Axle And Tread Drag

Two primary forces of friction oppose the motion of your car's wheels. One slips in at the wheel's center. The other pulls at its contact point with the track.

Shearing, Smearing and Pulling, Stretching

Axle friction shears metal and plastic molecules apart at the center of the wheel. Luricants can reduce the amount of shearing or replace the materials involved. It is even possible for a lubricant to change the nature of the friction from a sliding friction force to a rolling one. This happens if the lubricant shreds into microscopic balls that roll between the axle and wheel.

Tread or track friction provides traction for your wheels to spin. But traction comes with a cost. It is not a shearing force like axle friction, but a ripping or pulling force. As the wheel presses into the track, bonds form between the two. As it lifts from the contact point, the bonds between the two materials stretch like tacky glue. The pull from the stretch slows the wheel down.

There is another way rolling friction can manifest; rolling over imperfections in the track. Little ridges, bumps, dips and pits cause the entire wheel to lift and drop innumerable times over the course of the race. As the wheel jumps up and down, so does the axle it supports, changing both the effective tread and axle friction forces.

Torque

A torque is a force that acts through a distance rotated through. Your front axle supports Ff ounces of weight. The axle bearing down on the wheel bore produces a torque from axle friction. Its force is nfFfcos(O-q)/2 between the axle and wheel bore of one front wheel. This is written this way whether it is actually a shearing force or a rolling force. The value of nf will change, not its written expression.

Rolling Resistance

As a wheel rolls on a flat surface, it meets with all sorts of bumps and pits. On a track, these irregularities are typically much smaller than the radius of the wheel. But it still takes more force to roll over a bump.

At the point of just being able to roll over the bump, the sum of the torques about the contact point with the bump is:

0 = Ws-Fq since W pulls downward and F pulls horizontal.

As the size of the bump becomes smaller, the vertical separation q approaches R. Using R as the limit we can rearrange the expression as:

FR = Ws where the symbol s now becomes a length characterizing the resistance. It is called the coefficient of rolling resistance. Now F becomes,

F = Ws/R

It is easier to work with dimensionless numbers so let u = s/R be the coefficient of rolling friction that will be used in the models.

f = -uW, the minus sign is needed since the friction force is opposite in direction.

Rolling friction on a ramp

When the surface is tilted at an angle O, for O < 0 when the surface slopes downward, the weight of the wheel pulls downward at an angle to the track. This angle is O more than vertical (O+pi/2 in radians). As implicit in the above derivation, we only want to deal with the normal part of the weight. This is the part pushing at right angles to the track surface at the contact point. It's magnitude is WcosO. Substituting in the above for W, we get

f = -uWcosO

In general, when incorporating rolling friction into the models we will use the expression

f = -uN, for N the appropriate force normal to the track surface at the contact point.

This is equivalent to saying that rolling friction depends only on the normal force, just as sliding friction does!

Stickiness

Amazingly, the same result holds for stickiness on the track. The geometry is a bit different. The contact point on the track becomes the fulcrum. The torques are produced by the stickiness that is directed at right angles into the track and the forward force needed to pull the wheel from rest.

On a ramp there is an additional forward pulling force contributed by the part of the wheel's weight that is alligned with the track surface (WsinO). But for the friction force, we are only interested in the normal part of the weight which produces no torque about the contact point.

let G be the "gummy" normal force acting through a distance s. Then

FR = Gs and as before,

F = uG

Rolling friction from rest

When the surface is tilted at an angle O, for O < 0 when the surface slopes downward, the weight of the wheel pulls downward at an angle to the track. This angle is O more than vertical (O+pi/2 in radians). To see the force that must be overcome from rest on a ramp we consider the whole force of weight in the downward direction. The torque caused by the wheel's weight now acts through a lever-arm that is about RO+s long. Substituting this for s,

F = W(RO+s)/R

We get an approximation for small bumps (those less than about R/10)

F = W(O+s/R) = W(O+u)

So tilting the track decreases (O < 0) the force needed to roll from rest by the angle of tilt measured in radians.

Note that when O = u, no force is needed to move the wheel. From rest, the wheel is literally teetering on the top of the bump. The wheel must roll in one direction or the other. With a little geometry, we see the bump must have a height very near Rsin2O (that's RsinOsinO).

When O < -u, the surface slopes down enough so that the force is negative. The wheel "wants" to roll over the bump from rest. The negative force represents a restraining force necessary to keep the wheel from moving.

