Example Lesson: Buoyancy gravity buoyancy Why some objects float.
Experiment (2) BUOYANCY & FLOTATION (METACENTRIC HEIGHT) PART (2)
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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.
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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.
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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.
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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
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W
LWmWbWbWvm
OG
)2
790()190(1)35()125(
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3500
)2
790()35()190(300)125(3000L
WmWbOG
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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.
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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