```html Wind-as-“Sail” Estimate for 12 ft × 60 ft Cylinder Seastead

Rough Wind-Force / Thruster / Drift Estimates (20 mph wind)

Important: This is a first-pass, back-of-envelope estimate using simplified drag formulas. Real performance can differ a lot due to: wave/wind gusts, current, freeboard details, leg submergence, hull/leg interference, thruster efficiency vs speed, and yaw/roll coupling. For design decisions, you’ll want a naval architect / marine engineer to run stability + station-keeping + structural load cases.

Geometry & Assumptions Used

ItemAssumed valueNotes
Cylindrical “body” Diameter 12 ft (3.66 m), length 60 ft (18.29 m) Corrugated culvert, closed ends.
Exposed to wind Entire cylinder side exposed You wrote “about 8 feet above the water”; I treated the whole cylinder as above-water windage.
Projected area (broadside) Aair ≈ L·D ≈ 18.29·3.66 ≈ 66.9 m² Broadside to wind.
Wind speed 20 mph = 8.94 m/s Caribbean trade-wind scenario.
Air density ρair = 1.225 kg/m³ Sea-level standard.
Drag coefficient (air) Cd,air = 1.2 Cross-flow circular cylinder is often ~1.0–1.2; corrugation can increase drag. I used 1.2.
Thrusters 4 × units, each 720 lbf (3200 N) at 3.2 kW Total max thrust ≈ 2880 lbf (12.8 kN). Total max electrical ≈ 12.8 kW.
Underwater “lateral drag area” (very rough) Awater ≈ 17.8 m² Approximated from 4 legs: diameter 4 ft (1.22 m), submerged length ~12 ft (3.66 m) each, projected area ≈ 1.22·3.66 = 4.46 m² per leg → 17.8 m² total. (Cables & thruster pods ignored.)
Water density, drag coefficient ρwater = 1025 kg/m³, Cd,water = 1.0 Order-of-magnitude for bluff members in cross-flow.

Wind Force on the Cylinder (Broadside)

Using drag: F = 0.5 · ρ · C_d · A · V²

Scaling tip: wind force scales with V². If wind increases from 20→30 mph, force multiplies by (30/20)² = 2.25. So broadside force would be roughly 884 lbf × 2.25 ≈ 1990 lbf in 30 mph wind.

1) Can the thrusters hold the orientation you want? What power is needed?

1A) “Hold heading” (yaw control / pointing)

Purely from a force capacity standpoint, you have plenty of thrust:

But “holding orientation” is about moment (torque) as well as force. With four thrusters spread over a ~50 ft × 74 ft footprint, you can generate a yaw couple by pushing one side forward and the opposite side backward. A rough upper-bound for available yaw moment:

A crude “wind moment” depends on where the aerodynamic center of pressure sits relative to the system’s hydrodynamic pivot. If we assume an effective offset of, say, 3–9 m (highly uncertain), the wind yaw moment magnitude might be:

So at 20 mph wind, the thrusters are very likely capable of holding a chosen heading, with margin—assuming the thrusters can vector effectively and cavitation / ventilation isn’t limiting.

1B) “Hold orientation AND prevent drifting sideways” (station-keeping in wind)

If you attempt to keep near-zero drift relative to the water (ignoring current), thrusters must generate approximately the same horizontal force as the wind load:

Estimated electrical power (very rough): if thrust scales roughly with power for your operating point, then:

This power estimate is uncertain because propulsor efficiency and required power depend strongly on the vehicle’s speed through the water, propeller inflow, and thruster design. But it gives the right “kW scale” for 20 mph wind loads.

2) How fast would the wind blow it sideways if it was broadside to the wind?

A simple way to estimate “sideways drift speed” is to find the speed where wind force ≈ water drag force on the submerged structure. Using the same drag form in water:

F_water = 0.5 · ρ_water · C_d,water · A_water · V² → solve for V.

So a reasonable first estimate is:

If the legs/columns have more submerged length, added bracing, or if thrusters/frames add drag, the drift speed will be lower. If the underwater structure is “cleaner” than assumed, drift could be higher.

3) If you wanted to go 20–30° right/left of downwind, how well would that work?

Key point: a cylinder is mostly a drag device, not a lifting sail

A round cylinder (even corrugated) generally produces a force that is mostly downwind drag, not strong sideways lift like a wing/sail. That means:

Vector view (idealized): holding a course at an angle off downwind

Let the true wind push downwind with force F (here ~884 lbf when broadside). If you want your ground track (through-water track) to be at angle β from downwind, then to have no sideways slip you’d need:

Case (20 mph wind, broadside windage) β Along-course force F·cos(β) Side force to cancel F·sin(β) Implication
20° off downwind 20° 884 lbf × 0.940 ≈ 831 lbf 884 lbf × 0.342 ≈ 302 lbf Thrusters (or underwater lateral resistance) must supply ~300 lbf sideways to avoid leeway.
30° off downwind 30° 884 lbf × 0.866 ≈ 765 lbf 884 lbf × 0.500 ≈ 442 lbf Needs ~440 lbf sideways cancellation to truly track 30° off downwind.

What happens in practice with your configuration?

Estimated “downwind-drag propulsion” speed scale

If the wind drag is ~884 lbf and the underwater drag area estimate is in the right ballpark, the through-water speed that balances those forces was ~1.5 mph (Section 2). That suggests your wind-drag “sailing” mode might yield on the order of:

If you yaw the cylinder so it is not broadside, the projected area scales approximately with sin(φ), where φ is the angle between wind direction and cylinder axis: broadside (φ=90°) gives full force; 30° gives ~0.5 force; 20° gives ~0.34 force. So if you “feather” it to reduce loads, you also reduce available wind push.

Bottom-line Answers (20 mph wind)

  1. Orientation control: Likely yes at 20 mph. Broadside wind force estimate is ~884 lbf. With 2880 lbf total thrust available, you have margin for yaw control and for partial station-keeping. Rough electrical to fully cancel broadside drift is on the order of ~4 kW (high uncertainty).
  2. Broadside drift speed: Using a simple wind-force vs water-drag balance with the assumed submerged leg area, sideways drift is roughly ~1.5 mph (order-of-magnitude: 1–2 mph).
  3. 20–30° off downwind: A cylinder provides mostly downwind drag, not “sail lift,” so it won’t behave like a real sailboat. To actually track 20–30° off downwind without large leeway, you’ll likely need thrusters providing roughly ~300 lbf (20°) to ~440 lbf (30°) of sideways correction (or a dedicated underwater lateral plane / keel).

If you want, I can refine this

If you provide any of the following, the estimates can be tightened substantially:

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