0:07

Okay, so we've done a review of what these devices are.

Â There are certain benefits, everything has drawbacks,

Â now we need to get into the math of it.

Â How do we get the equations of motion of these things?

Â So we're going to be stepping through this carefully.

Â We are going to do it for spacecraft with a single VSCMG.

Â So that's a single variable-speed control-moment gyroscope.

Â That's a single axis control-moment gyroscope.

Â There have been a few that have flown this way.

Â Yes, Jordan?

Â >> Sorry, going back really quick to the CMGs,

Â do you know if there are any that use like flux spinning to do like, so

Â that could just levitate the bearing and rotate them?

Â >> There are, yeah.

Â because, especially, well, with the variable speed,

Â the Air Force had a nice program, Fred Levy was running that,

Â to look at energy storage, because as you spin up this wheels, as you de-spin them,

Â you can actually pull and you could turn mechanical energy in the dimo effect.

Â You can turn into electrical energy, so this could replace a battery.

Â But you have to spin to pretty high speeds to make them efficient.

Â So they're looking at 30, 40,000 RPM's, really fast and

Â mechanically with worrying about friction and little imbalances is quite tricky.

Â So there's some that are looking at magnetic bearings, levitation, and

Â those kind of things.

Â Not necessarily flux pinning, I haven't heard that one being used but, magnetic

Â bearings, absolutely, you will find some of those devices looking at that.

Â Yep, so there's a lot of mechanical challenges and how to do this well,

Â robust, it survives launch, where you get shaken to death, almost hopefully,

Â not slight.

Â So anyway, so think of VSCMG as a Hybrid, we'll have the math for

Â the CMG and for the reaction wheel all embedded in one.

Â So we can talk to this once.

Â I want you to understand this path, deal the parts of just specialties of it,

Â that's going to be our goal.

Â So goes with time, to wrap this up.

Â So very convenient because we can run both but,

Â this VSCMG's are devices being researched and

Â done actually phoned some small sats via CMG and tested those as well already.

Â So this is not just theoretical, this is starting to happen.

Â But this is stuff I did way back for my PhD, it's one of my chapters in the PhD.

Â So we're going to try and look at this.

Â One of the niceties I'm going to highlight is we talked about CMGs having

Â singularities.

Â What we'll find is VSCMG actually avoids this by move, by changing the wheel

Â speeds, I can actually produce torques about axis or to the tork direction.

Â And that can be useful, it's highly redundant, each device

Â if it's a variable speed you could use it as a reaction wheel, and as a CMG.

Â So that's two control modes that are for devices that gives you

Â eight control modes, and it's still a three degree problem.

Â So you've got a lot of redundancy built in which is sometimes nice, and

Â also combine power energy storage, something we just discussed.

Â That's something people have been researching and looking at and

Â would be nice, that we didn't need batteries as much.

Â 2:55

So how do we get equations of motion?

Â It all basically boils down to this one little, simple equation, all right?

Â How hard could that be?

Â It's kind of like, hey you're doing your whole dissertation f=ma, that's it.

Â Right, if you're doing orbits, trajectories, unstable manifolds.

Â All you're doing, and Daniel's laughing,

Â that's really that's all you're doing is f=ma, right?

Â How hard is that, in attitude it's h.equal to l how hard can that be?

Â And I'll show you here this is the hint of that.

Â If you want some more fun go chat with Cody and John from my lab,

Â they've been doing these imbalanced VSCMG equations that are quite interesting.

Â 3:35

Start in the right hand side, Andre?

Â >> [INAUDIBLE] >> Be more specific.

Â >> [INAUDIBLE] on the body.

Â >> Which body?

Â because we now have multiple.

Â >> Rigid body.

Â >> Only on the rigid bodies and not on the wheels?

Â No, external torque.

Â >> Exactly, right?

Â If this is a dynamical system,

Â this applied to blobs in space, jello flying through space.

Â But it was the net external torque acting on this dynamical system.

Â Now we don't have blobs but we will have systems of rotating,

Â it's a multi-rigid body problem essentially that we're solving here.

Â So that's just the net external torques.

Â We don't need motor torques.

Â Whatever that reactional spin torque is doesn't appear here,

Â because that's an internal torque.

Â The Gimbal access motor torque does not appear here it's also an internal torque.

Â So just the externals, gravity gradients, solar radiation pressure drag,

Â maybe you have thrusters to augment this.

