Cálculos de Estabilidade Naval

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The effect of shifting weight already on board
To find New DraftsExample 1, P.150 (HW)
T. moment = 120 ´ 45
= 5400 t-m (by the stern)
C. O. T = !.#$#%&'
()!)
= *+,,
-+,
= 22.5(cm) = 0.225 (m)
C. O. D. A = 
!
"
´ C.O.T 
= 
#$
%&#
´ 0.225 = 0.107 m
C. O. D. F = C. O. T – C. O. D. A
= 0.225 – 0.107 = 0.118 m
Present DF 5.50 m DA 6.50 m
C. O. D. F + 0.118 m C. O. D. A - 0.107 m
Final DF 5.618 m DA 6.393 m
A ship 126 m long is floating at drafts of 5.5 m F and 6.5 m A. The centre of flotation is 3 m aft of
amidships. MCT 1 cm = 240 tonnes m. Displacement = 6000 tonnes. Find the new drafts if a weight of 120
tonnes already on board is shifted forward a distance of 45 metres.
Ex 16-11, P.168
B.R = !
"#$
= %&
'&
C. O. T = ".)*)+,-
.$"$
C. O. D. A = /
0
´ C.O.T
C. O. D. F = C. O. T – C. O. D. A
Present DF 3.20 m DA 4.40 m
B.R ± - 0.07 m - 0.07 m
C. O. D. F - 0.012 m C. O. D. A + 0.010 m
Final DF 3.118 m DA 4.340 m
= 11&
'2&
= 3%
'&&
´ 0.022 
= 2.2 (cm) = 0.022 (m)
= 0.010 m
= 0.022 – 0.010 = 0.012 m
A ship 100m long has centre of flotation 3m aft of amidships and is floating at drafts 3.2
m F and 4.4 m A. TPC 10 tonnes. MCT 1 cm = 150 tonnes m; 30 tonnes of cargo is then
discharged from 20m forward of amidships and 40 tonnes is discharged from 12 m aft of
amidships. Find the final drafts.
Weight (t) distance f’m CF (m) (Head) moment t-m (Stern)
- 30 23 (fwd) 690
- 40 9 (aft) 360 .
- 70 330 (by the stern)
= 7.0 (cm) = 0.07 (m)
Ex 16-4, P.167
Weight (t) distance f’m CF (m) (Head) moment t-m (Stern)
+ 20 30 (fwd) 600
+ 45 25 (fwd) 1125
+ 60 15 (aft) 900
+ 30 3 (aft) 90.
+155 1725 990.
735 (by the head)
A ship is floating at drafts of 6.1 metres F and 6.7 metres A. The following cargo is then
loaded:
20 tonnes in a position whose centre of gravity is 30 metres forward of amidships.
45 tonnes in a position whose centre of gravity is 25 metres forward of amidships.
60 tonnes in a position whose centre of gravity is 15 metres aft of amidships.
30 tonnes in a position whose centre of gravity is 3 metres aft of amidships.
The centre of flotation is amidships, MCT 1cm = 200 tonnes m and TPC = 35 tonnes. Find
the new drafts forward and aft.
B.R = !
"#$
= '22
12
C. O. T = ".)*)+,-
.$"$
C. O. D. A = /
0
´ C.O.T
C. O. D. F = C. O. T – C. O. D. A
Present DF 6.100 m DA 6.700 m
B.S ± + 0.044 m + 0.044 m
C. O. D. F + 0.019 m C. O. D. A - 0.018 m
Final DF 6.163 m DA 6.726 m
= %12
4&&
= '
4
´ 0.037
= 3.7 (cm) = 0.037 (m) (by the Head)
= 0.018 m
= 0.037 – 0.018 = 0.019 m
= 4.4 (cm) = 0.044 (m)
CONTINUE…….Ex 16-4, P.167
Ex 16-5, P.167
A ship arrives in port trimmed 0.3 m by the stern and is to discharge 4600 tonnes of
cargo from 4 holds; 1800 tonnes of the cargo is to be discharged from No. 2 and 800
tonnes from No. 3 hold. Centre of flotation is amidships, MCT 1 cm = 250 tonnes m.
The centre of gravity of No. 1 hold is 45 m forward of amidships.
The centre of gravity of No. 2 hold is 25 m forward of amidships.
The centre of gravity of No. 3 hold is 20 m aft of amidships.
The centre of gravity of No. 4 hold is 50 m aft of amidships.
Find the amount of cargo which must be discharged from Nos. 1 and 4 holds if the ship is 
to sail on an even keel.
Cargo discharge from No.1 and 4 = total cargo – (No.2 +No.3)
= 4600 – ( 1800 + 800 ) = 2000 t
Initial trim= 0.30 m (by the stern)
Final trim = 0.00 m
C. O. T. = 0.30 (m) 30.0 (cm) (by the head)
T. moment = C. O. T.(cm) x MCTC
= 30 x 250
= 7500 ……….(1)
CONTINUE…….Ex 16-5, P.167
Let Amount of cargo discharged from No.1 = “w”
Amount of cargo discharged from No.4 = 2000 - w
Weight (t) distance f’m CF (m) (Head) moment t-m (Stern)
- w 45 (fwd) --- 45 w
- 1800 25 (fwd) --- 45000
- 800 20 (aft) 16000 ---
- (2000–w) 50 (aft) 100000-50w --- .
116000-50w 45000+45w .
