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. - PowerPoint PPT PresentationTRANSCRIPT
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ARO309 - Astronautics and Spacecraft Design
Winter 2014
Try LamCalPoly Pomona Aerospace Engineering
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Lecture 03: Numerical Integrations
Chapter TBD
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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:
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2nd Order Runge Kutta
• Given an 1st order ODE
• Applying Taylor Series expansion to 2nd order
• Since:
and
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2nd Order Runge Kutta
• Now we have
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2nd Order Runge Kutta
• Or
• The Coefficient in Butcher Tableaux form. Note:
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2nd Order Runge Kutta
• Graphically
to t1
slope=k1=f
slope=k2
yn+1 (Approximate Soln)
yoExact Solution
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Gernal Runge Kutta (explicit)
• The explicit RK method is given by
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Runge Kutta (explicit)
• Note that RK must satisfy:
Butcher Table
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Runge Kutta (explicit)• RK4
where
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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):
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RK4 Example
• From the problem:
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RK4 Example
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RK4 Example• RK4 fails for large time steps (range plot example)
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Lecture 04: Two-Body Dynamics:
Orbit Position as a Function of TimeChapter 3
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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
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Time versus True Anomaly
• Recall from Chapter 2
Since Then
Integrating from 0 (assuming tp = 0) to t and from 0 to θ
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Time versus True Anomaly
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Time versus True AnomalySimple Case: Circular Orbits (e=0)
If e = 0, then therefore
Since for a circular orbit we have then
FORCIRCULARORBIT OR
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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
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Time versus True Anomaly
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)
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Time versus True Anomaly
We need to fine out Me still ?
Let’s introduce another variable E = eccentric anomaly
Elliptical Orbits (0<e<1)
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Time versus True 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
Elliptical Orbits (0<e<1)
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Time versus True Anomaly
We need to fine out Me still
E
This is Kepler’s Equation
Elliptical Orbits (0<e<1)
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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 AnomalyElliptical Orbits (0<e<1)
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To find t given Δθ
• Fine Me:
• Find t:
Time versus True Anomaly
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.
Elliptical Orbits (0<e<1)
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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 AnomalyElliptical Orbits (0<e<1)
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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
Time versus True AnomalyElliptical Orbits (0<e<1)
For Me < 180 deg For Me > 180 deg
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To find θ given Δt
• After finding the converged E, then find θ
Time versus True AnomalyElliptical Orbits (0<e<1)
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Time versus True AnomalyParabolic Orbits (e = 1)
Then a = 1 and b = e, therefore we have b = a
MP = Parabolic Mean Anomaly
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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
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Time versus True AnomalyHyperbolic Orbits (e > 1)
Then a = 1 and b = e, therefore we have b > a
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Time versus True AnomalyHyperbolic 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.
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Time versus True AnomalyHyperbolic Orbits (e > 1)
Hyperbolic eccentric anomaly for the Hyperbola
Since:
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Time versus True AnomalyHyperbolic Orbits (e > 1)
We now have
Solving for F and since we now have
Using the following trig identities for sine and cosine
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Time versus True AnomalyHyperbolic 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.
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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
Time versus True AnomalyHyperbolic Orbits (e > 1)
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
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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
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Universal Variables
• Stumpff functions
or for z = αΧ2
,
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Universal Variables
• To use Newton’s method we need to define the following function and it’s derivative
• Iterate with the following algorithm
with
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Universal Variables
• Relation ship between X and the orbits
For t0 = 0 at periapsis
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Universal Variables
• Example 3.6 (Textbook: Curtis’s)
Find h and e
Since , then
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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
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Universal Variables
• Example 3.6 (Textbook: Curtis’s)
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Universal Variables
• Example 3.6 (Textbook: Curtis’s)
Thus we accept the X value of X = 128.5
where
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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
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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