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Troublshooting your pc for dummies - Bourg D.M.

Bourg D.M. Troublshooting your pc for dummies - Wiley publishing , 2002. - 350 p.
ISBN 0-596-00006-5
Download (direct link): fordummiestroublesho2005.pdf
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* In two dimensions its okay to leave the angular equation of motion as it is shown here, since the moment of inertia term is simply a constant scalar quantity
18 | Chapter!; Bask Concepts
In general, the time derivative of a vector, V, in a fixed (nonrotating) coordinate system is related to its time derivative in a rotating coordinate system by the following equation:
(d\/dt)flxed (d\/dt)mt + (w x V)
The (wxV) term represents the difference between Vs rime derivative as measured in the fixed coordinate system and Vs time derivative as measured in the rotating coordinate system. We can use this relation to rewrite the angular equation of motion in terms of local, or body-fixed, coordinates. Further, the vector to consider is the angular momentum vector Hcg. Recall that HCg = I to, and its time derivative is equal to the sum of moments about Lie bodys center of gravity. These are the pieces you need for the angular equation of motion, and you can get to that equation by substituting Hcg in place of V in the derivative transform relation as follows:
dHcg/dt I(duj/dt) + (cu x (Id)) ^ '
where the moments, inertia tensor, and angular velocity are all expressed in local (body) coordinates. Although this equation looks a bit more complicated than the one I showed you earlier, it is much more convenient to use, since I will be constant throughout your simulation (unless your bodys mass or geometry changes for some reason during your simulation) and the moments are relatively easy to calculate in local coordinates. Youll put this equation to use later, in'Chapter 15, when I show you how to develop a simple 3D rigid body simulator.
Inertia Tensor
Take another look at the angular equation of motion and notice that I wrote the inertia term, I, in bold, implying that it is a vector. Youve already seen that for 2D problems, this inertia term reduces to a scalar quantity representing the moment of inertia about the single axis of rotation. However, in three dimensions there are three coordinate axes about which the body can rotate. Moreover, in generalized three dimensions the body can rotate about any arbitrary axis. Thus, for 3D problems, I, is actually a 3 x 3 matrix, a second-rank tensor.
To understand where this inertia matrix comes from, you must look again at the angular momentum equation:
(rxux t)dm
where is the angular velocity of the body, r (see Figure 1-9) is the distance from the bodys center of gravity to each elemental mass, dm, and (ux r) is the angular momentum of each elemental mass. The term in parentheses is called a triple vector product and can be expanded by taking the vector cross products; r and w are vectors that can be written as follows:
r = xi + yj + zk u> tuxi + Uyj + u>z k
Inertia Tensor [ 19
Expanding the triple vector product term yields
HCg = J {[(y2 +12)* - xywy - XZOJJi -I- [~yxux + (z2 d:|J*:2)wy - yzwjj + [zxwx zyujy + (x2 + y2)wz]k} dm To simplify this equation, lets replace a few terms by letting
Ixx = J (?2 4- ?2) dm
l = j (z2 + x2) dm
Ilz 1 (x2 + y2) dm
lxy = lyx = J (xy) dm
I*Z = ^ZX J ixz) dm
Iyz Izy = I (yz) dm
Substituting these I -variables, some of which should look familiar to you, back into the expanded equation yields
HCg = [Ixx^x " IxyOJy Ixz^z]! T [ JyxWx + IyyWy iyz^Jj
+ [ izx^x IzyOJy + IzzWjk
Simplifying this a step further by letting I be a matrix:
Ixx fxy ixZ
I = Iyy Iyy iyz i'yx lzy fzz
yields the following equation:
Hcg = Iw
You already know that I represents the moment of inertia, and the terms that should look familiar to you already are the moment of inertia terms about the three coordinate axes, f tt , iyy, and L-. The other terms are called products of inertia:
ixy = iyx = J ?xy) dm
ixz = hr. = j (XZ) dm iyz = izy = J iyz) dm
20 j Chapter 1: Basic Concepts
Figure 1-9. Products of Inertia '
Just like the parallel axis theorem, theres a similar transfer of axis formula that applies to products of inertia:
f xy 1 o(>:y) + trtd^dy
?fxz fo(xz) + mdxdz .
Iy z Lj(yz) + mdy d,
where the I0 terms represent the local products of inertia, that is, the products of inertia of the object about axes that pass through its own center of gravity, m is the objects mass, and the d terms are the distances between the coordinate axes that pass through the objects center of gravity and a parallel set of axes some distance away (see Figure 1-10).
Figure 1-10. Transfer of Axes
Inertia Tensor { 21
Youll notice that I did not give you any product of inertia formulas for the simple shapes shown earlier. The reason is that the given moments of inertia were about the principal axes for these shapes. For any body there exists a set of axes oriented with respect to the body such that the product of inertia terms in the inertia tensor are all zero.
For the simple geometries shown earlier, each coordinate axis represented a plane of symmetry, and products of inertia go to zero about axes that represent planes of symmetry. You can see this by examining the product of inertia formulas, where, for example, all of the (xy) terms in the integral will be cancelled out by each corresponding ~{xy) term if the body is symmetric about the y-axis as illustrated in Figure 1-11.
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