Complete Ramp Model

Complete Model of A Grand Prix Race : Ramp

The starting ramp provides your car with all the energy it will get during the Grand Prix race. It is on the ramp that potential energy is converted to the kinetic energy that must overcome all sources of friction and drift that threaten to impede its progress. Each of these aspects of the physical race are integrated into a single model of the ramp.

Energy Relation

Everything significant on the ramp is included in this energy relation.

    ________Potential______ + _Kinetic Energy_ + _________Work By Drag__________
      Track         Car        Car     Wheels        Axles         Tread 
E = -mgLsinO = -mg(L-y)sinO + mv2/2 + Ifv2/Rf2 + (nfFfrf/Rf + 2uf(Ff/2+wf))ycosO
                                    + Irv2/Rr2 + (nrFrrr/Rr + 2ur(Fr/2+wr))ycosO

                                               + Work Done By Air
                                               + Wa

A few definitions will tame this energy relation and make its terms more meaningful.

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,

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

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

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

That makes the mechanical drag on the ramp, Dr, less than Df on the flat by a factor of cosO because the normal force is reduced by cosO.

Dr = DfcosO

The complete energy relation can now be written as,

E = -mgLsinO = -mg(L-y)sinO + Mv2/2 + Dry + Wa

This reduces to an equivalence between potential energy and kinetic energy plus the work of energy drains:

-mgysinO = Mv2/2 + Dry + Wa

We'd like to solve for v(y), but Wa has it locked away deep inside! But we can unlock it if we use the fact that E, the total energy, is a constant on the ramp. The individual terms change in their values but together they always add up to the value, E, at every position, y, on the ramp.

Physically, this constant condition is a result of the conservation of energy. Mathematically, it allows us to find the acceleration of car and from that the equations of motion on the ramp.

Passing Through A Black Hole

A black hole is a singularity in space and time, a place and period such that nothing beyond is knowable. There are two kinds; essential like the black hole and removeable like the one we have encountered in the energy relation above. We don't know what would happen if we passed through a real black hole, but we can pass through this one by taking a derivative.

Since E, the total energy, is constant, it does not change with time. It is eternally frozen according to your car's trajectory. In symbols we write,

0 = dE/dt = -d(mgLsinO)/dt = -d(mg(L-y)sinO + Mv2/2 + Dry + Wa)/dt

Apply the "d" operator of differentiation to each term on the right:

0 = -d(mg(L-y)sinO)/dt + d(Mv2/2)/dt + d(Dry)/dt + dWa/dt

Drop constant terms and take constants out of time dependent ones:

0 = mgsinOdy/dt + M(dv2/dt)/2 + Drdy/dt + dWa/dt

Apply the chain rule of differentiation to the more complicated terms and collect terms with common time dependencies:

0 = (mgsinO+Dr)dy/dt + 2Mv(dv/dt)/2 + dWa/dt

Note that dy/dt = v and dv/dt = a, so that

0 = (mgsinO+DR)v + Mva + dWa/dt

From the analysis of the force of air on your car, we can write

dWa/dt = kv3

Now a term with acceleration can be isolated on the left:

Mva = (-mgsinO-D1)v - kv3

Divide through by v to get

Ma = -mgsinO - D1 - kv2

But M = m + 2If/Rf2 + 2Ir/Rr2, so we can distribute the a and subtract 2Ifa/Rf2 + 2Ira/Rr2, the force needed to spin the wheels on the track, from both sides.

ma = -mgsinO - 2Ifa/Rf2 - 2Ira/Rr2 - D1 - kv2

This is the equation of motion along the surface of the ramp.

Equations of Motion

From studies of forces, we could have written the equations of motion directly. Expressions for weight, wheels, axle friction and air resistance were available. Deriving the equation of motion from the energy relation shows the tight connection between energy and force. It also verifies the peice-wise assembly of such an equation of motion from part forces. It is always good to verify model equations this way if possible. Finding an omittion or other mistake later can mean a lot of wasted effort.

The equation of motion along the ramp can then be written:

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

Some of our choices need further explanation.

The sign on mgsinO is negative because the angle of inclination, O, is negative, O < 0. That makes sinO negative. But the car is moved in the positive y direction (to the right) by the force of its weight. So we negate the term to keep the car headed in the right direction on the ramp.

If we had an uphill ramp, then the weight-force would be negative as we've expressed it, indicating movement to the left because of the way "y" is defined for the model.

The other forces are all positive measures and require a negative sign to make them oppose the direction of travel along y. The normal forces are only included in axle and tread friction. They do not directly contribute to movement in the y direction.

