Exploring Weighted Wheels

When raced by themselves, the wheels roll down the track more slowly than they do when mounted on the car! It is the weight of the car pushing the wheels forward via axles that speeds them up on the ramp. But that same weight focused at each axle pushes down on each wheel creating friction, axle friction.

The heavy wheeled car gets to the bottom of the ramp last

mgH = Mv2/2 => v2 = 2mgH/M so v is less as M is larger.

The heavy wheeled car can't win without reduced friction

mgH = mv2/2 + 2Iv2/R2 + nb(L1cosO+L2)r/R

mgH = mv2/2 + 2Iv2/R2 + nb(L1cosO+L2)r/R

v2(m+4I/R2)/2 = mgH - nb(L1cosO+L2)r/R

v = \[2(mgH - nb(L1cosO+L2)r/R)/(m+4I/R2)]

v = \[2R(mgHR - nb(L1cosO+L2)r)/(mR2+4I)]

When b, the body mass, is decreased the numerator increases making the speed, v, greater. But as I increases when that weight removed from the body is put into the wheels, it increases the denominator and makes the speed, v, lower. It is a tug-of-war. We want to know which wins, b or I.

How the moment of inertia changes when the body weight changes

When the body weight is reduced by Db and split among the wheels, assuming all four for now, the change in their moment of inertia is determined by considering the following model. It is the same model used to find the moment of inertia of AWANA and Boy Scout kit wheels.

I = sum of mi(Ri2+ri2)/2 for each cylinder indexed by i.

If we add another cylinder of mass, s, to the wheel where the outer, R, and inner, r, radii are known, then our moment of inertia model becomes,

I = s(R2+r2)/2 plus the sum of mi(Ri2+ri2)/2 for each original cylinder indexed by i.

Since the only cylinder affected by the weight change is the new one, we can immediately get the derivative of I with respect to s.

dI/ds = d(s(R2+r2)/2)/ds

Most of the expression is a factor without dependence on s, so it factors out

dI/ds = (ds/ds)(R2+r2)/2

s varies perfectly with itself, so ds/ds = 1

dI/ds = (1)(R2+r2)/2

And finally, the change in moment of inertia of the wheel brought about by a change in the wheel's mass by adding a ring of material is

dI/ds = (R2+r2)/2

Now note that if we call the change in body mass, Db, it has a negative value since it's being reduced from b, the body mass. Since Db is distributed to four wheels as an additional mass, s,

s = -Db/4

So the change in moment of inertia, DI, caused by a change in body mass, Db, is

DI = (dI/ds)s = -Db(R2+r2)/8

How the speed, v, depends on the body mass, b

dv/db = d(\[2R(mgHR - nb(L1cosO+L2)r)/(mR2+4I)])/db

dv/db = (d(2R(mgHR - nb(L1cosO+L2)r)/(mR2+4I))/db)/2\[2R(mgHR - nb(L1cosO+L2)r)/(mR2+4I)]

But the \[...] is just the speed, v, so

dv/db = (d(2R(mgHR - nb(L1cosO+L2)r)/(mR2+4I))/db)/2v

dv/db = 2R(d(mgHR)/db - d(nb(L1cosO+L2)r)/db)/2v(mR2+4I)

dv/db = 2R(0 - n(db/db)(L1cosO+L2)r)/2v(mR2+4I)

dv/db = -Rnr(L1cosO+L2)/v(mR2+4I)

How the speed, v, depends on the moment of inertia, I

dv/dI = d(\[2R(mgHR - nb(L1cosO+L2)r)/(mR2+4I)])/dI

dv/dI = (d(2R(mgHR - nb(L1cosO+L2)r)/(mR2+4I))/dI)/2v

dv/dI = 2R(mgHR - nb(L1cosO+L2)r)(d(1/(mR2+4I))/dI)/2v

dv/dI = -R(mgHR - nb(L1cosO+L2)r)(4dI/dI)/v(mR2+4I)2

dv/dI = -R(mgHR - nb(L1cosO+L2)r)(4)/v(mR2+4I)2

dv/dI = -2v2/v(mR2+4I)

dv/dI = -2v/(mR2+4I)

The battle between body mass, b, and moment of inertia, I, begins

The total derivative of speed with respect to body mass, b, and moment of inertia, I, is the sum of the derivatives multiplied by their changes, Db and DI.

Dv = (dv/db)Db + (dv/dI)DI

Substituting for the derivatives and the change in moment of inertia, DI, in terms of Db we get

Dv = -Rnr(L1cosO+L2)Db/v(mR2+4I) - 2v(-Db(Rs2+rs2)/8)/(mR2+4I)

Simplify

Dv = -Db(4Rnr(L1cosO+L2) - v2(Rs2+rs2))/4v(mR2+4I)

Since Db is negative, the change in speed at the end of the race, Dv, is positive if 4Rnr(L1cosO+L2) - v2(Rs2+rs2) is positive.

