MORE on Project 3

Ok, how do we deal with this thing? Hints: first consider the dependence on phi, then that on theta1, then that on theta2.

Dependence on phi
Figure 1: Effect of changing phi
See figure 1 to the right. Suppose that the elbow is initially at a point (x0,y0,z0) and angle phi0, and then moves through an angle phi. This causes a change in the position (x,y,z) as shown in the figure. Clearly the z location doesn't change. Consider, therefore only the x and y location. This is the projection onto the x-y plane; write

x0 = (projected-length) cos(phi0) and
y0 = (projected-length) sin(phi0).
Then consider moving through the angle phi:
x = (projected-length) cos(phi0 + phi)
and similarly for y. Using the trig identities
sin(a+b) = sin(a)cos(b) + sin(b)cos(a) and
cos(a+b) = cos(a)cos(b) - sin(a)sin(b),
and the fact that the z coordinate doesn't change, you should be able to find the transformation matrix for this linear transformation.

Dependence on theta1
Figure 2: Effect of changing theta1
Next, consider the effect of changing theta1, as shown in figure 2 to the right. In this case, there is no change in the position along an "x-prime" axis and there is change along a "y-prime" and the z axes, as shown in the figure. So, if we were to first twist the "y-prime" axis around to lie along either of the x or y axes, then apply the (two-dimensional) transformation, then twist back, we'd be set.

However, "twisting" the figure is just a transformation in phi, which we know how to do (see above!). So, let phi_tot be the current value of phi (which probably depends on x0 and y0, no?) -- then a transformation through the angle -(phi_tot) will move the "y-prime" axis to lie along the x-axis. In these coordinates, shown below, the transformation is strictly two dimensional.
Therefore, in these coordinates, you can (easily!) find a transformation matrix that accomplishes the rotation -- which will change the x and z coordinates but not the y coordinate.

Then, transform (using your phi transformation) back through the angle phi_tot to get the final point.

Dependence on theta2
Finally, what about theta2? This changes the position of the claw, but not the position of the elbow. You should be able to use a very similar matrix to that you found for theta1 here, with one difference: it changes the position of the claw relative to the elbow. The position of the claw relative to the elbow is just the difference in the locations of the two, so that you should finish with something like

(relative claw position) = Atheta2 (original relative claw position)
for some matrix Atheta2. You can apply this transformation after having first applied the effect of the phi and theta1 transformations to both the elbow and claw locations.

Last Modified: Thu May 6 01:36:13 CDT 1999
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