ARO309 - Astronautics and Spacecraft Design

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ARO309 - Astronautics and Spacecraft Design . Winter 2014 Try Lam CalPoly 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 Problem.

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### ARO309 - Astronautics and Spacecraft Design

Winter 2014

Try Lam

CalPoly 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 Problem

In Astodynamics Problems we usually deal with Initial Value Problem (IVP)

Mathematical Models (Equations)

Given:

Find:

Numerical Algorithm

2nd Order Runge Kutta
• Given an 1st order ODE
• Applying Taylor Series expansion to 2nd order
• Since:

and

2nd Order Runge Kutta
• Or
• The Coefficient in Butcher Tableaux form. Note:
2nd Order Runge Kutta
• Graphically

slope=k2

yn+1(Approximate Soln)

slope=k1=f

Exact Solution

yo

to

t1

Gernal Runge Kutta (explicit)
• The explicit RK method is given by
Runge Kutta (explicit)
• Note that RK must satisfy:

Butcher Table

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 fails for large time steps (range plot example)

### Lecture 04: Two-Body Dynamics: Orbit Position as a Function of Time

Chapter 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 Anomaly

Simple Case: Circular Orbits (e=0)

If e = 0, then

therefore

Since for a circular orbit we have

then

FOR

CIRCULAR

ORBIT

OR

Time versus True Anomaly

Elliptical Orbits (0<e<1)

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

Me = Mean anomaly for the ellipse

Time versus True Anomaly

Elliptical Orbits (0<e<1)

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

where n = mean motion = 2π/Te

OR

Time versus True Anomaly

Elliptical Orbits (0<e<1)

?

We need to fine out Me still

Let’s introduce another variable E = eccentric anomaly

Time versus True Anomaly

Elliptical Orbits (0<e<1)

OR

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

Therefore

or

Time versus True Anomaly

Elliptical Orbits (0<e<1)

We need to fine out Me still

E

This is Kepler’s Equation

Time versus True Anomaly

Elliptical Orbits (0<e<1)

To find t given Δθ

• Given orbital parameters, find e and h (assume θ = 0 deg)
• Find E:
• Find T (orbit period):
Time versus True Anomaly

Elliptical Orbits (0<e<1)

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.

Time versus True Anomaly

Elliptical Orbits (0<e<1)

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)
Time versus True Anomaly

Elliptical Orbits (0<e<1)

To find θ given Δt

• 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

Time versus True Anomaly

Elliptical Orbits (0<e<1)

To find θ given Δt

• After finding the converged E, then find θ
Time versus True Anomaly

Parabolic Orbits (e = 1)

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

MP = Parabolic Mean Anomaly

Time versus True Anomaly

Parabolic Orbits (e = 1)

STEPS:

Find h

Find MP

Find θ

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

Time versus True Anomaly

Hyperbolic Orbits (e > 1)

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

Time versus True Anomaly

Hyperbolic Orbits (e > 1)

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.

Time versus True Anomaly

Hyperbolic Orbits (e > 1)

Hyperbolic eccentric anomaly for the Hyperbola

Since:

Time versus True Anomaly

Hyperbolic Orbits (e > 1)

We now have

Solving for F and since

we now have

Using the following trig identities for sine and cosine

Time versus True Anomaly

Hyperbolic Orbits (e > 1)

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.

Time versus True Anomaly

Hyperbolic Orbits (e > 1)

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.

If α < 0, then orbit is hyperbolic

If α = 0, then orbit is parabolic

If α > 0, then orbit is elliptical

Where

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
• Example 3.6 (Textbook: Curtis’s)

Therefore

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

Universal Variables
• Example 3.6 (Textbook: Curtis’s)
Universal Variables
• Example 3.6 (Textbook: Curtis’s)

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
• Steps finding state at a future Δθ using Lagrange Coefficients
• Find r0 and v0 from the given position and velocity vector
• Find vr0 and α
• Find X
• Find f and g
• Find r, where r = f r0 + g v0
• Find fdot and gdot
• Find v

Where

and