TI 30Xa Algorithm: Acceleration, Velocity, Speed

TI 30Xa Algorithm: Acceleration, Velocity, Speed

Introduction and Algorithm

Given the acceleration (α), initial velocity (v0), and initial position (p0) of an object on a 2D plane, calculate velocity and position at time T.

Assuming that acceleration is constant, the equations are:

Acceleration: A = α

Velocity: V = α * t + v0

Position: P = α / 2 * t^2 + v0 * t + p0

With some simplification, we can simplify the work flow:

( I )

A = α

( II )

V = α * t + v0

V = A * t + v0 (II)

( III )

P = α / 2 * t^2 + v0 * t + p0

P = t * ( α / 2 * t + v0 ) + p0

P = t * ( α / 2 * t + 2 * v0 / 2 ) + p0

P = t * ( (α * t + 2 * v0) / 2 ) + p0

P = t / 2 * ( α * t + 2 * v0 ) + p0

P = t / 2 * ( α * t + v0 + v0 ) + p0

P = t * (V + v0) / 2 + p0

( I ): A = α

( II ): V = A * t + v0

( III ): P = t * (V + v0) / 2 + p0

Key Strokes:

1. Store the time T in Memory 1 ( T [ STO ] 1 )

2. Store the initial velocity in Memory 2 ( v0 [ STO ] 2 )

3. Enter the acceleration constant, α, press [ = ]

4. Calculate the velocity at time T: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ]

5. Calculate the position at time T: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] p0 [ = ]

Examples

Example 1:

α = 1.3

v0 = 0

p0 = 0

T = 10

1. 10 [ STO ] 1

2. 0 [ STO ] 2

3. Acceleration: 1.3 [ = ]

4. Velocity: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ] (Result: 13)

5. Position: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] 0 [ = ] (Result: 65)

Example 2:

α = -9.80665

v0 = 0

p0 = 10000

T = 30

1. 30 [ STO ] 1

2. 0 [ STO ] 2

3. Acceleration: 9.80665 [ +/- ] [ = ]

4. Velocity: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ] (Result: -294.1995)

5. Position: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] 10000 [ = ] (Result: 5587.0075)

Example 3:

α = -0.05

v0 = 10

p0 = 20

T = 15

1. 15 [ STO ] 1

2. 10 [ STO ] 2

3. Acceleration: 0.05 [ +/- ] [ = ]

4. Velocity: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ] (Result: 9.25)

5. Position: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] 20 [ = ] (Result: 164.375)

Example 4:

α = 4.26

v0 = 1.8

p0 = 0

T = 5

1. 5 [ STO ] 1

2. 1.8 [ STO ] 2

3. Acceleration: 4.26 [ = ]

4. Velocity: [ × ] [ RCL ] 1 [ + ] [ RCL ] 2 [ = ] (Result: 23.1)

5. Position: [ + ] [ RCL ] 2 [ = ] [ × ] [ RCL ] 1 [ ÷ ] 2 [ + ] 0 [ = ] (Result: 62.25)

Hope you find this useful.

Happy Halloween, Everyone!

Eddie

All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

HP Prime and Numworks: Plotting a Parametric Line of Motion

 HP Prime and Numworks: Plotting a Parametric Line of Motion

Plotting the Position of Motion

This program draws a 2D motion plot from an initial starting point (x, y) given initial velocity and acceleration. The rate and direction of both velocity and acceleration are assumed to be constant.

x(t) = ax * t^2 / 2 + vx * t + x0

y(t) = ay * t^2 / 2 + vy * t + y0

where:

t = number of seconds.

ax = acceleration in the x direction

ay = acceleration in the y direction

vx = initial velocity in the x direction

vy = initial velocity in the y direction

The HP Prime PPL program PLOTMOTION displays a traceable curve with a table of values.

The Numworks python script plotmtn.py uses math and matplot.pyplot modules. A scatter plot is laid on top of the path plot. The plot begins at the initial point where it is marked green.

HP Prime Program: PLOTMOTION

EXPORT PLOTMOTION()

BEGIN

// EWS 2024-07-22

LOCAL ch;

ch:=INPUT({C,D,A,B,V,U,N},

"Motion Plot Per Second",

{"x0:","y0:","vx:","vy:","ax:","ay:","n:"},

{"initial x position",

"initial y position",

"initial velocity x direction",

"initial velocity y direction",

"acceleration x direction",

"acceleration y direction",

"number of seconds"});

// user presses cancel

IF ch==0 THEN

KILL;

END;

// user presses OK

STARTAPP("Parametric");

'V*T^2/2+A*T+C'▶X1;

'U*T^2/2+B*T+D'▶Y1;

CHECK(1);

Parametric.Tmin:=0;

Parametric.Tmax:=N;

Parametric.Tstep:=1;

