0:09

We've already jumped ahead a little bit which is good.

Â Sometimes just rearranging your order helps things sink in more, right?

Â If it's rigid, what is the key differentiation?

Â So Sheila was on the right track earlier that it's all about these derivatives,

Â especially this little r dot.

Â And if it's rigid, then the derivative of

Â this vector as seen by the body frame has to be 0.

Â So you just end up with omega cross r, right?

Â If it's not rigid, then this little mass particle is maybe deploying, it's moving,

Â it's wobbling, it's doing something.

Â And you have extra turns.

Â That's, in fact, what you'd have to include.

Â Now you plug this in, and we only focus about this one.

Â You can separate this also for rigid body, like this plus this other term.

Â I'm not going to do that here.

Â But in the end, the angle momentum about

Â the center of mass ended up being this

Â minus r tilde, r tilde, dm times omega.

Â What did we call this big bracketed term here?

Â >> Is that the inertia tensor?

Â >> That's the inertia tensor.

Â And I like using the word tensor versus matrix, because there's a difference.

Â We have a position vector but

Â then we use a matrix to define the vector components with respect to some frame.

Â So the matrix represents a vector but the vectors you can fundamentally just add

Â vectors and add arrows and you don't have to assign frames.

Â Same thing here.

Â Now in this tilde notation, this is basically the cross product operator

Â which will be the vector equivalent of it, right?

Â You have to make sure that r tilde.

Â You have to take this vector, r, with respect to some frame, the B frame,

Â as an example.

Â This would be in the B frame, that's what you measure.

Â Now it's apples and apples and you can do all your matrix math.

Â You wouldn't have this in the C frame and this in the B frame.

Â And then you'd be mixing frames and it just wouldn't work out well, all right?

Â So we have to do the integration by picking a body frame and then we get this.

Â And also little r is defined relative to this body's center of mass.

Â That's how we defined little r.

Â What if we have a different point?

Â Maybe this thing isn't spinning about its center of mass, but

Â it's pivoted to an end of an arm, and it's pivoting like this.

Â All of a sudden,

Â I need the inertia about this endpoint, not about the center point, right?

Â Do we have to redo this integration over and over again?

Â 2:34

No, right?

Â So there's formulas on how to do this.

Â This, in essence, is my inertia tensor times omega.

Â So written like this,

Â treat this as a completely coordinate-independent formulation.

Â When you're actually numerically evaluated in some program like MATLAB or C or

Â something.

Â You have to make sure it's the same frames.

Â But when we derive controls,

Â we'll just write it this way and not assign frames yet.

Â That's the bookkeeping you do at the very end.

Â So good, that was the inertia tensor.

Â So let's wrap up some final things.

Â What if we have a, let's just flip the order.

Â Let's do attitudes first, just to mix it up.

Â What if we have a different orientation?

Â You have the inertia tensor given to you in B frames but

Â I want the inertia tensor in frame components.

Â I don't have to redo all those body intervals again.

Â Once you have them, you're good.

Â >> Do you have to left multiply and right multiply?

Â >> Okay, so give me the two letters of each DCM.

Â >> [INAUDIBLE] >> So it would be BE,

Â right, or no EB, sorry, backwards.

Â >> Won't let me erase.

Â Okay, EB, and the next one is?

Â >> [INAUDIBLE] >> Or B.

Â I like to remember it this way actually.

Â because this way if I know BE, great, then have to transpose this one.

Â If I know EB, fine, you have to transpose that one, right?

Â But it's basically if this is in the B frame, then it's a two-dimensional.

Â Think of tensors like two-dimensional vectors.

Â So instead of using the DCM once, it's a 2D thing, you have to use it twice.

Â And you're absolutely correct, it's a left and a right multiplication.

Â And when you bump them up like that the letters B should match up.

Â This is all on the B frame.

Â And then what's on the outside everything's being mapped to the E frame

Â and you get a inertia tensor in the E frame.

Â So a different rule of mnemonics trick that you can help to remember this stuff.

Â Definitely use it twice.

Â Good, so that's how we avoid, once we have this in some body fixed frame,

Â we can easily map it to any other body fixed frame.

Â I will use this trick several times, especially when deriving variable speed,

Â c and g equations of motion, and so forth.

Â So what do I mean by a principal coordinate frame?

Â Is it Tosh?

Â 5:36

>> They're 0.

Â >> So, they're 0.

Â If this is a principal frame,

Â you have a coordinate frame such that you get a nice diagonal form.

Â Chuck, can we always do this?

Â Or are there assumptions on the shape that allow us to do this?

Â It's a rigid system.

Â >> Yes, you can always do it.

Â >> You can always do this.

Â That's why for analysis very often you see papers or even our own stuff.

Â We need to start out with assuming it's a principal frame.

Â because instead of having nine elements to track and

Â now with symmetry it's actually six, but still it's about I can go down to three.

Â Three principal inertias fully define the mass distribution of a system.

Â Without loss of generality we can always do this.

Â In the code, though, we typically solve,

Â actually we do control problems that include the full tensor.

Â Because in real life they will never give you the mass balance.

