[BLANK_AUDIO]. Hi, this is module 21 of Three Dimensional Dynamics. Today's learning outcome is to conduct a translational transformation of inertial properties from one coordinate axis in a body to a different location. And we'll find that useful when we actually solve problems in the future. And so we need this skill. Here's what we came up with at the end of last module. We found the inertia matrix for a coordinate system that is fixed in a body with its origin at the mass center. And that's the inertial property matrix, symmetric. And, we said it includes, it's a tensor. It includes information about mass, shape and geometry. Now we're going to move the properties to another to another coordinate system, with it's origin up here at point B. And so this will be a translational transformation. We're not rotating the access, we're just translating it. And when we do that we're going to use something called the Parallel Axis Theorem. Now remember that the Parallel Axis Theorem relates a different point in the body always to the mass center, and so remember that. You always gotta go between the mass center and the other point in the body. I encourage you to look at other references, or the textbooks, regarding the Parallel Axis Theorem to get a good handle on it, but if you go into those references. This is how you calculate the new mass moments of inertia and the new products of inertia, using the Parallel Axis Theorem. So Ixx at point B is equal to Ixx at point C, plus m times the distance in the. And I'll show, show you what y-bar is and z-bar is in a second. So, often, in two dimension we say md-squared. And it's very similar in three dimensions. Here's the definition for the Parallel Axis Theorem for Iyy and Izz. And in this case as I said, I was going to tell you what x-bar, y-bar, and z-bar is. So x-bar, y-bar, and z-bar is the distance you go in the x, y and z direction between point B and point C. And again, here's the products of inertia, again you'll find these in, in a, in a, in any reference. The related to a dynamics course like this or other courses as well. And there's my three products of inertia. Again, where r from B to C is, is defined as x-bar in the I direction, plus y-bar in the j direction, plus z-bar in the k direction. So let's go ahead and do an example. we'll, we'll again find the mass moments of inertia and products of inertia with respect to this coordinate frame, shown with its origin at point B. For the uniform rectangular solid mass. And we start with the, inertia matrix at point C. Again, we found that at the end of last module. Here's our, our expressions for, or the definition for the, the mass, excuse me, the mass moments of inertia and the products of inertia transformed, or transferred, trans, in the, in the translational direction using the parallel axis theorem. So in this case, r from B to C for our problem is equal to. How far do we go? And, try to fill this out on your own, and then come on back. Okay, in the x direction, in going from point b to c, we go a distance, a over 2. So, the x-bar is going to be a over 2, in the, i direction. Then in the y direction we go a distance b over 2. So b over 2 in the j direction. And then going from point B to the point C in the z direction, we go down a distance d over 2. So that's not plus, but this should be minus. So we're going to erase that and say, okay, this is a minus. D over 2 in the k direction. Okay. And so now I can find Ixx about point B. So, I've got Ixx about point B, the mass moment inertia is Ixx about point C. Well we found that to be m over 12, times b squared plus d squared. And then plus m times y-bar. Well y-bar is b over 2 squared. Plus z-bar, and z-bar was minus d over 2. Squared. And if you do that math, you get Ixx about point B is equal to m over 3, b squared plus d squared. And you can do the same sort of approach for Iyy, about point B. And Izz about point B. And this is the results you'll get. Okay, now let's do the, products of inertia. And so here again, our definition for the Parallel Axis Theorem, I found rBC to be equal to these values, and so now I want to find I-xy-B. So I-xy-B. By the way, again, this is symmetric. So this is the same as I-yx-B. And you'll see some references switching them back and forth. It doesn't matter. Equals now. I-xy about point C, which we found was 0. And then minus m times x-bar, which was a over 2. And y-bar, which was b over 2. And that comes out to be minus m ab over 4. Same thing for I, same approach for I-yz. About point B, through point B, is I-zy, about point B equals, now, I-yz is 0. Here and here, are I-zy, same thing, and that's equal to 0 then, minus m times y-bar which is b over 2, and z-bar which is minus d over 2. So this equals m b d over 4, and that's positive. And finally I-xzb, which is the same as I-zxb equals 0 for I-zx or I-xzc minus m times z-bar, d over 2 minus and then x-bar, which is a over 2. And so this is, again, a positive mad over 4. Okay, so there's your products of inertia. We found our, moments of iner, mass moments of inertia. Here again is the mass moments of inertia we found, and the products of inertia, And so I can put those in matrix form, and now I found my inertia matrix for an axis at point B, with its origins at point B, oriented as shown. And we found that using the Parallel Axis Theorem, and we notice now that it's still symmetric, however it's not diagonal. We do have products of inertia now. And so, as a recap we did a translational transformation. We did this example, and this is the result we came up with.