// table and plot

STARTVIEW(10);

STARTVIEW(9);

END;

Numworks Python Code: plotmtn.py

from math import *
from matplotlib.pyplot import *

# 2024-07-23 EWS

print("Motion Plot per second from (0,0)")
c=eval(input("init. x position? "))
d=eval(input("init. y position? "))
a=eval(input("init. x velocity? "))
b=eval(input("init. y velocity? "))
v=eval(input("acceleration x? "))
u=eval(input("acceleration y? "))
n=int(input("number of seconds? "))

x=[v*t**2/2+a*t+c for t in range(n)]
y=[u*t**2/2+b*t+d for t in range(n)]

x0=min(x)-1
x1=max(x)+1
y0=min(y)-1
y1=max(y)+1
axis((x0,x1,y0,y1))
text(x0,y1-1,"Motion Plot")
plot(x,y,'gray')
scatter(x,y,color='blue',marker="h")
scatter(x[0],y[0],color='green',marker="h")
show()

Source

Tremblay, Christopher. Mathematics for Game Developers. Thomson Course Technology. Boston, MA. 2004. ISBN 1-59200-038-X.

Eddie

All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

Casio fx-991EX and fx-115ES Plus: Velocity Conversions

Casio fx-991EX and fx-115ES Plus: Velocity Conversions

The Casio fx-991EX Classwiz and Casio fx-115ES Plus calculators, among others, have 40 conversions in various areas. 

Keystrokes

Keystrokes: Casio fx-115ES Plus

[ SHIFT ] [ 8 ], enter the required two digit code

The conversions we'll be working with this blog entry:

ft → m, 03
m → ft, 04
yd → m, 05
m → yd, 06
mile → km, 07
km → mile, 08
km/hr → m/s, 19
m/s → km/hr, 20

Keystrokes:  Casio fx-991EX Classwiz

[ SHIFT ] [ 8 ].    The conversions will be grouped by category (length, area, volume, etc).

ft → m, 1, 3
m → ft, 1, 4
yd → m, 1,5
m → yd, 1,6
mile → km, 1, 7
km → mile, 1, 8
km/hr → m/s, ↓, 1, 1
m/s → km/hr, ↓, 1, 2

km/hr  (kilometers per hour) vs mph (miles per hour)

The calculators mentioned does not have a direct conversion between km/hr and mph.  However, because the time frame is the same (per hour), we can use the km/mile conversions (since 1 hour = 1 hour and no time conversion part is required, only the length).

Example 1:  150 km/hr to mph

150 km>mile returns 93.20567884

150 km/hr = 9320567884 mph

Example 2:  65 mph to km/hr

65 mile>km = 104.60736

65 mph = 104.60736 km/hr

More Examples

Example 3:  5.8 ft/s to m/s

Again, since the time portion is the same (second to second), we only have to worry about length.

5.8 ft>m returns 1.76784

5.8 ft/s = 1.76784 m/s

Example 4:  110 ft/s to km/hr

Now we are converting both length and time.   There are no direct time conversions on the calculators featured, so we have to use the manual conversion. 

Remember that 1 hr = 3600 s and 1 km = 1000 m. To convert from m to km, divide by 1000 and to convert from 1/s to 1/hr, multiply by 3600 (note: 1/s to 1/hr)

(110 ft>m) ÷ 1000 × 3600 returns 120.7008

110 ft/s = 120.7008 km/hr

Example 5:  40 m/s to km/hr

Here is another example where we both have to convert the length (m to km) and time component (1/s to 1/hr).   This is similar to the Example 3.  However, the Casio calculators featured on this blog has conversions between m/s and km/hr (velocity section).

40 m/s>kh/hr returns 144 km/hr

40 m/s = 144 km/hr

Example 6:  100 ft^2 to m^2

The calculators have two area conversions (acre/m^2).   However, we can use the same length conversion twice to make other areas.

100 ft>m ft>m returns 9.290304

100 ft^2 = 9.290304 m^2

Example 7:  1 mi to yd by using the conversions available

(1 mi>km × 1000) m>yd returned 1760

(1 km = 1000 m)

1 mi = 1760 yd

A Cautionary Tale

Example 8:  1 acre to ft^2

1 acre>m^2 m>ft m>ft returns 43559.99545

However, as defined, 1 acre = 43560 ft^2. 

Lesson:  use caution when chaining conversions, 100% accuracy isn't guaranteed.

Eddie

All original content copyright, © 2011-2020.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author

HP Prime and Casio fx-5800P: The Two Train Problem (updated 2/26/2018)

HP Prime and Casio fx-5800P:  The Two Train Problem

(updated 2/28/2018)

Introduction

Have you ever heard the infamous problem “two trains are heading towards each other…”? 

