And what we're going to do next is we're going to spend a little bit of time, at least two lectures, talking about torque free motion, because for spacecraft this is a pretty common situation. You're not always thrusting. You're not always reacting, because you're using lots of resources, energy, fuel. So how do we understand torque free motion. In fact there's many missions that exploit torque free motions to spin stabilize or do all kinds of stuff. And then we'll look at some disturbances like gravity gradients, we'll derive that from scratch using very much the same kind of notation. And dual spinners, that's another one. And then we'll jump into the control stuff afterwards. Okay, so I didn't put that one off. Let's talk about torque-free attitude motion. As we've been saying today, H dot is equal to L. If it's torque-free the L just becomes zero. That says the nerve, whatever this vector is, H is just a vector in space. And a vector has a magnitude times a direction. All we're saying with H dot equal to L is that as seen by an inertial observer, this vector is fixed. So that means, what you just seen before, H is equal to I omega, we compute all this stuff, we have I in the body frame, we have omega in the body frame. We very often write H is equal to something, something, something in the body frame. The body frame components of H for torque free motion are not zero. The inertias are fixed, but the omegas are going to vary, go back and forth. But if you reconstruct in the body frame, this is what the vector is, now you know what the body attitude is. So times BN, right. You map it back into the end frame. Those, the three inertia frame components have to be fixed. And that's essentially what we're doing here. This is in the body frame. This is typically how we write it. Here we've assumed what type of body frame to get this form. Principle, exactly. Now it's diagonal. We'll do this a lot for analysis. We can always do this for anybody. It just simplifies our basic stability analysis that we're looking at. So that we do that. Then times some DCM in whatever coordinates you wish. Some people that use something besides MRP. I just want to highlight that. That does happen. So you get this and now you have to take the derivative. This is only the N frame components. Now you have to take the derivative of all this stuff and set it equal to zero. And you get a lot of theta one dots and omega dots and yeah that has to be true but what do I do with this dots. It's not very insightful. So that's this is fundamentally what happens. So this is a good test if you're writing an integrator because you give it initial attitude. Give it a spin. No torque. Set the torque to zero and that equations of motion let it tumble. Let it spin and you watch the attitude history. And at every time step, if you compute this. From the attitude computed DCN, map your momentum coordinates from the body into the inertial frame, you should have three coordinates that give you perfect flat lines. If they do that, then you get really warm fuzzy feelings that your integrator is working. It's not a guarantee that you get warm fuzzy feelings something. There's always ways people can get creative and screw things up. But, if you preserve momentum in the inertial frame that's a really good indicator you're on the right track. It's good for numerical checks, it's not good for analysis. So the next step is now for analytical insight. How do we look at this stuff. So really, the key component is we want to do everything in the body frame. It helps us. And we want to look at, we want to make use that momentum is preserved. If we're doing torque free motion is there something else that preserved as well. [COUGH]. Any momentum. Yes, Spencer, energy actually. Do you remember the power equation T dot. Ended up being omega dotted with l, and at l0. That was one of the cases, how we can get 0 power on the system. So [INAUDIBLE] has to be preserved as well. That's a result we'll jump into next. So if you do everything in the body frame, H was equal to i omega. Omega, in the body frame is the classic omega one b one, omega two b two, and so forth. If you just take in the h vector and putting it in body frames, I'm just using h one, H two, H three. That's just my labels of the b frame components of the H vector. And I omega means H one must be I one omega one. That's that first vector component of H, because I'm using a principle frame now that has diagonalised my stuff. She is saying okay, good where we going with this. It's written age or in the body frame. Now instead of using just a re-written version of the prior slide what we're going to do next is instead of saying all three in actual vector component have to be constant, nice flat lines when you plot them out, we're just going to focus on the magnitude. because the magnitude is a scalar. And as we discussed in the first week, to take a derivative of a scalar doesn't have a framed dependency. So if H as a vector is constant in the frame, that must mean it's magnitude has to be constant as well. And therefore, the magnitude is something I can always compute even in body frame components. So that's a nice trick. We go, okay, how do we get the magnitude. Well, we're going to get the magnitude squared by just dotting H with itself. Or in a matrix form we do H transpose H. Same