Helical Mooring Screw – Installation Time & Lever‑Bar Design

This page gives a quick‑reference estimate for installing a single‑helix mooring screw with a dinghy‑pulled lever bar driven by a 10 hp outboard. It covers:

1. Assumptions & Basic Parameters

ParameterValueComments
Outboard power10 hp ≈ 7.5 kWTypical small‑boat outboard
Thrust available (low‑speed)≈ 800 N (≈180 lbf)Conservative for a 10 hp outboard moving a dinghy in a circle
Pull‑rope length (lever arm)10 ft (3.05 m) for 6‑in helix
12 ft (3.66 m) for 12‑in helix
Long enough to keep the dinghy clear of the anchor
Helix pitch (single‑start)0.33 ft (4 in) for 6‑in dia.
0.50 ft (6 in) for 12‑in dia.
Typical for these sizes
Soil (Caribbean sand)γ ≈ 18 kN m⁻³, φ ≈ 35°, c ≈ 0Medium dense to dense sand
Efficiency of rope‑lever system≈ 70 %Accounts for friction, rope stretch, & motion losses

2. Required Installation Torque

For a single‑helix anchor in cohesionless sand the ultimate torsional resistance is often approximated by an empirical “torque‑capacity” factor:

Treq ≈ k · D · S
where
D = helix diameter (ft)
S = embedment depth (ft)
k ≈ 150 lb·ft / ft² for dense sand (conservative 200 lb·ft / ft² can be used)

Applying the formula gives:

HelixD (ft)S (ft)Treq (ft·lb)Treq (Nm)
6‑in (0.5 ft) dia.0.57≈ 1 100 ft·lb≈ 1 490 Nm
12‑in (1 ft) dia.1.011≈ 2 500 ft·lb≈ 3 390 Nm

3. Pulling Force Needed at the Lever End

The effective torque after losses is

Teff = η·F·L  →  F = Teff / (η·L)

With η ≈ 0.70 and L = lever arm length:

HelixLever arm LRequired pulling force F
6‑in10 ft (3.05 m)≈ 340 N (≈ 77 lbf)
12‑in12 ft (3.66 m)≈ 680 N (≈ 153 lbf)

Both forces are well within the ≈ 800 N thrust the outboard can develop at low speed, so the system is mechanically feasible.

4. Rotation Speed & Installation Time

The dinghy’s forward speed translates into a tangential speed at the lever end:

vdinghy ≈ 2.5 m s⁻¹ (≈ 5 knots) → ω = v / L

Thus the theoretical revolutions per minute (rpm) are:

HelixL (m)ω (rad s⁻¹)RPM (≈ v/L)
6‑in3.05≈ 0.82 rad s⁻¹≈ 7.8 rpm
12‑in3.66≈ 0.68 rad s⁻¹≈ 6.5 rpm

Number of revolutions needed = depth / pitch:

HelixDepth (ft)Pitch (ft)Revolutions
6‑in70.33≈ 21
12‑in110.50≈ 22

Combining rpm and revolutions gives the minimum theoretical time:

HelixTheoretical time @ 7.8 rpm (6‑in) / @ 6.5 rpm (12‑in)Realistic time (allowing for higher soil resistance, rope slippage, & start‑up)
6‑in≈ 2.7 min3 – 5 min
12‑in≈ 3.4 min4 – 6 min
Practical tip: If the dinghy can maintain a higher cruising speed (up to 3 m s⁻¹) the times drop to ≈ 2 min for the 6‑in and ≈ 3 min for the 12‑in. In rough seas or loose sand, add 30 % to the estimates.

5. Lever‑Bar Design

5.1 Material

5.2 Geometry

Helix sizeRecommended bar lengthBar diameter / tube specTip reinforcement
6‑in helix 10 ft (3.05 m) 2 in (50 mm) solid or 2.5 in OD × 0.25 in wall tube Solid 6‑in long forged tip that fits the mooring‑eye bore
12‑in helix 12 ft (3.66 m) 2.5 in OD × 0.25 in wall tube (≈ 33 kg) or 2.5 in solid rod (≈ 75 kg) Solid 8‑in long tip, larger bore, and optional hardened sleeve for the eye

5.3 Weight

Bar configurationVolume (m³)Mass (kg)Mass (lb)
2 in solid, 10 ft≈ 0.0062 m³≈ 48 kg≈ 106 lb
2.5 in OD × 0.25 in wall tube, 10 ft≈ 0.0035 m³≈ 27 kg≈ 60 lb
2.5 in solid, 12 ft≈ 0.0116 m³≈ 91 kg≈ 200 lb
2.5 in OD × 0.25 in wall tube, 12 ft≈ 0.0042 m³≈ 33 kg≈ 73 lb
Recommendation: For the 6‑in helix a 10‑ft tube (≈ 27 kg) is light enough for two people to lift and maneuver, while still providing the needed torque. For the 12‑in helix a 12‑ft tube (≈ 33 kg) gives the best strength‑to‑weight ratio; a solid rod is unnecessarily heavy.

5.4 Tip‑Reinforcement (Eye‑End) Details

5.5 Handle & Quick‑Disconnect

6. Quick‑Connect / Re‑Use Tips

7. Summary of Results

Helix DiameterDepth RequiredEstimated Time (ideal)Estimated Time (practical)Suggested Lever BarApprox. Bar Weight
6 in (0.5 ft) 7 ft ≈ 2.7 min 3 – 5 min 10 ft, 2.5 in OD tube, solid tip ≈ 27 kg (≈ 60 lb)
12 in (1 ft) 11 ft ≈ 3.4 min 4 – 6 min 12 ft, 2.5 in OD tube, solid tip ≈ 33 kg (≈ 73 lb)

8. Putting It All Together – Example HTML Snippet

You can copy the table below directly into a web page:

<table>
  <tr>
    <th>Helix</th>
    <th>Depth</th>
    <th>Est. Time (ideal)</th>
    <th>Est. Time (realistic)</th>
    <th>Lever Bar</th>
    <th>Weight</th>
  </tr>
  <tr>
    <td>6‑in</td>
    <td>7 ft</td>
    <td>≈ 2.7 min</td>
    <td>3 – 5 min</td>
    <td>10 ft, 2.5 in OD tube, solid tip</td>
    <td>≈ 27 kg (≈ 60 lb)</td>
  </tr>
  <tr>
    <td>12‑in</td>
    <td>11 ft</td>
    <td>≈ 3.4 min</td>
    <td>4 – 6 min</td>
    <td>12 ft, 2.5 in OD tube, solid tip</td>
    <td>≈ 33 kg (≈ 73 lb)</td>
  </tr>
</table>

9. Safety & Operational Notes

Good luck with the prototype – the method described above should allow a small crew to embed a 6‑in helix in under five minutes and a 12‑in helix in under six minutes, using only a standard 10 hp outboard and a modestly heavy lever bar.