Today’s blog will cover the following question:

Two trains are heading towards each other, on their own separate train track.  Each train has going at their own speed (velocity).  The trains start out a distance apart.

1. When will the trains cross over the same spot, and

2. Where will the trains cross over the same spot?

According to the diagram above, we have two trains, labeled Train I and Train II, each going at velocity v and acceleration a.  The trains start at distance D apart.  We will call the point where the trains cross over the same spot, x. 

Notes:

1. To make our lives easier, let’s assume that Train I starts at position 0, while Train II starts at position D. 

2. Train I is going at velocity v and acceleration a. 

3. Train II is going at velocity –v and acceleration –a. Why negative?  Train II is traveling in the opposite direction of Train I.

Setting up the Equations

The general distance equation is:  x = x0 + v*t + a*t^2, where x0 is the initial position.

We are going to cover two scenarios:  one where there is no acceleration, that is the velocity of both trains is constant.  The other is where acceleration is present for at least one of the trains.  Acceleration is assumed to be constant. 

In general the distance equations for both trains are:

Train I:  x =        v1*t + 1/2*a1*t^2

Train II: x = D  - v2*t -  1/2*a2*t^2

What this boils down to are a system of two equations, solving for both t and x.

Case 1:  Constant Velocity

The trains are moving at constant velocity in this scenario.  Which means there is no acceleration, and hence a1 = 0 and a2 = 0.  Then:

Train I:  x =        v1*t

Train II: x = D -  v2*t

Subtracting equations gives us:

0 = -D + v1*t – -v2*t

0 = -D + (v1 + v2)*t

D = (v1 + v2)*t

t = D/(v1 + v2)

Once t is known, we can use substitution to find x.  Train I’s equation is simpler, so:

x = v1*t

Case 2:  Non-Zero Acceleration

Start with:

Train I:  x =        v1*t + 1/2*a1*t^2

Train II: x = D  - v2*t -  1/2*a2*t^2

Subtracting the equations and solving for t yields:

0 = -D + v1*t – -v2*t + 1/2*a1*t^2 - -1/2*a2*t^2

0 = -D + (v1 + v2)*t + 1/2*(a1 + a2)*t^2

Using the quadratic equation:

t = ( -(v1+v2) ± √( (v1+v2)^2 + 4*D*1/2*(a1+a2)) ) / (2*1/2*(a1 + a2))

t = ( -(v1+v2) ± √( (v1+v2)^2 + 2*D*(a1+a2)) ) / (a1 + a2)

Note that there are two roots for t.  However, given the problem, negative time does not make sense.  We can eliminate the root for t that is obviously negative.

Hence, our solution for time in this problem is:

t = ( -(v1+v2) + √( (v1+v2)^2 + 2*D*(a1+a2)) ) / (a1 + a2)

Again, we are going to use the simpler equation for Train I to figure x:

x = v1*t + a1*t^2

The Program TRAINS

The program TRAINS will solve this problem.  Below are the codes for both the HP Prime and Casio fx-5800p.  Enter each as a positive value as the directions are accounted for in the program.

In the program, Train I is considered the left train, while Train II is considered the right train.

HP Prime Program: TRAINS

EXPORT TRAINS()

BEGIN

// 2018-02-22 EWS

// 2 trains problem

LOCAL d,t,x;

LOCAL v1,a1,v2,a2;

INPUT({d,v1,a1,v2,a2},

"Two Opposing Trains",

{"Dist:","L Vel:","L Acc:",

"R Vel:","R Acc:"},

{"Distance between trains",

"Left Train: Velocity",

"Left Train: Acceleration",

"Right Train: Velocity",

"Right Train: Acceleration"});

// calculation

IF a1≠0 OR a2≠0 THEN

t:=(−(v1+v2)+√((v1+v2)^2+

2*d*(a1+a2)))/(a1+a2);

x:=v1*t+1/2*a1*t^2;

ELSE

t:=d/(v1+v2);

x:=v1*t;

END;

// results

RETURN {"Time",t,

"Position",x};

END;

Casio fx-5800P Program: TRAINS

“EWS 2018-02-22”

“DISTANCE”?→D

“LEFT-VEL.”?→A

“LEFT-ACC.”?→E

“RIGHT-VEL.”?→B

“RIGHT-ACC.”?→F

If E≠0 Or F≠0

Then

(-(A+B)+√((A+B)^2

+2*D*(E+F)))÷

(E+F)→T

A*T+1/2*E*T^2→X

Else

D÷(A+B)→T

A*T→X

IfEnd

“TIME:” ◢

T ◢

“POSITION:” ◢

X

Examples

Assume compatible units for each example.