The equations of motion reduce to

F = m a = -mgsinO - kv2 - 2a(If/Rf2+Ir/Rr2) - (nfFfrf/Rf + nrFrrr/Rr + 2uf(Ff/2+wf) + 2ur(Fr/2+wr))cosO

Acceleration

Rearrange to find the acceleration along the track and use Dr and M:

a(v) = (-mgsinO-Dr-kv2)/M

This is the acceleration along the track in terms of a changing velocity.

Let f = mgsinO + Dr, the constant part of the ramp force.

Now a(v) = -(f+kv2)/M, the force driving the car on the ramp divided by its inertial mass.

Velocity

Since acceleration, a = dv/dt, we can separate the velocity terms from the time terms by dividing dv/dt through by f+kv2 and forming an integral equality. The sum of small increases in velocity, dv, divided by the changing force at each step from 0 to 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 and speed from zero on this starting ramp. This means we won't carry around terms with the initial time or velocity in them.

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

If fk was greater than zero, friction, Dr, would overcome the car's forward force, mgsinO, and the car would stop. Remember, the ramp angle, O, is less than 0 because it is down sloping. From a reference table using fk < 0, we find

    /            
   | dv/(f+kv2) = arctanh(v\[-kf]/f)/\[-kf] for kf < 0
  /

This imposing arctanh function is pretty scarry at first sight, but can be understood pretty easily from its graph and a little algebra later on.

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

So -t/M = [arctanh(v\[-kf]/f)/\[-kf]](0,v) and since arctanh(0) = 0,

-t/M = arctanh(v\[-kf]/f)/\[-kf]

Multiply through by M:

t = -Marctanh(v\[-kf]/f)/\[-kf]

Solve for velocity, v, in terms of time, t:

v(t) = ftanh(-t\[-kf]/M)/\[-kf] which is

v(t) = -ftanh(t\[-kf]/M)/\[-kf] since tanh(-X) = -tanh(X)

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

    / y              / t
   | dy = -f/\[-kf] | tanh(t\[-kf]/M)dt.
  / 0              / 0

let X = t\[-kf]/M so dX = \[-kf]dt/M then from a reference table

    /
   | tanh(X)dX = ln(cosh(X))
  /

Where

cosh(X) = (exp(X) + exp(-X))/2 = Y and maps (-inf to inf) onto [1 to inf)

arccosh(Y) = +/-X mapping [1 to inf) onto (-inf to inf).

But to get dX = \[-kf]dt/M into our integrand, we use an old math trick:

multiply dt by 1 where 1 = \[-kf]/M/\[-kf]/M.

    / y              / t
   | dy = -f/\[-kf] | tanh(t\[-kf]/M) \[-kf]/M dt / \[-kf]/M so
  / 0              / 0

After simplifying (-f/\[-kf]) / (\[-kf]/M), the equality becomes:

    / y        / t
   | dy = M/k | tanh(t\[-kf]/M) \[-kf]dt/M
  / 0        / 0

Now using the reference relation from above,

y(t) = [Mln(cosh(t\[-kf]/M))/k[(0,t)

Evaluate t in the expression at both extremes, 0 and t to get:

y(t) = Mln(cosh(t\[-kf]/M))/k since cosh(t=0) = 1 and ln(1) = 0.

At t=0, hyperbolic cosine is 1 which gets mapped to 0 by the natural log. As t increases, the hyperbolic cosine increases so the natural log evaluates to an increasing positive value.

Solve for time, t, in terms of position, y.

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

exp(ky/M) = cosh(t\[-kf]/M)

Then use the arch hyperbolic cosine function, arccosh(), the inverse of the hyperbolic cosine, hcos(), to expose the time term in the same way.

arccosh(exp(ky/M)) = t\[-kf]/M

Clear away the multipliers around the time to get time in terms of position.

So t = Marccosh(exp(ky/M))/\[-kf]

What Does This Mean?

To see what we have, a few more definitions will help clear the smoke.

Lm = M/k is the distance needed to move an air mass equal to the car's inertial mass. Lm is on the order of half of a mile! It would take that much distance for your car to push its inertial weight in air out of its way.

Then, from above, velocity in terms of time is

v(t) = -ftanh(t\[-kf]/M)/\[-kf]

v(t) = \[-f/k]tanh(t\[-kf]/M) -- take the negative branch

Evaluate this velocity at an infinite time, inf, where a steady state occurs. This will help us understand the cryptic expression for speed at any time on the ramp.

v(inf) = vt = \[-f/k]tanh(inf\[-kf]/M) is the terminal velocity on the ramp. Terminal velocity is the speed reached but not surpassed by a car with drag Dr and A, allowed to slide down a long ramp with inclination O.

