6: Rigid Body Inertia Tensor

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来自 University of Colorado Boulder 的课程

Kinetics: Studying Spacecraft Motion

24 个评分

University of Colorado Boulder

Kinetics: Studying Spacecraft Motion

24 个评分

课程 2（共 4 门，Specialization Spacecraft Dynamics and Control）

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.

从本节课中

Continuous Systems and Rigid Bodies

The dynamical equations of motion are developed using classical Eulerian and Newtonian mechanics. Emphasis is placed on rigid body angular momentum and kinetic energy expression that are shown in a coordinate frame agnostic manner. The development begins with deformable shapes (continuous systems) which are then frozen into rigid objects, and the associated equations are thus simplified.

5: Rigid Body Angular Momentum6:55

6: Rigid Body Inertia Tensor3:24

6.1: Rigid Body Inertia about Alternate Points3:02

6.2: Rigid Body Inertia about Alternate Body Axes6:05

7: Rigid Body Kinetic Energy6:52

8: Rigid Body Equations of Motion13:13

8.1: Integrating Rigid Body Equations of Motion1:54

8.2 Example: Slender Rod Falling19:59

(Tips for Solving Spring Particle Systems)5:47

Optional Review: Rigid Body Properties14:15

Optional Review: Rigid Body Equations of Motion19:47

与讲师见面

Hanspeter Schaub
Glenn L. Murphy Chair of Engineering, Professor
Department of Aerospace Engineering Sciences

If you pick vector components, this is just written in a very corner frame,

independent way.

I'm just taking the cross product operator and representing it in a matrix form.

Now to actually numerically compute it, you'd have to take r1, r2, r3.

And then your first rule becomes 0 minus r3 plus r2, right, and you break it down.

That means you've picked a particular body frame.

So if you do that and that matrix with the minus sign, it gives you this definition.

If you replace r1, r2, r3 with just x, y, and z this should look very familiar.

And now you would have your y squared plus z squared, x squared plus z squared,

x squared plus y squared.

Those are your different moments of inertia definitions.

And then we have our off diagonal terms that appear as well.

So generally, to have a rigid body your inertia tensor, which is what I'm showing

here, will be a three by three matrix when you numerically evaluate it.

But to do so, there's two conditions.

One, we've picked about which point we're taking moments.

These r's are defined relative to the center of mass.

And then when you numerically evaluate it, you pick a particular coordinate frame and

say, okay, my one access point is forward.

The second one is to the left and then the third one is up.

If you pick a different coordinate frame, you would have different components.

And you get a different inertia matrix representation of the inertia tensor.

The inertia tensor is really an invariant,

that's it for this space craft has one inertia tensor.

But then how we numerically express it.

There's an infinity of answers, because I could pick a infinity of body frames

to break out these vectors and do the tildes and compute this.

Now this body integral is not very fun.

It gets tedious.

Cat programs love this.

They know where every nut and bolt and panel and fuel tank is, and

everything that goes on the spacecraft.

So it simply has to sum them up, know the location, do this math, times dn and

sum everything up, and it spits it out in a heartbeat.

Doing it analytically a little bit tedious, so

we will look at tricks that we don't have to redo this integral every time.

What if I am not taking moments about this point, but I am putting this at the end of

a robot arm and I need movement about this other point.

Do I have to redo all this math?

So angular momentum expression, as we said, was just hc.

You already identified that term as being nothing but the inertia tensor.

I am writing it as Ic times Omega.

There is now some notational subtlety here.

Here I picked a particular bonding frame.

Here I am not specifying body.

So I'm really equating matrix math is equivalent to this tensor vector math.

The inertia tensor, think of it as a two-dimensional vector, right?

I can write these quantities in a general way, but then when I evaluate any vector I

have to have a coordinate frame to break down the components and add and

subtract them.

But you can write them in a very coordinate agnostic way.

And that's what I'm having here.

I'm not saying, if this is the b frame, this better be in the b frame.

Otherwise, you're adding apples and oranges.

But, this can also both be in the q frame or the r frame, it doesn't matter.

In the vectorial way, we can just write the tensor operator like this.

So that's the Inertia Tensor.

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