Gravity doesn't matter for joint PID tuning

A placeholder, here to prove the pipeline works end to end: frontmatter, series filing, KaTeX, and code highlighting.

The claim

A robot joint under a PD loop sees gravity in two distinct ways, and they are ninety degrees out of phase with each other. Hanging straight down there is no gravity torque but maximum gravitational stiffness; held horizontally it is the reverse.

The plant is

$$P(s) = \frac{\Delta\theta}{\Delta\tau} = \frac{1}{Js^2 + bs + k_g}, \qquad k_g \equiv \frac{\partial g}{\partial \theta}$$

The constant torque $\tau_g$ never appears in $P(s)$ at all. It shifts the operating point and nothing else, so

$$\frac{\partial(\text{any stability margin})}{\partial \tau_g} \equiv 0$$

exactly, by linearity. Not approximately. Exactly.

Where $k_g$ goes

The stiffness term does survive, but only through a single dimensionless group:

$$\varepsilon \equiv \left(\frac{\omega_g}{\omega_c}\right)^2, \qquad \omega_g = \sqrt{\frac{|k_g|}{J}}$$

and $\omega_g$ is bounded by geometry alone. Because $J \ge m\ell^2$ by the parallel axis theorem while $|k_g|_{\max} = mg\ell$, the mass cancels:

$$\omega_g \le \sqrt{g/\ell}$$

For a human-scale limb this caps the gravity corner near 1 Hz regardless of how heavy the arm is.

Code renders too

import numpy as np

def gravity_corner_hz(k_g: float, inertia: float) -> float:
    """Gravity corner frequency in Hz, the only place gravity enters tuning."""
    return np.sqrt(abs(k_g) / inertia) / (2 * np.pi)

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