Okay, so that's the goal. We can have a single Richard spacecraft subject to a single VSCMG device. And the equations of motion are still h.equal to L. What has to be in H? What do you think? Henry? >> [INAUDIBLE]. >> True. >> [INAUDIBLE]. >> What else? The momentum of the pub? The gimbels? Because we have the Gimbels, that's actually some mechanical structure that holds it. It's going to move, and I'm just going to assume it's symmetric. So I have two arms, so that C G is in the center of the wheel as well. And then we have the wheel itself. So that's the total momentum of our system, okay? Good. Then what is L In this case? >> [INAUDIBLE] >> Which torque? Motor torques? >> [INAUDIBLE] >> No. Well, this is not the reaction. We're talking about spacecraft plus VSCMG. In that case, what is L? >> [INAUDIBLE] >> Give me an example? >> [INAUDIBLE] >> Yeah. External drag, gravity gradient torques. For any external torques acting on this system, right? because now we've got a two body system. Later on there'll be multi bodies and so forth. So still it's the same what we had before for single rigid body. We had an L, could have been granted great evening door thruster torques, whatever talks are producing onto the system. But the system is just now a spacecraft plus one of these gyroscopic devices, right? >> [INAUDIBLE]. >> No, we will get to that. So that will be next week. We'll get into the motor torque equations, where do those show up? Coz those are internal torques, right? You're a frame. Gimbal frame, you're the wheel. Those are two internal torques and they will not change the total momentum of the system. That's why they do not show up here. >> [INAUDIBLE]. >> So you're already asking motor torque questions, just hold those for next week. For now this is H. Motor torques do not show up in this H, right? This L is just the external torques acting on the system. We will break up the wheel and the wheel and gimbal system into H.equal to L to get the motor torques specifically. But right now, I'm just getting the overall equations of motion first. And then we'll get the other ones, okay? So we'll have to do this. And the key thing is we have to break up this system, because this gets really complicated pretty quickly. So I'm going to guide you through this. Here's some definitions. We will have the we mentioned the GSGTG. So that's the spin axis of the wheel, and this can vary as seen by the body because we're allowing it to gimbal. So whatever my spin axis is, there's a transverse axis. This is the one, two and then my three axis is GG. That's how we're defining the frame, spin, transverse and G is the gimbal axis. That's how we're gambling it. GG is the only one that's fixed as seen by the body, that's where you fit. Somehow you have to physically bolt the cmg to the spacecraft. Otherwise it's just going to go tumble all over the place, right? So that's where that connection is. Big omega is the wheel speed. That's the classic reaction will speed that you have relative to the body. If big omega zero then you stopped spinning about the Gs axis. And then the gimbal rate, that's how you're twisting it. So if gamma dot goes to zero, then just all of a sudden Gs and GT have become frozen. And it's just like a fly wheel, a reaction wheel, okay? So those are the coordinates that will use. Now with that we have the frames will need angular velocity vectors. So it's pretty straightforward. So from the body, the gimbal frame angular speed relative to the body is magnitude times direction. The magnitude will be gamma dot. This is how much we're twisting this gimbal frame. And the axis about which we're twisting is GG. Pretty straightforward, right? So that's gimbal frame relative to body. Let w be a will fixed frame and it will share Gs as its first axis, because it just makes it easier to go from G to W. So when you have this, that's fine. But now we're spinning about this axis. So I will have a WT and the W G. Those are my second and third axis in the wheel plane. And of course those spin and vary. We define them as a stepping stone, they will quickly vanish and we don't need them anymore. But just for now that defines a full wheel frame, and the angular speed of the wheel relative to the gimbal. Well the rate is omega and the axis is GS, right? Again Gs will vary with time that we need. And that's why I reused Gs across both frames because they're equal. So instead of having to remember that WS is equal to GS, I'm just calling it all Gs. And we'll put everything into the G frame quickly as we proceed, okay? So those are our angular velocities. The we're assuming a balanced system. So I have some structure here that holds it but I'm actually keeping the CG in the middle of the wheel. So the CG of the center of mass of this frame and wheel system is at the origin down there in the middle of that system. So the center of mass as your gimbling does not change. That's an important element. Then I'm assuming that this G S G T G G axis is a principal frame of your gimbal system. Usually is or it's pretty close, but for this analysis this is what we're deriving otherwise this would look way more complicated even more. So in that one that's just three. And I have a GS, and then you can see G T and G G. Three distinct inertias that I can assume the gimbal frame. But the wheel is different and in the wheel frame, the one Axis is a spin axis and then the two in plain. I'm just calling him I W T I W T, because the wheel is symmetric, right? They will have a unique spin axis and a unique transverse. And thetrans verses repeated here. Now due to