A Virtual Race Of Frictions

The task set forward by the question What is the worst source of friction? is carried out here. Using the parameter values listed on that page and the usual AWANA track and kit parameter values, the virtual race between the four cars is run. The resulting times and speeds are used to compute loss distances from the reference car.

Constants of this race

To make life easier and this race "shorter" some parameters are evaluated here that won't change for any of the cars in this race. Below, when each car is evaluated for time and speed, these parameters won't need recalculating. We'll just fetch them from here and speed the evaluation along. There are a number of measures that are not affected by friction. The trajectory, inertia and axle forces are constants in this race.

First, the trajectory length on the ramp:

L1 = Lr + (N+B-CMx)cos* + CMysin* + (Rr + CMycos* + CMxsin*)tan(O/2)

L1 = 110.5 + (1.5+3.875-1.9)(1) + 0 + (0.59 + 0.5(1) + 0)(-0.186947)

L1 = 110.5 + 3.475 + 1.09(-0.186947)

L1 = 113.7712 inches

The trajectory length on the flat:

L2 = Lf - (N+B-CMx)cos* - CMysin* + (Rr + CMycos* + CMxsin*)tan(O/2)

L2 = 249.5 - (1.5+3.875-1.9)(1) - 0 + (0.59 + 0.5(1) + 0)(-0.186947)

L2 = 249.5 - 3.475 + 1.09(-0.186947)

L2 = 245.8212 inches

Just for fun, compare the total trajectory length with the track.

L1 + L2 = 113.7712 + 245.8212 = 359.5924, 0.4 inches less than the track surface

Evaluate M, the inertial mass of the car,

M = m + 2If/Rf2 + 2Ir/Rr2

M = 0.01295 ozs2/in + 2(0.000048 ozins2)/(0.59 in)2 + 2(0.000048 ozins2)/(0.59 in)2

M = 0.01295 + 0.000275782 + 0.000275782

M = 0.01350 ozs2/in, yup, those spinning wheels make it seem heavier.

Evaluate the front axle force on the ramp,

Ff(ramp) = bg(CMx-CMytan(O+*))/B

Ff(ramp) = 4.666667(1.9-0.5tan(-0.369628))/3.875

Ff(ramp) = 4.666667(1.9-0.5(-0.387435))/3.875

Ff(ramp) = 2.5214664 oz

Evaluate the rear axle force on the ramp,

Fr(ramp) = bg(B-CMx+CMytan(O+*))/B

Fr(ramp) = 4.666667(3.875-1.9+0.5tan(-0.369628))/3.875

Fr(ramp) = 4.666667(3.875-1.9+0.5(-0.387435))/3.875

Fr(ramp) = 2.1452006 oz

Evaluate the front axle force on the flat,

Ff(flat) = bgCMx/B

Ff(flat) = 4.666667(1.9)/3.875

Ff(flat) = 2.2881722 oz

Evaluate the rear axle force on the flat,

Fr(flat) = bg(B-CMx)/B

Fr(flat) = 4.666667(3.875-1.9)/3.875

Fr(flat) = 2.3784948 oz

The Reference Car

A "typical" car is meant to represent one that could be assembled by anyone who has done nothing to make it faster. The friction coefficients are more guessed at than deduced from statistical studies of measurements. Yes, some measurements were made, but there is no way to determine how "typical" they are or if the term "typical" is really very meaningful in pinewood racing.

For a reference car, the important thing is that the values are not too extreme. Though it is nice to be able to say that "cutting axle friction in half gives you an advantage of X inches at the finish line", the results are best interpreted as indicators of the possibility of some advantage. But if you actually measured your car's coefficients and ran through these equations, you would be better able to quantify specifically an expected advantage. I don't know your car's measurements, so we'll use this reference car.

We start by determining an aerodynamic factor, k.

k = apA/2

k = 0.5(0.0000018677 ozs2/in4)2.5/2

k = 0.0000011673 ozs2/in2

Next is the drag due to axles and tread on the ramp.

