Complete Model of A Grand Prix Race : Flat

A good approximation of a Grand Prix track requires a long flat run following the starting ramp. This is where the real test comes. On the flat your Grand Prix car plays out the result of weight placement and friction reducing techniques as kinetic energy is frittered away.

Energy Relation

The energy relation on the flat is similar to that of the ramp. It has many of the same energy terms. But it differs from the ramp in two important ways.

First, the left hand side of the relation presents the amount of kinetic energy initially available on the flat. This amount is the same as the constant E = -mgLsinO on the ramp minus the amount used up in the work of friction, air resistance and loss at the join transition.

Second, because this part of the track is flat, theta is zero, O = 0, so sinO = 0 and cosO = 1. There is no potential energy to be considered on the flat. There is only kinetic energy and friction to deal with.

Using the index "0" for the initial velocity, written v0, is less by cosO (O the ramp angle) than the v0 derived on the ramp. The constant Ef is used for the energy. So the energy relation on the flat is:

     Init Kinetic Energy_ = _Kinetic Energy_ + _______Work By Drag_______
      Car      Wheels        Car      Wheels        Axles      wheels
Ef = mv02/2  +  Ifv02/Rf2 = mv2/2 + Ifv2/Rf2 + (nfFfrf/Rf + 2uf(Ff/2+wf))y
             +  Irv02/Rr2         + Irv2/Rr2 + (nrFrrr/Rr + 2ur(Fr/2+wr))y

                                             + Work Done By Air
                                             + Wa

A few definitions borrowed from the ramp will help reduce this energy relation and make its terms more comparable to those on the ramp.

M, the inertial mass of the car is,

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

Df, the mechanical drag on a flat track is defined with Ff and Fr evaluated at O=0, on the flat is,

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

There is no centrifugal force on the flat, so keep in mind that

Ff + Fr = 5 oz - 2wf - 2wr, where 5 oz is the AWANA weight limit.

The complete energy relation can now be written as,

E = Mv02/2 = Mv2/2 + Dfy + Wa

As on the ramp, we can't solve for v(y) because Wa has it locked away! But we can apply the same technique to find the equation of motion. Notice that the initial kinetic energy is a constant. So its time derivative is zero. We already evaluated the time derivatives of the remaining terms for the ramp, so we can already write the equation of motion.

Equations of Motion

Because the weight of the car bears straight down on the track, the tangental component of the weight disappears. All that remains are dissipating forces to work off the initial kinetic energy of your car. Because the kinetic energy of your car produces no force (Newton's First Law), though changes in it indicate forces are at work on it (Newton's Second Law), there is no force term to represent it. The energy relation on the flat is:

F = m a = GravTangent - Air - Wheels   - Axle Drag - Wheel Drag
F = m a = 0           - kv2 - 2Ifa/Rf2 - nfFfrf/Rf - 2uf(Ff/2+wf)
                            - 2Ira/Rr2 - nrFrrr/Rr - 2ur(Fr/2+wr)

Using the Df symbol defined above for mechanical drag, this reduces to:

F = m a = - kv2 - 2a(If/Rf2 + Ir/Rr2) - Df

Acceleration

Rearrange to find the acceleration along the flat and use M:

a(v) = -(Df + kv2)/M

This is the decceleration on the flat in terms of v because of the overall negative sign. Ignoring the sign, it is still called "acceleration" anyway.

On the flat, the velocity integral looks nearly the same as for the starting ramp except that the force is only that of mechanical drag, Df, and there is a new initial limit on the integral. This is the initial velocity, v0, the speed of your car upon reaching the flat. v0 is usually modeled as the speed at the bottom of the starting ramp. In the case of an abrupt join, it is less than the speed at the bottom of the starting ramp as mentioned above. It doesn't matter if the wheels of your car or track material are perfectly elastic or not.

Velocity

Since acceleration, a = dv/dt, we can separate the velocity terms from the time terms by dividing dv/dt through by Df+kv2 and forming an integral equality. The sum of small increases in velocity, dv, divided by the changing force at each step from the initial velocity, v0, to the final veloctiy, v, is equal to the sum of small steps in time, dt, needed to get to velocity, v, divided by the constant inertial mass of the car.