math, gives you the same stuff. So this times itself just gives you I one squared, omega one squared, I two squared, omega two squared. And you start adding them up. And all of this has to give you a constant. Because whatever spin you gave the system whatever gyrations are going on these things have to add up. Omega ones twos and threes can vary with time but they have to add up to give you the same number in the end same momentum. So good this gives me one equation. Kinetic energy was the other one that we talked about. If you look at kinetic energy that was one half omega transpose i omega, and if I is diagonal it really simplifies very quickly. And you end up with one half inertia rate squared for each of the principle axis, that's it. But energy is also preserved. So you notice in these two equations, the independent coordinates are the omegas. And giving it an initial spin, and I know the omegas, my rates, can change. But that gives me three degrees of freedom for the rates, subject to two constraints. Momentum magnitude has to be fixed. Energy has to be fixed. So in the end Jordan, two constraints, three degrees of freedom, how many degrees of freedom are you left with. One, right. So that means immediately, once we have top free motion, the Omega time history has become down to a single curve in three dimensional space. It's a one B sub space that we have to find. And so right now looking at this how do we intercept this, and this, and predict how that behaves. This is not quite geometrically intuitive. That's where the next stuff comes. Let me go back here. Could somebody tell me. Just replace omega one, two and three with X, Y and Z as your three coordinates that we're tracking. If you have AX squared plus BX squared plus Y squared plus CZ squared equal to a constant. What type of shape are you describing. >> Okay, not a circle because this is a three dimensional shape. It's not a sphere. >> [CROSSTALK] >> A, B, and C are different. If A, B, and C were the same it would ge a sphere. So what is the shape then. >> Ellipsoid. >> It's an ellipsoid that you actually, this is a classic equation for the surface of an ellipsoid. So that means, this is kind of like stuff, where we have this 4d hyper sphere, that's hard to visualize. A 3d surface, I can draw, so we'll see visualizations of this. But this equation actually means that your solutions for omega 1, 2, 3, reside on this ellipsoid. Does the ellipsoid have a constant shape, or does it change its shape with time? >> [INAUDIBLE] >> Why? >> Because its x,y, and z parameters [INAUDIBLE] over time? >> X, y, and z are your independent coordinates. Let's look at that. Too much fire, okay. If we have this, this is more of a classic way, you might see this ellipsoid and that's some constant, right Is this an ellipse? If a, b, and c are just constants is this an ellipse of a fixed shape, or is this ellipse varying with size? >> Fixed shape. >> Fixed shape, right? And so here we just have a lot of coordinates that. Where people get quickly confused is what is the independent stuff that we're looking at. So the omegas would be independent things, like x, y and z. You see in the classic ellipsoidal, Wiki notes or something. I1, 2 and 3, those are principle inertias that are fixed in time because we have a rigid body. If these weren't rigid, these would not be fixed things. And we'll see that actually with dual spinners, so you have multiple buttons. Same thing here, this also will make a squared y. So that's x squared y squared z squared times some constant equals to something that's constant. This is also an ellipse, so that that George was talking about right, that's the space we are actually looking for. We have two constraints, this is an ellipse, that's an ellipse. This two has to be intercepted and that's the answer, and that's not very intuitive, it's not easy to intercept two ellipsis to. So people came up with this other approach. Instead of treating omega one, two, three, as the independent coordinates we're trying to figure out what's happening. We use scaled versions of omega one, two, three. Basically you multiply in times of principle inertias so the new independent coordinates are simply your body frame angular momentum vector components. So I'm treating H1, 2, 3s. I can find H1, 2, 3 time histories, and then to map them back to omegas, all I do is divide them by the principal inertias and you get omegas again. But if you do this, and you get momentum, constraint looks like this, what type of geometric shape is this? This is a sphere, all right, it's basically constants equal to x square plus y square plus z square. So this is definitely a sphere, that's easy to draw, as you will see. And then the energy constraint, in terms of this, I will let you do this on your own. You just have to plug in that h1 is i1 omega1, yeah we arrange it you end up getting the energy constraint of divided by t gives you this. This is kind of a classic normalized forward where the ellipsoidal form is something old square equal to one. And then this these terms if you go look up ellipsoids which you could do easily. These become related to your semi-minor, semi-intermediate, and semi-major axes of that ellipsoid that you're going to have, right? This is just a unit, not a unit sphere. It's a sphere of size H, and then you get these energies