Dr = (nfFfrf/Rf + 2uf(Ff/2+mfg) + nrFrrr/Rr + 2ur(Fr/2+mrg))cosO

Dr = (0.2(2.5214664)0.049/0.59 + 2(0.01)(2.5214664/2+0.08333) + 0.2(2.1452006)0.049/0.59 + 2(0.01)(2.1452006/2+0.08333))0.932461

Dr = (0.068763248 + 0.058750752)0.932461

Dr = 0.118901832 oz

Now, the terminal velocity on the ramp. Watch out for the sign on the angle. If we get it wrong here, most of the equations will come out wrong as we go.

vt = \[-(mgsinO+Dr)/k]

vt = \[-(5(-0.361268)+0.118901832)/0.0000011673]

vt = \[1445590.8233]

vt = 1202.327253 in/s

Lm is an indicator of how much air must be moved as the car advances,

Lm = M/k

Lm = 0.01350/0.0000011673

Lm = 11565.15034 in

How long would it take to move a mass of air equal to car's mass at the terminal velocity? It's T.

T = Lm/vt

T = 11565.15034/1202.327253

T = 9.618970468 s

Next we evaluate some more exotic functions to get the speed at the bottom of the ramp, which is the same as the initial speed on the flat. We're ignoring the energy loss that would really happen at such an abrupt join in order to get results more closely matching a smooth transition race. All trigonometric functions are evaluated in radians, not degrees.

v0 = vt tanh(arccosh(exp(L1/Lm))) Note: tanh(X) = (e^X - e^-X)/(e^X + e^-X) Note: arccosh(X) = ln(X+\[XX-1])

v0 = 1202.327253 tanh(arccosh(exp(113.7712/11565.15034)))

v0 = 1202.327253 tanh(arccosh(exp(0.009837416)))

v0 = 1202.327253 tanh(arccosh(1.009885962))

v0 = 1202.327253 tanh(0.140497093)

v0 = 1202.327253 (0.13957989)

v0 = 167.8207054 in/s

Back to tread and axle friction, but this time on the flat.

Df = nfFfrf/Rf + 2uf(Ff/2+mfg) + nrFrrr/Rr + 2ur(Fr/2+mrg)

Df = 0.2(2.2881722)0.049/0.59 + 2(0.01)(2.2881722/2+0.08333) + 0.2(2.3784948)0.049/0.59 + 2(0.01)(2.3784948/2+0.08333)

Df = 0.062555250 + 0.064958750

Df = 0.127514000 oz

vi on the flat plays a similar role to terminal velocity, vt, on the ramp.

vi = \[Df/k]

vi = \[0.127514000/0.0000011673]

vi = \[109238.4134]

vi = 330.5123498 in/s

Tf on the flat is analogous to T on the ramp.

Tf = Lm/vi

Tf = 11565.15034/330.5123498

Tf = 34.99158306 s

The coasting time is used to find the time spent on the flat.

tc = Tfarctan(v0/vi)

tc = 34.99158306arctan(167.8207054/330.5123498)

tc = 34.99158306arctan(0.507759258)

tc = 34.99158306(0.469835729)

tc = 16.44029594 s

The coasting distance is also computed using the coasting time.

yc = -Lm ln(cos(tc/Tf))

yc = -11565.15034 ln(cos(16.44029594/34.99158306))

yc = -11565.15034 ln(cos(0.469835729))

yc = -11565.15034 ln(0.891642672)

yc = -11565.15034 (-0.114689818)

yc = 1326.404988 in

The most involved equation is that for the total time. If you follow carefully, you can see how it is actually made up of two parts; time on the ramp and time on the flat.

t = T arccosh(exp(L1/Lm)) + tc - Tf arccos(exp((L2-yc)/Lm)) Note: arccosh(X) = ln(X+\[XX-1])

t = 9.618970468 arccosh(exp(113.7712/11565.15034)) + 16.44029594 - 34.99158306 arccos(exp((245.8212-1326.404988)/11565.15034))

t = 9.618970468 (0.140497093 from v0) + 16.44029594 - 34.99158306 arccos(exp(-0.09343447825859))

t = 1.351437388407 (time on ramp) + 16.44029594 - 34.99158306 arccos( 0.9107976925226)

t = 1.351437388407 + 16.44029594 - 34.99158306(0.4255842173922)

t = 1.351437388407 + 16.44029594 - 14.8918654919

t = 1.351437388407 + 1.5484304481 (time on flat)

t = 2.8999 seconds

For the final speed of the car at the finsh line, a number is borrowed from evaluating the time to avoid duplication of effort.