Example 1:  Constant Velocity

Train I:  v1 = 40 (a1 = 0)

Train II: v2 = 35 (a2 = 0)

Distance:  300

Results:

Time:  4

Position:  160

Example 2:  Acceleration

Train I:  v1 = 40, a1 = 0.5

Train II: v2 = 35, a2 = 0.8

Distance:  300

Results:

Time:  3.87018761454

Position:  158.552092625

Thank you to Gianfranco Cazzaro, Dieter, and all who wrote to me pointing out the errors.  

Eddie

This is blog is property of Edward Shore, 2018.

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

Follow the Bouncing Ball

Blog Entry #67


Introduction

The goal is to find the velocity of a ball, and the angle it travels, given the initial velocity and angle.

The next two tables are for reference. Thanks to my cousin Gina for helping me with the Coefficient of Restitution! You are the best!

Variables:

Given:
v_i = initial velocity
θ = angle of impact before the bounce

Want to find:
v_f = final velocity
Φ = angle of travel after the ball bounces

Assumptions:
1. Gravity has negligible impact. We are talking about the moment just before the ball bounces to just after the ball bounces.
2. The coefficient of resistution (COR) is measured experimentally (see table). I thank my awesome cousin Gina Ramirez helped me with getting data. (see table above)
3. The x-direction is "forwards/backwards" and the y-direction is "up towards the sky/down towards the earth"
4. The coefficeint of friction only affects the y-direction (is this correct?)
5. Kinetic energy is not conserved.
6. The ground does not move and maintains a zero velocity.

Variables:
COR = coefficient of restitution, depending on the type of ball and surface the ball is being bounced on
μ = coefficient of friction, dependent on the surface the ball is being bounced on
v_ix, v_iy = initial velocity of the ball in the x-direction and y-direction, respectively
v_fx, v_fy = final velocity of the ball in the x-direction and y-direction, respectively

COR = √ ( (height of ball after the bounce) / (height of the ball before the bounce) )

Since momentum must be conserved in both the x and y directions:

m * v_i * cos θ = m * v_f * cos Φ
m * v_i * sin θ = m * v_f * sin Φ

m * v_i * cos θ - m * v_f * cos Φ = 0
m * v_i * sin θ - m * v_f * sin Φ = 0

m * v_i * cos θ - m * v_f * cos Φ = m * v_i * sin θ - m * v_f * sin Φ

m * ( v_i * cos θ - v_f * cos Φ ) = m * ( v_i * sin θ - v_f * sin Φ )

Note the definition of Impulse, I = m * Δv = Δp

Section A: Coefficient for Restitution

COR = v_fy / v_iy

v_fy = v_iy * COR

Hence: v_fy - v_iy = v_iy * COR - v_iy = v_iy * (COR - 1)

Section B: Angle After Bounce

tan Φ = v_fy / v_fx
tan Φ = (v_iy * COR) / v_fx
cot Φ = v_fx / (v_iy * COR)
COR * cot Φ = v_fx / v_iy

Section C: Angle Before Bounce

tan (-θ) = -v_iy / v_ix
- tan θ = - v_iy / v_ix
tan θ = v_iy / v_ix

Main:

Finding Final Angle After Bounce

Impulse, taking friction ( μ ) into consideration:

m * μ * (v_fy - v_iy) = m * (v_fx - v_ix)
μ * (v_fy - v_fi) = v_fx - v_ix

From Section A:

μ * v_iy * (COR - 1) = v_fx - v_ix
μ * (COR - 1) = v_fx/v_iy - v_ix/v_iy

From Section B:

μ * (COR - 1) = COR * cot Φ - v_ix/v_iy

From Section C:

μ * (COR - 1) = COR * cot Φ - tan θ
μ * (COR - 1) + tan θ = COR * cot Φ
(μ * (COR - 1) + tan θ) / COR = cot Φ
COR / (μ * (COR - 1) + tan θ) = tan Φ

which implies that:

Φ = arctan ( COR / (μ * (COR - 1) + tan θ) )

Finding Final Velocity

From Section B:

v_fx = (v_iy * COR) / tan Φ

We also know that:

cos Φ = v_fx / v_f

Then:

v_f = v_fx / cos Φ
v_f = (v_iy * COR) / (tan Φ * cos Φ)
v_f = (v_iy * COR) / sin Φ
v_f = (v_i * sin θ * COR) / sin Φ

Results

Final Angle:

Φ = arctan ( COR / (μ * (COR - 1) + tan θ) )

Final Velcoity:

v_f = (v_i * sin θ * COR) / sin Φ

Sources:

"Friction" http://physics.info/friction Retrieved 3/25/2012

"Coefficient of Restitution" Wikipedia. http://en.wikipedia.org/wiki/Coefficient_of_restitution Retrieved 3/22/2012

H. Brody "That's how the ball bounces" from "The Physics of Sports" edited by Armenti Angelo. American Institute of Physics: New York. 1992


This blog is property of Edward Shore. © 2012