Singularities and static forces

Why it matters

Near a singularity an industrial arm suddenly whips its wrist round or stops with a "singularity" alarm in the middle of a weld — a real safety and quality problem. The same Jacobian that predicts singularities also tells you the motor torques needed to push, press or hold a load, which is how motors and gearboxes are sized and how force control works.

Key ideas

Singularity. A configuration q where the Jacobian J(q) loses rank (det J = 0 for a square J). At and near it:

• the tool cannot move in one or more directions, whatever the joint speeds;
• tool motion in nearby directions needs very large joint speeds (q̇ = J⁻¹·ẋ blows up as det J → 0);
• the inverse-kinematics solutions merge, and there may be infinitely many;
• the arm can resist forces in the "lost" direction with zero joint torque — the load goes straight into the structure.

Types.

• Boundary (workspace) singularities — arm fully stretched or fully folded, on the edge of the workspace. Planar 2R: θ₂ = 0° or 180°. Easy to avoid by not using the edge of the envelope.
• Interior singularities — inside the workspace, from alignment of joint axes. In a 6-axis arm with a spherical wrist:
  • wrist singularity: joint 5 at 0°, so axes 4 and 6 line up — the most common in practice;
  • elbow singularity: arm fully stretched (θ₃ such that the wrist centre is at maximum reach);
  • shoulder singularity: wrist centre on the axis of joint 1, so the waist angle is undefined.
• Because of the wrist/arm decoupling, det J of a 6-axis arm factors into an arm part and a wrist part.

Measuring closeness. det J, the smallest singular value σ_min of J, or Yoshikawa's manipulability w = √det(J·Jᵀ). For a planar 2R arm w = |det J| = L₁L₂|sin θ₂|, largest at θ₂ = ±90°.

Handling singularities. Plan paths to stay clear; use joint-interpolated moves through the region; damped least squares q̇ = Jᵀ(JJᵀ + λ²I)⁻¹·ẋ; redundant (7-axis) arms use their extra joint to stay away.

Statics. In equilibrium, by the principle of virtual work (power in = power out, τᵀq̇ = Fᵀẋ = FᵀJq̇ for all q̇):

τ = Jᵀ·F

where F is the force (and moment) the tool applies to the environment, and τ the joint torques (forces for prismatic joints) needed. Gravity on the links adds a separate term g(q), handled in dynamics. Notice the duality: directions in which the tool cannot move at a singularity are exactly the directions in which forces need no joint torque.

Formulas

Singularity: rank J(q) < m; square J: det J(q) = 0

Planar 2R: det J = L₁·L₂·sin θ₂ (m²) — singular at θ₂ = 0°, 180°

Manipulability: w = √det(J·Jᵀ)

Static torques: τ = Jᵀ · F

• F — tool force in the base frame (N), with moments (N·m) for a 6-D wrench; τ — joint torques (N·m) or prismatic joint forces (N); J evaluated at the current q.

Planar 2R, tip force (Fx, Fy): τ₁ = −(L₁·s₁ + L₂·s₁₂)·Fx + (L₁·c₁ + L₂·c₁₂)·Fy τ₂ = −L₂·s₁₂·Fx + L₂·c₁₂·Fy

• equal to the moment of F about each joint axis (r × F).

Damped least squares: q̇ = Jᵀ(J·Jᵀ + λ²I)⁻¹ · ẋ

Worked examples

Example 1 (standard). Planar 2R, L₁ = 0.5 m, L₂ = 0.3 m, at θ₁ = 0°, θ₂ = 90°. The tool pushes on a surface with F = (10, 20) N. Find the joint torques.

1. Tool at (0.5, 0.3) m; J = [[−0.3, −0.3], [0.5, 0]] (from the 2R formula).
2. τ = Jᵀ·F: τ₁ = (−0.3)(10) + (0.5)(20) = −3 + 10 = 7 N·m.
3. τ₂ = (−0.3)(10) + (0)(20) = −3 N·m.
4. Check by moments: about joint 1 at (0, 0): x·Fy − y·Fx = 0.5·20 − 0.3·10 = 7 ✓; about joint 2 at (0.5, 0): 0·20 − 0.3·10 = −3 ✓.

Answer: τ₁ = 7 N·m, τ₂ = −3 N·m.

Example 2 (GATE level). A planar 2R arm (L₁ = 0.6 m, L₂ = 0.4 m) is held fully stretched and horizontal (θ₁ = θ₂ = 0°). (a) It holds a 5 kg payload at the tip (link weights neglected). (b) It instead pushes horizontally along its own length with 100 N. Find the joint torques in each case.

1. At θ = 0: J = [[0, 0], [1.0, 0.4]], det J = 0 — a boundary singularity.
2. (a) The tool must supply F = (0, +49.05) N (holding the 5 × 9.81 N load). τ₁ = 1.0 × 49.05 = 49.05 N·m; τ₂ = 0.4 × 49.05 = 19.62 N·m.
3. (b) F = (100, 0) N: τ₁ = 0·100 = 0; τ₂ = 0. The radial force goes entirely into the links.

Answer: (a) 49.05 N·m and 19.62 N·m; (b) 0 and 0 — the singular direction needs no torque.

Example 3 (near-singular speed). L₁ = L₂ = 0.5 m, θ₁ = 0°, θ₂ = 5°; the tool must move radially at 0.05 m/s. det J = 0.25·sin 5° = 0.0218 m². Solving J·q̇ = (0.05, 0) gives θ̇₁ = 1.14 rad/s and θ̇₂ = −2.29 rad/s — huge joint rates for a slow tool motion, which is why controllers slow down or fault near singularities.

Common mistakes

• Assuming any configuration with "nice" angles is singular — check det J; for 2R only θ₂ matters.
• Using J⁻¹ instead of Jᵀ for statics: τ = Jᵀ·F needs no inversion and holds even at singularities.
• Sign of F: τ = Jᵀ·F uses the force the robot applies to the environment.
• Forgetting gravity torques of the links themselves when sizing motors.
• Thinking singularities exist only at the workspace boundary — wrist singularities occur inside.

For GATE ME

Questions ask for the singular configurations of a planar arm (det J = 0), joint torques from a given tip force using τ = Jᵀ·F or moments, and conceptual consequences (loss of a direction, unbounded joint rates, zero torque for forces along the singular direction). Practise finding torques both by Jᵀ·F and by r × F as a check.

Quick check

1. At which θ₂ is a planar 2R arm singular?
2. A planar 1-link arm of length 0.4 m, horizontal, carries a 10 N downward load at its tip. Joint torque?
3. What does det J = 0 imply for q̇ = J⁻¹·ẋ?
4. Name the most common interior singularity of a 6-axis arm.
5. Does τ = Jᵀ·F require J to be invertible?

Answers: 1. 0° or 180°; 2. 4 N·m; 3. J⁻¹ does not exist — some tool velocities need unbounded joint rates or are impossible; 4. Wrist singularity (θ₅ = 0, axes 4 and 6 aligned); 5. No.