```html Seastead Scale Model & Full Scale Analysis

🌊 Seastead Wing-Float/Leg: Scale Model & Full-Scale Analysis

Table of Contents

Part 1: Mold Geometry & Foam Volume

Mold Description

The mold approximates a wing/airfoil cross-section:

CROSS-SECTION VIEW (not to scale) Leading edge (half-pipe, ~4" OD) ___----___ / \ / \ ← inside span ~3.75" | | \ / \ Plywood / Plywood \ (16" each) / (16" each) \ / \ / \ / \ / \ / \ / \ / Trailing edge

Cross-Section Area Calculation

We need to compute the area of the wing cross-section, which consists of:

  1. A semicircular leading edge (interior of the half-pipe)
  2. Two flat plywood sides extending back to a trailing point

Leading edge semicircle:

PVC pipe OD = 4.0 inches PVC pipe wall thickness (standard Schedule 40) ≈ 0.237 inches PVC pipe ID = 4.0 - 2×0.237 = 3.526 inches However, measured inside span = 3.75 inches. This suggests the pipe is closer to a thin-wall pipe or the measurement is across the inside of the cut half. For a 4" OD pipe cut in half, the inside chord would be close to the ID. The measured 3.75" makes sense for a thin-wall (SDR/DWV) 4" PVC pipe: → Wall thickness ≈ 0.125" (thin wall) → ID ≈ 3.75" Inside radius of half-pipe: r = 3.75 / 2 = 1.875 inches

Semicircle area:

A_semi = ½ × π × r² A_semi = ½ × π × (1.875)² A_semi = ½ × π × 3.5156 A_semi = 5.522 in²

Triangular portion from plywood sides:

Each plywood piece is 16″ long and hinges from an edge of the half-pipe. The two pieces meet at a trailing point forming a triangle. The base of this triangle equals the inside span of the pipe (3.75″) and the height is 16″.

A_triangle = ½ × base × height A_triangle = ½ × 3.75 × 16 A_triangle = 30.0 in²
⚠️ Note on geometry: The 16" plywood pieces likely don't extend perfectly perpendicular to the span. They angle inward from the pipe edges to meet at a trailing point. The triangle base = 3.75" (pipe span) and height ≈ 16" (plywood length ≈ hypotenuse, so actual height ≈ √(16² − 1.875²) ≈ 15.89"). The difference is negligible, so we use 16" as the effective height.

Total cross-section area:

A_total = A_semi + A_triangle A_total = 5.522 + 30.0 A_total = 35.52 in²

Volume of the mold:

Length = 3.5 ft = 42 inches V = A_total × Length V = 35.52 in² × 42 in V = 1,491.9 in³

Convert to more useful units:

1 gallon = 231 in³ V = 1,491.9 / 231 = 6.45 gallons 1 US cup = 14.4375 in³ V = 1,491.9 / 14.4375 = 103.3 cups In cubic feet: V = 1,491.9 / 1728 = 0.8634 ft³ In liters: V = 1,491.9 × 0.016387 = 24.45 liters
📐 Mold Volume ≈ 1,492 in³ ≈ 0.863 ft³ ≈ 6.45 gallons ≈ 24.5 liters

2-Part 2 lb/ft³ Foam Requirements

Two-part polyurethane pour foam (2 lb density) is mixed at a 1:1 ratio by volume (Part A : Part B). The liquid expands to fill the mold. The expansion ratio depends on the target density:

Target foam density = 2.0 lb/ft³ Mixed liquid density (typical) ≈ 68–72 lb/ft³ (≈ 9 lb/gal) Approximate expansion ratio = ~35:1 Volume of mixed liquid needed: V_liquid = V_mold / expansion_ratio V_liquid = 1,492 in³ / 35 = 42.6 in³ In cups: 42.6 / 14.4375 = 2.95 cups of mixed liquid Each part (A and B) at 1:1 ratio: Part A = 1.48 cups ≈ 1.5 cups Part B = 1.48 cups ≈ 1.5 cups
⚠️ Practical Advice:
🧪 Mix approximately 1.5 cups Part A + 1.5 cups Part B (3 cups total mixed liquid)
For safety margin in a constrained mold: ~1.75 cups each (3.5 cups total)
Verify with your foam product's stated expansion ratio. At 25× expansion you'd need ~2.1 cups each.

Weight of one foam float

Volume = 0.863 ft³ Density = 2.0 lb/ft³ Weight = 0.863 × 2.0 = 1.73 lbs per float

Part 2: Model Buoyancy & Maximum Total Weight

If 3 wing/legs are each 50% submerged in seawater, we need to find the buoyancy force.