Â Because maybe for momentum dumping people still use thrusters sometimes,

Â that could be good.

Â Now what is the h on the left hand side?

Â Robert.

Â 4:41

>> On the tip of my tongue, I'm just- >> The total angular momentum.

Â >> Total angular momentum, right?

Â All of this dynamic assistance.

Â So if it's a blob, it's just the entire thing, it's not quite a blob here.

Â For us it's a spacecraft, plus the wheels,

Â plus the frame whatever holds these things together.

Â Everything that has momentum, angular momentum, we have to account for.

Â And then what's this Baud up here, what type of time derivative are we taking?

Â 5:21

Center of mass, actually.

Â So this either has to be a center of mass point or taking moments and

Â torques about an inertial point.

Â Then h equal to l is true.

Â If you're talking about a body fix point, that's anything else than center of mass,

Â there was extra turnums that have to appear here.

Â So again these inbalance things makes things immediately much more complicated.

Â So h equal to l.

Â We can assume this is all about the center of mass of this system.

Â All the wheels are balanced so

Â even when I'm trying gyrating I'm not really changing.

Â When I'm putting the gimballing angles in I'm not changing the center of mass of

Â that CMG configuration, I'm just changing where the axis is pointing.

Â I'm changing the inertias but

Â not center of mass, so that makes huge simplifications.

Â So we need some frames to define this stuff, and I've got this gimbal frame.

Â So you will have gs is my spin axis, so this is your wheel,

Â if it's just the reaction wheel this is all you would have.

Â A reaction wheel in a gimbling frame, this is how we can twist that spinning wheel,

Â GG is the gimble axis, and that other.

Â And the CMG devices solves a pancake with the motor knob sticking out left and

Â right, it's actually the gg axis.

Â And so gs is the spin axis, gimble axis, and then gt is at the transverse axis,

Â it's the third axis that cannot complete a full frame.

Â GS crossed to GT gives you plus GG, okay?

Â So that's what these are, we've defined them.

Â Big omega here is going to be the wheel speed and

Â we're treating it here as time varying.

Â Later on we can freeze it to make it a CMG device, not a variable speed CMG.

Â Gimbel angles are defined through gamma, so the gimbel rate

Â at which I am twisting these things is gamma dot, and that's what that's defined.

Â So that's the rotation rate about this gg axis,

Â so kinematics is going to come back in.

Â And then so the full frame is gs, gt, gd, that's the g frame that we have.

Â This is the one we'll use a lot.

Â 7:42

There's a wheelframe, I'm going to introduce this quickly, and

Â then within a few steps, we're going to ignore it again.

Â We'll make some arguments why we can ignore it.

Â So the wheelframe needs three axis.

Â I'm picking principal axis, so the spin axis, ds, I can reuse it just to be lazy.

Â That's my first one and then I have a wt and

Â a wg which are two axis that are fixed relative to the wheel,

Â so as the wheel rotates, these axis are going to rotate with the wheel, right?

Â 8:12

So if you look at the wheel frame, you can go ahead and

Â get the angular velocity of the wheel system relative to the gimbal frame, and

Â that's simply going to be the wheel's spin rate times the wheel spin axis, that's it.

Â So if you sat on the wheel frame it's set all this gimbal frame, look at the wheel

Â you would see the wheel's spinning about that axis at that rate and vice versa.

Â So now we have our two.

Â If you need omega of w, relative to b, you would just add up the two of course,

Â and that's how, so we can get any combinations of omegas that we need.

Â Let's talk about inertias.

Â I'm assuming this thing is built in a way such that I have three principal inertias

Â of this gimbal contraption and they'd line up with the gs, gt, gt frame.

Â Recently good assumption, may not be perfect but

Â it's a good simplifying in the analysis that we're doing right now.

Â So I have IGS, IGG, GT and IGG.

Â So IG is the inertia's of the gimbal frame along the ST and G axis.

Â That's where the subscripts are coming from.

Â For the wheel, we can write that in the wheel frame, you're going to have

Â an inertia about here so that's going to be the wheel inertia about the s axis.

Â And then I have Iwt, because the wheel is perfectly symmetric, right?

Â No imbalances, no other warping, all these kind of stuff.

Â It's going to be the same inertia about the any

Â two orthogonal axis in the plane of the disc, of the wheel, right?

