Inverse kinematics

Why it matters

Tasks are specified in the workspace — "put the screw at (0.6, 0.2, 0.1) m pointing down" — but motors are commanded in joint angles. Inverse kinematics (IK) converts the desired tool pose into joint values. Every Cartesian move, every vision-guided pick and every offline program depends on it, and choosing the wrong one of several solutions can swing the arm into a fixture.

Key ideas

Problem. Given ⁰Tₙ (desired tool pose), find q such that FK(q) = ⁰Tₙ. Unlike forward kinematics, the equations are nonlinear and the answer may be:

• none — target outside the workspace, or the required orientation not achievable;
• finite and several — e.g. elbow-up/elbow-down for a 2R arm; up to 8 for a typical 6-axis arm with a spherical wrist (shoulder left/right × elbow up/down × wrist flip/no-flip); up to 16 for a general 6R arm;
• infinite — redundant arms (n > task DOF) or at singular configurations.

Solvability. A closed-form (analytical) solution exists for most industrial arms because they are designed for it: Pieper's condition — if three consecutive joint axes intersect at a point (a spherical wrist) or are parallel, the 6-DOF IK splits into position and orientation sub-problems.

Kinematic decoupling (6-axis arm with spherical wrist).

1. Find the wrist centre: p_w = p − d₆·a, where a is the approach (z) axis of the tool and d₆ the wrist-to-tool distance.
2. Solve the first three joints (θ₁, θ₂, θ₃) so the arm puts the wrist centre at p_w — geometry of a 2R arm in a vertical plane rotated by θ₁.
3. Compute ³R₆ = (⁰R₃)ᵀ·⁰R₆ and extract the wrist angles θ₄, θ₅, θ₆ as Euler angles.

Methods.

• Geometric — triangles and the cosine rule (planar arms, the first three joints).
• Algebraic — manipulate the matrix equation, use atan2 throughout.
• Numerical — iterate Δq = J⁻¹·Δx (Newton-Raphson), Jacobian transpose, or damped least squares Δq = Jᵀ(JJᵀ + λ²I)⁻¹·Δx near singularities. Needs a starting guess and returns only one solution, but works for any arm.

Use atan2(y, x), not tan⁻¹(y/x). atan2 returns the angle in the correct quadrant over (−180°, 180°].

Choosing among solutions. Prefer the solution nearest the current joint values (minimum joint motion), within joint limits and free of collisions. Robot controllers let you fix the "configuration" (e.g. elbow up, wrist no-flip) to keep it consistent along a path.

Formulas

Planar 2R arm reaching (x, y):

D = cos θ₂ = (x² + y² − L₁² − L₂²) / (2·L₁·L₂)

• Reachable only if −1 ≤ D ≤ 1, i.e. |L₁ − L₂| ≤ √(x² + y²) ≤ L₁ + L₂.

θ₂ = ±cos⁻¹(D) or more robustly θ₂ = atan2(±√(1 − D²), D)

• • and − give the two elbow configurations.

θ₁ = atan2(y, x) − atan2(L₂·sin θ₂, L₁ + L₂·cos θ₂)

Planar 3R with tool angle φ: wrist point x_w = x − L₃·cos φ, y_w = y − L₃·sin φ, solve 2R for (x_w, y_w), then θ₃ = φ − θ₁ − θ₂.

Articulated arm waist angle: θ₁ = atan2(y_w, x_w) (or θ₁ + 180° for the "reach-back" solution).

Wrist centre: p_w = p − d₆·a

Numerical IK step: Δq = J⁻¹(q)·(x_d − FK(q))

Worked examples

Example 1 (standard). Planar 2R, L₁ = L₂ = 1 m, target (1, 1) m. Find both solutions.

1. D = (1 + 1 − 1 − 1) / (2·1·1) = 0 → θ₂ = ±90°.
2. Elbow +: θ₁ = atan2(1, 1) − atan2(1·sin 90°, 1 + 1·cos 90°) = 45° − atan2(1, 1) = 45° − 45° = 0°.
3. Elbow −: θ₁ = 45° − atan2(−1, 1) = 45° + 45° = 90°.
4. Check (θ₁ = 90°, θ₂ = −90°): x = cos 90° + cos 0° = 1, y = sin 90° + sin 0° = 1 ✓.

Answer: (θ₁, θ₂) = (0°, +90°) or (90°, −90°).

Example 2 (GATE level). Planar 2R, L₁ = 0.5 m, L₂ = 0.3 m, target (0.6, 0.2) m.

1. Reach check: √(0.36 + 0.04) = 0.632 m, between 0.2 and 0.8 m ✓.
2. D = (0.40 − 0.25 − 0.09) / (2·0.5·0.3) = 0.06 / 0.30 = 0.20.
3. θ₂ = ±cos⁻¹(0.20) = ±78.46°.
4. atan2(0.2, 0.6) = 18.43°. For θ₂ = +78.46°: atan2(0.3·sin 78.46°, 0.5 + 0.3·0.20) = atan2(0.2939, 0.56) = 27.69°.
5. Elbow +: θ₁ = 18.43° − 27.69° = −9.26°. Elbow −: θ₁ = 18.43° + 27.69° = 46.13°.
6. FK check for (−9.26°, 78.46°): x = 0.5·cos(−9.26°) + 0.3·cos(69.20°) = 0.4935 + 0.1065 = 0.600 ✓; y = −0.0804 + 0.2804 = 0.200 ✓.

Answer: (θ₁, θ₂) = (−9.26°, 78.46°) or (46.13°, −78.46°).

Example 3 (wrist centre). A tool tip must be at p = (0.8, 0.2, 0.5) m pointing straight down, a = (0, 0, −1), with d₆ = 0.1 m. p_w = p − d₆·a = (0.8, 0.2, 0.5 + 0.1) = (0.8, 0.2, 0.6) m, and the waist angle θ₁ = atan2(0.2, 0.8) = 14.04°.

Common mistakes

• Using tan⁻¹(y/x): it loses the quadrant — use atan2.
• Reporting only one solution when the question asks for all, or forgetting the elbow sign when θ₂ is computed with cos⁻¹.
• Not checking reachability: |D| > 1 means no real solution.
• In 3R problems, solving the 2R part for the tool point instead of the wrist point.
• Subtracting the tool offset along the wrong axis when finding the wrist centre.
• Expecting a numerical method to find all solutions — it finds the one nearest the starting guess.

For GATE ME

Typical questions give a planar 2R arm and a target and ask for θ₂ (via the cosine rule), θ₁, the number of solutions, or whether the point is reachable. Practise the D-formula, the atan2 expression for θ₁, and always verify with forward kinematics.

Quick check

1. For a planar 2R arm, how many IK solutions exist for a point strictly inside the reachable annulus?
2. L₁ = L₂ = 1 m: is (2.5, 0) reachable?
3. What does |D| = 1 mean physically?
4. Why use atan2 instead of tan⁻¹?
5. What is Pieper's condition?

Answers: 1. Two (elbow up and elbow down); 2. No — it is beyond L₁ + L₂ = 2 m; 3. The arm is fully stretched or folded — one solution, on the workspace boundary (singular); 4. It gives the correct quadrant; 5. Three consecutive axes intersecting at a point (or parallel) — then a closed-form solution exists.