BUOYANCY & FLOTATION – METACENTRIC HEIGHT Report

1. Introduction: In this experiment we'll deal with floating body which a floating body is: a. 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. See Fig (1) b. Unstable: If it tends to a new equilibrium position after it had been tilted through a small angle. And a floating body to be unstable it is essential that the metacenter (M) is below the center of gravity. See Fig (1)

Fig (1): Stable and Unstable 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 → GM 

Px ………… (1) w. sin  1

Where, θ in radian * The metacentric height can be calculated as followed: MG = BM + OB – OG……………….......... (2) Where: 

BM 

I V

- BM is the metacentric 1 LW 3 12



radius , - I 



V: Total volume of displaced liquid.



V OB = 0.5 ( LxW )

Fig (2): Metacentric Height

2. Objective: 

To calculate the metacentric height of floating body.



To make sense of principle of buoyancy force and metacentric height of a flat bottomed vessel.

3. Apparatus:

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. See Fig (3) Fig (3): Set up

2

Pontoon measurement:    

4.

Pontoon dimension : Depth (D) = 170 mm Length (L) = 380 mm, Width (W) = 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.

Procedure:

At first part (1): Determination of floatation characteristic for unloaded and for loaded pontoon 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”), 8. Calculate GM practically where GM 

P(123) , 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), Where OG 

Wvm (OGvm )  Wb(OGb ) Wvm (OGvm )  Wb( x1)  Wvm  Wb Wvm  Wb

OGvm = 125 mm, OGb= x1: from table "1" .

3

Secondly 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. 6. Calculate GM practically where GM 

P (123) . 3500 sin 

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  ) 2 Where OG  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 (4): Weights & Dimensions

4

5. Results: Table (1): Results of Part (1) Bilge Weight Off balance wt.

Mean Def. θ (degree)

Exp. GM

Wb (gm)

P (gm)

0.00

50

2.2

53.4

100

4.5

52.25

150

7.2

49.06

200

9.9

47.69

2000.00

50

1.9

37.09

x1 = 30

100

3.9

36.16

150

6

35.30

200

8.3

34.08

4000.00

100

2.9

34.73

x1 = 37.5

150

4.7

32.16

200

6.4

31.52

250

8.2

30.79

GM at θ =0 from graph

(mm)

OG

OB

Theo. GM

(mm)

(mm)

(mm)

(mm)

55.297

164.93

125

15.789

55.719

37.994

98.95

87

26.315

38.265

36.236

70.38

75

36.842

32.222

Chart (2): GM for 2000 mast weight

Chart (1): GM for 0.0 mast weight 54

37.5

53

37 y = -0.7894x + 55.297

52

BM

y = -0.4649x + 37.994

36.5 36

51

35.5

50

35

49

34.5

48

34

47

33.5 0

2

4

6

8

10

0

12

5

2

4

6

8

10

Chart (3): GM for 4000 mast weight 35 34.5 34 33.5 33 32.5 32 31.5 31 30.5 30

y = -0.7092x + 36.236

0

2

4

6

8

10

Table (2): Results of Part (2) Off balance wt. P (gm)

Mean Def. θ (degree)

Exp. GM (mm)

GM At =0 From graph

BM

OG

OB

Theo. GM

(mm)

(mm)

(mm)

(mm)

M above Water level

30.643

141.36

125.4

18.421

34.36

159.781

17.95

141.36

136.28

18.421

23.42

159.781

3.771

141.36

147.28

18.421

12.501

159.781

4.2766

141.36

158.42

18.421

1.361

159.781

Mast Weight = 0.0 40

2.5

30.5

80

5.1

31.62

120

7.5

32.08

40

4.1

19.66

80

7.6

21.25

120

10.8

22.50

20

5.4

7.46

40

8

10.10

80

12.6

12.88

10

9.9

2.044

20

12.8

3.17

40

15.2

5.36

Mast Weight = 50.0

Mast weight = 100.0

Mast Weight = 150.0

Mast weight = 200.0 Unstable 6

Chart (4): GM for 0.0 mast weight 32.2 32.1 y = 0.1917x + 30.643

32 31.9 31.8 31.7 31.6 31.5 5

5.5

6

6.5

7

7.5

8

Chart (5): GM for 50 mast weight 23 22.5

y = 0.4244x + 17.954

22 21.5 21 20.5 20 19.5 19 3

5

7

9

11

13

Chart (6): GM for 100 mast weight 14 13

y = 0.7357x + 3.771

12 11 10 9 8 7 6 5 5

7

9

7

11

13

Chart (7): GM for 150 mast weight 6 5.5

y = 0.6175x - 4.2766

5 4.5 4 3.5 3 2.5 2 1.5 9

10

11

12

13

14

15

16

6. Comments & Recommendations:

Errors founded and may be affected on a results in the experiment due to many reasons listed below: 1. Zero error: If the setup reading isn’t zero at zero load. 2. Human error: if the experimenter red the outputs and calculate incorrectly.

It is recommended to make sure that there is no zero error.

8