Seawater density = 64.0 lb/ft³ (approximately) Volume of one float = 0.863 ft³ Submerged volume (50%) = 0.863 × 0.50 = 0.4317 ft³ Buoyancy per float = ρ_sw × V_submerged Buoyancy per float = 64.0 × 0.4317 = 27.63 lbs Total buoyancy (3 floats) = 3 × 27.63 = 82.9 lbs

For the model to float at 50% submersion, the total weight must equal the total buoyancy:

Total model weight = Total buoyancy force Total model weight = 82.9 lbs Weight of 3 foam floats = 3 × 1.73 = 5.19 lbs Remaining for structure/payload = 82.9 − 5.19 = 77.7 lbs
⚖️ Total model weight for 50% submersion on 3 floats ≈ 82.9 lbs
(Foam floats weigh ~5.2 lbs total, leaving ~77.7 lbs for the platform structure, tensegrity elements, and payload)
⚠️ Note: With 2 lb/ft³ foam, these floats are very light relative to the displaced water. The tensegrity platform structure above will need to be substantial (or ballasted) to bring total weight to ~83 lbs for proper 50% draft simulation. This is actually a nice design feature — you have huge reserve buoyancy for the model.

Part 3: Froude Scaling — Full-Scale Dimensions (1:6 Scale)

With a scale factor λ = 6, all linear dimensions scale by 6×:

Dimension Model (1:6) Full Scale (×6)
Float length 3.5 ft (42 in) 21 ft (252 in)
Leading edge diameter (inside) 3.75 in 22.5 in (1.875 ft)
Chord (nose to tail) ~17.875 in (1.49 ft) 107.25 in (8.94 ft)
Max thickness (at leading edge) 3.75 in 22.5 in (1.875 ft)
Plywood/side panel length 16 in 96 in (8 ft)

Froude Scaling Relationships

Quantity Scaling Factor Value for λ=6
Length λ 6
Area λ² 36
Volume λ³ 216
Mass / Force λ³ (same fluid) 216
Velocity √λ 2.449
Time √λ 2.449
Power λ3.5 529.1
📏 Full-scale float: 21 ft long × ~9 ft chord × ~1.9 ft max thickness
Cross-section area: 35.52 in² × 36 = 1,279 in² = 8.88 ft²
Volume per float: 0.863 ft³ × 216 = 186.5 ft³ per float

Part 4: Full-Scale Displaced Seawater (50% Submersion)

Method 1: Direct Calculation

Full-scale volume per float = 186.5 ft³ Submerged volume (50%) = 93.24 ft³ Seawater density = 64.0 lb/ft³ Displaced mass per float = 64.0 × 93.24 = 5,967 lbs ≈ 5,967 lbs (2.98 US tons) For 3 floats: Total displaced mass = 3 × 5,967 = 17,902 lbs ≈ 17,900 lbs (8.95 US tons)

Method 2: Froude Scaling from Model

Model buoyancy per float = 27.63 lbs Full-scale buoyancy = 27.63 × λ³ = 27.63 × 216 = 5,968 lbs ✓ (consistent)
🌊 Displaced seawater per float (50% submerged) ≈ 5,967 lbs (2.98 tons)
🌊 Total for 3 floats ≈ 17,900 lbs (8.95 tons)
This means the full-scale seastead platform could weigh up to ~17,900 lbs (structure + payload) and float at 50% draft.

Part 5: Drag Forces at Full Scale (1, 2, 3 MPH)

Setup & Assumptions

The floats are oriented with the leading edge forward (low Cd direction), moving through seawater. The submerged cross-section (50% of each float) presents a certain frontal area and wetted surface to the flow.

Drag Components

For a submerged streamlined body at low speeds, we consider:

Geometry of Submerged Portion

Full-scale float: Length (span, vertical when installed) = 21 ft Submerged length = 10.5 ft (50%) Chord = 8.94 ft (nose to tail, direction of travel) Max thickness = 1.875 ft (at leading edge) The float moves horizontally, chord-wise through the water. Frontal area per float (submerged portion, viewed from flow direction): A_frontal = submerged_length × max_thickness A_frontal = 10.5 ft × 1.875 ft = 19.69 ft² Wetted surface area per float (submerged): Approximate perimeter of cross-section: Semicircle arc = π × r = π × 0.9375 = 2.945 ft Two side panels = 2 × 8 ft = 16 ft (full scale) Total perimeter ≈ 18.95 ft Wetted surface = perimeter × submerged_length S_wet = 18.95 × 10.5 = 199.0 ft² per float Thickness-to-chord ratio: t/c = 1.875 / 8.94 = 0.210 This is a relatively thick "airfoil" (~21% t/c), similar to a NACA 0021. For a well-faired shape at low Reynolds numbers, a reasonable total drag coefficient based on frontal area: C_d (frontal) ≈ 0.08 to 0.12 for a streamlined strut/foil shape (much less than a cylinder at ~1.0, much more than a thin foil) We'll use C_d = 0.10 (based on frontal area) as a reasonable estimate for this blunt-leading-edge shape. This accounts for both form drag and skin friction.

