```html Seastead Backup Propulsion Analysis

Seastead Backup Propulsion Analysis

Platform data used in the calculations (all assumptions are explained in the text):

1. Primary propulsion redundancy

Two thrusters per side give differential thrust for steering. If one thruster fails the remaining three can still propel the platform, albeit at reduced speed and with a turning bias that can be compensated by angling the hull or using the kite/sea‑anchor lines.


2. Kedging with sea anchors (Question j)

Concept

Two large “parachute” sea anchors are deployed downstream. A dinghy pulls a line that runs through a block on the anchor and back to a winch on the seastead. While one anchor is pulling the other is hauled in, then the cycle repeats. The power quoted (2 kW) is the mechanical power the winch can apply to the line.

Speed estimate

The thrust is the drag of the sea anchor moving through the water. With a 10 m diameter parachute (projected area ≈ 78.5 m², Cd ≈ 1.2) the drag equation gives:

P = 0.5·ρw·Cd·Aanchor·v³ → v = ( 2P / (ρw·Cd·A) )1/3

Using P = 2000 W, ρw=1025 kg/m³, Cd·A ≈ 94 m² (anchor) + 14 m² (platform) = 108 m²:

v ≈ 0.33 m/s ≈ 0.74 mph

Thus a 10 m sea anchor at 2 kW would move the platform at roughly 0.7–0.8 mph. Larger anchors give more thrust but the speed actually drops because the drag rises faster than the power‑to‑thrust advantage.

Cost & weight


3. Kedging with ordinary anchors (Question j)

Concept

In shallow water a conventional Danforth or Bruce anchor is dropped, the line is winched in, then the anchor is retrieved and moved forward. The power is again limited to the 2 kW winch.

Speed estimate

Ignoring anchor‑slip, the speed is set by the platform drag (the same drag area as above). Using the same 2 kW:

v ≈ (2·2000 / (1025·14))1/3 ≈ 0.65 m/s ≈ 1.45 mph

In practice the anchor’s holding capacity will be the limit. A typical 30 kg (66 lb) Danforth can hold ≈ 2,000–3,000 lb (9–13 kN). If the anchor can only supply 9 kN of pull, the maximum steady thrust is ≈ 9 kN, and the speed that 2 kW can overcome is:

v = P / thrust = 2000 W / 9000 N ≈ 0.22 m/s ≈ 0.5 mph

Thus realistic speeds are in the 0.5–1 mph range, depending on seabed and anchor size.


4. Towing with a dinghy – Yamaha HARMO drives (Question 4)

Concept

A 14 ft dinghy carries three Yamaha HARMO rim‑drive motors. Each motor delivers 227 lbf (≈ 1 kN) of thrust; three together give ≈ 680 lbf (≈ 3,030 N). Power is drawn from the seastead’s batteries.

Speed estimate

Equating thrust to platform drag:

thrust = 0.5·ρw·Cd·A·v² → v = √( 2·thrust / (ρw·Cd·A) )

With thrust ≈ 3,030 N and Cd·A ≈ 14 m²:

v = √( 2·3030 / (1025·14) ) ≈ 0.64 m/s ≈ 1.4 mph

At lower motor RPM the thrust will be proportionally lower, giving ~1 mph. The method works well for short‑range “towing” to a dock or rescue vessel.


5. Kite propulsion (Question 5)

Concept

A stack of identical “sled” kites, each 6 ft × 2 ft (≈ 1.11 m²). The kite bridle is arranged so a single line pulls the platform forward. The wind creates lift that has a forward component (thrust). The analysis below assumes the kites are flown at a moderate angle (≈ 30° to the wind) giving a lift coefficient CL ≈ 0.5–0.8.

Speed down‑wind (0°)

Thrust = 0.5·ρa·CL·Akites·(Vwind – v)²

Set thrust = platform drag (0.5·ρw·Cd·Aplatform·v²). Using 20 kites (total A ≈ 22.3 m²) and CL = 0.5:

v ≈ 0.27 m/s ≈ 0.6 mph

With CL = 0.8, v ≈ 0.33 m/s ≈ 0.75 mph.