that come in there, and this is something we can start to put together. So for torque-free motion, to consider what happens to the rates over time as you're jumbling and tirading for anybody, this is nothing cylindrical or symmetric, it's any inertias. These are the two constraints, we've moved from Omega space to h space, and that gave us a simpler way to geometrically find these things. So that's a sphere, and this is the classic ellipsoidal equations, that you can find your a, bs and cs. Which are you're semi-axes for the three things, so yep, that's what you can find there. So we'll have three, if we have three distinct inertias, you will have three distinct axis, and a key element again is well let's see, can t vary on a spacecraft? I'm full of simple questions this morning. Torque-free motion, there's no external torque acting on a spacecraft, could t vary, ansel? >> Yeah. >> Could h vary? If it's torque-free, no external force acting on the spacecraft? You don't think so, you guessing, you feeling lucky? >> H = L, it goes back to that. Is that good for a rigid body? Yes, is it good for a continuum? Yeah, so if you have the spacecraft it's not rigid, it has fuel slosh. It has people wandering around in there, it's a space station. It has big panels that can flex, space station again, huge panels right? H dot equal to l still holds, if there's no external torque acting on the system, H has to be preserved. So for all the torque-free motions, we always treat H as a constant. That's religion, I mean, it doesn't change, t, though, the energy. We say this is rigid, but how rigid is any structure, everything has a little bit of flex. There's always something, maybe very stiff, but there is always some amount of flexing. So as soon as you flex, what happens mechanically? What happens to your energy state? Yeah, you're going to start dissipating energy. There's always a resistance to flexing and there's some damping coefficient. It might be really, really small but if internal friction there's stuff rubbing against ach other there. There's walls, there's always friction ways to dissipate energy, so even though it might be almost rigid and you now nothing is truly rigid. So in practice in this math we are assuming right now it's perfectly rigid but that's an artefact of our mathematics. In real life as an engineer we realize nothing is perfectly rigid, in fact what was the first satellite that went up for the U.S.? >> Explorer 1. >> Explorer. How did explorer 1 do? It was kind of in this shape, they go hey, we looked at this equations and if we spin about this axis. I heard spinning stabilizes things, so they spun about this axis right? Had big antenna sticking out, that's how they communicated. Those antenna were all flopping, doing stuff, anybody remember how long before it went unstable? It was less than an orbit, bery glorious. After that they started to listen to astrodynamicist they go to help if you do it by the right axis and the right stuff, and all right? So that was one because it is flexing it definitely lost energy and you will see here shortly why with all these constraints. This leads to something that can be very unstable, actually, and what does it move towards? There's definite stable equilibrium, but it's not this spin, where you're spinning about this nice symmetry axis, the skinny one, right? So energy can vary, that means, momentum doesn't, if it's torque-free, our momentum sphere is locked. But for that amount of momentum there is a finite range of energies that you can have, this ellipsoid has to intercept with momentum. So there's a minimum energy state you can have, and there's a maximum energy state you can have. So just because you have a fixed amount of momentum you cannot predict what the actual energy state is. There's a range of energies you can have as we dissipate energy we go from one to another on our own. What happens though as we change energy, you can see it's 2x the principle inertia x the energy it takes square root. As energy scales, it's going to scale all of my axes simultaneously. So this ellipsoid, basically, grows homogeneously or shrinks homogeneously with energy. As you change energy, you don't just make this ellipsoid longer in one direction and skinnier in another. It just grows and scales evenly. Right? That's the key thing you want to get out of this equation. Good, now, let's look at this. This is my first of many figures. We're going to assume, without loss in generality, that our principal in our body frame, the principal frame, it's lined up such that B1 gives Axis of maximum inertia. That would be like kind of like this axis here on this box. Then you have B2 is axis of intermediate inertia. That's the one in-between. That would be this kind of a skinnier side. And then axis of least inertia is here. As we said Principal frames is to find that permutation. I can always choose a frame where this is true. So, I don't have to do this math six different ways. I can just do it once. And you can flip your frames when if needed. And now we're plotting these things in momentum corner space. Not omega space but we realize that omega 2 omega H, sorry H3 H2 Are just omega 1 times I1 and so forth. The momentum shows up as a sphere, that's nice. The ellipsoid shows up here, and it has to intercept. So, we said earlier we've got three coordinates