Finding the Slowest Race Time

The task set forward by the question What is the slowest possible time? is carried out here. Using the parameter values listed on that page and the usual AWANA track and kit parameter values, the coasting distance, yc, is set to the length of the car's trajectory on the flat, L2. Then the yc is solved for the friction coefficient which is then used to find the time.

Evaluate the model

Let yc = L2

From the model of a car with wheels, but no air resistance we have,

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

Using the parameter values we can evaluate L2,

L2 = 249.5 - (0.6 + 5.8 - 5.8)(1) - 0 + (0.59 + 2.4(1) + 0)(-0.186947)

L2 = 249.5 - 0.6 - 2.99(0.186947)

L2 = 248.3410 inches

so yc = 248.3410 inches since it was assumed equal to L2

Working the other way to get n from the model, yc is:

yc = Mv02/2Df

so we have to evaluate v0.

v0 = \[2L1(-mgsinO-Dr)/M] without the cosO speed loss at the transition

and we have to evaluate L1, M and Dr.

L1 is the length of the car's trajectory on the ramp.

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

L1 = 110.5 + (0.6 + 5.8 - 5.8)(1) + 0 + (0.59 + 2.4(1) + 0)(-0.186947)

L1 = 110.5 + 0.6 - 2.99(0.186947)

L1 = 110.5410 inches

Note, L1 + L2 = 110.5410 + 248.3410 = 358.882 inches

That's 360.0 - 358.882 = 1.118 inches shorter 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

But the ramp friction is less than the flat friction, Dr = DfcosO so we have to evaluate Df.

Df = nfFfrf/Rf + 2uf(Ff/2+wf) + nrFrrr/Rr + 2ur(Fr/2+wr)

Df = n(4.5 oz)0.049 in/0.59 in + 0 + n(0 oz)0.049 in/0.59 in + 0

Df = n(4.5)0.049/0.59 - all the weight is on the front wheels!

Df = 0.373728n

Now evaluate Dr.

Dr = DfcosO = 0.373728n(0.932461) = 0.348486784n

Finally, we're ready to take a shot at v0, the initial speed on the flat.

v0 = \[2(110.5410)(-5(-0.361268)-0.348486784n)/0.01350]

v0 = \[221.082(1.80634-0.348486784n)/0.01350]

v0 = \[16376.4444(1.80634-0.348486784n)]

v0 = \[29581.426577-5706.974442n]

Now the coasting distance, yc = 248.3410 = Mv02/2Df

248.3410 = 0.01350(29581.426577-5706.974442n)/2(0.373728n)

Solve for the axle friction coefficient, n.

2(248.3410)0.373728n/0.01350 = 29581.426577-5706.974442n

13749.92374n = 29581.426577-5706.974442n

n(13749.92374+5706.974442) = 29581.426577

n = 29581.426577/(13749.92374+5706.974442)

n = 1.520357

To get the time from the time equation (using L1 and L2 for Lr and Lf):

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

t = \[2(0.01350)/0.373728n] ( 110.5410/\[248.3410] + \[248.3410] - \[248.3410-248.3410] )

t = \[0.072245055227/n] ( 110.5410/15.758839 + 15.758839 - 0 )

t = \[0.072245055227/n] ( 7.0145395863 + 15.758839 )

t(n) = 22.773378586\[0.072245055227/n]

We have a relation between time and the friction coefficient becuase the ramp angle was not involved dynamically in the determination of t. It was used to get L1 and L2, but it was not "tied" to a dynamic variable like speed, acceleration or position. But yc did it for us so, we have our n and can evaluate t(n).

t = 22.773378586\[0.072245055227/n=1.520357]

t = 22.773378586(0.217987341)

Confirmation

Running a similar car profile through "Race It" produces similar results, t = 5.065 and n = 1.3688. The difference is due to letting the car have a little air friction (to allow the algorithm to work at all) and running on a smooth transition. As a result, the coefficient was lower as air friction acted and the time was a bit slower because of some energy loss increase on the transition due to centrifugal force. In fact, had the air resistance truely been zero, the axle friction coefficient would still have been lower due to the centrifugal force on the transition.