Angle of Loll

Objectives    

Calculating the Angle of Loll Calculating the Effective GM at the Angle of Loll Correcting Angle of Loll Calculating Calculating the Angle of List Caused Cause d by a Transverse Transverse Shift of Weight when GM is Zero

A ship with a negative GM  A ship with a negative GM will not remain upright. It  will list, either to port or starboard, until the centre of buoyancy is able to attain a position vertically below the centre of gravity (G) at B2

Calculating the angle of loll 

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In this equation the values of GM and BM used are the original upright values. Because the upright GM is negative, the quantity  within the square root becomes positive.

Example In the upright condition a ship has KB 4.26 m, KG 7.15 m and BM 2.84 m. Calculate the angle of loll. KM = KB + BM GM=KM-KG KM = 4.26 + 2.84 = 7.10 m GM = 7.10 -7.15 = -0.05 m −2×−0.05

=

2.84

LOLL= 10.63° (to port or starboard)

=0.18765

GM at the Angle of Loll 



If the ship is heeled beyond the angle of loll righting levers become positive to act to right the ship back to the angle of loll. It follows that the ship must have acquired a new positive GM for this to happen

GM at the Angle of Loll 

This new metacentric height is the value shown as GM1 and is given by theformula:





 Where GM is the initial upright GM which is a negative  value and  is the angle of loll.

 It should be noted that the metacentre at this stage (M1) need not be on the centre line and will constantly move as the ship is heeled further beyond the angle of loll.

Example  An upright ship displaces 12500 t and has KG 7.84 m. 500 t is discharged from a position on the centre line Kg 6.00 m. Calculate the resulting angle of loll given that KB is 3.95 m and KM is 7.85 m in the final condition and the effective GM at the angle of loll

Example Calculate the KG and GM

Example

Correcting Angle of Loll    

G must be lowered below M

 Weight may be discharged from locations above G.  Weight may be transferred to a lower location. (derrick head?) Ballast may be loaded into a double bottom tank. 





If this is the case then a small tank on the low side must be filled first to prevent a roll to opposite side, with the danger of possible capsize due to gathered momentum.  A small tank will keep the FSE to a minimum and likewise the angle of list. Once the tank is full and no FSE then its opposite number may be filled.

Transverse Shift of Weight when GM is Zero 

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Consider a ship that is upright in a condition of neutral stability where GM = 0 with a weight 'w' on deck. The weight is moved transversely across the deck causing G to move off the centre line to GH The ship lists over and comes to rest when the centre of buoyancy attains a position below the centre of gravity.

Transverse Shift of Weight when GM is Zero

Transverse Shift of Weight when GM is Zero 

Calculating the Angle of List Caused by a Transverse Shift of Weight when GM is Zero

Example  A ship initially upright has displacement 12500 tonnes, KB 4.2 m, BM 4.6 m and a KG of 8.7 m has a weight of 50 t stowed on deck on the centre line at Kg 4.00 m. Calculate the list when the weight is lifted by the ship's crane, the head of which is 29.0 m above the keel, and then swung outboard 10 m from the centre line.

Example

Example

LIST WHEN GM=0 = 14.53°

SQA Stability  – July 2012  A vessel carrying timber on deck departs from port with a GM of 0.05m. The stability of the vessel deteriorates on passage and as a result the vessel settles to an angle of loll 12° to port. Even keel draught 5.700m in salt water. Investigate the effect of ballasting No.2 DB tanks, filling the tanks one at a time, in the following order: (1) centre, (2) port, (3) starboard. Using the Stability Data Booklet calculate EACH of the following: a)The initial negative GM prior to ballasting; b)The angle of loll on commencing to ballast the centre tank (assume weight negligible); c)The GM when the centre tank is full; d)The angle of heel when the port tank, TCG 5.00m, is full.

SQA Stability  – July 2012 a) Using E.K. of 5.700m, from hydrostatic particulars, Displacement = 11625.00 t KM = 8.63 m KB = 2.95 m BM = KM - KB = 8.63 – 2.95 = 5.68m

(Tan 12)2 = (-2 x GM) / BM 0.045180292 * 5.68 / 2 = - GM GM = - 0.128 m

SQA Stability  – July 2012 GM = - 0.128 m KG = KM – GM = 8.63 –(-0.128)= 8.758 m b) As soon as we start ballasting there will be free surface Calculate Moments about the Keel to determine the KG prior commencing to ballast  Weights (t) 11625 FSMSW 11625

KG (m) 8.758 1021 x 1.025 8.848

Moment (tm) 101811.75 1046.53 102858.28

SQA Stability  – July 2012 KG = 8.848 m GM = KM – KG = 8.63 - 8.848 = -0.218 m

TanθLOLL= √ 0.076760563 = 0.277056968 θLOLL= 15.5°

SQA Stability  – July 2012 c)The GM when the centre tank is full; Calculate Moments about the Keel to determine the KG when the centre tank is full

 Weights (t)

KG (m)

11625 271*1.025 277.775 Determine KM for Δ = 11896 t 11902.78

Moment (tm) 8.758

101811.75

0.59 8.567

163.88 101975.63

Displacement 11902.78 t KM = 8.580 m GM = 8.580 - 8.567 = 0.013 m

SQA Stability  – July 2012 d) Determine the Angle of Heel when the Port tank is full Calculate Moments about the Keel to determine the KG  Weights (t) KG (m) 11902.78 8.567 223*1.025 228.575 0.6 12131.36KM for Δ = 12119 8.417 t Determine

Moment (tm) 101971.12

Displacement 12131.36 t KM = 8.542 m GM = 8.542 - 8.417 = 0.125 m

137.15 102108.27

SQA Stability  – July 2012 Calculate Moments about the CL to determine the GGH  Weights (t)

Dist. from CL (m)

Moment (tm)

11902.78

0

0

228.575

5

1142.875

12131.355

0.094

1142.875

223*1.025

Tan θ List= GGH/GM = 0.094/0.125 = 0.753667 θ List =

37.0° to Port