As they tumble through space, objects like spacecraft move in dynamical ways. Understanding and predicting the equations that represent that motion is critical to the safety and efficacy of spacecraft mission development. Kinetics: Modeling the Motions of Spacecraft trains your skills in topics like rigid body angular momentum and kinetic energy expression shown in a coordinate frame agnostic manner, single and dual rigid body systems tumbling without the forces of external torque, how differential gravity across a rigid body is approximated to the first order to study disturbances in both the attitude and orbital motion, and how these systems change when general momentum exchange devices are introduced. After this course, you will be able to... *Derive from basic angular momentum formulation the rotational equations of motion and predict and determine torque-free motion equilibria and associated stabilities * Develop equations of motion for a rigid body with multiple spinning components and derive and apply the gravity gradient torque * Apply the static stability conditions of a dual-spinner configuration and predict changes as momentum exchange devices are introduced * Derive equations of motion for systems in which various momentum exchange devices are present Please note: this is an advanced course, best suited for working engineers or students with college-level knowledge in mathematics and physics.
3: Torque Free Motion General Inertia Case
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Del curso dictado por University of Colorado Boulder
Kinetics: Studying Spacecraft Motion
23 calificaciones
University of Colorado Boulder
23 calificaciones
Curso 2 de 4 en SpecializationSpacecraft Dynamics and Control
De la lección
Torque Free Motion
The motion of a single or dual rigid body system is explored when no external torques are acting on it. Large scale tumbling motions are studied through polhode plots, while analytical rate solutions are explored for axi-symmetric and general spacecraft shapes. Finally, the dual-spinner dynamical system illustrates how the associated gyroscopics can be exploited to stabilize any principal axis spin.
Conoce a los instructores
Hanspeter SchaubGlenn L. Murphy Chair of Engineering, Professor
Department of Aerospace Engineering Sciences
Torque free motion.
Momentum has to be constant.
The momentum magnitude has to be constant,
which is what I'm writing here in the first equation.
And kinetic energy has to be constant.
So we're not considering actual mechanical energy loss right now.
Like, before, like the Paul will plots, let's assume T and
H are perfectly preserved.
You can write the momentum at energy constraints in terms of your omegas.
H was I omega, so that for the magnitude is the component squared.
That's easy, magnitude squared.
And kinetic energy was 1/2, so the 2 comes to the left-hand side.
This one was just to make a transpose I Omega, that I is a diagonal.
Again, makes it very easy.
This is the form that you would have.
Now these are two constraints.
And we have three variables.
And quickly this illustrates again, that the omega curve that you're
going to get isn't going to be a plane or a volume, the solution space.
It's going to be one dimensional curve in three dimensional omega space, right?
That's what we're looking for.
We have three unknowns, omega one, two and three and two constraints.
So I can always use these two constraints and that's what I'm laying out here.
I say, okay well let me just pick to write everything in terms over omega 1.
Why? because it's the first thing one that
appears.
So I can use this and solve for omega 2 squared and
that's what you have in this equation.
And I can solve this for omega 3 squared which comes up here.
So you've got two equations and two terms, omega one and
two that I'm solving for, so just rule any algebra and you reduce it.
So, now because of these constrains if I have an omega one and I know my momentum
and I know my energy added at the beginning, I can quickly tell you what
the magnitude squared of omega two has and the same thing for omega three, all right?
So it's just the way of re-writing momentum and
energy constrains exclusively in omega terms.
Now what makes omega 1 special?
Nothing really.
There's nothing special about omega 1.
You could've just as easily solved everything in terms of omega 2 or
you could've used in two equations and say, well,
I'm going to solve everything for omega 1 and 2 in terms of omega 3, right?
You can solve this any which way you wish.
And substitute them again.
I will use all of them.
This was in the paper by Junkins, Jacobson,
Blanton that you've got a reference down here.
It's actually from 1973.
Most of you weren't even born.
But it's an eloquent result here and you will see,
this has some analogy to the axisymmetric cases.
But it gives you a more general formulation that you don't often see.
So this page, all we've done is we've taken momentum energy equations and
we've used it to express the omegas in terms of momentum and energy and
one of the other omegas, that's it.
So let's go back to the original problem.
This is the one we're trying to solve.
We've got these differential equations.
There's no L's on the right-hand side because there's no torque.
How do we solve this?
