Experiment (2) BUOYANCY & FLOTATION (METACENTRIC HEIGHT) PART (2)

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Experiment (2) BUOYANCY & FLOTATION (METACENTRIC HEIGHT) PART (2). By: Eng. Motasem M. Abushaban. Eng. Fedaa M. Fayyad. Purpose :. To determine the metacentric height of a flat bottomed vessel in two parts: PART (1) : for unloaded and for loaded pontoon. - PowerPoint PPT Presentation

Transcript of Experiment (2) BUOYANCY & FLOTATION (METACENTRIC HEIGHT) PART (2)

EXPERIMENT (2)

BUOYANCY & FLOTATION (METACENTRIC HEIGHT)

PART (2)

By:

Eng. Motasem M. Abushaban.

Eng. Fedaa M. Fayyad.

1

PURPOSE:

To determine the metacentric height of a flat

bottomed vessel in two parts:

PART (1) : for unloaded and for loaded pontoon.

PART (2) : when changing the center of gravity

of the pontoon.

2

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.

3

4

Remember:

- 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 PART (2) : 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. Calculate GM practically where

5. Draw a relationship between θ in degrees (x-axis) and

GM Practical (y-axis), then obtain GM when θ equals

zero. 5

.3500

)123(PGM

PROCEDURE

6. Move 50 gm bilge weight to the mast ahead,

then repeat steps 3,4&5.7. Repeat step 6 moving 100, 150 & 200 gm

bilge weight to the mast.

8. Determine the height of the center of

gravity for each loading condition according

to equation

6

W

LWmWbWbWvm

OG

)2

790()190(1)35()125(

7

3500

)2

790()35()190(300)125(3000L

WmWbOG

8

8. Calculate GM theoretically according to equation

GM (Th.) = BM + OB – OG

Notice: BM & OB are constants for all loading conditions, since the dimensions & the weight of pontoon do not alter.

9

Off balance wt. Mean Def. Exp. GM BM OG Theo. GM

P (gm) θ (degree) (mm) (mm) (mm) (mm)

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          

Table (2) \ Part (2)

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QUESTIONS