General Dynamics Models - Coasting Motion

When powered motion ends, a vehicle will generally coast to a stop. On this page, the assumptions used for the powered model are extended to work out the two main models for coasting.

The first is the more general case involving both sliding friction and resistance from fluid media. Vehicles constrained to a track or surface fit this model. Characteristic of the retarded motion described is a definite point at which the vehicle comes to a complete stop in finite time.

The second assumes no operating sliding friction. Vehicles immersed in a medium that are free to move fit this model if the medium is not too "thick" (ie., viscous). This motion is attenuated; never coming to a complete halt in theory. This attenuated motion model is presented on a separate page.

Notation

Mathematical models use mathematical notation. The various environmental, structural and performance factors are written using short one or two letter symbols. Each factor also has physical units that relate to physical measurements that might be measured given appropriate instruments when possible. The table below presents the notations of the factors modeled.

The various factors above are related to one another using physical laws and analysis. Some of these relationships have special meaning in understanding the model. These as well as kinematic factors (like time, acceleration, speed and distance) are recorded in the next table as a summary of the model.

Initial velocity is included for reference from the power model.

Terminal Motion Model

To model the deceleration of a vehicle we use a general form of Newton's second law, F = ma, that captures the forces that oppose most vehicular motion,

Ma = - Ds - kv2

The inertial mass, M, for this model must be coupled to deceleration along the line of motion. Wheels meet this criteria. Other parts like free spinning propellers may also. If there are no spinning parts, the mass of the vehicle is used for M. The inertial mass is a constant in the model, no mass is gained nor lost during travel.

Ds is a constant mechanical, sliding resistance. Generally, it is modeled as a scalar friction coefficient times a force that is at right angles to the sliding surface. It can be complicated by geometrical relationships to the surface on which the vehicle rests as with axle friction in wheels. This sliding friction is a constant in the model. Note that in some applications this sliding friction is different from that in the powered model! So it is given a different symbol.

All fluid media interactions that change in proportion to the speed of the vehicle squared, v2, are represented in the constant k. More than one media may be involved like air and water in the case of a boat. The terms expressing them are typically pressure drag terms (including lift) and skin friction terms. k is the total mass of media moved aside per inch of motion, hence it is called the "media coefficient" here.

Since the acceleration, a, is the change in speed over time, we can find an expression for the speed over time. Separate out the time and speed dependencies noting that a = dv/dt; an acceleration is a small change in speed with respect to a small change in time. Divide through by the right hand side above, multiply through by dt and swap sides to get,

dt = -Mdv/(Ds+kv2)

These will be integrated below, but first it is useful to define some constants to make the expressions derived for this model more meaningful.

Given that the media coefficient, k, is the effective mass of fluid media displaced per inch of movement, dividing the inertial mass of the vehicle, M, by it gives a distance.

Y = M/k is the distance needed to move an air mass equal to the vehicle's inertial mass. This is the same with or without powered motion.

P = \[kDs] is the square root of the mass of media moved per inch of motion multiplied by the non-media resistance forces on the vehicle. In terms of ROSI units, this is \[(ozs2/in2)oz] = ozs/in. ozs = (ozs2/in)(in/s) = mass times velocity = momentum. So P is the amount of momentum imparted by the vehicle to fluid media per inch of motion. It is called momentum transfer (per inch of motion) for this reason.

T = M/P is the inertial mass of the vehicle divided by the momentum transfer per inch of motion. Mass divided by momentum gives units of velocity. A distance divided by a velocity is a time. So, this displacement time, T, is the time needed to displace enough media mass to equal the vehicle's inertial mass at some speed.

V = \[Ds/k] is the square root of the sliding drag by the mass of air moved per inch of track. Work is force acting through a distance. In terms of units, ozin, represents work. So Ds/k can also be seen as the work or energy, e, expended to overcome drag per mass of fluid, j, displaced. Then expressing the energy as kinetic energy, e = jV2 for some velocity V. Hence Ds/k = e/j = V2. This V must be the speed of the displaced air mass matching its energy to work done by drag.

Time

We can use calculus to integrate the expression for small changes in time. The vehicle begins with an initial speed of v0, most often calculated from the powered motion model, while the clock starts at time zero. Any previous time can just be added to the time for terminated motion modeled here since there are no explicit time dependencies. But speed determines the media resistance.

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

The large 'S' shaped symbols indicate a special kind of addition (called integration) over very small steps according to some rather complicated rules.

From a table of integrals we find generally:

   /
  | dv/(Ds + kv2) = arctan(v\[k/Ds])/\[kDs] for kDs > 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.

So [t](0,t) = - M [arctan(v\[k/Ds])/\[kDs]](v0,v)

These indefinite forms must be evaluated on both extremes, (,), producing a term in initial velocity, v0,

t(v) = -M arctan(v\[k/Ds])/\[kDs] + M arctan(v0\[k/Ds])/\[kDs]

Both terms have units of time. Since v, the speed of the vehicle is decreasing, 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. Substituting the more meaningful symbols into the expression,

tc = T arctan(v0/V)

This is a constant that tells us how long the vehicle would travel before coming to a halt. The momentum transfer factor, P, is now \[kDs] and \[Ds/k] is a matching velocity, V. Using these symbols we get,

t(v) = tc - T arctan(v/V)

Velocity

Solve for v in terms of t and simplify. Start with the time expression above and add the last term on the right to both sides,

T arctan(v/V) + t = tc

Subtract off t and divide by T,

arctan(v/V) = (tc-t)/T

Apply the tangent function to both sides to liberate v/V,

v/V = tan((tc-t)/T)

Multiply through by V,

v(t) = V tan((tc-t)/T)

Position

Solve for position along the trajectory 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 = V | tan((tc-t)/T)dt.
  / 0       / 0