we're looking for, omega 1, 2, 3, subject to two constraints. The answer must be a curve. And that's where our curve is. Is that the only intersection? >> There's one I'm not showing. It's on the backside. Thank you. >> [LAUGH] >> So, even here, so if you have energy specified and momentum specified, there's actually two possible trajectories you could be on. They're related. They're kind of mirror images. But the shape isn't flat, this is kind of potato-shaped. Things, ellipsoid, but wrapped around a sphere, that's what the intersection curve looks like. And that's what you would have. And here you would wobble around in this direction, over there it's in this direction. So, this gives you something, this scale body inertia, you can tell right away, this is where my omegas have to reside. This can also be used as an integration check. So now we can play with energy levels and start to look at different spin conditions and also talk about stability of these. So, I know, those of you taking 3200, you've seen these slides before. So, then we'll quickly go to new stuff. With H, let's look at the smallest Ellipse you can do. If you shrink your energy then remember these axes are proportional to the principal inertias so this axis b1 that has the largest principal inertia. It's the longest elongation. If you shrink the ellipsis small as you can make it, the longest elongation is going to be the only point that just barely touches the momentum sphere. That's kind of the geometric interpretation. So that means, what kind of spin do you have? That's a single intersection point here or here. So that means, in this case, you have a spin that's doing either you're rotating positive or negative about your B1 axis, but you're doing a pure spin about a principle axis. And so in this case here, actually it would wobble and you'd be spinning this case that way. So H2 and 3 are 0 because you only have an H1 axis intercept, that's there. And you plug it back into the equations, in the kinetic equations. And in the end, if you have H and the principle inertias, you can compute with this what that minimum energy state is. So, there's a Clementine Mission that went to the moon, very successful. It was one of the first ones that found signs of potential water, ice kind of in the shadowy parts of the polar craters of the moon. Then after we was going to go off and visit some asteroid or comet or something. And along the way they had a slight glitch where all of the sudden the thrusters were told to be on and stay on just because until you run out of fuel. It wasn't just all the thrusters, maybe it was balanced, it was more like there's one thing went on, and so it spun up to a crazy momentum and the mission was over. Now, it's torp free It's still, probably whatever is flexing going on in the structure might still be dissipating energy but it will never ever come to rest unless you have some external force acting on the system. You need an external torque to change a total angle momentum, all right. So even with damping in the system that you have fuel slosh, it'll dissipate it but it can only dissipate it down. to this, that's the limiting case. Any lower energy that ellipsoid and the sphere don't intersect. You're violating momentum. Now you've changed momentum of the system and we must have applied an external force. Maybe SRPs could do some of this stuff. That's one thing we do with space debris tracking. SRPs Atmospheric drag. I'm looking at electrostatic drag forces, as well, and torques. They can change momentum over time, which really complicates the analysis of the objects. But, this is one of them. So, good. Do you think this is going to be stable? This is an equilibrium spin. That means, if I put my body in this condition my omega 2, and 3 are 0. If you go back to look at the other differential equations, what you end up getting is omega 1 data zero or omega 2 data zero, or omega 3 data zero. All right, let's actually look at, let me just go back a few. Here we go. If Omega 2 And omega 3 are 0, while 2 times 3, that's all 0. 0 times something is 0, something times 0 is 0, and torques are all 0. You can see a pure spin about b1 is going to be in equilibrium, where all the omegas are 0, right? So good. So we found that. Now we want to know, is this going to be a stable equilibrium. So if I take this object, and I do a pure spin about this, I'm never going to physically give it a pure spin. I'm just not that good. All right? I'm always off by a smidgen. So, stability is always well. This particular spin rate would be perfect to give you zero rates in everything. But if you're off by a little bit, are you going to stay close? Or is that going to drive it upside down? Gravity gradient is the easy example, right? Here, like this kind of a pendulum would be stable, if I do this, anyhow, it keeps wobbling, then have friction, it just keeps wobbling here to stay close versus this one. This is in equilibrium. But if I wouldn't hold it tight and I'm off by a little bit, it drives it away. That's an unstable one, all right? Here, if we have neighboring motion, we can't make energy any smaller, we wouldn't intercept. So we just make energy slightly bigger, what happens to the interception curves? Do they stay tight and bounded around that equilibrium? Louis? Yes, everything looks rounded. >> So, if you