Module 1: Locomotion & Biomechanics

Conservation of Angular Momentum (zero total):

\[ L_{\text{total}} = I_1 \omega_1 + I_2 \omega_2 = 0 \]

where \( I_1, \omega_1 \) are the moment of inertia and angular velocity of the front half, and \( I_2, \omega_2 \) for the rear half.

\[ I = I_0 + 2m_{\text{leg}} R^2 \]

where \( I_0 = \frac{1}{2}m_{\text{body}} r^2 \) is the body cylinder's moment and\( R \) is the leg extension distance. For a 4 kg cat:

• Tucked: \( R \approx 3 \) cm, \( I \approx 0.002 \) kg m\(^2\)
• Extended: \( R \approx 12 \) cm, \( I \approx 0.008 \) kg m\(^2\)
• Ratio: \( I_{\text{ext}}/I_{\text{tuck}} = 4 \)

\[ \frac{|\omega_1|}{|\omega_2|} = \frac{I_{\text{ext}}}{I_{\text{tuck}}} = 4 \]

If the rear rotates backward by angle \( \alpha \), the front rotates forward by \( 4\alpha \). Net rotation of the front relative to the ground: \( \Delta\theta_{\text{front}} = 4\alpha \). Net rotation of the rear relative to the ground: \( \Delta\theta_{\text{rear}} = -\alpha \).

In Phase 4, roles reverse: the rear gains \( 4\alpha \) while the front loses \( \alpha \). After one full cycle:

\[ \Delta\theta_{\text{front}} = 4\alpha - \alpha = 3\alpha \]

\[ \Delta\theta_{\text{rear}} = -\alpha + 4\alpha = 3\alpha \]

Both halves rotate by the same net amount \( 3\alpha \) per cycle. For \( \alpha = 60° \), one cycle gives \( 180° \) of total rotation — exactly what is needed to right the cat.

\[ \Delta\theta = \oint_{\gamma} \mathcal{A} \]

The net rotation \( \Delta\theta \) is the holonomy of the mechanical connection \( \mathcal{A} \)around a closed loop \( \gamma \) in shape space. This is mathematically identical to the Berry phase in quantum mechanics and the Hannay angle in classical mechanics — a deep connection between cat physics and gauge theory!

\[ \mathcal{A} = -\frac{I_1(\mathbf{s})}{I_1(\mathbf{s}) + I_2(\mathbf{s})} \, d\phi_1 - \frac{I_2(\mathbf{s})}{I_1(\mathbf{s}) + I_2(\mathbf{s})} \, d\phi_2 \]

where \( \mathbf{s} \) denotes the shape variables (leg extensions) and \( \phi_1, \phi_2 \)are the relative angles. The curvature of this connection determines the efficiency of rotation per unit shape change.

\[ \text{Fr} = \frac{v^2}{gL} \]

where \( v \) is speed, \( g \) is gravitational acceleration, and \( L \)is leg length. This dimensionless number represents the ratio of centrifugal to gravitational forces and universally predicts gait transitions across species (Alexander, 1989):

• Walk: Fr < 0.5 (pendular mechanics, always one foot on ground)
• Trot: 0.5 < Fr < 2.5 (diagonal limb pairs, spring-mass model)
• Gallop: Fr > 2.5 (asymmetric, suspended phases)

The vertical impulse must support body weight over one stride:

\[ \sum_{i=1}^{4} \int_0^{t_{c,i}} F_{z,i}(t) \, dt = MgT \]

Peak GRF for each limb during gallop (approximately sinusoidal contact):

\[ F_{\text{peak}} = \frac{\pi M g T}{2 \sum_i t_{c,i}} \]

At top speed, the duty factor (fraction of stride with a foot on ground) drops to ~0.2 per limb, meaning peak GRF can reach 3–4 times body weight per leg.

\[ E_{\text{elastic}} = \frac{1}{2} k \, \Delta L^2 = \frac{\sigma^2 V}{2E} \]

where \( \sigma \) is tendon stress, \( V \) is tendon volume, and \( E \approx 1.5 \) GPa is Young's modulus for tendon. The resilience (fraction of energy returned) of mammalian tendon is ~93%, making it an extremely efficient spring. For a 4 kg cat at full gallop, approximately 0.3 J is stored and returned per stride in each hind limb, representing about 30% of the total stride energy.

\[ f_{\text{stride}} \propto M^{-1/6} \]

A domestic cat at full gallop has \( f \approx 4 \) Hz (4 strides/sec), while a lion at top speed has \( f \approx 2.5 \) Hz. The predicted ratio:\( (190/4)^{1/6} = 2.4^{1/6} = 1.5 \), giving \( 4/1.5 = 2.7 \) Hz — close to the observed 2.5 Hz.

Force balance on P3:

\[ F_{\text{flexor}} \cdot r_{\text{tendon}} = F_{\text{ligament}} \cdot r_{\text{ligament}} + F_{\text{friction}} \cdot r_{\text{sheath}} \]

where \( r \) values are the moment arms of each force about the P2-P3 joint. At rest (\( F_{\text{flexor}} = 0 \)), the elastic ligament dominates and the claw is retracted.

Force balance on a cat climbing at angle \( \theta \) to horizontal:

\[ \text{Normal: } N = Mg\cos\theta \]

\[ \text{Tangential: } F_{\text{claw}} + \mu N \geq Mg\sin\theta \]

For the claw-as-hook model, the interlocking force depends on the claw tip radius \( r_c \)and the substrate roughness \( r_s \):

\[ F_{\text{claw}} = n_{\text{claws}} \cdot \sigma_y \cdot \pi r_c \cdot d_{\text{pen}} \]

where \( n_{\text{claws}} \) is the number of engaged claws, \( \sigma_y \) is the yield stress of the substrate, and \( d_{\text{pen}} \) is claw penetration depth.

\[ F_{\text{claw}} = 18 \times 5 \times 10^6 \times \pi \times 0.0002 \times 0.001 \approx 56 \text{ N} \]

Required force at vertical (\( \theta = 90° \)): \( Mg = 4 \times 9.81 = 39.2 \) N. Since \( F_{\text{claw}} > Mg \), the cat can climb vertically. Adding friction (\( \mu \approx 0.4 \)) provides additional margin.

\[ F_{\text{claw}} + \mu Mg\cos\theta = Mg\sin\theta \]

\[ \theta_{\max} = \arctan\left(\frac{F_{\text{claw}}/Mg + \mu}{1}\right) + \arctan(\mu) \]

More precisely, solving for \( \theta \):

\[ \sin\theta - \mu\cos\theta = \frac{F_{\text{claw}}}{Mg} \]

\[ \sqrt{1+\mu^2}\sin(\theta - \arctan\mu) = \frac{F_{\text{claw}}}{Mg} \]

\[ \theta_{\max} = \arcsin\left(\frac{F_{\text{claw}}}{Mg\sqrt{1+\mu^2}}\right) + \arctan(\mu) \]