From a reference table,

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

So let X = tc-t and a = 1/T 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 = V | tan((tc-t)/T) (-dt)/-1  simplifies to
  / 0      / 0

    / y        / t
   | dy = - V | tan((tc-t)/T) (-dt)
  / 0        / 0

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

[y](0,y) = - V [T ln(cos((tc-t)/T))](t,0)

Expand V and P (T = M/P) to see that V/P = 1/k and evaluate the indefinite forms at their limits to get,

y(t) = M ln(cos((tc-t)P/M))/k - M ln(cos(tcP/M))/k

Let yc = - Y ln(cos(tc/T))

In retrospect, this turns out to be the distance the vehicle has to travel before coming to a halt, the coasting distance. So,

y(t) = yc + Y ln(cos((tc-t)/T))

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 vehicle can travel.

Time

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

(y-yc)/Y = ln(cos((tc-t)/T))

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

exp((y-yc)/Y) = cos((tc-t)/T)

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

arccos(exp((y-yc)/Y)) = (tc-t)/T

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 - T arccos(exp((y-yc)/Y))

Coasting Time

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

t(v=0) = tc = T arctan(v0/V)

It is the displacement time multiplied by a regulating function. The regulator is controlled by the ratio of initial speed to the matching speed of vehicle mass to dispersed media. If this ratio is infinite, the arctan becomes pi/2 (it's evaluated in radians) and tc = piT/2. This is a time limit! No matter how fast your vehicle's initial speed is, it won't travel any longer than piT/2. (pi = 3.14159265.. is the ratio of a circle's circumference to its diameter.)

Coasting Distance

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

y(tc) = yc = - Y ln(cos(tc/T)) from above which is positive since tc is less than piT/2. Note that if tc = 0, yc = 0. Evaluate the cosine using radians.

Notice that if the friction was so high that the coasting distance had been less than the length of the expected trajectory, 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.

The coasting distance can be simplified for computation by expanding the coasting time term, tc, and reducing the resulting trigonometric function composition.

tc = T arctan(v0/V) and

yc = - Y ln(cos(tc/T))

Expand tc in the expression for yc,

yc = - Y ln(cos((T arctan(v0/V))/T))

Inverse terms cancel leaving the composition of three functions,

yc = - Y ln(cos(arctan(v0/V)))

The cos(arctan()) composition can be reduced by considering the definitions of the functions on a right triangle with side x and perpendicular y.

cosO = x/\[x2+y2], tanO = y/x so arctan(y/x) = O

Let y/x = v0/V

Rearrange the cosO expression to admit a substitution in terms of the new expression for y/x (or x/y as it turns out),

x/\[x2+y2] = (x/y)/\[(x/y)2+1] = (V/v0)/\[V2/v02+1]

Continue to simplify until each factor appears only once, if possible,

x/\[x2+y2] = V/\[V2+v02] = 1/\[1+v02/V2]

Substitute this for the composition in the coasting distance expression. Note that the minus sign cancels when the argument of the natural log function is inverted.

yc = Y ln(\[1+v02/V2])

Velocity via position

For the speed in terms of position, we compose the velocity in terms of the time in terms of position during deceleration, y.

v(t(y)) = V tan(arccos(exp((y-yc)/Y)))

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

1 = x2 + y2 (note this is a different "y")

and the relationship among trigonometric functions,

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

we can write

tan(arccos(exp((y-yc)/Y))) = tan(arccos(x)) = tan(O) where we have substituted x = exp((y-yc)/Y) 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(y)), 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(y-yc)/Y)-1] = \[exp(-2(y-yc)/Y)-1]

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

v(y) = V tan(arccos(exp((y-yc)/Y))) = V \[exp(-2(y-yc)/Y)-1]

This is a much easier expression for speed to work with.

To understand this we need to know that typical distances modeled for position along the deceleration trajectory, are much smaller than yc, the coasting distance. Also, yc is usually much smaller than the distance it takes the vehicle to move a mass of air equivalent to its inertial mass, M. This means that y-yc is negative so -2(y-yc)/Y is a positive number less than one. So the exponential evaluates to a number greater than one making the final speed positive in the typical application.

If y > yc, we know the vehicle will stop before going the entire expected distance; a final speed of zero. But if we use the equation for final speed, -2(y-yc)/Y 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.

Deceleration

To get an expression for the deceleration, start with the equation of motion and substitute in the expression for velocity in terms of time.

Divide the equation of motion by the inertial mass, M,

a(t) = - Ds/M - kv(t)2/M

Recognize that M/k = Y and expand the speed factor,

a(t) = - Ds/M - V2 tan2((tc-t)/T)/Y

Expand V and cancel the k by its inverse in the expression for Y,

a(t) = -(Ds + Ds tan2((tc-t)/T))/M

Collect terms in Ds,

a(t) = -Ds(tan2((tc-t)/T) + 1)/M

Since tan2O + 1 = sin2O/cos2O + cos2O/cos2O = (sin2O+cos2O)/cos2O = 1/cos2O

a(t) = -Ds/Mcos2((tc-t)/T)

But let's get this into a form more recognizable as an acceleration. Force is mass times acceleration, so multiply the right-hand-side by Mk/Mk,

a(t) = -MkDs/M2kcosh2((tc-t)/T)

Use the definitions of Y and P,

a(t) = -YP2/M2cosh2((tc-t)/T)

Use the definition of T,

a(t) = -Y/T2cosh2((tc-t)/T)

-Y/T2 looks much more like a deceleration, in/s2, though Ds/M is equivalent. cosh((tc-t)/T) is one when time is the coasting time. Note, that the sliding friction is independent of speed, so a(tc) will still have a negative value.