make it a little bit bigger, you would think, okay, you've got this little bit of a curve there. Okay, you're wobbling some. If you're not doing a fierce spin, you would wobble some. And we see that, right? If you take this object and I take it here and spin it, that spun pretty well, no biggie. Let's see So I've got some videos here. Same inertias, I'm showing an object that kind of scales with these inertias. I'm giving it a sphere spin, which would be in equilibrium. I should never see it wobble by other axis, it's just doing that pure B1 spin. Here I'm giving it some slight offsets with half a degree, and you can see, they look almost identical. They are not perfectly identical but they stay, these little wobbles they stay very small and bounded and life behaves well,right? Which kind of makes sense because we're doing deviations about a minimum energy states. And as you've often learned in physics, nature loves minimum energy. Everything kind of converges to minimum energy state. Right, that's the stable one That we would have. So, that's this spin. If we now look at intermediate energy, we increase our energy, right? Everything scales evenly, now of a sudden, the intermediate access which is along B2 intercepts You would have a pure spin about this point, or a pure spin about the other. That means when these object were spinning about this axis in a positive or negative set. Those are the two options if you have this specific energy state. And as before, you said H1 and H3 equal to 0. And then, you can quickly solve that intermediate energy state is momentum squared over 2 times the intermediate inertia. If you have that energy state, these are the two points you could be spinning. But it's not the intersection line isn't just 2 points, like with the minimum energy state. There's also this whole curve that happens, it's a saddle point that wraps around itself. That is a physical motion as well, and that's called the separatrix motion. So if you're spinning nearly about it, now, if I do a pure spin here, I'm good. That should give me just a spin, but is that one going to be stable? If you're off slightly, if you're not just on this point, but we increase energy. And all of a sudden you're at a point slightly above it up here. Are you going to stay close, With these things? What do you think? No, right? So this curve actually takes you from here, then back over here. Here it comes in, so the question is if you're on this curve, well, wait a minute. If I hit this point, do I go this way or that way, what happens? The answer is you never hit that point. Those spin conditions, you can put yourself on it mathematically perfectly. But if you're on the separatrix motion, you will get there with infinite amount of time. It's an asymptote, right? Otherwise mathematically you can see you have a decision point all of a sudden does the stuff go left or right? And that never happens, because you just asymptotically, it would take an infinite amount of time to actually reach that point. And who is that patient? So if you're off slightly, you will see interception lines that kind of look like this. But anytime I get close to this interception point, that's when I know this actually takes a long, long, long time. It'll hang out there, and then, eventually, it goes back over to the other side. This is why, if you try to flip an object about the axis of intermediate inertia, right? It flipped, and now it didn't, there we go, it flipped again, and it flipped again. In one twist and I'm not trying hard, it immediately does 180, 180, 180. It's very unstable in that sense, right? That becomes because of these kinds of behaviors. So, intermediate axis spins are never stable. They're highly unstable, but they may be deceiving. They may look stable for a short period of time. So, let's see if I can get this video to work. Went through the commercial. Let's see if YouTube is going to do this. This is on the space station, they have an object, they took it and they spun it. Why is it doing this motion? >> That's so cool [LAUGH]. It's perturbed very slightly and goes on. >> It's definitely a spin as they're doing it here, this is a spin about the axis of inner median inertia. But there's a little bit of a wobble, it's not a perfect spin. I guess astronauts aren't that well trained, all right? >> [LAUGH] >> So, this is hard to do, but you can see. For short periods of time, it almost looks stable, right? It gets close and it stays close, then crap. All hell breaks loose, right? And you flip upside down again, and then you hang out for awhile. No, YouTube, thank you, who knows what it's going to show us. >> [LAUGH] >> Okay, but it goes back and forth, yep, you can stop. So that's what's happening here. You can see, it looks stable for short periods of time, but it's getting close to here, hangs out for a while. But then, eventually, it's going to get over here and then hang out for a while and then come back. It never settles at one point, that's why it's not stable. You may look like you're staying close for a finite period of time. But as we go through, later on, the control proof, or the stable mean, and so forth. Once you enter a bounded region, you have to remain in that region forever, and you will not. because if you're off, you may stay there for 10,000 years, because you got really, really, really, really