Â So I just have Iwt, Iwt.

Â 10:02

I think this is really, that's an inertial tensor.

Â This is a matrix representation of the inertial tensor,

Â where I picked the w frame.

Â Here I'm claiming if I picked the W frame or

Â the G frame, they both give you the same answer.

Â Why do you think that is?

Â >> There's a degenerate Eigenvalue in the w inertia.

Â >> You're getting mathematical.

Â No, but you're right.

Â There is a repeated Eigenvalue in here.

Â That means that these axes are actually,

Â the Eigenaxes that generates these inertias are not unique.

Â If we go back and look at that.

Â We talked about Eigenvalue of an axis symmetric body.

Â Where we had IS and IT in some of the torque

Â 10:46

I could really, to get this diagonal from I just have to pick this axis and

Â I have to pick any two axis that are orthogonal to GS will do.

Â If any two axis will do, I can always pick GT and

Â GG if you wanted to, and they'll give you the same answer.

Â So actually I wrote up a quick mathematical thing.

Â So really I'm going to evaluate this now, I'm pulling in library.

Â I'm going to define now here this as a diagonal matrix

Â in this case the g frame, I've got this.

Â Now, since GS is my first axis, I'm doing what we call

Â m1 basically that was the one axis rotation that we needed with an angle.

Â So, you'd have 1, 0, 0 or I can do metrics form of wg, right?

Â So that's the classic note, that's the rotation because we're doing one axis

Â rotation where the GS is the first axis.

Â So if I translate here from the gimbal frame to the wheel frame,

Â this is what I could do, I get this answer and then you can

Â identify what does this cosine squared the same thing sine squared.

Â Those things are just going to drop out to b1, and

Â after all this math you get nothing but

Â exactly the same result again even though we did this code of transformation.

Â So if you don't believe the geometric arguments, you can quickly do the math and

Â validate and go, yeah.

Â Actually, this inertia tensor is diagonally both in the w frame and

Â in the g frame, that's all we need,

Â because that saves us a lot of transpose without then end up all you're

Â doing is creating a lot of trigs that become one in the end and not needed.

Â 12:29

Now I need to get some corner frames and attitudes.

Â We have the body attitude, we know how to describe that already,

Â MRP's, DCM, Cortanians whatever you want to wish.

Â We need to get the attitude of the gimball frame relative to the body frame.

Â And if you go back and look at how we defined DCM's, right?

Â It was basically the mapping from N to B and I'm just going to write that one out.

Â If your recall BN ended up being n1,

Â n2, n3 in B frame components,

Â that was one way to write it or each row,

Â 13:09

Was the b1, 2 and 3 in N frame components.

Â So if we go back and look at those definitions, if you have the three axis,

Â you can actually construct the DCN very quickly.

Â And in fact here, I'm using this one, so instead of having, what do i have here?

Â Let me go back.

Â Here we have BG instead of BM.

Â So instead of n1, 2, 3, I have the G first, second and third axis I put in.

Â And then the B frame component which is how we define them and

Â that's give you to DCM.

Â So that works.

Â So if we have BG.

Â Now if I need to get the inertia tensor which was diagonal in

Â the G frame which is great.

Â But I typically will need it in the end in the body frame, we'll do a lot of body

Â frame relative derivatives, I would have to reimposed multiply with the DCM, right?

Â To map it from one frame into another frame and you can do that.

Â The mathematics here become quite simple, IG is a diagonal and

Â this one I can write out like this.

Â So, let's just kind of do this by hand once.

Â 14:14

So, if I have BG is equal to my first

Â vector, my second, my third,

Â and then I need to compute BG times

Â I in G frame components times B,

Â now G transpose, right?

Â That would give me IG and the body frame.

Â Now, if you write it in the matrix form like this, it's going to be nothing but

Â gs, gt, gg that's the BG times

Â this diagonal which was Igs 0, 0, 0 Igt, 0,

Â 0, 0 Igg times the same matrix transposed.

Â So, this transpose becomes gs transposed, gt transposed gg transposed, right?

Â And now you can just start to carry out all this matrix math here.

Â If you look at the second part, this is going to be Igs times gs.

Â You can think of this like a vectrix,

Â matrix of vectors that we talked about earlier.

Â So out of this, all you'll going to get is Igs times gs hat transpose Igt,

Â the second row, 0 times this, 0 time this, drops out, gt

Â transpose plus Igg, gg transpose.

