Calculating  Force due to Wind

Most often ignored factor in shiphandling is effect of Wind, Ignoring which can lead to a serious incidents.

Now we will try to quantify the effect of wind i.e. the force in tons, So that we can have a fair idea and its effect in practical ship handling.

 

Understanding and able to calculate wind force is very critical to execute maneuvers in strong winds and avoid sudden accidents.

There are various formula/methods given to calculate force exerted due to wind:-

1. Use wind coefficients as per Mooring Equipment Guide - 4, Annex-A.

 

2. Force(T) = (v2/18)* Area/1000

    (Credit: From book “The Shiphandlers Guide” by Capt RW Rowe, Published by Nautical Institute)

    v=wind speed in m/s (1 m/s=2 kts)

    Area in Sq Mtrs

3. Force(T)=k X A X v2

    k=0.52 X 10-4 for beam wind

    k=0.39 X 10-4 for longitudinal wind

 

4. F(KN)=1/2 X d X v2 X A

    ​d=Density of air=1.229 kg/m3 

Any of the above method can be used to determine force due to wind.

 

Below is a comparative table:

Most of the tanker terminals suspend pilotage in winds about 26 to 28 knots.

 

As you can see from above table:-

  1. In case of VLCC in ballast, the force due to beam wind is 50 tons at 25 knots, i.e. it will take max power of 1 tug to just counter the effect of wind! So how many tugs do you think should be used?

  2. Hence vessel in ballast and vessels with large freeboard (Windage are 3000 Sq mtr or above) must properly plan their tug requirements if they intend to carry out pilotage in strong winds. If sufficient tugs are not available, movement must be delayed until wind has reduced. In no case movement should be attempted without thorough planning and tug requirements as at slower speeds and while maneuverings in Harbour or close to berths wind force will take over the vessel and in no time things can go out of control.

 

Exercise1.

A container vessel with LOA 294 mtr, freeboard 10 mtr, loaded with average 4 high GP containers along the length. Calculate the wind force in tons at for beam wind of 20kt and 25kt.

 

Solution:-

Height of 1 GP container=2.4 mtr

Height of 1 GP container=2.4 X 4 mtr = 9.6 mtr

Total freeboard = 10 + 9.6 =19.6 mtr

Windage area = 294 X 19.6 = 5762.4 Sq M

 

Wind Force (20 kt) = (10X10)X(5.762.4)/18 = 32 Ton ( May be just managed with 2 X 50 ton tugs)

Wind Force (25 kt) = (12.5X12.5)X(5.762.4)/18 = 50 Ton ( Advisable not to attempt berthing with just 2 X 50 ton tugs).

 

In case of wind gusts upto 35 kts this force will become 72 tons hence vessel may easily go out of control if sufficient powered tugs are not available.

 

So with above understanding I hope you will be able to make better decisions if you should carry out or postpone pilotage in strong winds. 

In next chapter lets discuss turning effect due to wind……………well beware of strong wind and be prepared!

Mumbai, India