Seastead Scale Model Calculations

Below are the calculations for your tensegrity scale model and the projected Froude-scaled figures for a full-size seastead implementation.

Part 1: The Scale Model

1) Mold Volume and Foam Mixture

Geometry Breakdown:

Length: 3.5 ft (42 inches).
Leading Edge Area: Half-circle with a 3.75" span (radius = 1.875"). Area = 0.5 × π × 1.875² ≈ 5.52 sq inches.
Trailing Edge Area: The 16" plywood pieces form a triangle. Base = 3.75", Sides = 16". Using the Pythagorean theorem, the height of this triangle is 15.89". Area = 0.5 × 3.75 × 15.89 ≈ 29.79 sq inches.
Total Cross Section: 5.52 + 29.79 = 35.31 sq inches.

Volume:

35.31 sq inches × 42 inches = 1,483 cubic inches (or 0.858 cubic feet).

Foam Mixture (2 lbs Density):

A 2 lb density foam yields 2 lbs of solid foam per cubic foot. You need 0.858 cu ft × 2 lbs = 1.716 lbs equivalent of liquid foam. Unmixed 2-part polyurethane liquid has a density slightly heavier than water (about 1.1 specific gravity).

Converting the required weight to liquid cups calculates out to roughly 3 US cups of pure liquid. However, accounting for expansion waste, liner crimping, and ensuring the mold packs out fully, you should mix slightly more.

Recommendation: Mix 1.75 to 2 cups of Part A and 1.75 to 2 cups of Part B (for a total mix of 3.5 to 4 cups). This ensures complete mold filling securely into the top hinges.

2) Total Weight of the Model in Equilibrium

If the model sinks exactly 50% into seawater, the total weight of the model must equal the weight of the sea water it displaces.

Total volume of 1 wing: 0.858 cu ft.
Submerged volume of 1 wing: 0.429 cu ft.
Submerged volume of 3 wings: 0.429 × 3 = 1.287 cu ft.

Seawater weighs approximately 64 lbs per cubic foot.
1.287 cu ft × 64 lbs = 82.4 lbs.

Answer: The total weight of your completed scale model (including the living area, foam legs, electronics, and ballast) must be 82.4 lbs to float halfway submerged.

Part 2: Full Scale Estimations (Using Froude 1:6 Scale)

1) Full Scale Dimensions

Using a scale factor of λ = 6, dimensions increase directly by a multiplier of 6.

Length/Draft: 3.5 ft × 6 = 21 feet long.
Maximum Thickness (Span): 3.75 inches × 6 = 22.5 inches (1.875 ft).
Chord Length (Front to back wing width): (1.875" + 15.89") × 6 = 106.6 inches = 8.88 feet.

2) Displaced Mass at Full Scale

By Froude scaling, volume and displacement weight scale by the cube of the scale factor (λ³).
6³ = 216.

One leg (50% submerged): 0.429 cu ft (model) × 216 = 92.66 cu ft.
92.66 cu ft × 64 lbs/cu ft = 5,930 lbs of displacement.
Total for 3 legs: 5,930 lbs × 3 = 17,790 lbs (Approx. 8.9 US tons).

Sanity check: Model weight (82.37 lbs) × 216 = 17,791 lbs.

3) Drag Force to Move at 1, 2, and 3 MPH

Assumptions:

Legs are vertical (Draft = 10.5 ft in the water).
Moving in the lowest drag direction (Wing leading edge forward, parallel).
Frontal Area of submerged sections: 1.875' (thickness) × 10.5' (draft) × 3 (legs) = 59.0 sq ft.
Drag Coefficient (C_d): Assumed at 0.15. While a perfect wing has a lower C_d, hinges, joints, surface imperfections, and hull skin-friction equivalent usually push working marine struts into the 0.12 - 0.18 range.
Formula: Drag Force = 0.5 × ρ × V² × C_d × Area

Speed (MPH)	Speed (ft / sec)	Estimated Total Drag Force (lbs)
1 MPH	1.47	19 lbs
2 MPH	2.93	76 lbs
3 MPH	4.40	170 lbs

4) Electrical Power (Watts) Required