close. But eventually you will, you give it enough time, you will bounce back. And it will go off to the other side. Hang out for a while and then come back. So that's kind of a really cool video, I like that one. It kind of illustrates what's happening, right? This is where you're going one direction, and then it floats around and comes back here. And, except for the air drag in that space station, it's really a torque-free motion. And it's just bouncing back and forth satisfying, this stuff. This is actually what leads to chaos as well. So with a single rigid body, he can proof completely rigorous mathematic that this is a chaotic system. If you're spinning nearly around a separatrix kind of energy state. because a infinitesimal change here has you spinning this way. And another one has you spinning this way, which is different orientations. And so infinite sensitivity to small changes is what redefines chaos. And so separatrix motions is a chaotic type of motion. So you see all kinds of papers published on it. So chaos is fun, here we look at it, I've just done a minor minor [INAUDIBLE], again 0.5 degrees. And even after less than one revolution, hugely different orientations, right? That's what you're seeing when you flip this thing and let it spin and its already tumbled. So definitely unstable, let's see what's the maximum energy state. If you scale this ellipse as big as you can, you end up with this kind of a huge ellipse, it still has to intercept. And that's going to be now at the polar points, the b3 axis. So this is a pure spin about to axis of least inertia, because that's how we define b3. And that would be this, and so if I try to spin this, it's reasonably stable, it's just an awkward way to spin it. Little bit of a tumble, but it looks stable. Again, you make your other Other 2, 0. Now from this term, you can derive energy is h squared over 2, minimum inertia. That makes sense, you divide by the smallest inertia to get the biggest energy state, right? So that's the maximum energy you could have with this amount of momentum. To have any more, you have to apply external torques and increase your sphere. But you intercept at 2 points, which is good. Now, here, if you have, we can't have more energy. That's a maximum energy state, if we have slightly less, Nathan, is that motion going to be stable or unstable? >> Unstable. >> Why? >> Because it'll go to the minimum. >> Why? >> because it's going to lose some energy, it's going to dissipate some energy. >> If you don't consider energy loss, and this was really Explorer 1, how bad could it be? Really, it's just a little dappy, nothing, it's very small, right? So without energy loss, if you're like this, that's exactly what Explorer 1 was doing, spinning about the skinny axis of the rocket. They just got it going, you would just be wobbling here. This looks perfectly stable, but as soon as you count the energy loss, you know this ellipse has to get smaller and smaller. And at some point, you're going to go through the separatrix. At that point, who knows what happens, and you might end up upside down, right side up. It's impossible to predict, it's a chaotic thing, right? And, then, at the end, with energy loss, where will we converge to? >> [INAUDIBLE] >> Yeah, this ellipse will go smaller and smaller and smaller. And the smallest one would be the motions here. So, in the end, everything in nature with rigid body converges to a flat spin. So for those of you doing space situational awareness, tracking debris objects and so forth, this is where things would converge assuming null external torques, right? So once you include SRP, and as a recent graduate we did this a lot. We're looking at the statics with Joe's work, things can go quite different all of a sudden. because now you need an external influence to pump up momentum and energy level, and it takes you out again. And things get way more complicated, much, much harder to track. But if it is torque free, these are the classic results. Good, and here's an example, there's no energy dissipation here. So this one, yes, it has a slight difference. You can see it a little bit more pronounced, but overall, nothing like the intermediate axis one. That one really went hog wild and went all over the place. So those are the three, you can look at all infinite [INAUDIBLE], that's what I'm looking at here. So minimum energy would be here, maximum energy would be here. Separatrix energies is here, and you can see the arrows how they go. So if you look at your H1, omega 1, 2, 3 coordinates spaces, you can actually quickly tell. Actually, we're wobbling about an axis of maximum inertia. Because I'm wobbling around in a positive sense. If I have my omegas or Hs move around this axis, and you can see that's actually orbiting in a negative. So this is a positive sense, this is a negative sense. I know I am spinning about an axis of least inertia, or near to it. And if it's flipping back and forth your closer to the separatrix, it's doing some crazy stuff. But you can look at all these interceptions, and this kind of gives you a quick geometric way to interpret stability rigid bodies. When we do the dual spinner, we'll do a complete mathematical way to talk about stabilities. And we'll rediscover the same results, just with a more of a linearization approach on how that works.