Exp. (2): BUOYANCY & FLOTATION METACENTRIC HEIGHT

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. GG1

= W. GM. sinθ

Px →GM  …………(1) W 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 , V 1 - I  LD 3 - I : Moment of inertia of pontoon 12 - V: Total volume of displaced liquid. V - OB = 0.5 ( ) LxD

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.

Fig.(3) Experimental set-up Fig.(4) Cross section Pontoon measurement: - Pontoon dimension : Depth (H) = 170 mm Length (L) = 380 mm, Width (D) = 250 mm.

- The height of the center of gravity of the pontoon is OGvm = 125 mm from outer surface of vessel base. - The balance weight is placed at x = 123 mm from pontoon center line.

- The weight of the pontoon and the mast Wvm = 3000 gm.

PART (1) : Determination of floatation characteristic for unloaded and for loaded pontoon Procedure: 1. Assemble the pontoon by positioning the bridge piece and mast. 2. Weigh the pontoon and determine the height of its center of gravity up the line of the mast. 3. Fill the hydraulic bench measuring tank with water and float the pontoon in it, then ensure that the plumb line on the zero mark. 4. Apply a weight of 50 g on the bridge piece loading pin then measure and record the angle of tilting and the value of applied weight. 5. Repeat step 4 for different weights; 100, 150, & 200 g, and take the corresponding angle of tilting. 6. Repeat the above procedure with increasing the bottom loading by 2000 gm and 4000 gm. 7. Record the results in the table ( Table " 1" ), P(123) 8. Calculate GM practically where GM  , W has three cases. W sin 9. Draw a relationship between θ (x-axis) and GM (y-axis), then obtain GM when θ equals zero. 10. Calculate GM theoretically according to equation (2), Wvm(OG ) Wb(OG ) Wvm(OG ) Wb(x1) where OG  vm b  vm Wvm Wb Wvm Wb

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. 2. Apply a weight of 300gm on a height of 190 mm from the pontoon surface. 3. Apply weights of 40, 80 &120 gms on the bridge piece loading pin, then record the corresponding tilting angle. 4. Move 50 gm bilge weight to the mast ahead, then repeat step 3. 5. Repeat step 3 moving 100, 150 & 200 gm bilge weight to the mast. P(123) 6. Calculate GM practically where GM  . 3500sin 7. Determine the height of the center of gravity for each loading condition. 8. Calculate GM theoretically according to equation (2), L Wvm(125) Wb(35) Wb1(190) Wm(790  ) where OG  2 W

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) Mean Exp. GM at Theo. Bilge Weight Off balance wt. BM OB Def. GM θ =0 GM θ from Wb (gm) P (gm) (mm) (mm) (mm) (mm) (degree) graph 0.00 50 2.13

100 4.45

150 6.90

200 9.23

2000.00 50 1.95

x1 = 30 100 3.98

150 6.10

200 8.25

4000.00 100 3.35

x1 = 37.5 150 5.10

200 6.90

250 8.75 Table "2": Part(2) Mean Theo. Off balance wt. Def. Exp. GM BM OG GM M above θ water P (gm) (degree) (mm) (mm) (mm) (mm) level Mast Weight = 0.0 40 2.40 80 4.88 120 7.50 Mast Weight = 50.0 40 3.45 80 7.23 120 10.50 Mast weight = 100.0 20 3.28 40 6.35 80 12.00 Mast Weight = 150.0 10 3.70 20 10.23 40 14.78 Mast weight = 200.0 Unstable