Jacobi Elliptical Functions

This blog entry is possible thanks to @mathematicsprof on Twitter. He posted an article "Jacobi Elliptic Functions from a Dynamic Systems Point of View", written by Ken Meyer Ph.D of the University of Cincinnati, which not only gave me something to read during lunch last Friday but also lead me on how to calculate Jacobi Elliptic Functions. I am so grateful!

By the way, I am on Twitter: @edward_shore.

Jacobi Elliptic Functions - An Introduction

Let k and t be parameters where 0 < k < 1 as the following system of differential equations is to be solved:

dx/dt = y(t) * z(t)
dy/dt = -z(t) * x(t)
dz/dt = -k^2 * x(t) * y(t)

which satisfy the initial conditions x(0) = 0, y(0) = 1, and z(0) = 1.

The Jacobi Elliptical Functions are defined to be the solutions to the above system.

The sine amplitude function is defined as sn(t, k) = x(t).

The cosine amplitude function is defined as cn(t, k) = y(t).

The delta amplitude function is defined as dn(t,k) = z(t).

A very interesting point, which has also managed to trip me up for years, is that there is no explicit closed formula for sn(t,k), cn(t,k), and dn(t,k). Yet mathematicians can find the derivatives and integrals of these functions.

Derivatives of the Jacobi Elliptical Functions

These derivatives, by looking at the system above, are:

d/dt sn(t,k) = cn(t,k) * dn(t,k)
d/dt cn(t,k) = -dn(t,k) * sn(t,k)
d/dt dn(t,k) = -k^2 * sn(t,k) * cn(t,k)

When k=0

By setting k=0 and finding solutions to the system equations becomes:

dz/dt = 0
z(t) = c
Since z(0)=1, conclude z(t)=1.

Then
dx/dt = y(t)
dy/dt = -x(t)
with initial conditions x(0)=0 and y(0)=1.

Since sin(0)=0, cos(0)=1, d/dt sin t = cos t, and d/dt cos t = -sin t, we can conclude that x(t) = sin t and y(t) = cos t. Meyer states that as k approaches 0, sn(t,k) approaches sin(t), cn(t,k) approaches cos(t), and dn(t,k) approaches 1.

A Very Familiar Identity

Similar to the famous identity:

sin²(θ) + cos²(θ) = 1

An identity of Jacobi Elliptical functions is:

sn²(t,k) + cn²(t,k) = 1

Why is this true? Take the derivative with respect to t and we find that:

2 sn(t,k) cn(t,k) dn(t,k) - 2 cn(t,k) dn(t,k) sn(t,k) = 0

How to Calculate

We can use the integral definition to assist us in calculating values for sn, cn, and dn. In this section, let the parameters be u and k, respectively. The integral definition stems from the incomplete elliptical integral:

where u and k are known and p is the value we are solving for.

The value p is known as the Jacobi Amplitude, am(u,k) for short. Then:

am(u,k) = p

sn(u,k) = sin(am(u,k)) = p

cn(u,k) = cos(am(u,k)) = p

dn(u,k) = √(1 - k² sin² (am(u,k)) = √(1 - k² sin²(p))

Advanced mathematics software or an online calculator, such as this one from Ke!san, are often used to calculate values of Jacobi Elliptical Functions. Fortunately, this task can be done with high-end graphing calculators, using their solve application.

I have been able to obtain accurate answers using the Hewlett Packard HP-50g, Texas Instruments TI-84+, and Casio Prizm fx-CG 10. The later two are relatively fast in finding values. Below I will show the setup for each of the three calculators. I am pretty sure the TI nSpire can handle this too. The following screens show how to obtain am(u,k), the Jacobi Elliptical functions easily follow.

In the solver screen, you can leave X blank (HP-50G only), or just assign any arbitrary value to X, such as 1, since X is the dummy variable in the integration and has no effect on calculations.

Here is a small table of values:

Resources

Meyer, Kenneth R. "Jacobi Elliptic Functions from a Dynamic Systems Point of View" The Mathematical Association of America. Monthly 108. October 2001 - retrieved 11/9/2012

Weinstein, Eric "Jacobi Elliptic Functions" - From MathWorld - A Wolfram Web Source, http://mathworld.wolfram.com/JacobiEllipticFunctions.html, retrieved 11/11/2012

Fun as always! Thank you as always. Until next time, Eddie




This blog is property of Edward Shore, 2012.

Projectiles - A Realistic Point of View (Linear Drag)

Projectile Motion

This blog is about projectile motion. Most resources give an introductory treatment of projectile motion. And yes, I will present the non-air resistance model in this section. But I am also going to present a more realistic model, taking into account air resistance, size of the projectile, and temperature.

All amounts are presented in SI (kilograms, meters, seconds) units.

Quick Conversions
1 ft = 3.280839895 m
1 lb = 0.4535924 kg

Simple Projectile Motion

A perfect projectile can be described by the following equations:

x(t) = v0 t cos θ
y(t) = v0 t sin θ - 0.5 g t^2

Where
v0 = initial velocity
θ = Intial angle, in degrees
g = gravitational acceleration constant = 9.80665 m/s^2 = 32.17404856 ft/s^2
And we assume the starting height is zero (on the ground)

Note that air resistence, features of the object, and other outside objects are ignored.

