# Seastead Scaling Analysis Here are the calculations for your 1/6th scale seastead model and its full-scale counterpart, using Froude scaling rules. ## Model Dimensions (1/6 Scale) ```html Seastead Scaling Analysis

Seastead Design: Scale Model to Full-Scale Conversion

Froude Scaling Principles

Froude scaling is used for hydrodynamic systems where wave effects are important. Key scaling relationships:

Since mass scales with λ³, the full-scale structure will have 216 times the mass of the model.

Model (1/6 Scale) Specifications

Main Structure

  • Triangular platform: 10 feet per side
  • Platform area above water
  • 3 columns/legs at each corner

Columns/Legs

  • Diameter: 5 inches (0.4167 feet)
  • Length: 6 feet
  • Submerged portion: 60% of length = 3.6 feet
  • Orientation: Angled 45° outward and downward from triangle corners
  • Support: 2 ropes from bottom of each leg to adjacent corners

Water Displacement Calculation (Model)

Volume displaced per leg = π × (radius)² × submerged length
Radius = 2.5 inches = 0.2083 feet
Submerged length = 3.6 feet
Volume = π × (0.2083 ft)² × 3.6 ft ≈ 0.490 ft³ per leg
Total volume displaced (3 legs) = 3 × 0.490 ft³ ≈ 1.470 ft³
Mass of water displaced = Volume × Density of water
Density of seawater ≈ 64 lb/ft³
Mass displaced ≈ 1.470 ft³ × 64 lb/ft³ ≈ 94.1 lbs

Model water displacement: ≈ 94.1 pounds

This represents the buoyant force supporting the structure's weight.

Parameter Model Value Units
Platform side length 10 feet
Leg diameter 5 inches
Leg length 6 feet
Submerged length 3.6 feet
Leg angle 45° degrees
Water displaced 94.1 lbs

Full-Scale Specifications

Length Scaling (λ = 6)

  • Platform side length: 10 ft × 6 = 60 ft
  • Leg diameter: 5 in × 6 = 30 in = 2.5 ft
  • Leg length: 6 ft × 6 = 36 ft
  • Submerged length: 3.6 ft × 6 = 21.6 ft

Water Displacement Calculation (Full-Scale)

Volume displaced per leg = π × (radius)² × submerged length
Radius = 15 inches = 1.25 feet
Submerged length = 21.6 feet
Volume = π × (1.25 ft)² × 21.6 ft ≈ 105.8 ft³ per leg
Total volume displaced (3 legs) = 3 × 105.8 ft³ ≈ 317.4 ft³
Mass of water displaced ≈ 317.4 ft³ × 64 lb/ft³ ≈ 20,300 lbs

Full-scale water displacement: ≈ 20,300 pounds

Mass/Weight Scaling (λ³ = 216)

Model displaced mass ≈ 94.1 lbs
Full-scale displaced mass ≈ 94.1 lbs × 216 ≈ 20,300 lbs
This buoyant force must equal the weight of the structure
Therefore, target weights:
Model target weight ≈ 94.1 lbs
Full-scale target weight ≈ 20,300 lbs
Parameter Full-Scale Value Units Scaling Factor
Platform side length 60 feet 6 (λ)
Leg diameter 2.5 feet 6 (λ)
Leg length 36 feet 6 (λ)
Submerged length 21.6 feet 6 (λ)
Leg angle 45° degrees 1 (no scaling)
Water displaced 20,300 lbs 216 (λ³)
Target structure weight 20,300 lbs 216 (λ³)

Summary

Key Results

  • Model (1/6 scale): Target weight ≈ 94 lbs
  • Full-scale: Target weight ≈ 20,300 lbs
  • Platform: 10 ft triangle (model) → 60 ft triangle (full)
  • Legs: 5" diameter, 6 ft long (model) → 2.5 ft diameter, 36 ft long (full)
  • Leg angle: Maintained at 45° in both scales

Design Implications: The full-scale seastead will be a substantial structure with a triangular platform 60 feet on each side, supported by three 2.5-foot diameter legs extending 36 feet downward at a 45° angle. The total weight should be approximately 20,300 pounds to match the buoyant force from displacement.

Notes:

1. Calculations assume seawater density of 64 lb/ft³. Fresh water would be slightly less (62.4 lb/ft³).

2. Froude scaling maintains hydrodynamic similarity between model and full-scale.

3. Rope dimensions should also scale by λ = 6 (diameter and length).

4. Model testing at 1/6 scale will have time scales reduced by √6 ≈ 2.45 (events happen faster in model).

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