Next, we're going to talk about angular momentum of the system. We said H dot equal to L. Right now we're working on the full spacecraft hub. That's the main rigid hub, plus the gimbal frame that can rotate, plus the wheel that can spin relative to the rotating gimbal frame, it's a three-body problem. Henry. What's in I_s? We'll get into that. That's the total momentum. What I should be able to do is H dot equal to L. That's the total angular momentum, but it has to be inertial angular momentum. Everything has to be momentum of the hub with respect to an inertial frame. The angular momentum of the gimbal, not relative to the body, but relative to an inertial frame. Then the wheel's momentum relative to the inertial frame. That's what has to get computed. Now that's the goal. We have three terms. Let's break them up one by one. The spacecraft hub, the rigid part of the hub, we can say, let's just take its inertia I_s, that's for the spacecraft. That's the classic rigid body you had in 5010 single rigid body about it. This is the angular momentum about its center of mass. We're not going to do about a general point because that's the orbit part. We know we can always write the momentum of a system as the angular momentum of the center of mass of the system plus the angular momentum about the center of mass. With that, we can just ignore, and without loss in generality, assume we can derive everything here about the center of mass of this system and therefore apply H dot equal to L. But the question now becomes what all goes into that I_s? Which is a question I actually have for you guys. If you have a blob and the center of mass of the system is C, and then you have some wheel, let's just pretend it's a wheel. I'm not going to draw a wheel and a gimbal. It's just too much drawing. It's too early in the morning. That is location O and that wheel is spinning. That's O relative to C. When we look at the wheel momentum, we will talk about the wheel momentum then about its center of mass, which will be at some location. If I compute the inertia of the spacecraft without a wheel about this point, I just account for that. Then later on, so H_w the momentum of the wheel is the wheel momentum about its center of mass. What are we missing in there if we do that? Henry? [inaudible]. There's mass and inertia of the system. If the wheel were having its center of mass at the center of mass of the system, then we're good. We're all happy. But typically we don't put all the wheels crunched in the middle. That's physically impossible. They're going to be attached somewhere else, by drawing it offset. How do these things relate? We're going to have to look at the momentum. I'm going to write this out. Do we have the momentum? What we need is the momentum relative to this point C. What we will typically write is the momentum of the wheel about point O, the inertial momentum of the wheel about that point, because that's where we have the classic inertia angular velocity, do all the stuff but those are not the same. What do we use to shift the point about which we take moments? Parallel Axis Theorem. Or in this case, we can also break it down and say, look, this has a certain distance, and that one, it's descent. The angular momentum of a system is the angular momentum of the center of mass of the system plus the angular momentum about the center of mass. I already have the angular momentum about the center of mass of the wheel. I still need to add the angular momentum of the center of mass and we derived that. That ended up just being plus the position vector of that center crossed with its inertial velocity times mass. We have different variables we used, but that's essentially what it was. We can do that and then you go, let's see, our dot is just going to be, this is a body-fixed vector. The wheel is not tumbling around and as it rotates, its CG does not move with respect, it's a balanced wheel. This time is just going to be o relative to c crossed with omega, crossed with o relative to c times mass. Now we can rewrite this. This omega, we can reverse this cross-product and I get minus r cross omega. Then we have r cross, r cross with a negative sign, which is equivalent to r tilde. R tilde transpose times mass and we put that in there. That's what we were missing. That's what we were missing before. To get the total momentum, that IS, is going to be the I of, how do we call this? Spacecraft minus reaction wheel. Or if we have VSCMGs, without all of those minus VSCMG. If you go to a cat problem, take out the VSCMGs, what's your inertia? But to be precise, you do need to include the center of mass parallel axis part of it, because when the CAD program gets the inertia it goes, "Okay, this particle here, that particle here." Yeah, the wheel has particles that move, so we handle that separately, but the center of mass doesn't move. It's nothing like a point source of inertia and mass times distance squared parallel axis theorem, so that's absorbed. The IS, is the basic default spacecraft without that plus parallel axis theorem that you have to do for every offset. You do need to include in there and that's what goes into that inertia. Now why did we do it? The spacecraft is rigid and we're taking advantage of that these vectors are constants as seen by the body. You could include them in the derivation, have a lot of terms floating around, but in the end it's the inertia tensor plus all these parallel axis, masters and square terms sum up. We might as well absorb them into one IS inertia tensor and say these are all the constant parts of the inertia of the spacecraft. That make sense? You can see already, we're getting into some subtleties here and what actually has to go to get the full momentum. That's what this IS is, is the rigid part of the spacecraft plus the center of mass contributions of the gimbal frame and the wheel frames are already all absorbed because they're all fixed. Those inertia contributions of fixed. Such a simple equation, so many questions already. But this now gives me, if I have that, that's like all your CMGs become point masses. You would have the total momentum, it becomes a rigid