Unstable
equilibrium and angle of loll were described in chapters 10 & 11 in Ship
Stability I of Capt. Subramaniam you may refer that. A vessel at the angle of loll is in an extremely precarious and
dangerous situation Wrong
action or no action on the part the ship's staff may cause the ship to capsize.
Even no action is dangerous because consumption of fuel and water from the
double bottom tanks would cause increase of KG making the vessel more unstable,
thereby increasing the angle of loll.
The angle of loll can be calculated by a simple formula derived from the wall-sided formula:
GZ = Sin θ (GM + ½BM tan²θ)
At the angle of loll GZ = Zero
So Sin θ (GM + ½BM tan²θ) = 0
So Sin θ = 0 or (GM + ½BM tan²θ) = 0
At the angle of loll, θ ≠ 0 so Sin θ ≠ 0
(GM + ½BM tan²θ) = 0 and Tan² θ = -2GM/BM
Tan θ = / -2GM
√ BM
Where θ = Angle of loll GM = The initial GM BM = BM when upright
Since this formula is derived from the wall-sided formula, it can be applied only when the immersed wedge and the emerged wedge are identical in shape.
Remedial action
1) Press up all slack tanks.
2) Run up SW into the DB tank which has
the smallest moment of inertia about its centre line. If this tank is not on the centre line of the ship, then on the lower side first, and after it is full, its counter part on the higher side.
3) Repeat action 2 with another tank and so on until the ship becomes stable.
4) If discharging or jettisioning deck cargo, do so from the higher side first, then from the lower side. If using,ship's own gear, due allowance must be made for the shift of COG, of each sling of cargo, from the UD to the derrick head during the operation.
Justification for such action.
At the angle of loll, any existing free surface effect must be eliminated/ minimized first. FWor HFO may require to be transferred internally such that the tanks finally remaining slack are those with the smallest moment of inertia about the tank's centre line.
While running up ballast into a DB tank, FSE would be created. This must be kept to a minimum. The necessity to fill up the tank with the smallest Ii' about its centre line is, therefore, vital. So also, the necessity to fill up only one tank at a time.
If the tank being ballasted is not on the centre line of the ship, but on either side like No:2 P and No:2 S, then fill up the lower side tank first f.e., if the ship is lolled to starboard, then fill up No:2 S first. After it is full, fill 'up No:2 P. The reasons for this:
Let 'Aº, be the angle of loll to starboard at first.
'pº, be the reduction in the angle of loll by completely filling up either No:2 P or No:2 S.
'qº' be the list caused by filling up either No:2 P or No:2 S.
'Rº, be the resultant inclination after completely filling up either No:2 P or No:2 S.
If No:2 S is run up first, R = A - P + q If No:2 P is run up first, R = A - P - q
In both cases, final R is to starboard.
It appears as if filling up the higher side tank would produce better results but it is not really so. The ship can loll to either side. If after the higher side tank is run up, wave action caused the ship to loll over to the other side (port side in this case), the ship would flop over to A - P + q to port and the momentum of flopping over will carry the inclination well beyond this. Since the GZ formed near the angle of loll is very small, the ship would heel over to port much more than A - P + q and take a very long time to return to this angle of inclination. If during this time (a) any openings went underwater &/or (b) a wave struck the ship adversely and/or (c) any cargo shifted, the ship may capsize.
By filling up the lower side tank first, the inclination would increase a bit to A - P + q at first, but this would be gradual and would would be gradual and would until No:2 P also is run up.
The same line of reasoning is applicable when considering discharge or jettison of cargo from the upper deck.
If the ship is in calm waters, such as inside a dock, the possibility of flopping over to the angle' of loll on the other side may not be there. In such a case, ballasting the higher, side tank or discharging deck cargo from the lower side may prove more effective and immediate.
The angle of loll can be calculated by a simple formula derived from the wall-sided formula:
GZ = Sin θ (GM + ½BM tan²θ)
At the angle of loll GZ = Zero
So Sin θ (GM + ½BM tan²θ) = 0
So Sin θ = 0 or (GM + ½BM tan²θ) = 0
At the angle of loll, θ ≠ 0 so Sin θ ≠ 0
(GM + ½BM tan²θ) = 0 and Tan² θ = -2GM/BM
Tan θ = / -2GM
√ BM
Where θ = Angle of loll GM = The initial GM BM = BM when upright
Since this formula is derived from the wall-sided formula, it can be applied only when the immersed wedge and the emerged wedge are identical in shape.
Remedial action
1) Press up all slack tanks.
2) Run up SW into the DB tank which has
the smallest moment of inertia about its centre line. If this tank is not on the centre line of the ship, then on the lower side first, and after it is full, its counter part on the higher side.
3) Repeat action 2 with another tank and so on until the ship becomes stable.
4) If discharging or jettisioning deck cargo, do so from the higher side first, then from the lower side. If using,ship's own gear, due allowance must be made for the shift of COG, of each sling of cargo, from the UD to the derrick head during the operation.
Justification for such action.
At the angle of loll, any existing free surface effect must be eliminated/ minimized first. FWor HFO may require to be transferred internally such that the tanks finally remaining slack are those with the smallest moment of inertia about the tank's centre line.
While running up ballast into a DB tank, FSE would be created. This must be kept to a minimum. The necessity to fill up the tank with the smallest Ii' about its centre line is, therefore, vital. So also, the necessity to fill up only one tank at a time.
If the tank being ballasted is not on the centre line of the ship, but on either side like No:2 P and No:2 S, then fill up the lower side tank first f.e., if the ship is lolled to starboard, then fill up No:2 S first. After it is full, fill 'up No:2 P. The reasons for this:
Let 'Aº, be the angle of loll to starboard at first.
'pº, be the reduction in the angle of loll by completely filling up either No:2 P or No:2 S.
'qº' be the list caused by filling up either No:2 P or No:2 S.
'Rº, be the resultant inclination after completely filling up either No:2 P or No:2 S.
If No:2 S is run up first, R = A - P + q If No:2 P is run up first, R = A - P - q
In both cases, final R is to starboard.
It appears as if filling up the higher side tank would produce better results but it is not really so. The ship can loll to either side. If after the higher side tank is run up, wave action caused the ship to loll over to the other side (port side in this case), the ship would flop over to A - P + q to port and the momentum of flopping over will carry the inclination well beyond this. Since the GZ formed near the angle of loll is very small, the ship would heel over to port much more than A - P + q and take a very long time to return to this angle of inclination. If during this time (a) any openings went underwater &/or (b) a wave struck the ship adversely and/or (c) any cargo shifted, the ship may capsize.
By filling up the lower side tank first, the inclination would increase a bit to A - P + q at first, but this would be gradual and would would be gradual and would until No:2 P also is run up.
The same line of reasoning is applicable when considering discharge or jettison of cargo from the upper deck.
If the ship is in calm waters, such as inside a dock, the possibility of flopping over to the angle' of loll on the other side may not be there. In such a case, ballasting the higher, side tank or discharging deck cargo from the lower side may prove more effective and immediate.
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