WALL SIDED FORMULA Learning Objectives 1. Unde Unders rsta tand nd the the dist distin inct ctio ion n betw betwee een n stability at small and large angles of heel by consideration of the Wall W all sided formula. 2. Calc Calcul ulat ate e the the angl angle e of of loll loll and and GM GM ang angle le of loll for a ship that is unstable. 3. Calc Calcul ulat ate e the the angl angle e of loll loll caus caused ed by a transverse shift of weight when GM is zero.
Stability at small angles of heel (initial stability) Fig. 16.1 shows a ship heeled to a small angle by an external force. The center of buoyancy has moved from B to B 1, which has a movement parallel to, and in the same direction as, the shift of the wedge from b to b1. The line of action of buoyancy buoyancy force acting upwards through B 1 passes through the initial tranverse metacenter (M). metacenter (M).
Stability at small angles of heel (initial stability) Fig. 16.1 shows a ship heeled to a small angle by an external force. The center of buoyancy has moved from B to B 1, which has a movement parallel to, and in the same direction as, the shift of the wedge from b to b1. The line of action of buoyancy buoyancy force acting upwards through B 1 passes through the initial tranverse metacenter (M). metacenter (M).
For small angles of heel (up to about 6 0) it is assumed that the movement of B to the low side follows the arc of a circle, BM being the radius of the arc known as the metacentric radius. BM = I V where I is the second moment of area ( or moment of inertia) of the water plane area. For boxed-shaped vessel BM BOX = LB3 12V
where L and B are the length and breadth of the water plane area respectively and V is the volume of displacement of the vessel (L x B x d). For small angles of heel the BM can be assumed constant, since there is no significant increase or change of shape of the water plane area as the ship heels, since the value of I must also be assumed constant.
Under these conditions righting lever maybe calculated using: GZ = GM x Sine 0 And the moment of statical stability or righting moment: RM = (GM x Sine 0 ) x Displacement This measure of stability is referred to as initial stability because it is related to the position of the initial transverse metacenter that is assumed to be at a fixed point within small angles heel.
Stability at large angles of heel for wallsided inclinations. When heeled to large angles it can no longer be assumed that the center of buoyancy moves in an arc. The transfer of buoyancy from high side to low is such that there is increasing vertical movement of B the vertical component of transfer of buoyancy increases at a faster rate than the horizontal component and B adopts a position at B2 rather than some position at B1 which it would have had if moving in an arc.
The water plane area at the larger angle of heel is larger, consequently BM is larger as a result of the greater value of moment of inertia of the water plane area (I). This causes the metacenter to move at larger angles of heel such that it is termed the prometacenter or moving metacenter. The righting lever arising from this higher position of the center of buoyancy (B 2) is: GZ = GX + XZ
which is greater than the lever GX that would have existed if the upthrust due to buoyancy had been applied at B1 and passed through M. The formula for this new GZ that applies for wall-sided inclinations is: GZ = (GM x Sine0 )+(½ BM x Tan20 x Sine0) GZ =
GX
+
XZ
This simplifies to: GZ = (GM + ½ BM Tan20 ) Sine 0
where GM and BM are the values for the ship in upright condition. When using: GZ = GM x Sine0 it should be noted that for a ship that has a large GM, the error in using this formula would remain small up to a larger angle of heel than for a ship having a small initial GM value. Example: A box-shaped vessel length 120 m, breadth 18 m floats on a even keel draft of 8.0 m in salt water. KG is 6.4 m. Calculate the righting lever (GZ) when the vessel is inclined by an external force to: a) 50 b) 300
Solution: KM = KB + BM KBBOX = draft = 8.000 = 4.000 m 2
2
BMBOX = LB3 = 120 x 183
= 3.375 m
12V 12 x (120 x 18 x 8) therefore KM = 4.000 + 3.375 = 7.375 m GM = KM ± KG = 7.375 ± 6.400 = 0.975 m
a) GZ value at 50 Because this is a small angle of heel: GZ = GM x Sine0 GZ = 0.975 x Sine 50 = 0.0849768 m (ans) Alternatively using wall-sided formula: GZ = (GM + ½ BM Tan20 ) Sine0 GZ = (0.975 + 3.375 x Tan 250) x Sine0 2 = 0.0861026 m (ans)
Even within small angles of heel there will be difference in answers, because the water plane area of a vessel will increase in reality. The small angle formula for GZ should only be used for angles of heel up to 50 or 60. b) GZ values at 30 0 (The wall-sided formula must be used for this large angle of heel). GZ = (GM + 1/2 BM Tan 20 ) Sine0 GZ = (0.975 + 3.375 Tan2300)Sine0 = 0.769 m (Ans)
