Soapbox Racer Physics

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The force F acting to move a soapbox cartie down a slope of angle θ is m g sin(θ), where m is the mass of the cartie and g is the acceleration due to gravity.

In order to accelerate the cartie, this force has to overcome drag, rolling friction and wheel inertia, and whatever is left over accelerates the cartie. So;

m g sin(θ) = Ft + Fr + Fd + FR


Ft = Translational Force (i.e. moving the whole machine down the hill)
Fw = Force rotating the wheels
Fd = Aerodynamic drag force
FR = Rolling friction

Aerodynamic drag Fd is given by;

Fd = -(1/2)ρCdAv2   (see )

Rolling friction is linear in both velocity and vehicle mass, and has a CR (rolling resistance coefficient) of about 0.0055. It is proportional to the force (Newtons) normal to the track. See For all four wheels, we get;

FR = -CRmg cos(θ)

Ft is pretty straightforward - for a cartie of mass m accelerating at a;

Ft = m a

The Fw term is a little more fiddly; the angular acceleration ω of the wheel with moment of inertia I is caused by the torque T

T = I ω

I for a hoop of mass mw is mw r2, and ω is given by a / r, so the above equation becomes;

T = mw r2 (a / r)
T = mw r a

T also equals Fw r, so

Fw r = mw r a
Fw = mw a

i.e. The radius of the wheel has no effect - the extra inertia of the larger wheel is cancelled out by the larger torque applied to rotate it.

Substituting these two terms back into the initial equation we now have;

m g sin(θ) = m a + mw a + Fd + FR
m g sin(θ) = (m + mw) a + Fd + FR

which can be rearranged for a as;

a = (m g sin(θ) - Fd - FR) / (m + mw)

Or, in full (substituting the expressions for Fd and FR

a = (m g sin(θ) - (1/2)ρACdv2 - CRmg cos(θ)) / (m + mw)

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scottishcarties's picture

gravity racer terminal velocity

The equations above can be rearranged to calculate the maximum speed of a soapbox on any given slope (provided the course is long enough).

Maximum speed is achieved when the force on the cartie due to gravity is equal to the forces due to rolling resistance, moment of inertia and drag. i.e;

Ft = Fd + FR + Fw


Fd = Ft - FR - Fw

(1/2)ρACdv2 = Ft - FR- Fw

or, rearranging for V,

V = √ (2(Ft - FR - Fw) / ρCdA)

Substituting the equations above in for the forces gives;

V =  √ (2(m g sin(θ) - CRm g cos(θ) - mw a) / ρCdA)

But when V is at the maximum, acceleration a = 0, so the mw a term disappears;

V =  √ (2(m g sin(θ) - CRm g cos(θ) ) / ρCdA)

So. That's all very clever, but what does it mean? Well - if you plug in some typical values for the variables, you can plot a set of curves to show how mass and drag affect your terminal velocity.

For instance,on a slope of gradient 1 in 10 (10%), with a rolling resistance of .0055, air at standard temperature and pressure, you get the following graph;

Soapbox Maximum Speed

I.e. Heavier carties go faster, but the increase in top speed per Kg decreases as overall weight increases. And it doesn't really matter how much weight you chuck into your cartie if it has the aerodynamics of a warehouse as you'll just never catch a slick and streamlined machine no matter how hard you try.

scottishcarties | January 21, 2013 - 14:23
scottishcarties's picture

wheel size and terminal velocity

The other interesting thing that drops out of this analysis is the slightly surprising observation that the mass and diameter of your wheels has no effect on your terminal velocity.

Well - sort of. There are still frictional losses through tyre deformation, bearings etc, and of course your wheels have aerodynamic drag too. But when you consider your wheels in terms of the force used to increase the speed at which they are rotating, at terminal velocitythere is no force required to do this.

scottishcarties | January 21, 2013 - 15:07

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