't see mentioned here. I'm not sure how you get the cart going, but if you start out the race by pushing the car, then consider the following. If the car is heavier, it will take longer to accelerate, because it will require a greater force. This could be crucial.

Also, I wanted to make a comment about the discussion of "rolling friction". First of all, the friction between the wheels and the ground is essentially a good thing. You need that friction for the wheels to transmit torque and move your vehicle whatsoever. The punchline is that "rolling friction" is essentially negligible compared to sliding friction, which is the more commonly considered type. I just wanted to mention this because it seems like the comments may be exaggerating this effect just slightly.

Insofar as it is a negative effect, it is a fairly complex effect mainly due to elastic deformation. Values for the coefficient of rolling friction according to Wikipedia look to be very small, on the order of .001-.03 for driving surfaces. For sliding friction, coefficients are generally much higher.

Ref:
http://en.wikipedia.org/wiki/Rolling_resistance
http://en.wikipedia.org/wiki/Coefficient_of_friction#Coefficient_of_friction

Also, I wanted to make a comment about the discussion of "rolling friction". First of all, the friction between the wheels and the ground is essentially a good thing. You need that friction for the wheels to transmit torque and move your vehicle whatsoever.
Actually not. If a mass m is sliding down a slope of angle θ, the downward accelerating force is mg sin(θ). If a wheel of mass m and radius R (with all the mass in the rim) is rolling down the slope, its moment of inertia is mR2, and the downhill accelerating force is only (1/2) mg sin(θ), because the other half of the downward force is accelerating the rotational inertia. So a wheel sliding w/o friction has twice the downward accelerating force of a wheel rotating w/o slipping.

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω

The last force, other than friction, is the effect of the moment of inertia of the wheels. A solid disk wheel of mass mw has a moment of inertia I = (1/2) mwr2, where r is the radius of the wheel. If the wheel is rolling without slipping, r2 becomes v2. If all of the mass is in the rim (rubber), it is mwr2. When all of the mass is in the rim, half the total kinetic energy is in the rotation, and half in the translation. The best way to minimize this is to use light wheels. This creates a slowing down of the soapbox acceleration by an effective force (for a wheel with all the mass in the rubber). For all four wheels we get (recall that mw is included in m).

Fw = -(1/2)mwg sin(θ)

Thus if all of the soapbox mass were in the wheels, the net downward effective translational accelerating force would be +(1/2)mg sin(θ). Putting all these together we get

Fsum = mg sin(θ) - (1/2)ρACdv2 - CRmg cos(θ) - (1/2)mwg sin(θ)
So assuming that the CR term isn't ridiculous, then the best thing to do is load it up with weight, but you want the wheels as light as possible. However, I don't quite understand how you calculated the effective force due to the angular momentum of the wheels.

For wheels with all the weight in the rim, I=(1/2)mwr2, so L=Iω=(1/2)mwrv and F=d(L)/dt = (1/2)mwrg sin(θ) . Where am I going wrong?

Intuitively, I'm thinking a heavier car starts off with a lower acceleration, but ends up with a higher velocity.

(No that can't be right.
I had the same thought process....

The coefficients of friction are not constant with load.....Has anyone questioned if the coefficient of friction (both rolling and static) depends upon weight?
That's the thought I had when reading posts...seems like it should but the net result may be insignificant....

So a wheel sliding w/o friction has twice the downward accelerating force of a wheel rotating w/o slipping.

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
Even if what you say is true, this is impossible. A wheel cannot slide without friction. It's irrelevant.

Well, it was in response to someone else's post that a wheel requires friction to roll. So it's relevant to that. :)

Also, I wanted to make a comment about the discussion of "rolling friction". First of all, the friction between the wheels and the ground is essentially a good thing. You need that friction for the wheels to transmit torque and move your vehicle whatsoever.
Bob S's response to this might be true, too, but I think a better argument is that this rolling "friction" (I think rolling resistance is more accurate) is different from and unrelated to the friction required to transmit driving/braking/cornering forces to the road. Rolling resistance is always a bad thing. Also, I'm not sure why we're discussing sliding friction here... did I miss a post? Unless we're looking at cornering/braking forces, I don't see a need to consider any sliding friction - I think assuming there is no slipping between the wheels (undriven in a soapbox racer) and the ground is safe.

-Kerry

Yea, rolling resistance is probably better terminology, because that's really what we're talking about. But my point was that rolling resistance is a (relatively) minor effect, and you are always going to have some of that no matter what wheels you use.

So assuming that the CR term isn't ridiculous, then the best thing to do is load it up with weight, but you want the wheels as light as possible. However, I don't quite understand how you calculated the effective force due to the angular momentum of the wheels.

For wheels with all the weight in the rim, I=(1/2)mwr2, so L=Iω=(1/2)mwrv and F=d(L)/dt = (1/2)mwrg sin(θ) . Where am I going wrong?
For a wheel with all the mass in the rim, the moment of inertia is I = mwr2. [Note: no factor of 1/2]. The rotational energy with the wheel rolling w/o slipping is
WR = (1/2) I w2 = (1/2)mw(rw)2 = (1/2)mwv2
since for a wheel rolling w/o slipping rw = v, where v is the translational velocity.
But the translational energy of the wheel is also WT = (1/2)mwv2
So the total kinetic energy at any time is WR + WT = mwv2
So the effective translational force on the wheel is only (1/2) mg sin(θ). The other half is used to spin the wheel (moment of inertia).

In your equations, F·r = d(L)/dt. since abs( F x r) = F·r is a torque.

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
± − · × ÷ √

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