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Attenuated motion happens in the absence of sliding friction when powered motion ends. On this page, the assumptions used for the powered model are extended to work out a the attenuation model for coasting.
This attenuated model 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.
The other coasting model 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. This terminated motion model is presented on a separate page.
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.
|M||ozs2/in||Inertial mass of vehicle|
|k||ozs2/in2||Aero/hydrodynamic coefficient (due to media)|
|L||in||Length of powered run|
|t||s||Time of travel|
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.
|Equations of Motion||oz||Ma = - kv2|
|Displacement Distance||in||Y = M/k|
|Initial Velocity||in/s||v0 = vt \[1-exp(-2(y+y0)/Y)]|
|Momentum Transfer||ozs/in||P = v0k|
|Displacement Time||s||T = t0 = M/P|
|Time||s||t(y) = T exp(y/Y) - t0|
|Velocity||in/s||v(y) = v0 exp(-y/Y)|
|Velocity||in/s||v(t) = v0/((t+t0)/T)|
|Distance||in||y(t) = Y ln((t+t0)/T)|
|Acceleration||in/s2||a(t) = - kv02/M((t+t0)/T)2|
Initial velocity is included for reference from the power model.
To model the deceleration of the vehicle we use a general form of Newton's second law, F = ma, that captures the forces that oppose most vehicular motion,
Ma = - kv2
The inertial mass, M, for this model must be coupled to deceleration along the line of motion. Wheels generally don't meet this criteria because they would introduce various types of sliding friction. Passive paddle wheels may meet this criteria. If there are no spinning parts, the mass of the vehicle is used for M. The inertial mass is a constant in the model.
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.
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, 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/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 = v0k; In terms of ROSI units, this is (in/s)(ozs2/in2) = 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.
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/k | dv/v2 / 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.
The result is an expression of the indefinite forms
[t](0,t) = - M/k [-1/v](v0,v)
The result of evaluating these forms is
t(v) = Y (1/v - 1/v0)
We'd like to set v to zero to find out how long it would take to stop, but 1/v is infinite! According to this physical model, the boat never really stops. In practice, one takes some measurements of vehicle speed to set a nominal speed at which the vehicle can be considered effectively stopped. Knowing what that speed is, you can get an idea of how long it takes to stop using the expression above.
Solve for v in the time expression by dividing through by Y,
t/Y = 1/v - 1/v0
Add the initial speed term to both sides and swap sides,
1/v = t/Y + 1/v0
v(t) = 1/(t/Y + 1/v0)
To get a more standard form, normalize the denominators,
v(t) = 1/((v0t+Y)/Yv0)
Note the term Yv0 in the denominator of the denominator. It "flips" up to the numerator to get,
v(t) = Yv0/(v0t+Y)
This shows that as t becomes large, the denominator gets large compared to Yv0 and v gets closer to zero. Since v0t gets large compared to Y as time progresses, v(t) is very near Y/t for large enough times, t.
To see the effect on acceleration at "large" values of t, we can take the time derivative of v(t) to get the deceleration near the "end" of the motion.
a(t) = dv/dt = -Y/t2
You can use this expression to find a time past which you can consider the vehicle to have stopped by setting a minimum deceleration, as, and solving for t. Remembering that as is negative, we get,
ts = \[-Y/as] where the "subscript", "s", on t and a means "stopped"
Separate time from position variables in the velocity expression using v = dy/dt,
dy = Yv0dt/(v0t+Y)
Integrate starting the distance and time at zero. Any distance and time traveled prior to the attenuated motion is just added back in later.
/ y / t | dy = Yv0 | dt/(v0t+Y) / 0 / 0
The result is [y](0,y) = Yv0 [ln(v0t+Y)/v0](0,t)
Evaluate the limits to get,
y(t) = Yv0 (ln(v0t+Y) - ln(Y))/v0
Simplify by recognizing the pattern ln(A) - Ln(B) = Ln(A/B).
y(t) = Y ln((v0t+Y)/Y)
Simplify by distributing the divisor Y in the log function,
y(t) = Y ln(1 + v0t/Y)
With the 1 in the log argument, y(t) will always be positive. But if v0 = 0, then the distance will always be zero as it should be.
Note that y(ts) = Y ln(1 + v0ts/Y) is the maximum distance the vehicle can travel before being considered effectively stopped.
Solve y(t) for t to get time in terms of distance.
ky/M = ln(1 + v0kt/M)
Swap sides and apply the exponential function to both sides,
1 + v0kt/M = exp(ky/M)
Simplify to isolate t,
t(y) = M(exp(ky/M) - 1)/v0k
This time result can be plugged into the velocity we got earlier to get velocity in terms of position.
v(t(y)) = Yv0/(v0t(y)+Y)
Make the substitution and simplify, some terms cancel,
v(t(y)) = v0/exp(y/Y)
Invert the exp function and multiply its argument by a negative,
v(y) = v0 exp(-y/Y)
Starting with the equation of motion, solve for acceleration,
a(v(t)) = - v(t)2/Y
Plug the expression for speed in terms of time in to it and simplify
a(t) = - Yv02/(v0t+Y)2
Now we can see that when v0t is much greater than Y, we can consider Y = 0 and get the same expression as above for deceleration at "large" times,
a(t) = - Y/ts2
To get a form that is more like that of the coasting model, continue to simplify the full expression by dividing through by Y2,
a(t) = - v02/Y((v0t+Y)/Y)2
With P = v0k and T = t0 = M/P
a(t) = - kv02/M((t+t0)/T)2 where kv02 has units of force, ounces.
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|Copyright © 1998, 2002 by Michael Lastufka, All rights reserved worldwide.|