# Homework Help: Serial link chain manipulator with constrained geometry

1. Nov 23, 2011

### nucloxylon

1. The problem statement, all variables and given/known data

I have a three link revolute manipulator at the origin. I know all the link lengths. The joint angle for the third link is coupled to the second such that Theta3 = k*Theta2. I want to determine the joint angles (thetas) of the manipulator given that the third link should lie on a line an angle Alpha from the x - axis at a known position.

I believe the solution should be unique but I can't seem to wrap my hands around an equation that proves it.

To summarize:
Knowns:
Angle of line: Alpha
And it's position (I forgot to label that in the diagram, but say we know the equation of the line y = mx + c)
Coupling function: Theta3 = Theta2 * k

Unknowns
Theta1, Theta2, Theta3
Joint positions, X1,Y1, X2, Y2, X3, Y3

2. Relevant equations
Theta3 = Theta2 * k
Alpha = Theta1 + Theta2 + Theta3 = Theta1 + Theta2(1+k)
X1 = L1cos(Theta1)
Y1 = L1sin(Theta1)
X2 = L1cos(Theta1) + L2cos(Theta1+Theta2)
Y2 = L1sin(Theta1) + L2sin(Theta1+Theta2)
X3 = L1cos(Theta1) + L2cos(Theta1+Theta2) + L3cos(Alpha)
Y3 = L1cos(Theta1) + L2cos(Theta1+Theta2) + L3sin(Alpha)

3. The attempt at a solution

I tried assuming I knew one of the joint end positions, i.e X2, Y2, because my intuition tells me there's got to be only one solution. Then maybe I could find another set of equations and somehow cut it out. Here's my work so far:

By law of cosines:
X2^2 + Y2^2 = L1^2 + L2^2 - 2L1L2cos(beta)
where beta is 180 - Theta2 (see diagram)
beta = acos((L1^2 + L2^2 - (X2^2 + Y2^2)) / (2L1L2))
Theta2 = 180 - beta
Theta3 = (180 - beta) * k
Theta1 = alpha - Theta2 - Theta3

But now I'm stuck because I can't come up with another good set of equations to remove X2 and Y2 since I don't actually know them. Any insight or advice would be great.

Thanks!

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Last edited: Nov 23, 2011