body. Nothing else is going on and we're good. If you just have not a CMG, but just a little point-mass there with the same mass, you would be rigorous with that. That's this one. Now we have to look at the gimbal frame and we need its angular momentum about the inertial angular momentum. What you're going to get here is the angular momentum of the gimbal about the gimbal center of mass which is okay. Because the momentum of the center of mass is already absorbed into IS omega term, right? This is what we just argued. But it's still the inertial momentum so that often the errors people quickly make is they put somehow omega G relative to B or something. It needs to be the inertial angular momentum about the gimbal frame center of mass, so this has to be omega g relative to n in here. Henry. [inaudible] They vary with time. Then you have to get all the momentums right. The question is, would it be easier to just include in your spacecraft stuff? You could do it. That was that h is equal to h of the hub plus h of the wheel plus h of the gimbal. You could combine all that and that's essentially what we're getting. All the math we're doing better be equivalent to your approach to. I would argue this is easier because I'm breaking it up into manageable smaller parts and putting them carefully together. I'm making sure I'm not missing anything in here. But in the end it should all be equivalent. I have the total momentum of the system however you got there. I like just breaking complex problems into small problems that I know how to do and then the rest is complexity. Here though the key thing is I needed the angular momentum about the center of mass of the gimbal. I need omega g relative to n, omega g. We had omega G relative to B. That was simply Gamma dot times GG, right? That was the gimbling of the frame that we do about the GG axes. But I also need omega B relative to n and the inertia tensor, I can write out in a coordinate agnostic way. Sorry. Could take a dry throat now. Too much talking. I_G_s, I_G_g, I_G_t, say that fast times these outer products times Omega B/N. Then the other part here is this inertia tensor times something g_g will just pull out this last term g^T dotted with g_g will go to 0. Those are orthogonal vectors. This part times the inertia tensor will just give me this component. Do this yourself, there's many ways you could write this algebra, whatever makes you comfortable. Now, we can simplify this because you will find these terms appear a lot with VSC and gs. Will have a g_s transposed or dotted with Omega vector, a g_t Omega and g_g Omega. When we write Omega, we typically don't write Omega like this. We write Omega as Omega 1 B1, Omega 2 B2, and Omega 3 B3, because that's what the rate gyro gives you. That's what you're going to be sensing. But those are taking the Omega vector, I'm breaking it down into body frame components. Here in the math will find a lot of Omega projected onto gimbal frame components. This Omega here, with the left superscript G is the same Omega that's Omega B/N. I've just now taking a coordinate frame transformation and projected it onto g_s, g_t and g_g. That make sense? With these definitions, that's the projection of this onto the first, second, and third gimbal frame axis. GS transposed Omega will just be shorthand, omega S, that's the spin axis component of my body angular velocity. Omega T is the transverse axis component of the angular velocity and Omega G is the gimbal frame axis component of the same vector. This is the same Omega not a new Omega vector, just different vector components and this would just simplify my math because it makes it more compact versus having these terms all over the place. [inaudible] Exactly. The DCM is embedded in that projection stuff that we're doing. Left and right multiply times the DCM so that DCM times Omega is really just Omega transformation into that frame. Exactly, so it all hopefully starts clicking. If you do that, then this part simplifies because it's just going to be I_G^s times the Omega s, times the g_s axis. We're left with the gs on the map here. The same for the other two, but in the g_g axis, we will also have the gimbal rate plus that. We get a reasonably compact answer of the inertial angular momentum of the gimbal frame with respect to the inertial frame, but it's about the gimbal center of mass. The rest is already absorbed elsewhere. Now the wheel is basically ditto. Here, H_W is the inertial angular momentum of the wheel about its center of mass. In inertial means, I need to take omega W with respect to N. We had G with respect to N by just adding these two, and to get W, you have to add Omega W with respect to G, and that was big Omega times g_s. The wheel speed times the spin axis. It's another vector. We can write that out. If I do this, I_W are broken up into three ways. I'm just showing you different ways to the algebra. I'm not saying one is better than another, different people will click with different algebra or sometimes they don't see it one way, then they see this, now they prefer the first. That's why I'm trying to mix it up a little bit. I'm going to do this term first, then this term and this term. Here, I'm breaking up this like before, as the principal inertia times these outer plane projections of the eigenvectors of that system. Those were the principal axis that we had, times Omega B/N and again, g_s transpose Omega B/N is just going to be Omega s. This term could be written compactly like this. That's one way. Let's look at this one. I'm going to use a matrix representation. I_W in the gimbal frame was the spin axis and the two transverse and everything else. It's the principle. The gimbal frame was diagonal. The angular velocity was Gamma. about the third axis. It was g_s, g_t, g_g, g_g is the gimbal axis. It's 000 Gamma. You carry this out, you would get 00 I_W_t Gamma. which is equivalent to I_W_t Gamma. times the g_g, the third axis. We're going from matrix representation back to vector representation. You don't have to do this back and forth between vectors and matrices, but sometimes people do it this way and go huh, that clicks to me. That's where I get it.