0 =< 4Rnr(L1cosO+L2) - v2(Rs2+rs2)

Note that this says that at a given speed, v, the ring of added mass, Rs2+rs2, must be smaller than a certain size, say Rsc2+rsc2 in order for the speed to be increased. We can solve for this maximum ring size, Rsm2+rsm2.

v2(Rsm2+srm2) = 4Rnr(L1cosO+L2)

Rsm2+rsm2 = 4Rnr(L1cosO+L2)/vc2

Using some typical values for the parameters in this equation we can solve for a typical maximum ring size.

Rsm2+rsm2 = 4*0.59*0.2*0.049*(110.5*0.9325+249.5)/(150)2

Rsm2+rsm2 = 0.000362 in2

The smallest the inner radius could be for an AWANA wheel is about 0.118 inches. Using this as the value of rsm2, we can solve for the outer radius Rsm.

Rsm2 = 0.000362 - rsm2 = 0.000362 - (0.118)2 = 0.000362 - 0.013924 = -0.013562

But that's negative! So no ring of shifted weight can be used to speed up the typical car! How can the parameters be changed to raise Rsm? The outer wheel radius, friction coefficient and track lengths can be increased or the given finish speed decreased. Of course, increasing the friction coefficient and track lengths will lower the finish speed too! But we don't want to raise the friction coefficient and we can't change the track lengths, so we can't increase the speed of our cars by shifting weight from the body to the wheels unless we have bigger wheels!

How big do the wheels need to be?

Let's solve for R in the condition above and evaluate with typical parameters and the 0.118 radial limit imposed by the shape of the AWANA wheels.

Rsm2+rsm2 = 4Rnr(L1cosO+L2)/vc2 from above.

R = vc2(Rsm2+rsm2)/4nr(L1cosO+L2)

Now evaluate to get a ball park guess.

R = 150@*(0.118@+0.118@)/(4*0.2*0.049*(110.5*0.9325+249.5)) = 45.34 inches

This value is very large. Keep in mind that the wheel must keep all of its properties like moment of inertia and weight the same. A wheel like that would be very thin and probably wouldn't be able to hold up the car! Besides, we evaluated at the minimum radius for Rsm, 0.118 inches. So whatever weight is shifted from the car body would have to fit inside that raduis.

Proof that dt/dyc = 0 is not attainable

Looking at the total race time equation from the model, we can see that longer coasting distance, greater inertial mass and less friction on the flat, make the race time longer.

t = \[2M/Df] ( Lr/\[yc] + \[yc] - \[yc-Lf] )

The only term that could decrease the time is Lr/\[yc]. But finding the value of yc where it could make the time better shows that yc would have to be only a couple of feet (high friction or low ramp) and the flat would have to be less than a quarter the length of the ramp! These conditions are not met in typical Grand Prix races, so for our purposes, the time is always made worse by increasing the wheel weight.

Proof:

dt/dyc = \[2M/Df] ( -Lr/\[yc]3 + 1/\[yc] - 1/\[yc-Lf] )/2

0 = dt/dyc = 1/\[yc] - Lr/\[yc]3 - 1/\[yc-Lf]

1/\[yc] = Lr/\[yc]3 + 1/\[yc-Lf]

1 = Lr/yc + \[yc/(yc-Lf)]

1/(1 - Lr/yc) = \[(yc-Lf)/yc]

yc/(yc-Lr) = \[1-Lf/yc]

yc2/(yc-Lr)2 = 1-Lf/yc

yc2 = (yc-Lf)(yc-Lr)2/yc

yc3 = (yc-Lf)(yc-Lr)2

yc3 = yc3-2yc2Lr+ycLr2-yc2Lf+2ycLrLf-Lr2Lf

0 = yc2 - yc(Lr2+2LrLf)/(2Lr+Lf) + Lr2Lf/(2Lr+Lf)

0 = yc2 - ycLr(Lr+2Lf)/(2Lr+Lf) + Lr2Lf/(2Lr+Lf)

yc = Lr(Lr+2Lf)/2(2Lr+Lf) +/- \[Lr2(Lr+2Lf)2/4(2Lr+Lf)2-Lr2Lf/(2Lr+Lf)]

yc = Lr(Lr+2Lf)/2(2Lr+Lf) +/- Lr\[(Lr+2Lf)2-4Lf(2Lr+Lf)]/2(2Lr+Lf)

yc = Lr(Lr+2Lf)/2(2Lr+Lf) +/- \[Lr2+4LrLf+4Lf2-8LrLf-4Lf2]/2(2Lr+Lf)

yc = Lr(Lr+2Lf)/2(2Lr+Lf) +/- \[Lr(Lr-4Lf)]/2(2Lr+Lf)

yc = ( Lr(Lr+2Lf) +/- \[Lr(Lr-4Lf)] )/2(2Lr+Lf)

0 < Lr-4Lf => Lf < Lr/4, the flat must be less than a quarter the length of the ramp for the critical coasting distance to exist.

yc = ( 100*(100+40) +/- \[100*(100-80)] )/(2*(200+40))

yc = ( 14000 +/- \[2000] )/480

yc = ( 14000 +/- 44.721 )/480 = 29.26 or 29.07 - would need very high friction