WALL SIDED FORMULA

Learning Objectives

1. Understand the distinction between stability at small and large angles of heel by consideration of the Wall sided formula.

2. Calculate the angle of loll and GM angle of loll for a ship that is unstable.

3. Calculate the angle of loll caused 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 B1, 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 force acting upwards through B1 passes through the initial tranverse metacenter (M).

For small angles of heel (up to about 60) 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 BMBOX = 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 wall- sided 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 (B2) 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 Tan250) 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 300 (The wall-sided formula must be used for this large angle of heel).

GZ = (GM + 1/2 BM Tan20) 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

5.7m

9.8m 8.0 m

Fig 16.4

0

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 Tan2 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 wall-

sided 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 B2.

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 formula-

GZ = (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 Tan0LOLL =

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.

-2 x GM BM

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 0LOLL =

Tan 0LOLL =

Tan 0LOLL =

Tan 0LOLL = 0.18765 = 10.630 to port

or starboard (Ans)

-2 x GM BM

-2 x – 0.05 2.84

0.102.84

• 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 0LOLL = -2 x GM-2 x GM BM

Tan 0LOLL = -2 x -0.067 3.90

0LOLL = 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 = 0.136 m (Ans) Cos 0 Cos 10.500

• 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 ‘w’ 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 (1) W

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 GGH), and the formula for GZ at large angles of heel being the wall-sided formula; then:

GX = (GM + ½ BM Tan20) 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

Rearranging gives: 2 x w x d = Tan30

BM x W

Therefore:

Tan 0LIST WHEN GM = 0 =

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

22 x w x d BM x W

3

Solution: Tan 0LIST WHEN GM =0 =

Tan 0LIST WHEN GM =0 =

= 0.20274

Tan 0LIST WHEN GM = 0 = 11.460 (Ans)

In any calculation that ask for the list to be calculated and a former part of the calculation gives a GM of exactly 0, then the above formula must be used.

2 x w x d BM x W

3

2 x 20 x 104.80 x 10000

3