v = vi tan(arccos(exp((L2-yc)/Lm)))

v = 330.5123498 tan(0.4255842173922 from t)

v = 330.5123498 (0.4532872221905)

v = 149.8170 in/s

Other kinematic variables like accelerations and positions at certain times and speeds etc. can be determined, but for present purposes, we just really need time and speed.

The Car with no Tread Friction

This car is the same as the reference car except it has no tread friction.

THe equations are computed the same way, but in a few places results from an evaluation above can be used instead of recomputing it.

Dr = (nfFfrf/Rf + 2uf(Ff/2+mfg) + nrFrrr/Rr + 2ur(Fr/2+mrg))cosO

Dr = (0.2(2.5214664)0.049/0.59 + 0 + 0.2(2.1452006)0.049/0.59 + 0)0.932461

Dr = 0.07227890301589 oz

vt = \[-(mgsinO+Dr)/k]

vt = \[-(5(-0.361268)+0.07227890301589)/0.0000011673]

vt = \[1485516.380355]

vt = 1218.817615706 in/s

T = Lm/vt

T = 11565.15034/1218.817615706

T = 9.488827689203 s

v0 = vt tanh(arccosh(exp(L1/Lm))) Note: tanh(X) = (e^X - e^-X)/(e^X + e^-X)

v0 = 1218.817615706 tanh(arccosh(exp(113.7712/11565.15034)))

v0 = 1218.817615706 (0.13957989 from reference car v0)

v0 = 1218.817615706*(0.13957989)

v0 = 170.1224287303 in/s

Df = nfFfrf/Rf + 2uf(Ff/2+mfg) + nrFrrr/Rr + 2ur(Fr/2+mrg)

Df = 0.2(2.2881722)0.049/0.59 + 0 + 0.2(2.3784948)0.049/0.59 + 0

Df = 0.07751412983051 oz

vi = \[Df/k]

vi = \[0.07751412983051/0.0000011673]

vi = 257.6909670153 in/s

Tf = Lm/vi

Tf = 11565.15034/257.6909670153

Tf = 44.87992138007 s

tc = Tfarctan(v0/vi)

tc = 44.87992138007arctan(170.1224287303/257.6909670153)

tc = 44.87992138007(0.5834983962903)

tc = 26.18736215089 s

yc = -Lm ln(cos(tc/Tf))

yc = -11565.15034 ln(cos(26.18736215089/44.87992138007))

yc = -11565.15034*(-0.1808742083673)

yc = 2091.837412393 in

t = T arccosh(exp(L1/Lm)) + tc - Tf arccos(exp((L2-yc)/Lm)) Note: arccosh(X) = ln(X+\[XX-1])

t = 9.488827689203 arccosh(exp(113.7712/11565.15034)) + 26.18736215089 - 44.87992138007 arccos(exp((245.8212-2091.837412393)/11565.15034))

t = 9.488827689203 (0.140497093 from reference car v0) + 26.18736215089 - 44.87992138007 arccos(0.852468629626)

t = 9.488827689203 (0.140497093) + 26.18736215089 - 44.87992138007(0.5501069235238)

t = 1.333152706311 (ramp) + 1.498606672546 (flat)

t = 2.831759378857 s

v = vi tan(arccos(exp((L2-yc)/Lm)))

v = 257.6909670153 tan(arccos(exp((245.8212-2091.837412393)/11565.15034)))

v = 257.6909670153 tan(0.5501069235238)

v = 257.6909670153 (0.6132523387953)

v = 158.0295882085 in/s

The Car with no Axle Friction

This car has no axle friction, but is otherwise the same as the reference car.