Before with axis symmetry, one of them was 0.
So we could say omega 3 is fixed and therefore we took a derivative of this,
substitute into the others, right?
And came up with nicely decoupled ones.
Now everything varies with time and I can no longer do that trick directly.
But there's still similar steps.
We're going to start out with the same kind of a step.
We have three fully coupled and now non-linear differential equations,
whereas before I had two coupled linear differential equations.
I'm still going to take a derivative just because it worked so well for the axis and
metric let's try it here, right.
And you do this, and then with chain rule, Omega is Omega 2's 3's Omega 1, everything
varies with time, and here you go, to make a 1, 2, 3 double outs divided by inertia.
That's basically it.
So you're going okay, great, where is this going?
So you took a derivative again.
Now similar as to with the axis symmetric case, omega 1 dot for
the axis symmetric only depended on omega 2 because omega 3 was constant.
Well, what we're going to do now is we plug in the second and
third differential equation into here.
And rewrite this all and
you end up with this expression here that you going to have.
So omega 1.., is equal to omega 1 omega 2 squared.
And omega 1 omega 3 squared.
As you can see there's still quite coupled.
There's still non-linear Just I went from three non linear first order differential
equations to three non linear second order differential equations.
Seems like we're going backwards, we're not making much progress.
But here's the elegant thing now, from the prior slide,
I have for example omega 2 squared In terms of just a constant and
something times omega 1.
And the same thing for omega 3 squared.
So if you look at the second order differential equations,
here is omega 1, omega 1.
I can write omega 2 squared in terms of a constant and omega 1.
And the same thing for omega 3 squared.
So by using the momentum and the kinetic energy constraints,
I can substitute those in and instead of having time varying quantities,
I have everything in terms of constants of energy and consular momentum magnitude.
And what you can do for one, you can follow the same process here.
We use here the expressions where omega 1 and 3 are written in terms of omega 2.
And here we use the third set that writes omega 1 and 2 in terms of omega 3, right.
There's three different ways we could take two constraints and solve for
two to three variables.
If you do that and plug them in, so that's the process,
those equations we found, and you plug them in here.
And then you do the same thing for the other two look what happens in the end.
This is omega 1, just a bunch of constants.
Omega 1 times constant something times omega 1 squared.
So you're going to get omega 1 term and
then something times omega 1 squared will give you omega 1 cubed essentially.
So the final form for everyone of these omegas, I can write it as a,
what's called a homogeneous undamped Duffing equations.
Homogeneous just means the right hand side is 0.
Undamped, there's no omega dot.
Just omega double dots and omegas.
And the Duffing equation, is this classic, this would be a spring mass equation.
The Duffing equation is a spring mass with a cubic Stiffness term added to it, so
like a cubic spring instead of a linear spring.
It's a common modeling thing to see as well.
So, all three of the omegas actually have an equation of this form.
But you would have these As and Bs which depend on omega 1, 2 and 3, and
you can see, the inertias are computed differently, and
when you plug it in, energy and momentum come in with different terms so
we have formulas for the As and Bs have to be.
So this is actually very elegant because we often see systems that are,
we end up with x double dot plus kx, right?
If it's x double dot plus k equal to 0,
it should be trivial to talk about linear stability, at least.
You've linearized the system for small motions locally.
If k is positive, we know we'll have a restoring force let me just draw
that out because we're about to mute that.
So if you, If I have M.
Forget M, we'll just do normalized.
That means x double dot = kx.
So if x is here, you can look at what happens to x double dot.
If x is positive, the x double dot is going to be negative.
And it's a linear relationship.
So if you're too far ahead, the force is going to restore you back to the origin.
If you're too far behind, and x is negative,
-k be the positive value that then makes it positive, you can catch up.
This gives you the typical oscillator equation that you'd have, but
it's a stable motion.
So that works.
But this result is typically due to a linearization that you get.
Here have we linearized anything when we derived this?
No, we just used full nonlinear differential equations,
differentiated them, substituted in energy momentum, and
you end up with three equations, we've a first order term and a cubic term.
The cubic term 2, if you do structures, you often see this cubic term.
But it's the result of an approximation.
Once you go beyond the first order stiffness,
the next term people typically include is the cubic term.
If the fractions get really big like strain hardening on a spring or something.
You get this extra stiffness that comes in, that's the cubic term.
But again,
in structures, that's an approximation of a much more complex behavior.