Â So that's that part times this, so,

Â 16:15

This would have to go here, and this goes here.

Â This should be a three by one.

Â It's a three by three times a three by one.

Â That should be a three by one, one by three times three by one.

Â Now all I have to do is gs times this term, gt times that term,

Â gg times this term, and that's where you get the final answer.

Â So from here, if I go backwards, IGs, Gs,

Â Gs transpose plus these other terms.

Â That's all we're doing.

Â So it's a nice analytic way that we can do these projections

Â from the inertia of one frame to another.

Â And you will see this is a handy result to what we'll use several times as we go

Â through that.

Â So that's what was done here.

Â This one also in the gimbal frame, I can just do the same thing, but

Â instead of IGs, IGT, I have IWs, IWT and that's it.

Â It's a little bookkeeping.

Â So you get exactly the same result.

Â So now we have the inertia's written in two different frames.

Â Good, next step, Angular Momentum.

Â You guys said H had to be the angular momentum of the total system.

Â The system will be composed of a single VSCMG, so

Â it's a spacecraft, that's part B, that's the body.

Â Then I have G that denotes the inertia and

Â momentum of the gimbal frame that has some mass and it rotates and does stuff.

Â And then you have HW which is the momentum of just the wheel

Â itself spinning about that.

Â So we have to add up all these momentums.

Â And each one of these momentums is of these bodies relative to

Â the inertial frame.

Â So for the body this is dead simple, you've done this.

Â I'm using IS now, that is my inertia tensor of the spacecraft.

Â You'll see a different I appearing later on.

Â So this one right now is just of the spacecraft itself.

Â It has the center of mass of the reaction wheel,

Â CNG devices already accounted for in that.

Â But what we're not doing is the momentum of the wheel-disk part spinning about

Â the center of mass of those disks.

Â So a lot of it is already lumped in there.

Â And this part will be easy to go forth.

Â The other ones get more complicated.

Â So for the gimbal frame, it's the same math at the first step.

Â We have IG times, now I need omega G relative to N and

Â omega G relative to N is omega G relative to B plus B relative to N, right.

Â Earlier we kept calling this just omega, here and being a little explicit

Â because we've got lots of different frames and lots of different omegas.

Â So it's a little bit easier to keep track of.

Â So we need both.

Â This one was simply gamma dot times Gg and that's put in here.

Â And the other one is omega BN.

Â The inertia one I've written with this projections into G frame

Â components already.

Â So, IG times omega BN gives me this answer, and

Â IG times this one will give me this answer.

Â Now, again, I'm jumping through a few steps directly.

Â So let me just look at it once.

Â This is a tricky one to do yourself in the homework and different ways to solve this.

Â One thing you could say is, you know you have IG like this,

Â and omega G relative to B is just gamma dot times Gg.

Â 19:40

Zero, they're orthogonal, right?

Â So that drops out.

Â And same thing with the Gg times Gs.

Â They're also orthogonal, they drop out.

Â So that's one way to do it.

Â I think on the next slide I show you another way with the matrix method.

Â That will give you the same answer hopefully.

Â It's the same physics.

Â Now these terms, gs transposed with omega BN will happen a lot.

Â If you look at, if we were showing the definition,

Â let us recall how we typically write omega.

Â 20:09

So omega BN, we typically write it as omega 1b1 + omega 2b2 + omega 3b3.

Â So we have a reframe, the omega 1, 2,

Â 3s are nothing but the 1, 2 and 3 access components.

Â Omega 1 is really defined as b1 dotted with omega BN,

Â or in matrix form this is equivalent to b1 transpose omega BN, all right?

Â Now, here we don't have b1 dotted with omega, we have gs dotted with omega.

Â So if you have omega BN,

Â gs hat, this is really what is going to give you, if you think of this,

Â if this is your omega vector.

Â And here is your gs vector.

Â It's giving you the gs component of that omega vector essentially.

Â And we're just going to call that, instead of omega 1, 2, 3, we've already used 1, 2,

Â 3 for the body frame stuff.

Â So I can't reuse the same names, I'm just going to use omegas, omegat and

Â omegag for the gs, gt, gg axis of that stuff.

Â 21:19

So a slight notational thing but it helps us.

Â So that's what you see defined right here.

Â Where we have Omega s is gs hat transpose with omega.