symmetry of the disk, I'm going to say the inertia tensor expressed in wheel frame components is the same as the inertia tensor expressed in G frame components. Why is that? Why does symmetry cause that? No, the W frame rotates relative to the gimbal frame. >> Because the only access [INAUDIBLE] always a wine [INAUDIBLE]. Yeah the the first one Gs that on right? The difference about which axis one two or three are we rotating to go from G to W. >> [INAUDIBLE] >> Is it one two or three? One, Right? The G S. The way we defined the G frame, it was G S G T G G. And then the W frame was Gs omega T, omega G. So we're rotating about the one axis. So actually yeah because if you look at it, if you look down at the mass distribution and you have this one thing rotating another one not, it's like having a spinning disk beneath you. A blade, circular saw blade or a cd player or some of those disks. And if you just look at it, even though you're not rotating you cannot tell that the mass distribution is changing. Everything is perfectly balanced. To look at the math of that, I can just bring this up quickly. Mathematica did this earlier. We can compute a one axis rotation. So if the theta is the angle that the wheel has rotated now relative to the gimbal frame, that would be from gimbal to the wheel, the dcm. That's what I would have. And I'm defining my inertia tensor and F W means in the W frame. The way I can write that here, I can't do underscores unfortunately in mathematica. So therefore to compute that you would pre imposed multiply with the dcm and the right transpose. And you get this. And you can see cosine squared times these transverse inertia. Science squared times the transverse inertia, which becomes what? Just the inertia, right? Those things cancel out. And sure enough, you get back the same thing if you simplify it, right? So it's a quick thing that you could actually prove to yourself if you want to. I just wanted to show you the math. So no magic here, no hand waving. So for us rigorously the inertia matrix, in respect of W frame and the G frame are actually the same. That's something we can take advantage of as we derive this. Now, what is this mapping from the gimbal frame to the body frame? If you remember the basic dcm definition, it turns Bn was defined such that it's a three by three matrix all the direction cosines. And the first row was simply B one expressed in end frame components, right? So much in the N one, N two, N three. And the second row was B two, third row was BN. Or the first column was N one expressed in the frame components. And the second column was N two, and then N three. Same thing holds here, I'm looking at B G. So the first column must be the first base vector of G, which we call GS. And it will be in the frame components GT and then GG. So if we have to construct that and we'll do this analytically here. Actually not just numerically, we can solve and simplify this would be convenient. These rotations, we have these inertias. And I have the gimbal frame inertia and the wheel inertia and everything now is assumed to be in the G frame. We're taking advantage that the wheel inertia in the G and the B and the w the same. That's where we can drop the W frame really. If you're going to pre and post multiply times this stuff and you end up with these projections. So we'll do this math by hand quickly. But what it really means if the gs gs transposed gives you a three by three sub plane really. And that's the primary plane from inertia tensor perspective of IGS, that inertial term. That was the principal inertia. And so if Gs is going to be one zero zero for example, and you would multiply this out. You would just extract the three by three, that was one in the 11 element and everything else goes to zero, right? That's it. That's like the sub plane where I G s. But if this is any other axis, it's not 100010, but just three general numbers, unit vectors, you would get the sub plane, that is for that principle inertia. And the same thing for G T and G G. But let's derive that quickly, because that's a good exercise. So we said BG the inertia of the gimbal frame in the G frame, times B G transpose. B G we said is Gs gt gg. And I'm going to drop the left superscript Bs here. I'm assuming you know those, with respect to which frame to do it. I'm just going to write them as a essentially. Or one by three vector of matrix of vectors. I can write it like that. And then here if you plug this in, you would have gs gt gg times Igs Igt Igg, right? We said it was principle in the gimbal frame zero times this transposed, which gives you Gs hat transpose Gt hat transpose Gg hat transpose, right? So, if you carry this out, this times this will just give you my gs transposed Igt. So that simplifies to Igs gs hat transpose, Igt gt hat transpose, Igg gg hat transpose. And now this times this will just give you my IGs Gs Gs transpose plus the others, right. So you can just carry out this algebra like that. And that's actually a really handy results that we have, because this is true. And now in the end, I'm saying this isn't B frame components because I'm assuming these have been decomposed in the B frame. But if they needed to be in another frame, you just have to put all the Gs Gt Gts into the queue frame, and then you get the inertial sensor in the queue frame with these projections. So it's another way to actually express your inertia tensors in terms of the principal vector sub planes and the Eigen values of the system. The principle inertias. So we can do that for the wheel and for the gimbal and we're good.