Let's do some examples on Lagrangian. We're having x-coordinates, y-coordinates. This was the classic kinetic energy minus potential energy. X doesn't appear, so I'm taking the partial of this with respect to x dot. This is my integration constant there. You can see if I invert that, I get the relationship, px was Beta that we had earlier, and that B matrix would really be just 1 over m times identity operator, in this case, very trivial. But now I have a relationship there. If you plug the Routhian, you take this relationship and plug it into your Lagrangian, my kinetic energy now is written this way, where I don't have x dot appearing because x dot has to satisfy this condition. That's cool. If you expand this, you will find the Routhian. The Routhian is the part that depends on the non-cyclic coordinates. You would have y in here. There's no more on R, there's nothing else R dot, there's nothing else left. The extra term, if you carry this out, this term goes into the Routhian, this goes into the Routhian, and this term carried out m over m squared just becomes 1 over m, the 1/2 and the px squared is here. This is the term that is just a constant for a particular dynamical system. Again, px depends on initial conditions. Different initial conditions, you'd have different constants. You can't just compute it once and then apply it to 60 different cases like how you're throwing the ball. You'll recompute it every time. But once you have it, that's there. Therefore, taking the gradient of the Lagrangian, this is always going to drop out and you can just do it off the Routhian, which is this one. If you then apply this, you get these equations of motion directly. [inaudible] [inaudible]? This one? Yes. Does look like it. Thanks. I'll go fix that for next time. That's the example there. If you did the same thing here, and I'll let you do this on your own, you would get the same kind of stuff. Here we got these equations which happened to have Theta dot in them. They're a little less trivial than the one example I just did. You'd find the Routhian instead of the Lagrangian. The Routhian would be in terms of p theta. You do the Lagrange's classic equations for r, you would end up with this immediately. There's not that much difference really, but either way. That's what we're talking about with Routhians there. It helps you solve for the ones that really matter. It doesn't mean Theta. I could throw a ball this way, or I could throw a ball this way, bouncing across the plane with no obstacles, and which way I throw it doesn't really matter on the bouncing motion. It matters on the position. Were you aiming here or here? But for the rest of it, it doesn't. My biggest thing with cyclic coordinates is invariants of motion. That's a nice thing to check. Once you find that, that's something you can take advantage off. I have a simpler system. I have a single differential equation. Numerically, I can propagate that, and that I am moving out to the constant angular rate or direction is not going to be corrupted versus if I had XYZ coordinates and then integration errors, I might have little wobbles in that. Here, you're going to have dead on perfect. It's moving in one straight line, always. There's nice advantages there if you can do cyclic coordinates and take advantage of things that are not things you have to numerically solve for essentially.