Question 4 of 7: Inverse position kinematics — all closed-form solutions (20 marks)

Question 4: Inverse position kinematics — all closed-form solutions (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The forward kinematics established in Question 3, and a desired endpoint position with no joint limits imposed.

Find. Every closed-form solution $(q_{1},q_{2},q_{3})$ that places the end point at $(x_{e},y_{e},z_{e})$.

Approach. Take the three position equations from $^{0}_{3}T$, notice that $q_{1}$ decouples, and solve the remaining pair as a polar-coordinate problem in the $y_{0}$–$z_{0}$ plane.

1. Write the three scalar position equations. The translation column of $^{0}_{3}T$ gives$$x_{e}=2+q_{1},\qquad y_{e}=-(2+q_{3})\sin q_{2},\qquad z_{e}=(2+q_{3})\cos q_{2}.$$Three equations in three unknowns, and the first one contains only $q_{1}$, so the system splits immediately — a direct consequence of the first joint sliding along an axis that no later joint can reorient.
2. Solve the decoupled prismatic variable. Rearranging the first equation,$$\boxed{q_{1}=x_{e}-2}$$This has exactly one solution for any commanded $x_{e}$, so it multiplies the solution count by one and every branch below shares it.
3. Eliminate $q_{2}$ by summing squares. Squaring and adding the second and third equations removes the trigonometry entirely, because $\sin^{2}q_{2}+\cos^{2}q_{2}=1$:$$y_{e}^{2}+z_{e}^{2}=(2+q_{3})^{2}\;\Longrightarrow\;2+q_{3}=\pm\sqrt{y_{e}^{2}+z_{e}^{2}}.$$The $\pm$ is the whole source of multiplicity in this arm. Since the question states there is no joint limit, the negative root is admissible: the slider simply passes back through its own origin and extends the other way.
4. Recover $q_{2}$ with a two-argument arctangent. Dividing the second equation by the third,$$q_{2}=\mathrm{Atan2}\!\left(\frac{-y_{e}}{2+q_{3}},\;\frac{z_{e}}{2+q_{3}}\right),$$where the sign chosen for the radius in the previous step feeds through. Using $\mathrm{Atan2}$ rather than $\arctan(y/z)$ is essential: the single-argument form collapses the four quadrants onto two and would silently discard half the workspace.
5. Collect the two complete solution branches. Writing $\rho=\sqrt{y_{e}^{2}+z_{e}^{2}}$,$$\boxed{\text{Branch 1:}\quad q_{1}=x_{e}-2,\quad q_{3}=\rho-2,\quad q_{2}=\mathrm{Atan2}(-y_{e},\,z_{e})}$$$$\boxed{\text{Branch 2:}\quad q_{1}=x_{e}-2,\quad q_{3}=-\rho-2,\quad q_{2}=\mathrm{Atan2}(y_{e},\,-z_{e})=\mathrm{Atan2}(-y_{e},\,z_{e})+180^\circ}$$There are exactly two, and no more: the arm has no elbow to flip and no shoulder to swing behind itself, so the sign of the radius is the only discrete choice available.
6. Work a numerical example to confirm both branches close. Command the tip to $(x_{e},y_{e},z_{e})=(2.6,\,-1.3,\,2.1)$. Then $\rho=\sqrt{1.3^{2}+2.1^{2}}=2.470$ and $q_{1}=0.600$ in both branches, with$$\text{Branch 1: } q_{2}=31.76^\circ,\ q_{3}=0.470;\qquad\text{Branch 2: } q_{2}=211.76^\circ,\ q_{3}=-4.470.$$Substituting either triple back into $^{0}_{3}T$ returns the commanded point to full machine precision, which is the only check that actually tests the algebra end to end. Note that the two branches differ by exactly $180^\circ$ in $q_{2}$ and carry radii equal in magnitude and opposite in sign, as expected.
7. Identify the degenerate case. If $y_{e}=z_{e}=0$ then $\rho=0$, the two branches merge, $q_{3}=-2$ and $q_{2}$ becomes arbitrary — the tip sits on the joint-3 axis and rotating the revolute joint moves it nowhere. This is a genuine kinematic singularity of the position problem and is worth stating explicitly, because a controller inverting these equations numerically will otherwise divide by zero there.