Â Omega T is gT transpose with omega and gg is the projection of g onto omega.

Â So the omega vector, this is again without any subscripts, we imply it's omega BN.

Â At some point it just gets old to write all that stuff over and over again.

Â So this is omega BN is equal to

Â these three vector components of that along the g axis.

Â So if you use that definition then you have here gs transpose times omega and

Â that becomes nothing but this little omega s.

Â So, the same momentum expression here can be rewritten a little bit more compactly

Â in this form and that will help us in same label keeping that we do.

Â So, now we have to angular momentum vector off the g frame relative to

Â the inertial frame.

Â What we need next is the angular momentum of the wheel.

Â >> [INAUDIBLE] >> Itself.

Â Then it's the same formula.

Â You need the angular velocity of the wheel relative to inertial, and

Â we know the angular velocity of the wheel relative to gimbal.

Â That was big omega times gs.

Â That was the spin axis side.

Â Gimbal relative to body,

Â that was the gamma dot times Gg, that's the gimbaling rate.

Â And then of course, this is the classic spacecraft body rate that you'd have.

Â If you plug that in you now have three of these terms to evaluate.

Â And I will just to show you different ways to do it, I'll do it different ways here.

Â This first term I've written out, and

Â it's going to be this inertia tensor which we rewrote earlier as a result of

Â the principal inertias times these outer products times omega.

Â And in here quickly you can see again gs hat transposed omega.

Â That's what we defined to be omega s.

Â The projection of the gs axis onto omega was the omega s component.

Â Same thing with omega t and omega g.

Â So that's one way we can get that first term.

Â That gives us this one.

Â 23:19

Now we want to get the other two.

Â I'm showing you a matrix representation, for

Â some people this might be a easier way to think.

Â Both of them should all give you the same stuff.

Â This term I can write in G frame components as this diagonal matrix.

Â G relative to B, that was gamma dot times Gg.

Â So, in g frame component it's 00 gamma dot, all right?

Â And then you just do your typical matrix math with which will give you 00 IWt that

Â along the third axis, as the vector representation of that.

Â That's the third axis is Gg, so you always get back the same answer.

Â You could do the same thing also for this one, just here omega w relative to g

Â was big omega times gs, so it's big omega along the first axis of the wheel frame.

Â This inertia tensor in the wheel frame.

Â You carry out the matrix math,

Â you get just something in the first axis, which is the gs representation.

Â So, this ties back to the, you know, week one stuff that we did.

Â Vector matrix representation.

Â You should be able to go back and forth and

Â whatever's most convenient It'll get you there, it should be the same answer.

Â Now, we have these, we add them up to the other term, you combine them,

Â this gives us now the angular momentum of the wheel that we need.

Â 24:34

A few more things, as we take derivatives, we'll want inertial derivatives and,

Â of course, we'll use transport theorems because a lot of these inertia tensors

Â are fixed by the body frame.

Â And also some of these other axis have benefits.

Â So, the gs, gt,

Â gg axis you could write analytically if you know their orientation at launch.

Â I locked them down and gs was pointing here and gt was pointing here, and

Â there's no gimble angle, right?

Â And after launch, you rotated 15 degrees,

Â you can actually compute with your classic cosines and sines.

Â What are the current orientations of those axes?

Â gg stays always the same, hopefully, that means you bolted it down correctly,

Â all right?

Â The gimbal axis is supposed to stay body fixed, always.

Â But for a VSCMG, the GSGT, we have to treat generally as time varying.

Â So, you can write this and in code in fact, it would simulate this.

Â If I know my gimbal angle I can't compute at any instant.

Â If I know my initial orientations at launch what are my current orientations of

Â this frame of the wheel?

Â How far has it twisted?

Â All right, so we will need derivatives of this stuff.

Â If we take body frame derivatives, different ways you can do it.

Â If you want to, you could just use this expression here and take derivatives.

Â because these are actually fixed as seen by the body.

Â And cosine becomes minus sine, which is right here.

Â Sine becomes cosine, and you can see the derivative of this is

Â going to be the gt direction and vice versa.

Â And the derivative of this one is just common sense, is going to be zero.

Â That's one way.

Â Or you can also use a transport theorem and say, well wait a minute,

Â 26:42

Different ways you can approach it.

Â I like the transport, it's very compact, you can get there very,

Â very quickly without writing it out.

