```html Seastead Leg Side-Load Estimate

Side-Load Estimate for One Seastead Leg

Below is a rough engineering estimate for how much sideways distributed force one of your marine-aluminum foil legs might withstand before structural failure, based on the dimensions you gave. This is not a final structural design and should not be used for construction without review by a licensed naval architect / structural engineer.

Important: The real failure risk is probably not simple shell yielding of the 1/2-inch aluminum skin. More likely critical issues are:

1. Geometry used

You asked: if waves from the side push on the leg, how much evenly distributed side force along the leg could it take before it breaks?

2. Simplified structural model

I modeled one leg as a vertical cantilever beam, fixed at the top where it joins the main triangle structure, with a sideways distributed load over the submerged half.

To keep the estimate simple, I assumed:

Approximate section properties

Treating the foil very approximately as a thin rectangular closed section:

Approximate second moment of area for bending sideways: 

I ≈ 2.1 ft4

Section modulus: 

S = I / c ≈ 2.1 / 5 = 0.42 ft3

Convert to cubic inches: 

S ≈ 0.42 × 1728 ≈ 726 in3

Yield bending moment: 

Myield ≈ σy S ≈ 35,000 psi × 726 in3 ≈ 25.4 × 106 in-lb
Myield ≈ 2.12 × 106 ft-lb

3. Distributed load that would reach yield

For a cantilever with a uniform load w over submerged length a = 9.5 ft, the top moment is: 

M = w a² / 2

So: 

w = 2M / a² = 2 × 2.12×106 / 9.5² ≈ 46,900 lb/ft

Total sideways force over the submerged portion: 

F = w a ≈ 46,900 × 9.5 ≈ 446,000 lb
Very rough answer:
One leg, if idealized as a 1/2-inch-thick closed aluminum foil shell and loaded evenly along the submerged 9.5 ft, reaches first-yield bending at about:

4. More realistic allowable load

For actual marine structures, you do not want to operate near yield. Also, this simple calculation ignores stress concentrations, weld weakness, buckling, fatigue, and dynamic amplification. A crude safety factor of 3 to 5 would be more reasonable for concept screening.

Basis Distributed load Total side force on one leg
First yield (very rough) ~46,900 lb/ft ~446,000 lb
Yield / 3 ~15,600 lb/ft ~149,000 lb
Yield / 5 ~9,400 lb/ft ~89,000 lb
Practical concept-level range: If there is no substantial internal framing inside the foil, a more believable allowable design side load may be on the order of only 90,000 to 150,000 lb total per leg, and possibly less if panel buckling or weld details govern.

5. What wave height might create that much force?

This part is much more uncertain than the beam calculation. Wave force depends on:

For a rough side-force estimate on one submerged foil, we can use drag: 

F = 0.5 ρ Cd A V²

Take:

Then: 

F ≈ 4520 V² Newtons

or in pounds-force: 

F ≈ 1016 V² lb, with V in m/s

Velocity needed to create certain forces

Total side force on one leg Required crossflow velocity Velocity in knots
89,000 lb ~9.4 m/s ~18 kt
149,000 lb ~12.1 m/s ~24 kt
446,000 lb ~20.9 m/s ~41 kt

These are very high water-particle or relative crossflow speeds. Ordinary non-breaking swell usually does not create such huge broadside forces on a body this size unless there is strong dynamic interaction.

Relating wave height to water velocity

A rough deep-water estimate for maximum horizontal particle velocity near the surface is: 

umax ≈ πH / T

where:

Solving for wave height: 

H ≈ uT / π

Using the velocities above:

Force target Velocity needed Wave height if T = 8 s Wave height if T = 10 s Wave height if T = 12 s
89,000 lb 9.4 m/s ~24 m (79 ft) ~30 m (98 ft) ~36 m (118 ft)
149,000 lb 12.1 m/s ~31 m (101 ft) ~39 m (126 ft) ~46 m (151 ft)
446,000 lb 20.9 m/s ~53 m (175 ft) ~66 m (218 ft) ~80 m (262 ft)
Interpretation: If we use simple drag physics, the wave heights needed to generate those loads as steady side force are unrealistically large. That suggests the leg probably does not fail first from ordinary broadside swell drag alone.

6. What probably matters more than steady drag

For your concept, the dangerous loads are more likely from:

  1. Roll-induced bending: the whole platform tilts, and the buoyancy/hydrodynamic forces shift hard onto one or two legs.
  2. Breaking-wave impact: short-duration slamming pressures can be much higher than steady drag.
  3. Connection loads: even if the leg shell is strong enough, the joint into the deck truss may fail first.
  4. Local shell buckling: a 1/2-inch skin spanning large unsupported foil panels may buckle far below material yield.
  5. Fatigue: repeated cycles can crack aluminum weldments at stresses far below one-time yield.

7. Bottom-line answer

Estimated one-leg side load capacity, very roughly:

Wave height needed?

If you estimate force from simple steady drag, the required wave-induced crossflow velocities correspond to extremely large wave heights—much larger than ordinary swell. So the critical case is probably not regular side-wave drag, but rather:

8. Recommended next step

For this concept, a useful next step would be a preliminary structural model with:

If you want, I can next produce any of these in HTML too:

This estimate is conceptual only and should not be used as a final engineering basis for human-occupied offshore structures.

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