Seastead Active Stabilizer Analysis Report

1. Additional Buoyancy Force per Foot of Immersion

Given Leg Geometry:
• NACA 0035 foil, 8.5 ft chord (effective), 21.5 ft span
• 35% thickness ratio → max thickness = 0.35 × 8.5 = 2.975 ft
• Trailing edge cut 0.5 ft short → effective span for buoyancy ≈ 21.0 ft
• Foil area (planform) ≈ 8.5 ft × 21.5 ft = 182.75 ft²
• Average waterplane width varies with immersion depth

Waterplane Area Calculation:
For a NACA 0035 at mid-chord (max thickness), half-breadth y = ±1.4875 ft
At 50% immersion (design waterline at mid-chord), waterplane width ≈ 2.975 ft
Waterplane area per leg ≈ 2.975 ft × 21.5 ft = 63.96 ft²

Buoyancy per inch of immersion:
ΔF = ρ × g × A_wp × Δh
ρ (seawater) = 64 lbf/ft³ | A_wp = 63.96 ft² | Δh = 1/12 ft
ΔF = 64 × 63.96 × (1/12) = 341 lbf per inch per leg
ΔF = 64 × 63.96 × 1 = 4,093 lbf per foot per leg

Result

341 lbf/inch 4,093 lbf/foot per leg

Three legs combined: 1,023 lbf/inch or 12,280 lbf/foot

Context: Total displacement = 27,500 lbf → 1 ft extra immersion = 44.6% of total buoyancy

Important: Waterplane area changes with immersion depth (foil shape). At deeper immersion, width increases toward max chord; at shallower, it decreases nonlinearly. The 341 lbf/in is valid only near the design waterline (50% chord). For large motions, use nonlinear hydrostatics.

2. Wave Height Reduction Analysis

2.1 The "4 ft → 3 ft" Claim

Physics Check:
Wave height reduction requires force to counteract wave excitation.
Wave force on leg (Morison equation, inertia-dominated for large body):
F_wave ≈ ρ × C_m × V × a_water
C_m ≈ 2.0 (foil), V = displaced volume per leg = 27,500/3/64 = 143 ft³
For 4 ft wave, 8 sec period: a_max = ω² × A = (2π/8)² × 2 = 2.47 ft/s²
F_wave ≈ 64 × 2.0 × 143 × 2.47 = 45,300 lbf peak per leg

Stabilizer Force Required:
To reduce 4 ft wave to 3 ft (25% reduction), need to counteract ~25% of wave force:
F_stab_needed ≈ 0.25 × 45,300 = 11,300 lbf per leg

Stabilizer lift: L = ½ × ρ × V² × A_stab × C_L
At 6 knots (10.1 ft/s), ρ = 1.99 slugs/ft³:
For A_stab = 4 ft² (2 ft chord × 2 ft span), C_L = 1.0:
L = 0.5 × 1.99 × 10.1² × 4 × 1.0 = 405 lbf — INSUFFICIENT

Conclusion: A small "airplane-style" stabilizer on each leg cannot reduce a 4 ft wave to 3 ft at 6 knots. The forces are 28× too small. Wave height reduction of this magnitude requires either:
• Much larger stabilizer surfaces (≈ 100+ ft² per leg)
• Active ballast transfer between legs (SWATH-style)
• Much higher speeds (>20 knots) for foil lift to scale with V²
• Resonant-frequency cancellation (only works at specific wave periods)

2.2 What IS Achievable: Resonant Motion Suppression

Realistic Value Proposition: The stabilizer can effectively suppress resonant heave/pitch/roll at the natural frequency of the platform (typically 8-15 sec for this displacement). This prevents the "resonant amplification" where 1 ft waves produce 3-5 ft platform motions. This is the primary benefit — not reducing individual wave crests.

Metric	Value	Notes
Heave natural period (estimate)	9-12 sec	Depends on waterplane area & mass
Pitch natural period	7-10 sec	Coupled with heave
Resonant amplification factor (no damping)	3-5×	Can turn 1 ft waves into 3-5 ft motions
With active stabilizer (critical damping)	1.0-1.5×	Eliminates resonance peak
Typical motion reduction in sea state 3-4	40-60%	RMS motion, not peak wave height

3. Stabilizer Performance at 4-8 Knots

3.1 Assumptions for "Airplane" Stabilizer per Leg

Parameter	Value
Stabilizer planform area (per leg)	4.0 ft² (2 ft span × 2 ft chord)
Aspect ratio	1.0 (low, thick foil for strength)
Max lift coefficient (C_L_max)	1.2 (with flap/elevator)
Drag coefficient at zero lift (C_D0)	0.015 (NACA 0015-ish section)
Actuator force (peak)	2,000 lbf (electric linear, 6 in stroke)
Actuator speed	12 in/sec
Control bandwidth	2-3 Hz

3.2 Performance Estimates

Speed	Dynamic Pressure (psf)	Max Lift Force (lbf)	Drag at Zero Lift (lbf)	Drag at Max Lift (lbf)	Power at Zero Lift (HP)	Power at Max Lift (HP)	Heave Authority (inches @ 1 Hz)
4 kn (6.75 ft/s)	45.4	218	2.7	18.5	0.03	0.20	±0.8
5 kn (8.44 ft/s)	70.9	340	4.3	28.8	0.06	0.39	±1.3
6 kn (10.13 ft/s)	102.1	490	6.1	41.4	0.10	0.67	±1.8
7 kn (11.81 ft/s)	138.9	667	8.3	56.5	0.16	1.06	±2.5
8 kn (13.50 ft/s)	181.4	871	10.9	74.1	0.24	1.59	±3.2

Heave Authority Calculation:
Stabilizer force F_stab opposes heave acceleration: F = m × a
Added mass in heave ≈ 1.5 × displaced mass = 1.5 × (27,500/3/32.2) = 427 slugs/leg
a_max = F_stab / m_added → displacement amplitude = a_max / ω²
At 6 kn, F_stab = 490 lbf, ω = 2π rad/s (1 Hz):
a = 490/427 = 1.15 ft/s² → amp = 1.15/(2π)² = 0.029 ft = 0.35 inches
But at resonant frequency (ω_n ≈ 0.6 rad/s, 10 sec period):
amp = 1.15/0.6² = 3.2 ft — this is where authority matters!
Table shows inches of motion suppression at resonance, not single-wave crest reduction.

Key Limitation: At 4-8 knots, stabilizer forces (200-900 lbf) are small vs. wave excitation forces (10,000-50,000 lbf). The stabilizer cannot significantly reduce individual wave-induced motions. Its value is resonance suppression and steady trim control.

Component	Spec	Unit Cost (USD)	Qty	Subtotal
Marine aluminum wing (6061-T6, CNC + weld)	4 ft², 0.5" skin, internal ribs	$450	1	$450
Aluminum actuator mount & fairing	CNC machined 6061	$280	1	$280
Electric linear actuator	2,000 lbf