Grand Prix Racing - | The Science of Fast Pinewood Cars |

The task set forward by the question Can a slower car win? is carried out here. Using the parameter values listed on that page and the usual AWANA track and kit parameter values, the time equation of the model of a car with axles but no friction is evaluated. Since that model was not developed as a part of the derivation of the simple race model, it will be done here.

We begin with the equations of motion. Note the wheels provide a resisting force while they are spinning up on the ramp and because the acceleration on the flat is negative, a pushing force on the flat as they spin down!

Equations of Motion on ramp

F = ma = -mgsinO - 2Ifa/Rf2 - 2Ira/Rr2

Equations of Motion on flat

F = ma = - 2Ifa/Rf2 - 2Ira/Rr2

These equations can be written more compactly using the inertial mass, M.

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

Equations of Motion on ramp

F = Ma = -mgsinO

Equations of Motion on flat

F = Ma = 0

Now, these have the same form as a block on a ramp. So we can use those equations if we remember to use M instead of m for the mass. Since the flat equations weren't derived there, it'll be done here. Note that the energy in the wheels simply make the car seem a bit heavier than it really is standing still! Sounds a bit like what happens when a spaceship approaches the speed of light - it gets heavier! That's one hint why some physicists think matter is just space spinning in on itself.

Time to the end of the ramp is, (L1 has been used for y)

t(L1) = \[-2L1/gsinO]

Speed at the end of the ramp is, (t(L1) is used for t)

v(t(L1)) = -gsinOt(L1)

We start with the simple equation of motion.

F = Ma = 0

Since M is a constant, a, acceleration is the only quantity that can change. But 0 is also a constant! So acceleration must be zero, a = 0. This doesn't mean that the speed is zero, only that it does not change. So to model this over time, we can just write the linear equations,

v(t) = v0 (no time dependence) and

y(t) = v0t so

t(y) = y(t)/v0

Using L2 for the trajectory length on the flat, t becomes,

t = L2/v0

It's time to evaluate these equations and get the best time. We'll also get the speed to compare with the fastest speed on our question page Can a slower car win? But before we can start, we need to find the trajectory lengths on the ramp and flat. For that, we use the equations from the race summary.

The length of the trajectory on the ramp is,

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

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

L1 = 110.5 + 6.4 - 0.55897153

L1 = 116.3410 inches

The length of the trajectory on the flat is,

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

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

L2 = 249.5 - 6.4 - 0.55897153

L2 = 242.5410 inches

To see how much "warp" we have in the path, add the two parts of the trajectory and subtract from the length of the track surface which is Lr + Lf = 360 inches.

L1 + L2 = 358.8820, 1.118 inches shorter than the surface of the track!

Now we can evaluate the time to the end of the ramp.

t(L1) = \[-2L1/gsinO]

t(L1) = \[-2(116.341)/(386.088)(-0.361268)]

t(L1) = \[1.6681956706]

t(L1) = 1.29159 seconds

To get time on the flat, we have to know the speed at the end of the ramp.

v(t(L1)) = -gsinOt(L1)

v(t(L1)) = -386.088(-0.361268)1.29159

v(t(L1)) = v0 = 180.15257423 in/s

Letting y = L2 in the equation for time on the flat found above,

t(L2) = L2/v0

t(L2) = 242.5410/180.15257423

t(L2) = 1.34631 seconds

Adding these two times together, we get:

t(L1) + t(L2) = 1.29159 + 1.34631 seconds

t = 2.6379 seconds |
---|

v = 180.1526 in/s |

Using the low position of the center of mass, -0.59 inches, the length of the trajectory on the ramp is,

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

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

L1 = 110.5 + 6.4 - 0

L1 = 116.9 inches

The length of the trajectory on the flat is,

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

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

L2 = 249.5 - 6.4 - 0

L2 = 243.1 inches

Add the two parts of the trajectory and subtract from the length of the track surface which is Lr + Lf = 360 inches. This time, we get zero as we must since the car's center af mass is on the track surface.

Now we can evaluate the time to the end of the ramp.

t(L1) = \[-2L1/gsinO]

t(L1) = \[-2(116.9)/(386.088)(-0.361268)]

t(L1) = \[1.6762110854]

t(L1) = 1.29469 seconds

To get time on the flat, we have to know the speed at the end of the ramp.

v(t(L1)) = -gsinOt(L1)

v(t(L1)) = -386.088(-0.361268)1.29469

v(t(L1)) = v0 = 180.5850 in/s

Letting y = L2 in the equation for time on the flat found above,

t(L2) = L2/v0

t(L2) = 243.1/180.5850

t(L2) = 1.34618 seconds

Adding these two times together, we get:

t(L1) + t(L2) = 1.29469 + 1.34618 seconds

t = 2.64087 seconds

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