Note, we start the clock over again from zero on the flat. This means we won't have to carry around terms with the initial time in them. But we will need to add the ramp time to the flat time to get the total race time.

   / v               / t
  | dv/(Df+kv2) = - | dt/M
 / v0              / 0

From a table of integrals we find generally:

   /
  | dv/(Df + kv2) = arctan(v\[k/Df])/\[kDf] for kDf > 0
 /

This arctan function can be understood from its graph and a little algebra.

On the other side of the equality, the time integral evaluates to -t/M.

So -t/M = [arctan(v\[k/Df])/\[kDf]](v0,v)

Unlike the similar expression on the ramp, this one must be evaluated on both extremes, (v0,v), producing a term in initial velocity.

t(v) = -Marctan(v\[k/Df])/\[kDf] + Marctan(v0\[k/Df])/\[kDf]

Both terms have units of time. Since v, the speed of the car at the finish line is less than it is at the beginning of the flat, the negative term is less in magnitude than the positive one. Having worked out the details previously, we can name the positive term "coasting time", tc, to make this derivation more concise and meaningful.

tc = Marctan(v0\[k/Df])/\[kDf] This is a constant that tells us how long the car would roll on a flat surface before coming to a halt.

Now t(v) = tc - Marctan(v\[k/Df])/\[kDf]

which is more manageable and easier to solve for v(t).

Solve for v in terms of t and simplify:

v(t) = \[Df/k]tan((tc-t)\[kDf]/M)

Position

Solve for position on the track in terms of time. To do this, separate variables with v = dy/dt and integrate to get y in terms of time:

    / y            / t
   | dy = \[Df/k] | tan((tc-t)\[kDf]/M)dt.
  / 0            / 0

From a reference table,

    /
   | tan(aX)dX = -ln(cos(aX))/a
  /

So let X = tc-t and a = \[kDf]/M then dX = -dt

But to get dX = -dt into our integrand, we use an old math trick:

multiply dt by 1 where 1 = -1/-1.

    / y            / t
   | dy = \[Df/k] | tan((tc-t)\[kDf]/M) (-dt)/-1  simplifies to
  / 0            / 0

    / y             / t
   | dy = -\[Df/k] | tan((tc-t)\[kDf]/M) (-dt)
  / 0             / 0

Now use the reference relation, to get position, y, in terms of time, t.

y(t) = M/k [ln(cos((tc-t)\[kDf]/M))](0,t)

Evaluation at t=0 does not produce a 0, but another term:

y(t) = Mln(cos((tc-t)\[kDf]/M))/k - Mln(cos(tc\[kDf]/M))/k

Let yc = -Mln(cos(tc\[kDf]/M))/k

In retrospect, we know this is the distance the car would have to travel before coming to a halt. So,

y(t) = yc + Mln(cos((tc-t)\[kDf]/M))/k

This looks funny with the "+" in the middle. However, the coasting time, tc, is always greater than the measured time, t. The cosine (evaluated using radians) is then always positive and less than 1. This insures that the natural log function, ln(*), will always evaluate to a negative number. So the term on the right of the "+" represents the maximum distance the car can travel.

Time

Solve for time, t, in terms of position, y, t(y). First, clear the natural log function of multipliers to get:

k(y-yc)/M = ln(cos((tc-t)\[kDf]/M))

Use the exponential function, exp(), the inverse of the natural log, ln(), to expose the cosine function by applying it to both sides.

exp(k(y-yc)/M) = cos((tc-t)\[kDf]/M)

Then use the arch cosine function, arccos(), the inverse of cosine, cos(), to expose the time term in the same way.

arccos(exp(k(y-yc)/M)) = (tc-t)\[kDf]/M

Clear away the multipliers around the time and subtract the coasting time, tc, from both sides to get time in terms of position.

t(y) = tc - Marccos(exp(k(y-yc)/M))/\[kDf]

What Does This Mean?

A few definitions can help us interpret the details of these expressions.

Lm = M/k is the distance needed to move an air mass equal to the car's inertial mass. This is the same on a flat or ramp.

vi = \[Df/k] is the square root of the division of axle and tread drag by the mass of air moved per inch of track. Work is force acting through a distance. In terms of units, oz x in, represents work. So Df/k can also be seen as the work or energy, e, expended to overcome drag per mass of air, j, displaced. Then expressing the energy as kinetic energy on the flat, e = jvi2 for some velocity vi. Hence Df/k = e/j = vi2. This vi must be the speed of the displaced air mass matching its energy to work done by drag.