Drag Force Calculation

Drag equation: Fd = ½ × ρ × V² × Cd × A

Seawater density: ρ = 1.99 slugs/ft³ (= 64 lb/ft³ ÷ 32.174 ft/s²) Speed conversions: 1 MPH = 1.467 ft/s 2 MPH = 2.933 ft/s 3 MPH = 4.400 ft/s Frontal area per float (submerged) = 19.69 ft² C_d = 0.10 Number of floats = 3 Total frontal area = 3 × 19.69 = 59.06 ft²
Speed (MPH) Speed (ft/s) ½ρV² (psf) Drag per Float (lbs) Total Drag, 3 Floats (lbs)
1 1.467 2.14 4.2 12.6
2 2.933 8.56 16.9 50.5
3 4.400 19.27 37.9 113.8
Sample calculation for 1 MPH: q = ½ × 1.99 × (1.467)² = ½ × 1.99 × 2.152 = 2.141 lb/ft² F_per_float = 2.141 × 0.10 × 19.69 = 4.22 lbs F_total = 3 × 4.22 = 12.6 lbs
🚢 Total drag force (3 floats, oriented streamlined):
  @ 1 MPH: ~13 lbs
  @ 2 MPH: ~51 lbs
  @ 3 MPH: ~114 lbs
Drag scales with velocity squared. These are low speeds — the Froude number Fr = V/√(gL) ranges from 0.05 to 0.15, so wave-making resistance is minimal.
⚠️ Additional drag sources not included above:

A practical design factor of 1.5× to 2× on these bare-hull numbers is prudent.

Part 6: Motor & Propeller Power Requirements

Assumptions

On propeller efficiency: At these very low speeds (1–3 MPH) for a ~9-ton vessel, you're in a regime where propeller design matters enormously. A well-matched large-diameter, slow-turning prop could achieve 50–55% efficiency. A poorly matched small prop might only manage 30–35%. We'll show both a conservative and optimistic case.

Power Calculations

Effective (towing) power: Pe = Fdrag × V

Shaft power: Pshaft = Pe / ηprop

Electrical input power: Pelec = Pshaft / (ηmotor × ηdrive)

1 HP = 550 ft·lbs/s = 745.7 watts

Conservative Case (ηtotal = 35%)

Speed (MPH) Total Drag (lbs) Effective Power (watts) Electrical Input (watts) Electrical Input (HP)
1 12.6 25 72 0.10
2 50.5 201 573 0.77
3 113.8 680 1,935 2.60
Sample calculation for 2 MPH (conservative): P_effective = 50.5 lbs × 2.933 ft/s = 148.1 ft·lbs/s = 200.9 watts P_electrical = 200.9 / 0.352 = 570.7 watts

Optimistic Case (ηtotal = 48% — well-designed large prop system)

Speed (MPH) Total Drag (lbs) Effective Power (watts) Electrical Input (watts) Electrical Input (HP)
1 12.6 25 53 0.07
2 50.5 201 419 0.56
3 113.8 680 1,416 1.90
Electrical power for 3 full-scale wing-floats at various speeds:

Speed Conservative (watts) Optimistic (watts) Practical Estimate (watts)
1 MPH 72 W 53 W ~60–75 W
2 MPH 573 W 419 W ~420–575 W
3 MPH 1,935 W 1,416 W ~1,400–1,950 W
Power scales with V³ — doubling speed requires 8× the power!

Practical Motor Recommendations (Full Scale)

Target Speed Suggested Motor Size Battery Considerations (for 8 hours)
1 MPH (station-keeping) Single trolling motor, ~100W ~600 Wh = modest 12V battery
2 MPH (repositioning) 1–2 trolling motors or small pod drive, ~750W total ~4.6 kWh = small solar + battery bank
3 MPH (transit) 2–3 kW electric outboard or pod drive ~15.5 kWh = substantial battery bank
💡 Design Insight: At just 1 MPH, a single ~100W solar panel could potentially provide continuous station-keeping power during daylight hours. This is very favorable for a seastead that primarily needs to hold position with occasional repositioning. A 1 kW solar array could sustain ~2 MPH cruising during peak sun.

Summary of All Results

Model Scale (1:6)
Float volume1,492 in³ (0.86 ft³, 6.45 gal)
Foam per float (2 lb/ft³)~1.73 lbs
Foam mix: Part A + Part B~1.5 + 1.5 cups (add 15-25% margin)
Buoyancy per float (50% submerged)27.6 lbs
Total model weight (3 floats, 50% draft)82.9 lbs
Full Scale (×6)
Float dimensions21 ft long × 8.9 ft chord × 1.9 ft thick
Float volume186.5 ft³ per float
Displaced seawater per float (50%)5,967 lbs
Total displacement (3 floats)17,900 lbs (8.95 tons)
Drag @ 1 MPH (3 floats)~13 lbs → 60–75 W electrical
Drag @ 2 MPH (3 floats)~51 lbs → 420–575 W electrical
Drag @ 3 MPH (3 floats)~114 lbs → 1,400–1,950 W electrical

Analysis prepared for seastead design evaluation.
All calculations use standard naval architecture methods and reasonable engineering assumptions.
Values should be validated with physical model testing and CFD analysis for final design.

Key References: Principles of Naval Architecture (SNAME), Froude scaling laws, ITTC recommended procedures for model testing.

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