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Unformatted text preview: E—102 Transforming Unit Vectors Between
Two Reference Frames Differing by a Single Angle Consider the two reference frames R and R1 with their respective unit vectors i, j, k and i1, j1 and k1 shown below. Their orientation differs only by the angle of rotation 8 , which can
be constant or a function of time. Assume we wish to express i, j and k in terms of i], jl, and
k1. ’ Y 2’2 u
‘ Fig. 1 Let us begin with the relation for i. We begin by breaking up the unit vector i into
components parallel to and perpendicular to the directions of i1 and jl as shown below. [i [cosG Fig. 2 Now, the component of i in the direction of i1 is simply lijcose. Since the component
is in the direction of i1, we can create a vector component as [Moose1‘1 = coseil. Now the component of i in the direction of j1 is simply lilsinB. Since the component is in the negative
direction of j I, we can create a vector component as ~ilsin6 j1 = —sin6 jl. Thus finally, i : costii1 ~ sinGj1 (1) Now the process described above can be time consuming There is, however a short cut.
Consider Figs 1. Let B : 0. Obviously, in this case, i = i1. The trig function which is maximum when 9 = O is cosine. Thus, one component for the case when 6 t 0 is coseil. Including the cosine automatically handles the component when 6 s 0. Now let 6 : 90° .
Obviously, in this case i = —jl. The trig function which is maximum when 8 = 90° is sine. Thus, one component for the case when 6 at 0 is —sir16 j1. Including the sine automatically handles the component when 6 at O. The final result is obtained by adding the two components
just obtained as i goosei1 — sinej1 (2)
the same result as given in Eq. 1‘ This procedure can be continued to express j in terms of i1 and jx. A little practice with
this technique will allow you to do it quickly and easily. ...
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 Winter '08
 Eke
 Trigonometry, Dot Product, component, 90°, Standard basis

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