Component Forces

As the weight of the axle presses on the wheel bore on a ramp, the total force Ff is split up into two components. One part of the force presses through the contact point of the wheel on the track. This part of the force is the normal force that produces axle friction and tread friction. cos(O-q) is the factor that splits this normal force out from the total. The other pushes the axle parallel to the track surface accelerating the car. It gets multiplied by sin(O-q).

The distance axle friction acts through is the radius of the wheel bore, rf, not the radius of the axle. It is the wheel's rotation that is affected, not the axle's. Torque due to axle drag in one front wheel is written as nfFfrfcos(O-q)/2. Note, Ff also depends on the car's angle on the track.

Your car's wheels also produce a torque from the friction caused by the normal part of the axle force at the point of contact with the track. This force for a front wheel is uf(Ff/2+wf)cos(O-q), tread friction. again, cos(O-q) selects the part of the force at right angles to the track. wf is the weight of the wheel which also bears down on the point of contact. Rf, the outer radius, provides the leverage. This torque is expressed as uf(Ff/2+wf)Rfcos(O-q).

The force that propells the axle forward is (Ff/2+wf)|sin(O-q)|. This includes the part of the axle support force that pushes along the track. It also includes the part of the wheel weight that pushes along the track. This combined force acts through the outer wheel radius, Rf, also. (Ff/2+wf)Rf|sin(O-q)| is the torque. q is the shift in the contact angle between the axle and wheel caused by centrifugal force. On a flat surface, q = 0.

Total Torque

The total torque for each front wheel is (Ff/2+wf)Rf|sin(O-q)| - uf(Ff/2+wf)Rfcos(O-q) - nfFfrfcos(O-q)/2 or

(Ff/2+wf)Rf(|sin(O-q)|-ufcos(O-q)) - nfFfrfcos(O-q)/2 or

(Ff/2+wf)Rf|sin(O-q)| - (uf(Ff/2+wf)Rf+nfFfrf/2)cos(O-q)

Your wheels rotate about 100 times as your car speeds down the track. A small change in the rotation angle multiplied by the distance of the lever through which work is performed gives a small distance on the track. It can be added up over the track surface.

Let Rf be the outer wheel radius, dO the small angle rotated through and dy the small distance covered along the track by means of the small rotation. Then

RfdO = dy so dO = dy/Rf

The work done by these forces written as small rotations over the track surface (small torques) is

  / O
 | ( (Ff/2+wf)Rf|sin(O-q)| - (uf(Ff/2+wf)Rf+nfFfrf/2)cos(O-q) ) dO
/ 0
   / y
= | ( (Ff/2+wf)Rf|sin(O-q)| - (uf(Ff/2+wf)Rf+nfFfrf/2)cos(O-q) ) dy/Rf
 / 0
   / y
= | ( (Ff/2+wf)|sin(O-q)| - (uf(Ff/2+wf)+nfFfrf/2Rf)cos(O-q) ) dy
 / 0

Since q = 0 and sinO is constant on the ramp (again no centrifugal force), this evaluates to

( (Ff/2+wf)|sin(O-q)| - (uf(Ff/2+wf)+nfFfrf/2Rf)cos(O-q) )y

which is (Ff/2+wf)y|sin(O-q)| - (uf(Ff/2+wf)+nfFfrf/2Rf)ycos(O-q), the amount of work done to spin the wheel minus the amount of work done by friction to oppose spinning.

If the track were curved, O and q would not be constant and the solution of the integral would depend on knowing what O(y), the slope at each point along the surface, was and how the centrifugal force reduced it through q(y) at each point, y.

On the ramp, the part of the work done by friction is (uf(Ff/2+wf)+nfFfrf/2Rf)ycosO. These terms stand for the work of tread friction and axle friction. Without the "y", each term is the instantaneous force of friction. uf(Ff/2+wf)cosO at each tread, nfFfrfcosO/2Rf at each axle.

On the flat, the part of the work done by friction is (uf(Ff/2+wf)+nfFfrf/2Rf)y. Since O = 0, cosO = 1. These terms stand for the work of tread friction and axle friction. Without the "y", each term is the instantaneous force of friction. uf(Ff/2+wf) at each tread, nfFfrf/2Rf at each axle.

Summing the forward force of all four wheels, 2(Ff/2+wf)|sin(O-q)| + 2(Fr/2+wr)|sin(O-q)| we find (Ff + Fr + 2wf + 2 wr)|sin(O-q)| which is just mg|sin(O-q)|. On the ramp this is mg|sinO| and 0 on the flat.