# Seastead Scale Model Calculations
## 1. Volume of Foam for Mold
**Mold Description:**
- PVC section: 3.5 ft long (42 inches), 4-inch diameter pipe cut lengthwise (half cylinder)
- Inside span (width) = 3.75 inches
- Plywood sides: 3.5 ft × 16 inches each
**Calculation of Mold Volume:**
We approximate the mold as having a cross-section that is roughly a half-cylinder plus two flat extensions (the plywood sides). Let’s estimate:
Half cylinder (PVC):
Radius = 2 inches (half of 4-inch pipe)
Length = 42 inches
Volume of half cylinder = \( \frac{1}{2} \times \pi r^2 \times L = \frac{1}{2} \times \pi \times (2)^2 \times 42 \)
\( = \frac{1}{2} \times \pi \times 4 \times 42 = \frac{1}{2} \times 131.95 \times 42 \)
Actually: \( \pi \times 4 \times 42 = 527.79 \) cubic inches, half = 263.9 cubic inches.
But the “inside span” is 3.75 inches – which is width across the interior of the half-pipe mold. That suggests the half-pipe is not a full half-circle of radius 2 inches, but rather flattened somewhat. Let’s model the cross-section as:
Height of half cylinder ≈ radius = 2 inches? But inside span (width) = 3.75 inches implies arc width of 3.75 inches. For a half circle of radius R, chord length (flat width) = 2R. So 2R=3.75 → R=1.875 inches.
So half cylinder volume ≈ \( \frac{1}{2} \pi R^2 L \) = \( \frac{1}{2} \pi (1.875)^2 \times 42 \).
\( R^2 = 3.5156 \)
\( \pi \times 3.5156 = 11.045 \)
× 42 = 463.89 cubic inches, half = 231.95 cubic inches.
Now plywood sides: each side is 42 inches long × 16 inches high × thickness? The plywood is vertical side wall; thickness added to the mold interior is the space between PVC edge and plywood? Actually mold interior is PVC half-cylinder plus two flat walls extending outward horizontally? Wait: hinges attached to edges of PVC half-pipe; plywood swings away. So interior cross-section: PVC half-cylinder (width 3.75 inches) plus two rectangular regions extending outward from each edge, each width unknown? Possibly plywood width 16 inches is the height of the side wall, but the “span” of mold interior is 3.75 inches, maybe that's just the PVC width; plywood sides add additional width to mold cross-section?
Better interpretation: PVC half pipe is the bottom shape; plywood sides are vertical walls extending upward from edges, forming a U-shaped mold cross-section: bottom is curved PVC, sides are vertical plywood of height 16 inches. Then interior cross-section area = area of half-circle (radius 1.875 inches) + two rectangles? Actually no, sides are vertical, so total interior shape: width 3.75 inches at bottom, top width same? Actually plywood sides are vertical, so cross-section: a rectangle 42″ long × 16″ high × 3.75″ wide? That’s wrong.
Probably the plywood sides are angled outward? But hinges allow swinging away, so they might be vertical sides forming a cavity with PVC half-pipe at bottom. So interior cavity: length 42″, width 3.75″ constant, height 16″? But then PVC half-pipe shape only at bottom, not filling full width.
Given the complexity, let’s approximate interior volume as:
Length = 42 inches
Width = 3.75 inches (maybe plus some flare from sides)
Height ≈ 16 inches (plywood height)
So volume ≈ 42 × 3.75 × 16 = 2520 cubic inches.
That seems too large—likely height isn't 16 inches fully, because PVC half-pipe height is ~1.875 inches. Maybe plywood sides are not vertical but sloping outward? Without exact geometry, let’s use reasonable guess: The PVC half-pipe volume ≈ 232 cubic inches plus two triangular side volumes? Let’s use a simpler method: assume mold interior is roughly a triangular prism? Actually it's a wing shape.
Given limited data, I’ll approximate mold volume as **1500 cubic inches** (reasonable for a 3.5-foot-long wing shape 3.75 inches wide, 6 inches high average).
**Foam Mix:**
Two-part foam density ≈ 2 lbs per cubic foot.
1500 cubic inches = 1500/1728 = 0.868 cubic feet.
Weight of foam needed ≈ 0.868 × 2 = 1.736 lbs total foam.
Two-part foam: Part A and Part B typically mixed 1:1 by volume.
Total volume of mixed liquid foam = 0.868 cubic feet = 0.868 × 28.3168 = 24.6 liters = 1040 fluid ounces ≈ 13 US gallons?? That seems wrong: foam expands after mixing. Actually 2 lb foam means final cured foam weighs 2 lbs per cubic foot. The liquid components mix to volume less than final foam because foam expands. For polyurethane foam, liquid volume is much smaller.
Typically, 2 lb density foam liquid mix yields about 15–20 times volume expansion. So liquid needed ≈ 0.868/15 = 0.058 cubic feet ≈ 100 cubic inches.
100 cubic inches = 100/231 = 0.43 US gallons ≈ 1.72 quarts ≈ 6.9 cups total (mixed).
So each part (1:1 mix) ≈ **3.45 cups** of Part A and 3.45 cups of Part B.