Â So, we can get the body frame derivatives,

Â the key is the third one, that's where it's locked down.

Â That one doesn't vary, the other two are a function of the gimbal rate.

Â If you need inertial derivatives, which we can do it here,

Â we've chosen to do b-frame derivatives.

Â We've already got these answers as a sub-result,

Â plus the omega crossed these axes again.

Â And you can do that, and carry out, if you write omega.

Â The easy trick here is to get this answer in your homework I would say.

Â Write omega node as omega 1 b1, omega 2 b2 because you're crossing the gs.

Â Instead write omega as omega s gs, omega t gt, right?

Â Use the g frame components of omega and those definitions and

Â that's where these terms come in.

Â And then, it's just a classic cross products between vectors in

Â the same frame.

Â So, 1 crossed 2 gives you 3, 3 crossed 1 give you minus 2, all the usual rules.

Â And that's how you can get there then as well.

Â So, that would work.

Â So then, you should derive this yourself then.

Â I'm just kind of showing you the highlights and how we go after it.

Â Since we have omega s, omega t and

Â omega g, the vector components of the body rate in the g frame components.

Â We will also have derivatives of this.

Â Omega s, omega t, omega g are scalars.

Â So, what type of derivative am I taking on the left hand side?

Â 28:13

Tebow.

Â >> Time derivative it's- >> As seen by what frame?

Â >> Doesn't matter.

Â >> Doesn't matter, right?

Â Just a scalar, so that stuff comes back again.

Â If it's just a scalar it's just a time derivative.

Â Over here, Tebow, what type of derivative have I chosen to use here?

Â >> An inertila derivative.

Â >> Inertial derivative.

Â But that was purely a choice, I didn't have to use an inertial derivative.

Â I could have used any, as long as I'm consistent,

Â you can't use these derivative chain rules into one part in a and one part in b.

Â If you've taken a, make everything a, if you've taken b, make everything b.

Â The math will work itself out.

Â So, you can do that, so that term is just okay,

Â the angular acceleration dotted in the the gs direction, great.

Â This term here, gs hat transposed omega.

Â If you go back and look at gs hat is here,

Â so we have all this math times, what did we do,

Â times omega, gs hat transposed omega.

Â So, if you can see here, gs hat, this stuff, transposed omega,

Â gt transposed with omega, what does that give me?

Â >> Omega t.

Â >> Omega t, exactly.

Â That was the projection.

Â Gg transposed omega gives you omega g.

Â And you're going to have an omega g, omega t with that product and here omega t,

Â omega g with the product.

Â 1 plus, 1 minus, they both cancel and all you're left with is here times gt times

Â that term, that second term, which was gt transposed omega which gives you omega t.

Â 29:47

Let's say we did body frame.

Â How is that going to look different?

Â Let me do that by hand.

Â So, just so I can prove, so if we use this formula, I don't need to fire here, okay.

Â If we use this formula as Tebow says, that's just a time derivative.

Â Now, let's say, I'm going to use body frames.

Â I'm going to use primes as a body frame instead of writing ddt and b's and

Â all that stuff.

Â You would have gs transposed prime,

Â right, times omega plus gs hat omega BN prime.

Â Well, one thing, omega BN prime, how does that relate to the initial derivative?

Â It's the same thing, right?

Â Omega, the B and the N frame is the same.

Â So, that's kind of cool.

Â We get right away gs hat transposed omega BN dot, the inertial derivative.

Â So, that's good.

Â This one, we need the body frame derivative of gs, and

Â if you go back, we did that actually and I had showed you the results.

Â Here we go.

Â The body framed derivative of that, different ways you can derive it, but

Â I'll use the transport theorem, was nothing but gamma dot gt.

Â And if I plug that one in, gamma dot gt transposed omega BN,

Â this whole thing is nothing but the definition of that and

Â you get the same answer.

Â So, as expected, hopefully, we took here, b frame derivatives, in the slides I'm

Â taking n frame derivatives, I get the same answer both ways.

Â So, that's always a good thing to practice then you know you're doing all your

Â math correctly.

Â 32:01

And now the fun, you need to do this yourself, I'm showing you the highlights.

Â [LAUGH] But this is now where you apply your excellent skills on

Â transport theorem and differentiation, we need H dot = L, so

Â I would say since we've broken it up as H of the hub B H of the gimbal and

Â H of the wheel, do each separately.

