Exploring Lifted Wheels

When a wheel is lifted, not only does the total rotational inertia of the wheels decrease but the weight on the other wheels increases. The car accelerates more with the decreased wheel inertia but slows with the increased axle friction. Beginning with an energy equation modified for number of wheels lifted, this balance between inertia and friction is explored. In the process, we find some limiting cases in which lifting wheels won't help.

A Model

The model energy equation can be changed to directly reflect the number of wheels touching the track. j is used to indicate how many wheels are touching the track by making these assignments:

For none lifted, j = 2; for one lifted, j = 3/2; for two lifted, j = 1.

Now the energy realation with no air resistance or tread friction can be written.

E = mgH = mv2/2 + jIv2/R2 + nr(m-2jw)(L2+L1cosO)/R

Let's separate out the amount of axle friction due to the extra weight of the lifted wheels (the friction work term with j in it).

E = mgH = mv2/2 + jIv2/R2 + nrm(L2+L1cosO)/R - 2nrjw(L2+L1cosO)/R

The reduction in inertial energy from the four wheels on track configuration is (2-j)Iv2/R2. The increase in work done by axle friction is 2nr(2-j)w(L2+L1cosO)/R. Which is greater? If the inertial energy reduction is greater, then the car will be faster.

The inequality of interest is

(2-j)Iv2/R2 > 2nr(2-j)w(L2+L1cosO)/R

In order to find limiting conditions or evaluate this inequality, we need to find the velocity, v, from the energy relation above. Rearanging it we have,

(m + 2jI/R2)v2/2 = mgH - nr(m-2jw)(L2+L1cosO)/R

Let M = (m + 2jI/R2) to simplify a few steps; it's the inertial mass of the car.

Mv2/2 = mgH - nr(m-2jw)(L2+L1cosO)/R

Now we isolate v2 on the left

v2 = 2(mgHR-nr(m-2jw)(L2+L1cosO))/MR

Getting back to the inequality, we can now substitute for v2 to get

2(2-j)I(mgHR-nr(m-2jw)(L2+L1cosO))/MR3 > 2nr(2-j)w(L2+L1cosO)/R

Simplify

I(mgHR-nr(m-2jw)(L2+L1cosO)) > MR2nrw(L2+L1cosO)

Expand M and distribute terms in m and w.

ImgHR - Inrm(L2+L1cosO) + 2Inrjw(L2+L1cosO) > mR2nrw(L2+L1cosO) + 2Inrjw(L2+L1cosO)

Cancel out the jw terms (the result does not depend on how many wheels are lifted)

ImgHR - Inrm(L2+L1cosO) > mR2nrw(L2+L1cosO)

Divide through by m

IgHR - Inr(L2+L1cosO) > R2nrw(L2+L1cosO)

Divide through by the track length terms

IgHR/(L2+L1cosO) - Inr > R2nrw

Simplify more

IgHR/(L2+L1cosO) > nr(wR2 + I)

IgR/nr(wR2 + I) > (L2+L1cosO)/H

This expression pits wheel parameters against track parameters. The effect of lifting wheels, depends on;y on the wheels/axles and how long the track is. At some point, the track will be long enough so a particular set of wheels can not benefit from lifting. How long?

For kit wheels, the left hand side evaluates to 0.000048*386.088*0.59/(n*0.049*(0.0002158*0.59*0.59+0.000048))= 1812.40/n inches. Using typical values for the track parameters, the right hand side is (259.5+110.5*0.9325)/39.2 = 352.54/39.2 = 8.99 inches.

1812.40 > 8.99n since n, the friction coefficient is most always less than one. You'd have to intentionally get the wheel stuck to have a coefficient of 1.

Dividing 1812.40 by 8.99 gives

n < 201.60. So lifting wheels works, unless you glue your wheels to your axle! In that case, it doesn't matter how long the track is.

For n = 0.01, a fast car, the track flat length, L2, would need to be,

1812.40/0.01 > (L2+110.5*0.9325)/39.2

39.2*181240-103.04 > L2

L2 < 7104504.96 inches = 112.13 miles. In order for a pinewood car to go that far, the ramp would likely have to run down a mountain slope! This is not feasible for most races.