Saturday, June 14, 2014

HP Prime: Equations of Motion Under Constant Power


Introduction
 
When considering the motions of automobiles, the calculating acceleration, velocity, and distance can be increased if the power output of the vehicle and the vehicle’s mass are considered. 

It is necessary to go beyond the simple equations of motion when considering vehicles.  For reference, I list the simple of equations of motion here:

v = a*t + v0

x = a*t^2/2 + v0*t + x0

where a is acceleration, v is velocity, and x is position.  v0 and x0 are initial velocity and initial position, respectively.  Acceleration is assumed to be constant.

Considering Power

In the paper “Constant powers equation of motion”, Roger Stephenson (see Sources below) explained that the automobile’s engine can produce constant torque at a certain RPM (revolutions per minute).  However, for the car to accelerate requires either the engine’s rpm or gear ratio linking the engine to the wheels must change.  Drivers aim to change gears to engine rpm at a level where the most power is produced.  This causes torque at the wheels to change.  However, power is kept at a constant level and can be used to modify the equations of motion.

Let P be the power of the vehicle and m it’s mass.  The kinetic energy of the automobile at t=0 is:

KE = ½ * m * v^2 = P * t

Solving for velocity, we get

‘v = √(2*P*t/m)

Stephenson introduced the variable Z, zip, defined as:

Z = √(P/m)

Z nicely combines the vehicle’s constant power and mass.  Deriving the equations of motion started with the recognition of relationship between work and power:

Power = Work / Time

Since Work = Force * Displacement and by Newton’s Second Law, Force = Mass * Acceleration, we arrive at:

P = W /t

P  = m * a * x/t

P/m = a * x/t

Since Z = √(P/m) and v = x/t, and recognizing that a = dv/dt,

Z^2 = dv/dt * v

From there, Stephenson derived the following equations of motion under constant motion in Appendix A in his paper:

 a = Z * (2*t + (v0/Z)^2) ^ (-1/2)

v = Z * (2*t + (v0/Z)^2) ^ (1/2)

x = (Z/3) * ( (2*t + (v0/Z)^2) ^ (3/2) – (v0/Z)^3) ) + x0

Where Z = √(P/m)

Using standard US units, power is stated in ft*lb/s and mass is stated in slugs.   The following conversion factors may be necessary:

1 hp (horsepower) = 550 ft*lb/s

1 slug = 32.17404 lbs (mass * gravity)

1 mph = 22/15 ft/s

1 ft/s = 15/22 mph

The Program CARFORCE

The following HP Prime program, CARFORCE, takes the following for arguments: 

P = power in horsepower (hp)

M = mass in pounds (lb)

I = initial velocity (mph)

T = time to evaluate (seconds)

CARFORCE will return a list of four answers:

Zip (numerical value only (ft/s^1.5))

Acceleration in miles per hour per second

Velocity in miles per hour

Position in feet


This program illustrates the use of units in an HP Prime program.  A way to separate a value from a value_unit couple is to divide such value by 1_unit.  Example:

15_ft / 1_ft = 15

The program listed is shown below.  The program assumes that x0 = 0. 

EXPORT CARPOWER(P,M,I,T)

BEGIN

// 2014-06-14 EWS

// Roger Stevenson, 1980

// Lloyd W. Taylor, 1930

// power (hp),mass (lb),

// initial velocity (mph), time (sec)

LOCAL Z,W,A,V,X,L0;

// convert

P:=CONVERT(P*1_hp,1_(ft*lbf/s))/1_(ft*lbf/s);

I:=CONVERT(I*1_mph,1_(ft/s))/(1_(ft/s));

M:=CONVERT(M*1_lb,1_(slug))/(1_slug);

Z:=√(P/M);

// temp

W:=2*T+I²/Z²;

A:=Z/√W;

V:=Z*√W;

X:=Z/3*(W^1.5-(I/Z)^3);

L0:={Z,A*15/22,V*15/22,X};

L0:=L0*{1,1_(mile/(h*s)),1_(mile/h),1_ft};

 

RETURN L0;

 

END;

 Update:  If you have trouble running the program, here is an alternative program which uses numeric conversions instead of the CONVERT function.

EXPORT CARPOWER(P,M,I,T)
BEGIN
// 2014-07-25 EWS

// Roger Stevenson, 1980
// Lloyd W. Taylor, 1930
// power (hp),mass (lb),
// initial velocity (mph), time (sec)
LOCAL Z,W,A,V,X,L0;
// convert
. use decimal conversions

P:=550*P; // hp to ft*lbf*s^-1
I:=22/15*I; // mph to ft/s
M:=M/32.1740485564;

Z:=√(P/M);
// temp
W:=2*T+I²/Z²;
A:=Z/√W;
V:=Z*√W;
X:=Z/3*(W^1.5-(I/Z)^3);
L0:={Z,A*15/22,V*15/22,X};

RETURN L0;
END;

Example:

Data:  Vehicle has a 100 hp engine, has a mass of 4,000 pounds.  Figure the equations of motion after 10 seconds elapsed.  Assume the vehicle is initially stopped.

CARPOWER(100, 4000, 0, 10) returns

{21.0331445022, 3.2066959696_mile/(h*s), 64.1339193918_mile/h, 627.087211833_ft}

Sources
Lloyd, Taylor W. “The Laws of Motion Under Constant Power”  The Ohio Journal of Science, v30 n4 (July, 1930), 218-220.  https://kb.osu.edu/dspace/bitstream/1811/2458/1/V30N04_218.pdf

Revised 6/12/2014
 
Stephenson, Roger  “Constant power equations of motion”  (Published January 29, 1982) from the compilation book The Physics of Sports edited by Angelo Armenti, Jr.  American Institute of Physics: New York  1992, pp. 284-289
   

Considering power, calculating acceleration, velocity, and position become more accurate.
 
Until next time, have a great day!  Happy Birthday Mom and Dad!  Happy Father’s Day!


Eddie

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