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worksheet4.20100126.4b5f93ea568e15.47128035

worksheet4.20100126.4b5f93ea568e15.47128035 - Worksheet on...

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Worksheet on Euler angles On the orientation of a rigid body A proper orthogonal matrix A general formula Let Θ ˛ @ 0, Π D and j , Ψ ˛ @ 0, 2 Π L . The matrix In[1]:= H H = 88 Cos @ Ψ D , Sin @ Ψ D , 0 < , 8 - Sin @ Ψ D , Cos @ Ψ D , 0 < , 8 0, 0, 1 << . 88 Cos @ Θ D , 0, - Sin @ Θ D< , 8 0, 1, 0 < , 8 Sin @ Θ D , 0, Cos @ Θ D<< . 88 Cos @ j D , Sin @ j D , 0 < , 8 - Sin @ j D , Cos @ j D , 0 < , 8 0, 0, 1 <<L MatrixForm Out[1]//MatrixForm= Cos @ Θ D Cos @ j D Cos @ Ψ D - Sin @ j D Sin @ Ψ D Cos @ Θ D Cos @ Ψ D Sin @ j D + Cos @ j D Sin @ Ψ D - Cos @ Ψ D Sin @ Θ D - Cos @ Ψ D Sin @ j D - Cos @ Θ D Cos @ j D Sin @ Ψ D Cos @ j D Cos @ Ψ D - Cos @ Θ D Sin @ j D Sin @ Ψ D Sin @ Θ D Sin @ Ψ D Cos @ j D Sin @ Θ D Sin @ Θ D Sin @ j D Cos @ Θ D is proper orthogonal, since In[2]:= Transpose @ H D .H Simplify Out[2]= 88 1, 0, 0 < , 8 0, 1, 0 < , 8 0, 0, 1 << In[3]:= Det @ H D Simplify Out[3]= 1 Finding Θ , j , and Ψ when | h 33 | 1 Consider a general transition matrix H with entries h ij . In[4]:= H = Table @ Symbol @ "h" <> ToString @ i D <> ToString @ j DD , 8 i, 3 < , 8 j, 3 <D Out[4]= 88 h11, h12, h13 < , 8 h21, h22, h23 < , 8 h31, h32, h33 << Necessary conditions Suppose that h 33 1. Since each column of H corresponds to a unit vector, it follows that h 33 < 1. Then, there exists a unique Θ ˛ H 0, Π L such that cos H Θ L = h 33 and sin H Θ L 0. Since each column of H corresponds to a unit vector In[5]:= H @@ 3, 1 DD 2 + H @@ 3, 2 DD 2 == 1 - H @@ 3, 3 DD 2 . h33 fi Cos @ Θ D Simplify Out[5]= h31 2 + h32 2 Sin @ Θ D 2

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It follows that h 31 , h 32 £ sin H Θ L . Given this value of Θ it follows that there exists a unique j ˛ @ 0, 2 Π L such that cos H j L sin H Θ L = h 31 and sin H j L sin H Θ L = h 32 .
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