30° off‑wind

The component of wind acting along the boat’s direction is V = 20 mph · cos30° ≈ 17.3 mph (7.74 m/s). Repeating the calculation gives:

v ≈ 0.20–0.25 m/s ≈ 0.45–0.55 mph

Could we reach 2 mph?

Solving for the required kite area to achieve v = 2 mph (0.894 m/s) yields:

Even with a very efficient lift coefficient you would need > 100 kites, weighing many hundreds of kilograms and costing many thousands of dollars. Reaching 2 mph with kites alone is not realistic for this size of platform.


6. Towing by a second seastead (Question 6)

If a second platform is nearby, a light rope bridge can be used to pass power from the “towing” platform’s batteries to the “towed” platform’s thrusters. Assuming each platform can devote 2 kW to the thrusters (the primary design), the combined available power is 4 kW (or more if both run their thrusters).

Using the same drag area (Cd·A ≈ 14 m²):

v = ( 2·P / (ρw·Cd·A) )1/3 = ( 2·4000 / (1025·14) )1/3 ≈ 0.82 m/s ≈ 1.8 mph

If each platform can run 4 kW (both thrusters), total 8 kW gives:

v ≈ ( 2·8000 / (1025·14) )1/3 ≈ 1.04 m/s ≈ 2.3 mph

Thus a paired tow can comfortably achieve 2 mph, limited mainly by the capacity of the rope bridge and the battery energy available.


7. Single‑thruster operation (Question 7)

With only one functional thruster (or two on the same side) the platform can still make progress by “sailing” at an angle to the wind. The thruster provides a forward component while the kite or sea‑anchor lines can be used to counteract side‑slip. Using a single 2 kW thruster:

v ≈ ( 2·2000 / (1025·14) )1/3 ≈ 0.65 m/s ≈ 1.45 mph

By pointing the hull about 20–30° off the wind, the net motion can be directed down‑wind with a modest lateral offset. The exact heading is a trade‑off between speed and side‑drift; a 30° heading gives a forward component of cos30 ≈ 0.86, i.e. about 1.2 mph forward, plus a small sideways drift that can be corrected by the kite or a trailing line.


Summary Table

Method Power used (W) Estimated Speed (mph) Pros Cons Approx. Cost / Weight
Sea‑anchor kedging (10 m parachute) 2 000 W (winch) 0.7–0.8 Simple, no moving parts, works in open water Slow, needs a dinghy to reset anchors, anchor drag limits speed $800–$1,500 (new)
Weight ≈ 50–100 kg
Regular anchor kedging 2 000 W (winch) 0.5–1.0 Uses conventional ground‑anchor, cheap Anchor must hold, limited by seabed, slow reset Anchor $100–$400
Weight ≈ 20–40 kg
Dinghy + 3 × HARMO (681 lb thrust) ≈ 3 × 1 kW (motor input) 1.0–1.5 Good for short‑range tow, easily stowed Requires a dinghy, limited battery capacity Motors ≈ $3,000 (each)
Weight ≈ 30 kg total
Kite stack (20 kites) 0 W (wind only) 0.6–0.8 (down‑wind) No fuel, silent, simple hardware Very low speed, needs many kites for higher speed, wind‑dependent ~$50–$100 per kite → $1,000–$2,000 for 20
Weight ≈ 5 kg each → 100 kg total
Neighbor‑tow (combined power) 4 000 W (2 × 2 kW thrusters) ≈ 1.8 Redundant power, can maintain 2 mph Requires second platform, coordination Only the cost of extra cable/rope bridge
Single‑thruster operation 2 000 W ≈ 1.4 (straight) – 1.2 (30° heading) Simple, works with existing hardware Reduced speed, limited steering, need to use wind/kite for side‑force balance No extra cost

Note: All numbers are order‑of‑magnitude estimates. Real performance will depend on actual drag coefficients, anchor holding capacity, wind gusts, sea state, and battery state. They are intended only for preliminary planning.

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