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Bollard Force

Reference data and engineering information about bollard force for mechanics applications.

bollardforceCalculator

Overview

Engineering reference data for Bollard Force in mechanics.

Key Formulas

Newton's Second Law

F=maF = ma

Force = mass × acceleration.

Work

W=FdcosθW = Fd\cos\theta

Work = force × displacement × cos(angle).

Kinetic Energy

Ek=12mv2E_k = \frac{1}{2}mv^2

Energy of motion.

Potential Energy

Ep=mghE_p = mgh

Gravitational potential energy.

Variables

SymbolDescriptionUnit
FFForceN
mmMasskg
aaAccelerationm/s²
vvVelocitym/s

Interactive Charts

Bollard - quay ship mooring

References

Angle-Turns Conversion Reference

The relationship between turns, degrees, and radians when wrapping rope around a bollard:

TurnsDegreesRadians
1/490°π/2
1/2180°π
1360°
2720°

Worked Examples

Example 1: Single Turn Around a Bollard

For one complete turn (360° = 2π radians) with friction coefficient μ = 0.5:

S=Fe(0.5×2π)=Feπ0.043FS = F \cdot e^{-(0.5 \times 2\pi)} = F \cdot e^{-\pi} \approx 0.043F

Result: A single turn reduces the required effort force to approximately 4.3% of the load force — less than 5% of the original load.

Example 2: Mooring a Ship

Given:

  • Ship velocity: 0.05 m/s
  • Stopping time: 2 seconds
  • Ship mass: 20,000 kg
  • Rope angle: 180° (half turn, π radians)
  • Friction coefficient: μ = 0.4

Step 1 — Calculate retardation: a=ΔvΔt=0.05 m/s2 s=0.025 m/s2a = \frac{\Delta v}{\Delta t} = \frac{0.05 \text{ m/s}}{2 \text{ s}} = 0.025 \text{ m/s}^2

Step 2 — Calculate load force (F): F=ma=20,000 kg×0.025 m/s2=500 N=0.5 kNF = m \cdot a = 20{,}000 \text{ kg} \times 0.025 \text{ m/s}^2 = 500 \text{ N} = 0.5 \text{ kN}

Step 3 — Calculate effort force (S): S=500e0.4π=142 N=0.14 kNS = 500 \cdot e^{-0.4\pi} = 142 \text{ N} = 0.14 \text{ kN}

Result: With a half turn around the bollard, the dock worker only needs to exert 142 N to hold a 500 N load — a reduction of over 70%.

Practical Force Reduction

Rule of Thumb: One complete turn (360°) around a bollard with typical rope friction (μ ≈ 0.5) reduces the required effort force to approximately 10% of the load force. Additional turns provide further dramatic reductions following the exponential decay relationship.

The exponential nature of the Capstan equation means that:

  • Friction coefficient (μ) has a significant impact — rougher ropes/finishes increase holding power
  • Each additional wrap multiplies the force reduction exponentially
  • The system is self-locking at sufficient wrap angles, requiring minimal effort to hold large loads

Load Force Determination

The load force FF in bollard applications is often derived from the dynamics of the moored vessel or object. Common calculations include:

Retardation Force:
F=maF = m \cdot a
where mm is mass (kg) and aa is acceleration (m/s²).

Retardation from Velocity Change:
a=ΔvΔta = \frac{\Delta v}{\Delta t}
where Δv\Delta v is the change in velocity (m/s) and Δt\Delta t is the stopping time (s).

Friction Coefficient Reference

The friction coefficient μ\mu between a rope and a typical steel or cast iron bollard is generally in the range of *0.3 to 0.5. This value is crucial for calculating the required effort force and must be selected based on rope and surface materials.

Radian Angle Conversion Formula

For calculations, the contact angle in degrees must be converted to radians: αrad=2παdegrees360=παdegrees180\alpha_{\text{rad}} = \frac{2\pi \cdot \alpha_{\text{degrees}}}{360} = \frac{\pi \cdot \alpha_{\text{degrees}}}{180}

This conversion is essential for applying the capstan equation S=Fe(μαrad)S = F e^{-(\mu \alpha_{\text{rad}})}.

Effort Force Ratio Table

5 rows
Effort force to load force ratio (S/F) for various wrap angles and friction coefficients
Angle(°)
Radians(rad)
Turns
S/F (μ=0.3)
S/F (μ=0.5)
901.5710.250.6220.456
1803.1420.50.3870.208
2704.7120.750.240.095
3606.28310.1490.043
72012.56620.0220.002

Source: engineeringtoolbox.com

Bollard Force Calculator

Interactive calculator based on the Capstan equation. Input load force, friction coefficient, and contact angle to compute the required effort force.

Practical Applications

Ship Mooring

The text provides a specific mooring scenario:

  • Ship mass: 20,000 kg
  • Arrival velocity: 0.05 m/s
  • Stop time: 2 seconds
  • Retardation: a=0.052=0.025 m/s2a = \frac{0.05}{2} = 0.025 \text{ m/s}^2
  • Load force: F=ma=500 N=0.5 kNF = ma = 500 \text{ N} = 0.5 \text{ kN}
  • Effort force (half turn, μ=0.4): S=500e0.4π=142 N=0.14 kNS = 500e^{-0.4\pi} = 142 \text{ N} = 0.14 \text{ kN}

This demonstrates how a half turn reduces the holding force to ~28% of the mooring load.

Engineering Considerations

  1. Friction Coefficient Range: The text specifies μ = 0.3-0.5 is common for rope around steel or cast iron bollards
  2. Force Reduction: One full turn (360°) reduces effort to ~5-10% of load (depending on μ)
  3. Safety Factor: Multiple turns should be used for critical applications
  4. Rope Material: Natural fiber ropes typically have higher μ than synthetic ropes
  5. Bollard Condition: Worn or greased bollards will have lower μ values

Multi-Turn Applications

For heavy loads or high-security mooring:

  • Two turns (720°) reduces effort to less than 0.5% of load (μ=0.5)
  • Three turns (1080°) provides essentially zero effort required
  • Practical limit: Rope becomes difficult to retrieve if too many turns are used

Capstan Equation Verification

The core equation S=FeμαS = Fe^{-\mu\alpha} can be verified against the chart data:

  • For μ=0.5, 360°: S/F=e0.5×2π=eπ0.043S/F = e^{-0.5 \times 2\pi} = e^{-\pi} \approx 0.043
  • This matches the 4.3% value from the text's example

Safety Notes

  1. Always use additional safety factors beyond calculated values
  2. Consider dynamic loads (wave action, wind gusts) not in static calculations
  3. Regularly inspect ropes and bollards for wear
  4. Train personnel in proper belaying techniques
  5. Never stand in the bight of a loaded rope