Â It just helps manage all the algebra.

Â Otherwise you're bound to get one letter crossed with another.

Â Do one by at the time.

Â So if I take H dot using the transport theorem.

Â I take the derivative as seen by the gimbal frame,

Â because I'm using a lot of gimbal frame components.

Â But then I need to add plus omega of the gimbal frame,

Â relative to the inertial, right?

Â because dot is still the inertial derivative, cross with that H expression,

Â carry all that algebra.

Â This is what you get, I'll let you do that.

Â You guys are very good at this now.

Â You do the same thing for the gimbal stuff and this is what you get,

Â the gimbal frame stuff.

Â And then for the body itself, well that's pretty trivial, we've done this before for

Â a single rigid body, you regain the same derivative expression, right?

Â Now we pack it together H dot = L is a sum of all three.

Â 33:10

We're going to use this definition, J was the inertia of the gimbal frame and

Â the wheel, the combined VSCMG device.

Â This is the inertia of the rigid part of the spacecraft.

Â So this one is actually constant, as seen by the body frame.

Â This one isn't, because inertia, everything is spinning and twisting.

Â And I now is the combined spacecraft inertia, that's the full thing.

Â The rigid part plus all these spinning, twisting wheel parts, all combined.

Â That means I in this notation actually becomes time varying.

Â If you simulate these things, every time step,

Â you have to recompute where are these axes, where are these inertias and

Â sum up all these inertia contributions again.

Â So it becomes even as seen by the body, I is time varying but IS is the rigid part.

Â That's what you took advantage of when you did the earlier differentiation of HB,

Â that it was fixed as seen by the body frame.

Â 34:04

Yes, Louis? >> How do you guys account for

Â the center of mass of the gimbal system, in the body or no?

Â >> So all of these things have been taken in account for.

Â IS here actually already accounts for the parallel axis part.

Â That if the CG of the wheels aren't mounted in the center of the spacecraft,

Â center of mass of the spacecraft, but they're offset somehow.

Â That is a fixed offset, and that's already accounted for

Â actually here when we derive this stuff.

Â So here these are just the wheel parts that that's the momentum and

Â inertia is spinning about their own perspective center of masses.

Â 34:37

So center of mass is all subtlety.

Â I'm glossing over somewhat, but this is rigorous, this works,

Â it's just you have to account for depending on where you place them, okay.

Â There's this much mass out there,

Â that M distance squared term is already put into this part.

Â because that part is fixed, relative, it's a balanced wheel.

Â If it's inbalanced, your question would be very good, because then with an imbalance,

Â as I'm rotating, then the center of mass

Â of the wheel is actually orbiting inside the craft and then this wouldn't work.

Â You'd have to be much more careful about which points you do it so.

Â Being balanced allows us to, as complicated as it is, and I'm sure

Â you'll agree in the homework, being balanced makes this much, much easier.

Â So that's a key assumption, so now we throw this all together.

Â H dot = L, and I combine terms, which you will do as well,

Â you can come up with these kind of a terms.

Â Now this is the I inertial tensor of the hub,

Â the rigid plus all the spinning stuff.

Â You can write it out into this form, which is our equation of motion of

Â the spacecraft with a single variable speed CMG.

Â L is still the net external torque we were talking about earlier, so

Â that could be your gravity gradient torque acting on this or deviation pressure,

Â if we have thrusters acting for generating pure torques.

Â But it's all about the system center of mass.

Â And then inside you have all these terms.

Â So the first thing I always do is let's make sure these equations make sense if we

Â simplify them.

Â If we bolt down like on launch, we have no gimbling, we have no wheel acceleration,

Â everything is locked down for launch, I should get back the equations of motion of

Â a single rigid body, which was this part and this part and the L.

Â Everything else should vanish.

Â So if you look at that and you see, okay, if I have gamma dot, if I'm locked down,

Â there's no gimbling going on.

Â So gamma dot has to be 0.

Â This vanishes, this vanishes, let's see,

Â gimbal acceleration vanishes.

Â That's 0 and this is 0, so no gimbling makes a lot of them vanish.

Â If you take the remaining equations you've got one here,

Â IWS omega dot, and you've got this is all vanished, that vanished.

Â You got this term that's non-zero, and this term that's non-zero.

Â Go back and look at your dual spinner equations of motion.