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

Get back to the velocity expression to see how the terminal velocity is encoded in it twice!

v(t) = \[-(mgsinO+Dr)/k]tanh(t\[-k(mgsinO+Dr)]/M)

Simplify:

v(t) = vt tanh(tvt/Lm)

Better yet, use T = Lm/vt, the time needed to move an air mass equal to car's inertia at the terminal velocity on the ramp. Then,

v(t) = vt tanh(t/T)

We now see the velocity at time, t, on the ramp as a fraction of the terminal velocity. The hyperbolic tangent never quite reaches one, even when t is very large compared to T. The hyperbolic tangent regulates the fraction of the terminal velocity expressed.

With similar reductions,

t(y) = T arccosh(exp(y/Lm)) - T is "regulated" is this expression but the time, t(y), is unlimmited.

For a long distance, y, t(y) is nearly Ty/Lm = y/vt. For smaller ones, the arccosh makes the time larger than y/vt since the terminal veloctiy, vt, has not yet been achieved.

y(t) = Lm ln(cosh(t/T)) - likewise, Lm is "regulated" but y(t) unlimited.

To get the acceleration in terms of time, start with the acceleration in terms of velocity:

a(v) = (-mgsinO-Dr-kv2)/M

Substitute the expression for velocity in terms of time:

a(t) = a(v(t)) = (-mgsinO-Dr-k(vt tanh(t/T))2)/M

Simplify. Distribute the square:

a(t) = (-mgsinO-Dr-kvt2 tanh(t/T)2)/M

Substitute the expression for terminal velocity, vt:

a(t) = (-mgsinO-Dr-k(-(mgsinO+Dr)/k) tanh(t/T)2)/M

Cancel k:

a(t) = (-mgsinO-Dr+(mgsinO+Dr) tanh(t/T)2)/M

Collect terms to get another "regulated" expression.

a(t) = (mgsinO+Dr) (tanh(t/T)2 - 1)/M

The constant part of the ramp force is divided by the constant inertial mass. This is a constant acceleration on the ramp, the same as it would be without air resistance! Air resistance adds a scaling factor that regulates the acceleration. At the starting line, t=0, the acceleration is a(0) = -mgsinO-Dr as expected. If the ramp were long enough and time very long, the tanh term gets close to 1 so the acceleration is close to zero. At that point the air resistance stops acceleration and the car's speed becomes constant at the terminal velocity.

To get the speed at the bottom of the ramp, v0, we use the expressions for velocity in terms of time and time in terms of distance, y. Using a distance equal to the trajectory length on the ramp, L1, we get:

v(t(L1)) = v0 = vt tanh(Marccosh(exp(kL1/M))kvt/Mkvt)

Simplified, the expression becomes:

v0 = vt tanh(arccosh(exp(L1/Lm)))

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

1 = x2 - y2

and the relationship among hyperbolic trigonometric functions,

tanh(u) = sinh(u)/cosh(u) = y/x

we can write

tanh(arccosh(exp(L1/Lm))) = tanh(arccosh(x)) = tanh(u) where we have substituted x = exp(L1/Lm) and u = arccosh(x)

Since tanh(u) = y/x all that needs to be done is to solve for y in terms of x, plug this new expression for tanh(u) in terms of x in the expression for v0, expand x and simplify.

y = \[x2-1]

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

tanh(u) = y/x = \[x2-1]/x = \[(x2-1)/x2] = \[1-1/x2]

Expand x,

tanh(u) = \[1-1/exp(2L1/Lm)] = \[1-exp(-2L1/Lm)]

Substitute this into the expression for v0 to get,

v0 = vt tanh(arccosh(exp(L1/Lm))) = vt \[1-exp(-2L1/Lm)]

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

When L1 = Lm, v0 = 0.92987vt. For L1 > Lm, v0 gets closer to vt, but can never reach it since the hyperbolic tangent can never reach 1. But typically, Lm is much larger than L1 so that the car never gets close to reaching its terminal velocity.

Verification

When a, the aerodynamic drag coefficient is so low it is almost zero, these expressions ought to become the same as the ones for the simple car with axles. They indeed do.

If you try to find these limits yourself, you will need to use the expression of the hyperbolic functions in terms of natural log, ln, natural exponentiation, exp, etc. and their derivatives. Most can be found in the limit by applying L'hopital's Rule two or three times in turn.

Check the limit for t(y) squared since applying L'hopital's Rule to t(y) results in a square root in the numerator and one in the denominator leading to infinite application of L'hopital's Rule! Taking the square removes the square roots.

Note: arccosh(X) = ln(X+\[XX-1])

Note: tanh(X) = (e^X - e^-X)/(e^X + e^-X)