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Exp. (2): BUOYANCY & FLOTATION – METACENTRIC HEIGHT Purpose : To determine the metacentric height of a flat bottomed vessel. Introduction: A floating body is stable if it tends to return to its original equilibrium position after it had been tilted through a small angle. For a floating body to be stable it is essential that the metacenter ( M ) is above the center of gravity; metacentric height ( MG ) should be positive. Fig. (1) Stable & unstable equilibrium The greater the metacentric height, the greater is the stability, however, very large metacentric heights causes undesirable oscillations in the ships and are avoided. Theory: If a body is tilted through an angle θ , B1 will be the position of the center of buoyancy after tilting. A vertical line through B1 will intersect the center line of the body at (M) (Metacenter of the body), MG is the metacentric height. The force due to buoyancy acts vertically up through B1 and is equal to W, the weight of the body acts downwards through G. The resulting couple is of magnitude Px
Px = W. GG = W. GM. sinθ → W Px GM …………(1) θ in radian Fig.(2) Metacentric height
- The metacentric height can be calculated as followed: MG = BM + OB – OG ……………….......... (2) Where:
I
BM - BM is the metacentric radius ,
I LD - I : Moment of inertia of pontoon
- V: Total volume of displaced liquid.
- OB = 0.5 ( LxD
V
Experimental Set-up: The set up consists of a small water tank having transparent side walls in which a small ship model is floated, the weight of the model can be changed by adding or removing weights. Adjustable mass is used for tilting the ship, plump line is attached to the mast to measure the tilting angle.
- Record the results in the table ( Table " 1" ),
- Calculate GM practically where sin
W
P
GM , W has three cases.
- Draw a relationship between θ (x-axis) and GM (y-axis), then obtain GM when θ equals zero.
- Calculate GM theoretically according to equation (2), where Wvm Wb WvmOG Wb x Wvm Wb WvmOG Wb OG OG vm b vm
OGvm = 125 mm, OGb= x1: from table "1". PART (2) : Determination of floatation characteristic when changing the center of gravity of the pontoon. 1.Replace the bilge weights by 4x 50 gm weights.
- Apply a weight of 300gm on a height of 190 mm from the pontoon surface.
- Apply weights of 40, 80 &120 gms on the bridge piece loading pin, then record the corresponding tilting angle.
- Move 50 gm bilge weight to the mast ahead, then repeat step 3.
- Repeat step 3 moving 100, 150 & 200 gm bilge weight to the mast.
- Calculate GM practically where 3500 sin
P ( 123 )
GM .
- Determine the height of the center of gravity for each loading condition.
- Calculate GM theoretically according to equation (2), where W
L
Wvm Wb Wb Wm OG
Where : In case of 50 gm, L = 10 mm. In case of 100 gm, L = 20 mm. In case of 150 gm, L = 30 mm. In case of 200 gm, L = 40 mm. Fig.(5) Weights & Dimensions
Tables of results: Table "1": Part(1) Bilge Weight Off balance wt. Mean Def. Exp. GM GM at θ =
BM OB
Theo. GM Wb (gm) P (gm) θ (degree) (mm) from graph (mm) (mm) (mm) 0.00 50 2. 100 4. 150 6. 200 9. 2000.00 50 1. x1 = 30 100 3. 150 6. 200 8. 4000.00 100 3. x1 = 37.5 150 5. 200 6. 250 8.