Dr = (nfFfrf/Rf + 2uf(Ff/2+mfg) + nrFrrr/Rr + 2ur(Fr/2+mrg))cosO

Dr = (0 + 2(0.01)(2.5214664/2+0.08333) + 0 + 2(0.01)(2.1452006/2+0.08333))0.932461

Dr = 0.04662292878007 oz

vt = \[-(mgsinO+Dr)/k]

vt = \[-(5*(-0.361268)+0.04662292878007)/0.0000011673]

vt = 1227.80721423 in/s

T = Lm/vt

T = 11565.15034/1227.80721423

T = 9.419353629758 s

v0 = vt tanh(arccosh(exp(L1/Lm))) Note: tanh(X) = (e^X - e^-X)/(e^X + e^-X)

v0 = 1227.80721423 (0.13957989 from reference car v0)

v0 = 171.3771959034 in/s

Df = nfFfrf/Rf + 2uf(Ff/2+mfg) + nrFrrr/Rr + 2ur(Fr/2+mrg)

Df = 0 + 2(0.01)(2.2881722/2+0.08333) + 0 + 2(0.01)(2.3784948/2+0.08333)

Df = 0.04999987 oz

vi = \[Df/k]

vi = \[0.04999987/0.0000011673]

vi = 206.9632305647 in/s

Tf = Lm/vi

Tf = 11565.15034/206.9632305647

Tf = 55.88021750745 s

tc = Tfarctan(v0/vi)

tc = 55.88021750745arctan(171.3771959034/206.9632305647)

tc = 55.88021750745(0.6916158364678)

tc = 38.64764337337 s

yc = -Lm ln(cos(tc/Tf))

yc = -11565.15034 ln(cos(38.64764337337/55.88021750745))

yc = -11565.15034 ln(cos(0.6916158364669))

yc = -11565.15034 (-0.2610836755661)

yc = 3019.471959241 in

t = T arccosh(exp(L1/Lm)) + tc - Tf arccos(exp((L2-yc)/Lm)) Note: arccosh(X) = ln(X+\[XX-1])

t = 9.419353629758 (0.140497093 from reference car v0) + 38.64764337337 - 55.88021750745 arccos(exp((245.8212-3019.471959241)/11565.15034))

t = 1.32339180292 (ramp) + 38.64764337337 - 55.88021750745 arccos(exp(-0.2398283357933))

t = 1.32339180292 (ramp) + 38.64764337337 - 55.88021750745 arccos(0.7867629085054)

t = 1.32339180292 (ramp) + 38.64764337337 - 55.88021750745 (0.6652493313148)

t = 1.32339180292 (ramp) + 1.473366042858 (flat)

t = 2.796757845778 s

v = vi tan(arccos(exp((L2-yc)/Lm)))

v = 206.9632305647 tan(0.6652493313148 from t)

v = 206.9632305647 (0.7845506058671)

v = 162.3731279317 in/s

The Car with no Air Drag

For this car, we need to use a different model because setting a to zero makes k become zero and singularities appear everywhere in the equations. So we use equations from the model of a car with wheels, but no air resistance. We make the same changes to the model as above reguarding the energy loss at the join and use L1 and L2 for Lr and Lf.

v0 = \[2L1(-mgsinO-Dr)/M]

v0 = \[2(113.7712)(-5(-0.361268)-0.118901832)/0.01350]

v0 = \[28441.7578221]

v0 = 168.6468434988 in/s

yc = Mv02/2Df

yc = 0.01350(168.6468434988)2/2(0.127514000)

yc = 0.01350(28441.7578221)/0.255028

yc = 1505.574801976 in

t = \[2M/Df] ( L1/\[yc] + \[yc] - \[yc-L2] )

t = \[2(0.01350)/0.127514000] ( 113.7712/\[1505.574801976] + \[1505.574801976] - \[1505.574801976-245.8212] )

t = 0.4601537306598 ( 113.7712/38.80173709998 + 38.80173709998 - 35.49300778993 )

t = 1.349224185161 (ramp) + 1.52252413576 (flat)

t = 2.871748320921 s

tc = \[2ycM/Df]

