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Unformatted text preview: f link 4 is determined. In summary, all we need to
ﬁnd are the two angular velocities ω3 , ω4 .
The actual computation is quite straight forward. We start again from the vector loop equation:
R2 + R3 + R4 − R1 = 0, Rj = rj eiθj , j = 1, . . . , 4. Recalling that the derivative of a complex number
Z (t) with ﬁxed modulus, |Z | = const, i.e., Z (t) = |Z |eiθ , is Z = iθZ , we ﬁnd upon diﬀerentiating
the loop equation
i θ 2 R2 + i θ 3 R3 + i θ 4 R4 = 0 .
The real and imaginary parts of this complex equation yield two real equations which can be written
in matrix form as
r2 cos θ2 ˙
r3 cos θ3 r4 cos θ4
r2 sin θ2
r3 sin θ3 r4 sin θ4
Assuming the position analysis has already been done, angles θ3 , θ4 are known (and of course, θ2
and θ2 are given). Thus we have two linear equations in two unknowns: θ3 , θ4 . We can make a small
simpliﬁcation in these equations – essentially by dividing out the given quantity θ2 – the advantage
being that the resulting equations will no longer depend on time. Thus, we deﬁne
k3 = ˙
θ2 k4 = 1 ˙
θ2 and we have the equations
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- Fall '13