Hydraulics Lab - ECIV 3122
Experiment (2): Buoyancy
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
Hydraulics Lab - ECIV 3122
Experiment (2): Buoyancy
Px = W. GG1
= W. GM. sinθ
→ GM 
Px
…………(1)
w. sin 
Where, θ in radian
Fig.(2) Metacentric height
* The metacentric height can be calculated as followed:
MG = BM + OB – OG………………..........(2)
Where:
- BM 
- I
I
V
- BM is the metacentric radius ,
1
LW 3
12
- I : Moment of inertia of pontoon
- V: Total volume of displaced liquid.
- OB = 0.5 (
V
)
LxW
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.
Hydraulics Lab - ECIV 3122
Fig.(3) Experimental set-up
Experiment (2): Buoyancy
Fig.(4) Cross section
Pontoon measurement:
- 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.
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.
Hydraulics Lab - ECIV 3122
Experiment (2): Buoyancy
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".
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.(5) Weights & Dimensions
Hydraulics Lab - ECIV 3122
Experiment (2): Buoyancy
Tables of results:
Table "1": Part(1)
Bilge Weight Off balance wt.
Wb (gm)
P (gm)
0.00
50
100
150
200
2000.00
50
x1 = 30
100
150
200
4000.00
100
x1 = 37.5
150
200
250
Mean
Def.
θ
(degree)
Exp.
GM
(mm)
GM at
θ =0
from
graph
BM
OB
Theo.
GM
(mm)
(mm)
(mm)
Hydraulics Lab - ECIV 3122
Experiment (2): Buoyancy
Table "2": Part(2)
Off balance wt.
P (gm)
Mast Weight = 0.0
40
80
120
Mast Weight = 50.0
40
80
120
Mast weight = 100.0
20
40
80
Mast Weight = 150.0
10
20
40
Mast weight = 200.0
Unstable
Mean
Def.
θ
(degree)
Exp. GM
BM
OG
Theo.
GM
(mm)
(mm)
(mm)
(mm)