ARO309 - Astronautics and Spacecraft Design

ARO309 - Astronautics and Spacecraft Design

Winter 2014

Try LamCalPoly Pomona Aerospace Engineering

Lecture 03: Numerical Integrations

Chapter TBD

Introductions

• In this lecture we will look at how dynamical system problems are solved numerically

Real Life ProblemReal Life Problem

Mathematical Models (Equations)

Mathematical Models (Equations)

Numerical AlgorithmNumerical Algorithm

In Astodynamics Problems we usually deal with Initial Value

Problem (IVP)

Given:

Find:

2nd Order Runge Kutta

• Given an 1st order ODE

• Applying Taylor Series expansion to 2nd order

• Since:

and


• Now we have


• Or

• The Coefficient in Butcher Tableaux form. Note:


• Graphically

to t1

slope=k1=f

slope=k2

yn+1 (Approximate Soln)

yoExact Solution

Gernal Runge Kutta (explicit)

• The explicit RK method is given by

Runge Kutta (explicit)

• Note that RK must satisfy:

Butcher Table

Runge Kutta (explicit)• RK4

where

RK4 Example

• Goal: Integrate the 2-Body Problem using RK4 for a single step with step size = 60 sec

• Equations of Motion (EOM):

• State Equation (around Earth):

RK4 Example

• From the problem:

RK4 Example

RK4 Example• RK4 fails for large time steps (range plot example)

Lecture 04: Two-Body Dynamics:

Orbit Position as a Function of TimeChapter 3

Introductions

• Chapter 2 (Lection 1 and 2) relates position as a function of θ (true anomaly) but not time

• Time was only introduced when referring to orbit period

• Here we attempt to find the relations between position of the S/C and time Kepler’s Equation

Time versus True Anomaly

• Recall from Chapter 2

Since Then

Integrating from 0 (assuming tp = 0) to t and from 0 to θ


Time versus True AnomalySimple Case: Circular Orbits (e=0)

If e = 0, then therefore

Since for a circular orbit we have then

FORCIRCULARORBIT OR

Time versus True AnomalyElliptical Orbits (0<e<1)

Then a = 1 and b = e, therefore we have b < a

Me = Mean anomaly for the ellipse


Therefore we have

From the orbit period of an ellipse we know (or can derive) that

Therefore we can solve for me as function orbit period as

OR where n = mean motion = 2π/Te

Elliptical Orbits (0<e<1)


We need to fine out Me still ?

Let’s introduce another variable E = eccentric anomaly



This relates E and θ, but it leaves the quadrant of the solution unknown and you get two values of E for the equation. To eliminate this ambiguity we use the following identity

OR

Therefore or



We need to fine out Me still

E

This is Kepler’s Equation


To find t given Δθ

• Given orbital parameters, find e and h (assume θ = 0 deg)

• Find E:

• Find T (orbit period):


To find t given Δθ

• Fine Me:

• Find t:


Question: What if you are going from a θ = θa to θ = θb?

Answer: Find the time from θ = 0 to θ = θa and the time from θ = 0 to θ = θb. Then subtract the differences.


To find θ given Δt

• Given orbital parameters, find e and h (assume θ = 0 deg

• Find T (orbit period):

• Find Me:

• Find E using Newton’s method (or a transcendental solver)



• Using Newton’s Method:

– Initialize E = Eo:

– Find f(E):

– Find f’(E):

– If abs( f(E) / f’(E) ) > TOL, then repeat with

– Else Econverged = En


For Me < 180 deg For Me > 180 deg


• After finding the converged E, then find θ


Time versus True AnomalyParabolic Orbits (e = 1)

Then a = 1 and b = e, therefore we have b = a

MP = Parabolic Mean Anomaly

Time versus True AnomalyParabolic Orbits (e = 1)

Thus given t or Δt we can find MP

To fine θ we can find the root of the below equation

Which has one real root

STEPS:Find hFind MP

Find θ

STEPS:Find hFind MP

Find θ

Time versus True AnomalyHyperbolic Orbits (e > 1)

Then a = 1 and b = e, therefore we have b > a


Where the Hyperbolic mean anomaly is

Thus we have

Similar with Ellipse we will intro a new variable, F, the hyperbolic eccentric anomaly to help solve for the Mean Hyperbolic anomaly, Mh.


Hyperbolic eccentric anomaly for the Hyperbola

Since:


We now have

Solving for F and since we now have

Using the following trig identities for sine and cosine


We now have

Therefore we now have:

This is Kepler’s Equation for Hyperbola

Similar to Elliptical orbits we can solve for F as a function of θ, which is found to be. Thus given θ we can find F, and Mh, and finally t.

STEPS TO FIND θ (given t)• Set initial F0 = Mh where

• Find f and f’

• If abs( f / f’ ) > TOL, repeat steps with updated F

• Else, Fconverged = Fi. Now find θ


If time, t, was given and θ is to be found then we have to solve for Kepler’s equation for hyperbola iteratively using Newton’s method

Universal Variables

• What happens if you don’t know what type of orbit you are in? Why use 3 set of equations?

• Kepler’s equation can be written in terms of a universal variable or universal anomaly, Χ, and Kepler’s equation becomes the universal Kepler’s equation.

WhereIf α < 0, then orbit is hyperbolicIf α = 0, then orbit is parabolicIf α > 0, then orbit is elliptical

Universal Variables

• Stumpff functions

or for z = αΧ2

,

Universal Variables

• To use Newton’s method we need to define the following function and it’s derivative

• Iterate with the following algorithm

with

Universal Variables

• Relation ship between X and the orbits

For t0 = 0 at periapsis

Universal Variables

• Example 3.6 (Textbook: Curtis’s)

Find h and e

Since , then

Universal Variables


Therefore

So X0 is the initial X to use for the Newton’s method to find the converged X

Universal Variables


Universal Variables


Thus we accept the X value of X = 128.5

where

Lagrange Coefficients II• Recall Lagrange Coefficients in terms of f and g coefficients

• From the universal anomaly X we can find the f and g coefficients

Lagrange Coefficients II

andWhere

• Steps finding state at a future Δθ using Lagrange Coefficients

1. Find r0 and v0 from the given position and velocity vector

2. Find vr0 and α

3. Find X4. Find f and g5. Find r, where r = f r0 + g v0

6. Find fdot and gdot7. Find v

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ARO309 - Astronautics and Spacecraft Design