Â You may start out with a frame that's perfect, and at the very last minute

Â astronaut A and B switch places, and next thing you know your CG's offset.

Â Crap, my inertia tensor's now fully populated, darn astronauts!

Â So there's always something.

Â So in the code, we want to have formulations and

Â controls that handle any inertia tensor.

Â But we want to also realize for analysis and

Â we design it typically they're diagonal or nearly diagonal.

Â The off diagonal terms might be very small.

Â We want to have solutions that work for everything.

Â So good, this is symmetric matrix from the definition.

Â Could you have inertias, I1 = 800,

Â I2 = 20, and I3 = 50.

Â I'm just using generic units.

Â I'm not throwing in anything particular, they're just consistent numbers.

Â 8:07

Here we go.

Â If you look in this, this is the I1, I2, I3.

Â These off diagonals are all 0, right?

Â So I1 i r2 squared and 3 squared.

Â This one is r1 squared and 3 squared.

Â So summed up, I get r1 squared 2 squared and 2r3 squared.

Â Compared to the third, it's just r1 squared and 2 squared plus an r3 squared.

Â That squared term can at best be 0, which means they're equal.

Â 8:35

But if anything else, it's going to have to have something bigger, all right?

Â So when you set up inertias, be careful that you actually use a realistic example

Â where the sum of 82 principal inertias has to be bigger than or

Â equal in a limiting case to the third one.

Â So there's different kind of properties that you can pick up here.

Â So good, we have that.

Â Now we know how to do this.

Â How do we go from, what's the relationship?

Â If you have a regular 3x3 matrix and

Â you're trying to find the principal inertias.

Â How do you go from the 3x3 tensor matrix representation to the three principals?

Â 9:14

What was back row scratching your head.

Â What's your name?

Â >> Mark. >> Mark, thank you.

Â >> Can you repeat the question please?

Â >> Sorry? >> Can you repeat the question please?

Â >> How do I go from a 3x3 inertia tensor representation to a diagonal?

Â How do I find the principal values of- >> You have to find the eigenvalues and

Â the eigenvectors.

Â >> Exactly, right?

Â So you have to find eigenvalues, eigenvectors of the system.

Â And the eigenvalues,of the 3x3, they are all of the principal inertias,

Â that's cool.

Â And then the associated eigenvectors are the axes of that principal inertia.

Â And are relative to the body, it turns out.

Â This is a principal axis, and

Â this is a principal axis, then you'd have a third one, right?

Â Now are these eigenvectors always going to be unique?

Â 9:58

Jordan, think of a cylinder.

Â If you get this math,

Â I can give you a nice 3x3 matrix of a cylinder with some weird skewed frame.

Â You find principal inertias.

Â If it's a cylinder, what should you expect now, all of a sudden?

Â >> Expect one to be up the axis and two to be in that, in any plane.

Â >> Right. >> Anywhere in that plane.

Â >> So you will have a repeated eigenvalues set orthoganol to the symmetry axis.

Â The inertia of a cylinder about this axis or this axis is just going to be the same.

Â So you get a repeated eigenvalue.

Â And the plane is unique in that case, but not necessarily the values, right?

Â So this is how these things all relate again.

Â And these matrices are symmetric, turns out they're also symmetric

Â definite because the principal inertias have to be positive.

Â You would never have a negative principal inertia.

Â It just doesn't make physical sense.

Â 11:23

Just like the vector between the two points?

Â >> Yeah, so let me just say

Â that c relative to q,

Â they're close.

Â It's usually a parallel axis there, right?

Â If you do it for 1d rotation it was always easy,

Â it was something the inertia about the center of mass plus the distance.

Â So if I have the inertia about the center of mass,

Â I want the inertia about this endpoint.

Â I do mass times distance squared, it's squared I think, that I get there.

Â This is essentially mass times distance squared,

Â and this is the distance between those two points that you'd have.

Â It doesn't have to be an inertia point,

Â it could be another point as well, like base point.

Â Now what are some the trickeries in here?

Â This is typically given to you in the Q.

Â Let's say this is your spacecraft queue, Star Trek, right?

Â Anybody remember Q?

Â That was one of the characters in there.

Â So in his spacecraft, Q lives in a very Q-centric world.

Â So this Rc relative to Q is usually given to you in a Q frame.

Â This stuff is probably given to you in the body frame.

Â Now before I actually can do this math,

Â what do you have to do with these quantities?

Â 13:18

Tony, what would you do?

Â >> Do a rotation matrix on the R.

Â >> You can also just say, well, so I know this in the Q frame or

Â I can also find c relative to q in the B frame.

Â Which is BC times R C relative to Q in the Q frame, right?

Â If you did this now you plug these in here, and then this.

Â Just make sure it's consistent.

Â I didn't say we have to have the answer in B or Q components.

Â You can do either.

Â Well, you can do one and then map the answer back into the frame you want.

Â That works as well, if it's a third frame that's not either C or Q or B.

Â So you can see lots of different ways this goes.

Â This is the general math but if this is the coordinate independent way,

Â a coordinate frame independent way to write it.

Â When you evaluate it, still make sure you have all the right stuff.

Â Otherwise, nothing works.

Â Okay, good.

Â