134
2 T3 + I203C3 = 0 (A5)
03S3 = 0 (A)
be simultaneously satisfied. We now examine the consequences of having all four conditions satisfied. Condition (A6) requires that 03=0 or S3=0.
Case 1: 03=0 and S3+0, then r3=l and frora condition
(A5), we see that 02 must be equal to 0 (and 72=1). Condition (A3) then requires that a3=0 but 3=a3=0 means that joint axes 3 and 4 coincide and the arm is degenerate.
Case 2: c3=$3=0, then C3=l and, as in the previous case, 02=0. Condition (A4) then becomes a2+a3=0 if C3=1, or a2-a3=0 if C3=-l. When C3=1 (E3=0), then a2+a3=0 means that a2=a3=0 since both link lengths a2 and a3 are non-negative numbers. Therefore, this case means that o2=a2=0 and o3=a3=0 so that joint axes 2, 3 and 4 coincide and the arm is degenerate.
When C3=-I (03=7r), then 2-a3=0 and a2=a3. Since C3=02=0, axes 2, 3, and 4 are parallel. The values a2=a3 and 83=7 force joint axes 2 and 4 to be aligned, thereby forcing the arm in a degenerate configuration.
Case 3: S3=0 and 03+0. This case occurs when 83=0 or when e3=7r. When 83=0, then C3=1 and condition (A4) yields a2=a3=0 as in the previous case. Condition (A3) yields 02=0 or d3=0.
Since a2=0, the case c2=0 means that joint axes 2 and 3 are aligned and the arm is degenerate. Assuming d3=0 and
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