By Guo D., Ji L.

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**Sample text**

The angular velocity (ω1 , ω2 , ω3 ) is finite but does not correspond to the time derivative of an angular coordinate in the body-fixed system. To obtain the corresponding equations in body-fixed (embedded) coordinates, we have to explicitly compute the transformations to factor in the inertia forces due to rotation. 23): ωs = A−1 ωb = AT ωb . Just because the coordinate system rotates with the body does not mean that the angular velocity in the body-fixed coordinate system is zero. 10. Nevertheless, this angular velocity does not correspond to the change of an angle in the body-fixed coordinate system.

The sign of the result is reversed if the order in the product is reversed); this is reminiscent of the cross product for vectors. Additionally, for quaternions there is a ‘unit operation’ 1, such that I · 1 = 1 · I = I, J · 1 = 1 · J = J, K · 1 = 1 · K = K. 48) q = w1 − xI − yJ − zK. In the following, we will drop 1 when it is not necessary. e. for two quaternions q1 and q2 , q1 · q2 = q2 · q1 in general. Nevertheless, the quaternion product is associative: (q1 · q2 ) · q3 = q1 · (q2 · q3 ). The dot ‘·’ for quaternion multiplication is often omitted.

So one has to be much more careful with rotations than with translations. 6 Rotations of a book by 90◦ around two axes: in the sequence from (a) to (c), the book is rotated first around the z-axis and then around the x-axis; in the sequence from (d) to (f), the book is rotated first around the x-axis and then around the z-axis. 7 Rotation around two of three orthogonal axes (x and y) on a globe gives a rotation around the third axis (z). 7, when we rotate the positive x-axis by 90◦ around the y-axis from the equator to the north pole, and then down again by 90◦ around the x-axis, we get the same result as if we had simply rotated it by 90◦ around the z-axis.