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Differential Equations Solutions 59

# Differential Equations Solutions 59 - θ and sin θ to...

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Chapter 12 Solutions: Case Study: Achieving a Common Viewpoint: Yaw, Pitch, and Roll (coauthored by David A. Schug) CHALLENGE 12.1. (a) First we rotate the object by an angle φ in the xy -plane. Then we rotate by an angle θ in the new xz -plane, and fnish with a rotation o± ψ in the resulting yz -plane. (b) We will use the QR decomposition o± a matrix; any nonsingular matrix can be expressed as the product o± an orthogonal matrix times an upper triangular one. One way to compute this is to use plane rotations to reduce elements below the di- agonal o± our matrix to zero. Let’s apply this to the matrix Q T . Then by choosing φ appropriately, we can make Q yaw Q T have a zero in row 2, column 1. Similarly, by choosing θ , we can ±orce a zero in row 3, column 1 o± Q pitch Q yaw Q T (without ruining our zero in row 2, column 1). (Note that since we require π/ 2 <θ<π/ 2, i± cos θ turns out to be negative, we need to change the signs on cos
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Unformatted text preview: θ and sin θ to compensate.) Finally, we can choose ψ to ±orce Q roll Q pitch Q yaw Q T to be upper tri-angular. Since the product o± orthogonal matrices is orthogonal, and the only upper triangular orthogonal matrices are diagonal, we conclude that Q roll Q pitch Q yaw is a diagonal matrix (with entries ± 1) times ( Q T ) − 1 . Now convince yoursel± that the angles can be chosen so that the diagonal matrix is the identity. This method ±or proving this property is particularly nice because it leads to a ±ast algorithm that we can use in Challenge 4 to recover the Euler angles given an orthogonal matrix Q . CHALLENGE 12.2. A sample MATLAB program to solve this problem is available on the website. The results are shown in Figure 12.1. In most cases, 2-4 69...
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