Â 37:18

Now, so this is good for both devices, but

Â you can see the gimbling adds a lot of terms.

Â There's gamma, gamma, gamma, all the stuff that happens here.

Â What if you have a reaction wheel mode, we talked about that.

Â We basically get a double spin equations of motion, way simpler.

Â What if we make it a classic CMG?

Â In that case, the wheel's speed has to be constant, not variable speed.

Â 37:45

One term, woop-de-doo.

Â You got all this other stuff, so

Â the CMGs really aren't that much easier than the variable speed CMGs,

Â you just now accounted for all the other stuff spinning and gyrating.

Â >> Why is it that they haven't flown VSCMGs?

Â >> They have, and in fact every CMG is in fact a variable speed

Â CMG because it launches spun down and they have to slowly spin it up.

Â But what they do is a simple control that they, it's complexity of the stuff.

Â Here, let me show you what generates the control that you have.

Â We have body rates are pretty small.

Â I mean, we're talking small.

Â People freak out when this craft goes more than two degrees per second,

Â that's nothing.

Â The wheels will be going way, way faster.

Â So CMGs, we're looking at 10,000 rpm.

Â So the big omega is going to be the big honking elephant.

Â So this is small, gimbal rates are small, this stuff is all small.

Â What's going to give us the big torque?

Â It's going to be this term, right here, that's the wheel speed, which is huge.

Â IWs, which is your biggest inertia of this disk because we move the mass out about

Â the GS axis, that give us that ring, the pancake shape.

Â And gamma dot is directly proportional to that.

Â So, if you hold the fixed wing speed, this is the control torque that we would use

Â in a control application, and that's what they're after.

Â You could make everything else variable, but that's really more in a later

Â research, we've been finding in how we can avoid singularities.

Â But the trick is, with the reaction wheel, what generates the torque is this one.

Â I'll show you the motor torque equation in the next class.

Â But as you accelerate a wheel you get, with a minus sign,

Â an opposite torque back onto the craft.

Â But you only get one Newton meter in, you get one Newton meter out.

Â Here though, if I gimble, with a little effort, the faster the wheel is going,

Â the bigger the torque I get.

Â And that's what they've been focusing on.

Â So, devices could do this.

Â The issue is the motor torque about the spin axis typically are not that strong.

Â If you have to muscle through a singularity acting like a reaction wheel,

Â you're simply going to saturate that little motor very, very quickly.

Â So the way they're designing it, they're doing the minimal effort required,

Â which means it tracks it within a simple bound, but it's not very tight bound, and

Â it doesn't have much.

Â It may take, it just needs you to spin it up and keep it roughly at the same speed,

Â that's all they need.

Â So it would be a hardware change.

Â So some have flown them like this and actually demonstrated that you can do it,

Â it's just a little bit more complexity in the control and people, air force in

Â particular have been looking at these form of energy storage devices and

Â to account for them.

Â There's extra complexity, they feel like I can do it.

Â Or some CMG manufacturer who was chatting with me during my PhD too, and

Â they're saying well, we could also take a reaction wheel off the shelf and a CMG

Â off the shelf, and combine both devices to get the same mathematical benefits.

Â 40:35

Off the shelf is always cheaper than trying to custom

Â design specifically for something.

Â So until these become common, there's different ways you can do it.

Â But for here, this is kind of, where are we going to go?

Â So what we can see, my final comments on this is, with these equations of motion,

Â you see now there that as the wheel speed grows, we can get a very large torque.

Â But also there's limits in each reaction wheel, how fast you can spin it and

Â that's that big omega dot.

Â The gimbling is a nice mode, but you will also have singularities.

Â Actually, I think I could show those as well.

Â This will be something about the gt axis, that's one axis.

Â If I have four devices, I'm going to have four gt axes.

Â And depending on how you've lined it up,

Â you can have all those axes become coplanar.

Â And if your control is outside of that, which will happen commonly with external

Â disturbances, that's a gimbal lock situation.

Â Because these axis about which I get these nice generous torques vary with time,

Â it just makes a lot of things more complicated.

Â But that was the singularity that we have so, anyway so you can see,

Â CMGs, definitely a lot of benefits, the amplification but

Â also mechanical challenges, more expensive.

Â But you get a lot of bang for your buck,

Â way bigger torques than what you get with a reaction wheel, but

Â it comes at a cost and the controls get more challenging, too.

Â