Example: Hitting a Golf Ball

Use the simplistic model to plot a trajectory with an initial velocity of 150 mph (67.056 m/s) at a 35° angle.

The trajectory lasts for approximately 7.844 seconds, with a range of approximately 430.86 meters and a height of approximately 75.4 meters.

A More Realisitc View of Projectile Motion

The equations above describe simple projectile motion. However, if we want a realistic picture, we have to consider factors such as temperature, air pressure, and the size of the projectile.

So of the variables we will consider are the drag coefficient, the density of air, and the object's terminal velocity.

Note: Linear drag is assumed. This is best for objects moving at slow speeds. No lift is used.

Drag Coefficient

The drag coefficient is a dimensionless quantity that quantifies the resistance of an objects's movement, taking into consideration factors such as the object's size, shape, and smoothness.

Approximate Drag Coefficients:
Perfect Smooth Baseball: 0.1
Baseball: 0.3
Golf Ball: 0.4
Tennis Ball: 0.6

Density of Air

The density of air can be determined using the Ideal Gas Law:

P V = N R T

where
P = absolute pressure = 101325 kPa
R = Specific Gas Constant = 287.058 J/kg K
T = temperature in Kelvins
N = number of moles (mass)
V = volume of the gas
ρ = air density = N/V in kg/m^3

Solving for ρ gives:

ρ = P / (R T) ≈ 352.9774471/T

Note: To convert degrees Celsius to Kevlins, add 273.15.

Terminal Velocity

When an object falls, it reaches a terminal velocity when the drag becomes equal to the object's weight. At terminal velocity, the object no longer accelerates.

Terminal Velocity is measured by:

V = √ ( ( 2 × mass of object × g ) / ( drag coefficient × ρ × surface area of object ) )

TROJECT
Calculator: CASIO fx-CG10 (Prizm)

Note: Let [triangle] be the symbol for the "stop and display triangle"


{.3, .6, .4} → List "C"
{.145, .057, .045} → List "M"
{.041043, .003318, .001385} → List "S"
Deg
ClrText
"Init Velocity="?→ V
"Init Angle="? → θ
9.80665 → G
Menu "Object","Baseball",1,"Tennis Ball",2,"Golf Ball",3
Lbl 1: 1 → I : Goto Y
Lbl 2 : 2 → I : Goto Y
Lbl 3 : 3 → I : Goto Y
Lbl Y
Menu "Temperature","104°F/40°C",A,"95°F/35°C",B,
"86°F/30°C",C,"77°F/25°C",D,
"68°F/20°C",E,"59°F/15°C",F,
"50°F/10°C",G,"41°F/5°C",H,
"32°F/0°C",I
Lbl A : 1.1839 → P : Goto Z
Lbl B : 1.2041 → P : Goto Z
Lbl C : 1.2250 → P : Goto Z
Lbl D : 1.2466 → P : Goto Z
Lbl E : 1.2690 → P : Goto Z
Lbl F : 1.2920 → P : Goto Z
Lbl G : 1.3163 → P : Goto Z
Lbl H : 1.3413 → P : Goto Z
Lbl I : 1.3943 → P : Goto Z
Lbl Z
List "M"[I] → M
√ (( 2 × M × G) ÷ (List "C"[I] × P × List "S"[I])) → W
M × G ÷ W → K
Solve(-MGX ÷ K + M ÷ K × ( 1 - e^(-KX ÷ M)),100) → E
MV cos θ ÷ K × ( 1 - e^(-KE ÷ M)) → R
"Range="
R [triangle]
-M ÷ K × ln ((MG) ÷ K × (V sin θ + MG ÷ K)^-1) → F
-MGF ÷ K + M ÷ K × (V sin θ + MG ÷ K) × (1 - e^(-KF ÷ M)) → H
"Height="
H [triangle]
ClrGraph
ViewWindow -.5, R + .5, 1, -1, H+1, 1, 0, E, .05
θ → A
ParamType
G SelOff
G SelOn 1
"MV cos A ÷ K × (1 - e^(-KT ÷ M))" → Xt1
"-MGT ÷ K + M ÷ K × (V sin A + MG ÷ K) × (1 - e^(-KT ÷ M))" → Yt1
DrawGraph

-----------

Variables:

M = mass of object, in grams
V = initial velocity, in metes/seconds
A = θ = initial angle, in degrees
C = drag coefficient
S = radius of the object
P = gas density
K = terminal velocity

Revisiting the Example: The Golf Ball

Use the proposed model to plot a trajectory with an initial velocity of 150 mph (67.056 m/s) at a 35° angle. Assume it is 77°F outside (25°C).

The trajectory has a range of approximately 163.46 meters and a height of approximately 45.02 meters. It lasts about 6.177 seconds.

References:

Article: Erlichson, Herman. "Maximum Projectile range with drag and lift, with particular application to golf." The College of Staten Island, 1982

From:

Armenti, Angelo Jr. (editor) The Physics of Sports American Institute of Physics 1983 pg. 71-78

Reference Links:

Terminal Velocity (NASA.gov)

Drag Coefficient (NASA.gov)

Drag on a Baseball (NASA.gov)

Air Pressure