A box ± shaped vessel length 116 m, breath16 m and depth 9.8 m and is upright floating on an even keel draft of 5.7 m in salt water. KG is 6.0 m. Calculate the moment of statical stability when the vessel is heeled to the angle of deck edge immersion. Solution: KM = KB + BM KBBOX = draft = 5.700 = 2.850 m 2
2
BMBOX = LB3 =
116 x 163
= 3.743m
12V 12 x (116 x 16 x 5.7) therefore KM = 2.85 + 3.743 = 6.593 m GM = KM ± KG GM = 6.593 ± 6.000 4.1m 0 9.8m
8.0 m 5.7m
Fig 16.4
To calculate the angle of heel at which deck edge immersion takes place, consider Fig. 16.4 Tan 0 = OPP = 4.1= 0.51250 = 27.140
ADJ
8
GZ =(GM + 1/2BM Tan20 ) Sine0 = (0.593 + 3.743 x Tan 2 27.140) Sine 27.140 = 0.495 m RM = GZ x Displacement = 0.495 x (116 x 16 x 5.7 x 1.025) = 5368 t-m (Ans)
ANGLE OF LOLL Calculating the angle of loll using the wallsided formula A ship with a negative GM will not remain upright. It will capsize, either to port or starboard , until the center of buoyancy is able to attain a position vertically below the center of gravity (G) at B 2. At an angle of loll GZ is zero. The accurate formula for calculating GZ for wall sided inclinations is: GZ =(GM+1/2BM Tan20 )Sine0
1. A ship upright with negative GM; G above M
2. Ship starts to capsize as a result of negative GZ, B moves outward
3. Ship will settle at an angle of loll with B below G. GZ is zero
A formula for calculating the angle of loll value can be derived from this as follows: Expanding the above formulaGZ = (GM x Sine0 ) + (1/2BM Tan20 x Sine0 ) At the angle of loll GZ is zero 0 =(GM x Sine0 )+(1/2BM Tan20 x Sine0) -(GM x Sine0 ) = (1/2BM Tan20 x Sine0 ) Dividing both sides by Sine0 gives- GM = 1/2BM x Tan20 or - GM = BM x Tan20 2
-2 x GM = Tan20 Tan0 LOLL =
-2 x GM BM
BM 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 a KG 7.15m KB 4.26 m, BM 2.84 m. Calculate the angle of loll. Solution: KM = KB + BM = 4.26 + 2.84 = 7.10 m GM = KM - KG = 7.10 ± 7.15 = - 0.05 m Tan 0 LOLL = Tan 0 LOLL = Tan 0 LOLL = Tan 0 LOLL = 0.18765 =
-2 x GM BM
0.10 2.84
-2 x ± 0.05 2.84
10.630 to port or starboard Ans
Calculating the effective 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 as shown in Fig. 16.1
This new metacentric height is the value shown as GM1and is given by the formula: GM at the angle of loll = -2GM Cos 0 where GM is the initial upright GM which is a negative value and 0 is the angle of loll. It should be noted that the metacenter at this stage (M1) need not be on the center 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 center 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. Solution: Calculate KG GGv = w x d = 500 x (7.84 ±6.00) = 0.077m W±w
12500 - 500
Final KG = 7.840 + 0.077 = 7.917 m Final GM = KM ± KG = 7.85 ± 7.917 = - 0.067 m BM = KM ± KB
BM = 7.850 ± 3.950 = 3.900 m
Tan 0 LOLL =
-2 x GM BM
Tan 0 LOLL =
-2 x -0.067 3.90
0 LOLL = 10.500 (to port or starboard) (Ans) Calculate the effective GM at the angle of loll Gm at the angle of loll = 2GM = -2 x ± 0.067 Cos 0 Cos 10.500
= 0.136 m (Ans)
Calculating the angle of List caused by a transverse shift of weight when GM is zero. Consider a ship that is upright in a condition of neutral stability where GM = 0 with a weight µw¶ on deck. The weight is moved transversely across the deck causing G to move off the center line to GH. The ship list over and comes to rest when the center of buoyancy attains a position below the center of gravity.
GGH = w x d W
(1)
W
In the listed condition the horizontal component of GGH is represented by GX (Fig. 16.8).
Had the ship been heeled by an external force to the same angle of heel, instead of being listed by moment of weight, a righting lever GZ would have existed , being the same length as GX. In triangle GGHX:
Cosine 0 = ADJ HYP Cosine 0 = GX GGH
GX = Cos 0 x GGH
(2)
Combining formula (1) and (2): GX = Cos 0 x w x d W Since GZ (for the ship heeled) = GX (the horizontal component of GG H), and the formula for GZ at large angles of heel being the wall-sided formula; then: GX = (GM + ½ BM Tan 20 ) Sine 0 Cos 0 x w x d = (GM + ½ BM Tan2 0 ) Sine 0 W
Since the GM = 0: Cos 0 x w x d = ½ BM Tan20 x Sine 0 W Dividing both sides by Cos 0 gives: w x d = ½ BM Tan30 W Because: Tan 0 = Sine 0 Cos 0 w x d = BM Tan2 0 W
2
2 x w x d = Tan 30
Rearranging gives:
BM x W Therefore: Tan 0 LIST WHEN GM = 0 =
3
22 x w x d BM x W
Example: Calculate the list of a ship displacing 10000 tons when a weight of 20 tons is shifted transversely through 10 m given that BM is 4.80 m, GM is 0.00m