It's like the next order term that's included.
Here this is a perfectly rigorous equation.
So in structures, this is often found as an approximation.
Here there's no approximation.
Is it exact thing for tumbling general inertia, motions.
So good, so we want to study this now.
A little bit more detail.
We have three equations.
When we only had these parts,
that was the axisymmetric case it was easy, spring mass.
It was a sine and a cosine response, right?
Now here it's going to get a little more complicated.
Unfortunately I don't have analytic answers as we have with the axisymmetric
codes but there are some arguments we can talk about like global stability of these
motions and departures that you can write it in this form is kind of a nice thing.
So yeah, then I'll lead departure from Hooke's law.
That's the linear part, this is an exact differential equation.
So we can do this.
So this is actually three equations.
Are these three equations decoupled?
>> [COUGH] >> Sorry, what was your name?
>> Mark. >> Mark, are they decoupled?
>> Yes.
>> Okay. Are they independent?
>> Yes.
Well, not because of the constants.
>> You said it with such conviction you almost had me convinced.
Why are they not independent?
Everybody agrees they're decoupled.
If you look at this each equation has only omega 1s in there.
And the next term only has omega 2's and the other one has only omega 3's there's
no omega 1 times omega 3's as we had originally.
So we definitely made them independent.
Once you have all the initial conditions and these parameters you can solve three
independent differential equations numerically and get the answer.
But independent is something different that means,
lets say I have a spin about omega 1 only and
the next simulation is a spin about omega 1 and tumble about B2 or something.
It is the same omega 1 but it has a different B2.
What will be different in these differential equations?
because you're right, they aren't independent, Matt?
>> They are energy momentum terms?
>> Exactly.
The energy momentums are actually what cross couples them, so
they're not independent.
I can't just go well, if omega 1 is one radium per second,
I have a fixed answer that's not true.
Because the coefficients that go here depend on momentum and energy.
And so it's a little bit more complicated if you also tumbling about another access,
your energy state is different, momentum states are different, and
that will give you different A's and B's.
So you have to solve it for that particular case.
So there are three uncoupled equations but not independent equations.
Is that make sense?
They're coupled through this coefficients through energy and momentum, precisely.
So, the three uncoupled oscillators and that's there.
If I write them out to As and Bs, this is what they are, as you can see, T and
H has appeared.
The inertia's always the same, it's the same with your body.
But as your math was saying, if H is different and
T is different, I would have different coefficients, and all of a sudden and
then we get a different response with these omegas, right?
So that's where it does matter.
But it's very elegant so for the general case you can always write this as
a harmonious, homogenous on them to Duffling equation.
Now with these parameters we'll look at this a little bit more carefully but
you can see, for
example, the cubic stiffness terms Bs, there's differences of inertias again.
So we know right away if it's axis symmetric,
some of these things will go away, some insight you get.
But at the same time, I'd say if it's a cube or
a sphere, all of these terms go away.
For a sphere, none of that matters.
Even in here there's always differences of inertia so for sphere again, As and
Bs go away which we talked about.
A sphere you can tumble about any axis and it'll just continue to tumble,
your mega 1, 2, 3s are all flat lines, always, or for a cube the same response.
But for general bodies, if I1 is your largest inertia.
You can see this is going to be a positive term times a positive term,
this is going to be a positive cubic stiffness.
Down here again, if I1 is the largest, so this is going to be negative,
that's negative, negative times negative is positive again.
Again we'll have a positive stiffness whereas here
you'll have a negative stiffness.
So is positive stiffness good or bad for a stability for a cubic one?
>> Good.
>> It's good.
It's pretty much analogous.
Instead of a liner response, now you have something that's kind of cubic.
But you need the same kind of a sign, right?
If you're positive, you want something pulling you back to the origin.
The origin is the equilibrium that we'll be studying here.
And if you're negative, you want something positive to pull you back.
It's just got to be more aggressive for big departures, but
way less aggressive for small departures.
So there's a difference in the cubic terms.
That's why for a linearization, this is the linear part it's going to dominate.
But if you go really big departures, the cubic one is always going to win.
At some point, x cubed is going to be bigger than x,
regardless of what these constant coefficients are, these As and Bs.
So good.
So you can see just looking at the signs of this we can get
some insight into what must be happening and
we'll cover this a little bit more in another slide as we go through it.
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