Tf = Lm/vi, the time needed to displace an air mass equal to the car's inertial mass at the speed needed to match its kinetic energy to the work of drag.

Using these definitions, we revisit each of the above expressions.

The coasting time is the time when the car's velocity is zero:

t(v=0) = tc = Tfarctan(v0/vi)

It is the air dispersal time at the speed matching the air's energy to mechanical drag multiplied by a regulating function. The regulator is controled by the ratio of initial speed on the flat to the matching speed of air energy to mechanical drag. If this ratio is infinite, the arctan becomes pi/2 (it's evaluated in radians) and tc = piTf/2. This is a time limit! No matter how fast your car starts on the flat, it won't roll any longer than piTf/2.

The coasting distance is the distance reached at the coastig time:

y(tc) = yc = -Mln(cos(tc\[kDf]/M))/k from above.

Use the definitions of Lm and TF to simplify to:

yc = -Lm ln(cos(tc/Tf)) which is positive since tc is less than piTf/2. Note that if tc = 0, yc = 0. Evaluate the cosine using radians.

Continuing to make similar use of the definitions, more of the above expressions become simpler. The trigonomic functions should all be evaluated using radians.

t(y) = tc - Tf arccos(exp((y-yc)/Lm))

Notice that if the friction was so high that the coasting distance had been less than the length of the track, the equations would "break down" and the "real" car would stop on the track. The equations break down in the sense that at some point in the evaluation of the time function, the arccos function receives an argument greater than 1 because y-y0 is positive.

y(t) = yc + Lm ln(cos((tc-t)/Tf))

v(t) = vi tan((tc-t)/Tf)

a(v) = -(Df + kv2)/M

a(v(t)) = -(Df + k(vi tan((tc-t)/Tf))2)/M

a(t) = -(Df + kvi2tan((tc-t)/Tf)2)/M

a(t) = -(Df + Dftan((tc-t)/Tf)2)/M

a(t) = -Df(tan((tc-t)/Tf)2 + 1)/M

For the speed at the finish line, we compose the velocity in terms of the time in terms of the length of the car's trajectory on the flat.

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

But we can do a lot better! By using the definition of a unit circle,

1 = x2 + y2

and the relationship among trigonometric functions,

tan(O) = sin(O)/cos(O) = y/x

we can write

tan(arccos(exp((L2-yc)/Lm))) = tan(arccos(x)) = tan(O) where we have substituted x = exp((L2-yc)/Lm) and O = arccos(x)

Since tan(O) = y/x all that needs to be done is to solve for y in terms of x, plug this new expression for tan(O) in terms of x in the expression for v(t(L2)), expand x and simplify.

y = \[1-x2]

Substitute y into the expression for tan(O) where we left off two lines above,

tan(O) = y/x = \[1-x2]/x = \[(1-x2)/x2] = \[1/x2-1]

Expand x,

tan(O) = \[1/exp(2(L2-yc)/Lm)-1] = \[exp(-2(L2-yc)/Lm)-1]

Substitute this into the expression for v(t(L2)) to get,

v(t(L2)) = vi tan(arccos(exp((L2-yc)/Lm))) = vi \[exp(-2(L2-yc)/Lm)-1]

It's like collapsing a telescope. This is a much easier expression for race end speed to work with.

To understand this we need to know that typical distances for L2, the flat length, are much smaller than yc, the coasting distance. Also, yc is usually much smaller than the distance it takes the car to move a mass of air equivalent to its inertial mass, Lm. This means that L2-yc is negative so -2(L2-yc)/Lm is a positive number less than one. So the exponential evaluates to a number greater than one making the end speed positive in the typical race.

If L2 > yc, we know the car will stop on the flat part of the track before crossing the finish line; a final speed of zero. But if we use the equation for final race speed, -2(L2-yc)/Lm is a negative number less than one. Since it is less than zero, the exponential evaluates to a positive number less than one so the number inside the squareroot is negative! This leads to purely "imaginary" results that indicate the expression is not valid in this case.