tc = \[2(1505.574801976)0.01350/0.127514000]

tc = 17.85476408264 s

v(t) = (tc-t)Df/M

v(t) = (17.85476408264-2.871748320921)0.127514000/0.01350

v(t) = 141.5217979141 in/s

Summary of the Friction Race Standings

From the evaluations above, we found that the reference car finished the race with a time of 2.8999 seconds and a speed of 149.8170 inches per second. Using the time differences from this reference time to each improved car, we multiply by the slower speed, the reference car's, to get a good estimate of how far ahead of the reference car each improved car was as it passed the finish line. This distance represents the limit of advantage that might be gained if the reference car was worked on to reduce each type of friction.

Aerodynamic Drag advantage = (2.8999-2.8717)*149.8170 = 4.22 in

Tread Friction advantage = (2.8999-2.8318)*149.8170 = 10.2025377 in

Axle Friction advantage = (2.8999-2.7968)*149.8170 = 15.4461327 in

This does not mean that the reference car could be improved to beat itself by 4 + 10 + 15.5 = 29.5 inches! The effects of reduced friction are a bit more than additive. Using the best time for a "four wheeler" with no friction, we estimate the reference car loses by just over three feet!

Sensitivity of Wheel Friction to Axle and Tread Friction Coefficients

To find a relation between the critical values of n and u, start with the general expression for axle and tread friction. When they both produce the same amount of friction, the car's time will be dominated by both parameters, not just one of them. By going futher and taking derivatives, we can see which parameter will be the most advantageous to work on to reduce race time.

Wheel Drag (Dw)= Axle Friction + Tread Friction = nFr/R + 2u(F/2+w)

Let nFr/R = 2u(F/2+w)

Then n/u = 2R(F/2+w)/Fr = 2R(1/2r+w/Fr)

From this analysis, we see that the critical ratio between n and u is different on the ramp and flat and for front and rear wheels since the axle force, F, is different in all those cases. Using the values from the reference car race, let's see how they differ.

For the ramp front wheels, Ff(ramp) = 2.5214664 oz

n/u = 2(0.59)(1/2(0.049)+0.08333/F(0.049)) = 1.18(10.20408163265+1.700612244898/F)

n/u = 12.83667164666

For the ramp rear wheels, Fr(ramp) = 2.1452006 oz so by the same evaluation method, n/u = 12.9762637849

For the flat front wheels, Ff(flat) = 2.2881722 oz, n/u = 12.91781432912

For the flat rear wheels, Fr(flat) = 2.3784948 oz, n/u = 12.88451060283

For a Boy Scout wheel, a typical value would be

n/u = 2(0.59)(1/2(0.059)+0.11111/F(0.059)) = 1.18(8.474576271186+1.883220338983/F)

For F = 2.28 oz, about half the weight of the body of a Boy Scout car (accounting for the heavier wheels)

n/u = 10.97464912281

If we measure the tread friction coefficient on the track, then the critical value for n is n = (n/u)u, so we can present the following table to help us compute the critical values for n if we can find u. (Of course it works the reciporcal way too if you measure n, u = n/(n/u).)

If only a small change is made in axle friction or tread friction how much does that affect the wheel friction? To find this answer, we take derivatives of the wheel friction with respect to the friction coefficients then evaluate them for the Boy Scout car to see what the effect is.

dDw/dn = d(nFr/R + 2u(F/2+w))/dn = d(nFr/R)/dn + 0 = Fr/R

dDw/du = d(nFr/R + 2u(F/2+w))/du = 0 + d(2u(F/2+w))/du = F+2w

Both depend on directly on F, but dDw/dn als depends on the ratio of inner to outer wheel radius, a number that is about 1/10.

For example, say we change the tread friction by du = -0.005. Then the change in wheel friction would be about dDw = (F+2w)du = (2.28+0.22222)*(-0.005) = -0.0125111 oz. But for the same amount of change in n, dn = -0.005, the change in wheel friction would be about dDw = Frdn/R = 2.28*(0.059)*(-0.005)/0.59 = -0.00114 oz, a much smaller improvement.