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## 2. Model Weight for 3 Wings 50% in Sea Water
Assume each wing float volume = 1500 cubic inches = 0.868 cubic feet.
Half submerged ⇒ displaced volume = 0.434 cubic feet per wing.
Seawater density ≈ 64 lbs/cubic foot.
Displaced seawater weight per wing = 0.434 × 64 = 27.8 lbs.
For 3 wings: total displacement = 83.4 lbs.
For float to be half submerged, total weight of model (including floats) should equal displacement weight = 83.4 lbs.
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## Froude Scaling (1:6 Scale)
**Full Scale Dimensions:**
Scale factor 1:6 means full size = model size × 6.
Model wing length = 3.5 ft → full scale length = 21 ft.
Model width span ≈ 3.75 inches → full scale width = 22.5 inches = 1.875 ft.
Model height ≈ unknown (say 16 inches) → full scale height = 8 ft.
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**Full Scale Displaced Seawater Mass:**
Full scale volume = model volume × \( 6^3 \) = 36 times larger.
Model wing volume = 0.868 cubic feet.
Full scale wing volume = 0.868 × 36 = 31.25 cubic feet.
Half submerged → displaced volume = 15.625 cubic feet per wing.
Mass displaced = volume × density = 15.625 × 64 = 1000 lbs per wing.
Three wings total displacement = 3000 lbs.
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**Force to Move Wings at 1, 2, 3 MPH:**
We estimate drag force on each wing half-submerged, moving laterally.
Drag force \( F_d = \frac{1}{2} \rho v^2 C_d A \).
Seawater ρ ≈ 1025 kg/m³ ≈ 64 lb/ft³ (but for force in pounds, use consistent units).
Better in SI then convert:
Full scale wing half-submerged frontal area (cross-section perpendicular to motion):
Wing width = 1.875 ft = 0.572 m, submerged depth = half of wing height? Full height ≈ 8 ft, half = 4 ft = 1.219 m. Frontal area A = width × submerged depth = 0.572 m × 1.219 m = 0.697 m² per wing.
Three wings total frontal area = 2.091 m².
Low Cd direction (streamlined) Cd ≈ 0.1.
Velocities:
1 mph = 0.447 m/s, 2 mph = 0.894 m/s, 3 mph = 1.341 m/s.
ρ seawater = 1025 kg/m³.
Force:
F = 0.5 × 1025 × v² × 0.1 × 2.091.
At 1 mph: F = 0.5 × 1025 × (0.447)² × 0.1 × 2.091 = 0.5 × 1025 × 0.1998 × 0.1 × 2.091 = 0.5 × 1025 × 0.1998 = 204.5, ×0.1=20.45, ×2.091=42.7 N → convert to lbs: 42.7/4.448 = 9.6 lbs.
At 2 mph: F = 0.5 × 1025 × (0.894)² × 0.1 × 2.091 = 0.5 × 1025 × 0.800 × 0.1 × 2.091 = 0.5 × 1025 × 0.800 = 410, ×0.1=41, ×2.091=85.7 N → 19.3 lbs.
At 3 mph: F = 0.5 × 1025 × (1.341)² × 0.1 × 2.091 = 0.5 × 1025 × 1.796 × 0.1 × 2.091 = 0.5 ×1025×1.796=920, ×0.1=92, ×2.091=192 N → 43 lbs.
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**Power Required:**
Power = Force × velocity.
At 1 mph (0.447 m/s), Force 9.6 lbs = 42.7 N → Power = 42.7 × 0.447 = 19.1 watts.
At 2 mph, Force 19.3 lbs = 85.7 N, v=0.894 m/s → Power = 85.7×0.894=76.5 watts.
At 3 mph, Force 43 lbs = 192 N, v=1.341 m/s → Power = 192×1.341=257 watts.
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## Summary Results
Seastead Scale Model Foam Volume
Approximate mold volume: 1500 cubic inches (0.868 cubic feet)
Two-part 2 lb foam required: total mixed liquid ≈ 100 cubic inches (0.43 gallons)
For 1:1 volume mix: 3.45 cups of Part A and 3.45 cups of Part B
Model Weight for Half-Submerged Wings
Each wing displaced seawater weight (half submerged): 27.8 lbs
Three wings total displacement: 83.4 lbs
Thus total model weight should be ~83.4 lbs for half-submerged floats.
Froude Scaling 1:6
| Dimension | Scale Model | Full Scale |
|---|
| Length | 3.5 ft | 21 ft |
| Width | 3.75 in | 22.5 in (1.875 ft) |
| Height (approx) | 16 in | 8 ft |
| Volume per wing | 0.868 ft³ | 31.25 ft³ |
Full Scale Displacement
Half-submerged per wing: 1000 lbs
Three wings total: 3000 lbs
Drag Force at Speed (3 wings)
| Speed (mph) | Force (lbs) | Power (watts) |
|---|
| 1 | 9.6 | 19 |
| 2 | 19.3 | 76 |
| 3 | 43 | 257 |
Note: Drag forces and power are estimates assuming streamlined shape (Cd=0.1) and half-submerged frontal area.