Wing/Leg Mold, Buoyancy, and Full-Scale Estimates

1) Foam volume needed for the mold

Cross-sectional area

Let:

• Half-circle radius: r = 3.75 / 2 = 1.875 in
• Flat side length: s = 16 in
• Chord from top pipe edge to bottom pipe edge: c = 3.75 in

The two plywood sides form an isosceles triangle behind the half-circle. Triangle height:

h = sqrt(s² - (c/2)²) = sqrt(16² - 1.875²) ≈ 15.8898 in

Triangle area:

A_triangle = (1/2) × c × h = 0.5 × 3.75 × 15.8898 ≈ 29.793 in²

Half-circle area:

A_halfcircle = (1/2) π r² = 0.5 × π × 1.875² ≈ 5.522 in²

Total cross-sectional area:

A_total ≈ 29.793 + 5.522 = 35.315 in²

Volume of one molded float/leg

Length = 42 in

V = A × L = 35.315 × 42 ≈ 1483.2 in³

Convert to cubic feet:

1483.2 / 1728 ≈ 0.8588 ft³

Convert to gallons:

1483.2 / 231 ≈ 6.42 gallons

Convert to liters:

1483.2 × 0.016387 ≈ 24.3 liters

Foam mixing amount for 2-part 2 lb foam

Most 2-part expanding polyurethane foams are mixed 1:1 by volume, and the quoted density (2 lb/ft³) is the final cured foam density, not the mixed liquid density. So the required liquid mix depends on the foam’s expansion ratio / yield, which varies by product.

Using the final foam volume above:

Required finished foam volume = 0.8588 ft³

If your product yields exactly its stated final volume, then you need enough mixed liquid to produce 0.8588 ft³. A common rule of thumb for 2 lb foam is roughly 8 to 10 times expansion by liquid volume. That would mean total mixed liquid around:

• At 8× expansion: 6.42 gal / 8 ≈ 0.803 gal total mixed liquid
• At 10× expansion: 6.42 gal / 10 ≈ 0.642 gal total mixed liquid

In cups (16 cups per gallon), total mixed liquid is roughly:

• 0.803 × 16 ≈ 12.85 cups total
• 0.642 × 16 ≈ 10.27 cups total

Since it is 1:1 by volume, each part would be:

• Part A: about 5.1 to 6.4 cups
• Part B: about 5.1 to 6.4 cups

A practical target would be:

Mix about 6 cups of Part A + 6 cups of Part B per leg, then adjust based on actual product yield and whether you want slight overfill.

Summary for one mold

2) Model weight if 3 wing/legs are each 50% in seawater

For flotation, the total model weight should equal the buoyant force from the displaced seawater. If each leg is half submerged, then each leg displaces:

0.5 × 0.8588 = 0.4294 ft³ of seawater

For 3 legs:

V_displaced,total = 3 × 0.4294 = 1.2882 ft³

Take seawater weight density as:

ρ_seawater ≈ 64 lb/ft³

So supported weight is:

W = 1.2882 × 64 ≈ 82.44 lb

Answer: If the model has 3 such legs and each is 50% submerged in seawater, the total model weight should be about 82.4 lb.

Froude Scaling

Scale = 1:6 model, so full scale linear dimensions are 6× the model.

1) Full-scale dimensions

Cross-sectional area scales by 6² = 36, and volume by 6³ = 216.

Full-scale volume of one wing/leg:

V_full = 0.8588 × 216 ≈ 185.49 ft³

2) Full-scale displaced seawater if half submerged

One full-scale leg, half submerged:

V_disp,1 = 185.49 / 2 ≈ 92.74 ft³

Weight of displaced seawater:

W_1 = 92.74 × 64 ≈ 5935 lb

For 3 legs:

W_3 = 3 × 5935 ≈ 17,805 lb

3) Force to move 3 full-scale legs at 1, 2, or 3 MPH

For one full-scale leg:

• Length = 21 ft
• Maximum thickness (pipe span) = 1.875 ft
• Half submerged depth ≈ 0.9375 ft

So projected frontal area per leg is approximately:

A_1 ≈ 0.9375 × 21 = 19.6875 ft² = 1.829 m²

For 3 legs:

A_total ≈ 59.06 ft² = 5.488 m²

Speeds:

• 1 mph = 0.447 m/s
• 2 mph = 0.894 m/s
• 3 mph = 1.341 m/s

Using F = 0.5 ρ V² Cd A for all 3 legs combined:

4) Approximate electrical power needed at 1, 2, or 3 MPH

Mechanical towing power is:

P_mech = F × V

Electrical input power depends on propeller, drivetrain, and motor efficiency. A reasonable overall efficiency for a small marine electric propulsion system is:

η_overall ≈ 0.5 to 0.7

I will show results using 60% overall efficiency. So:

P_elec = P_mech / 0.60

So, under these streamlined assumptions, a rough estimate for electrical input power for the 3 full-scale legs is:

• 1 MPH: about 40 W
• 2 MPH: about 340 W
• 3 MPH: about 1.1 kW

Final Answer Summary