116000 - 50w – (45000 + 45w) (by the head) (2)
Trim moment1 = Trim moment 2
7500 = 71000 – 95w
95w = 63500
w = 𝟔𝟑𝟓𝟎𝟎
𝟗𝟓
= 668.42 t discharge from No.1
2000 – w = 1331.5 t discharge from No.4
Ex 16-9, P.168
A ship floats in salt water on an even keel displacing 6200 tonnes. KG = 5.5 m, KM = 6.3
m, and there is 500 tonnes of cargo yet to load. Space is available in No. 1 ‘tween deck
(KG 7.6 m, centre of gravity 40 m forward of the centre of flotation) and in No. 4 lower
hold (KG 5.5 m, centre of gravity 30 m aft of the centre of flotation). Find how much
cargo to load in each space to complete loading trimmed 0.6 m by the stern, and find also
the final GM. MCT1cm = 200 tonnes m.
Initial trim = 0.00 m (even keel)
Final trim = 0.60 m (by the stern)
C. O. T. = 0.60 m 60 (cm) (by the stern)
T. moment = C. O. T.(cm) x MCTC
= 60 x 200 = 12000 ……….(1)
Let Amount of cargo loaded to No.1 = “w”
Amount of cargo loaded to No.4 = 500 - w
Weight (t) distance f’m CF (m) (Head) moment t-m (Stern)
+ w 40 (fwd) 40 w ---
+ (500-w) 30 (aft) --- 15000-30w.
40 w 15000-30w.
15000 - 30w – 40w (by the stern) (2)
CONTINUE…….Ex 16-9,168
Trim moment1 = Trim moment 2
12000 = 15000 – 70w
70w = 3000
w = 𝟑𝟎𝟎𝟎
𝟕𝟎
= 42.86 t loaded to No.1 TD
500 – w = 457.14 t loaded to No.4 LH
Weight x KG = moment about keel
6200 x 5.5 = 34100
42.86 x 7.6 = 325.736
457.14 x 5.5 = 2514.27
6700 = 36940.006
KG = 𝒎𝒐𝒎𝒆𝒏𝒕
𝒘𝒆𝒊𝒈𝒉𝒕
= 𝟑𝟔𝟗𝟒𝟎.𝟎𝟎𝟔
𝟔𝟕𝟎𝟎
= 5.513 m
GM = KM – KG 
= 6.3 – 5.5 = 0.787 m
Ex 16-2, P.167 (OOW BOE)
Trim = DA – DF
= 7.4 – 7.2
= 0.2 m (by the stern)
Initial trim= 0.20 m (by the stern)
Final trim = 0.00 m (even keel)
C. O. T. = 0.20 (m) 20.0 (cm) (by the head)
T. moment = C. O. T.(cm) x MCTC
= 20 x 200
= 4000 ……….(1)
An oil tanker 150 m long, displacement 12500 tonnes, MCT 1 cm 200 tonnes m, leaves
port with drafts 7.2 m F and 7.4 m A. There is 550 tonnes of fuel oil in the foward deep
tank (centre of gravity 70 m forward of the centre of flotation) and 600 tonnes in the
after deep tank (centre of gravity 60 m aft of centre of flotation). The centre of flotation
is 1 m aft of amidships. During the sea passage 450 tonnes of oil is consumed from aft.
Find how much oil must be transferred from the forward tank to the after tank if the
ship is to arrive on an even keel.
CONTINUE…….Ex 16-2, P.167
Let Amount of cargo shifted from fwd to aft = “w”
Weight (t) distance f’m CF (m) (Head) moment t-m (Stern)
- w 130 (fwd to aft) --- 130 w
- 450 60 (aft) 27000 --- .
27000 130w .
27000 – 130w (by the head) (2)
Trim moment1 = Trim moment 2
4000 = 27000 – 130w
130w = 23000
w = 𝟐𝟑𝟎𝟎𝟎
𝟏𝟑𝟎
= 176.9 t shifted to Aft
Ex 16-23, P.170
Let w = amount of water run into FPT.
B.S = 𝒘
𝑻𝑷𝑪
= 𝒘
𝟑𝟎
(cm)
New DA = 6.30 (m) + 𝒘
𝟏𝟓
(cm)
Req. DA = 6.20 (m)
C. O. D. A (1) = N. DA – Req. DA
= 0.10 + 𝒘
𝟑𝟎
(m)
= 10 + 𝒘
𝟑𝟎
(cm) = 𝟑𝟎𝟎 7𝒘
𝟑𝟎
…(1)
C. O. D. A (1) = C. O. D. A (2)
𝟑𝟎𝟎 7𝒘
𝟑𝟎
= 0.08w
300 + w = 2.39w
300 = 1.39w
w = 300/1.39 = 215.8 ts (run into FPT)
C.O.T = 𝒘×𝒅
𝑴𝑪𝑻𝑪
(Head) = 𝒘× 𝟕𝟎
𝟒𝟐𝟎
= 𝟕𝟎𝒘
𝟒𝟐𝟎
C. O. D. A (2) = 𝟔𝟕
𝟏𝟒𝟎
x 𝒘
𝟔
(cm)…(2)
A ship 140 m long arrives off a port with drafts 5.7 m F and 6.3 m A. The centre of
flotation is 3 m aft of amidships. TPC 30 tonnes. MCT 1 cm = 420 tonnes m. It is required
to reduce the draft aft to 6.2 m by running water into the forepeak tank (centre of
gravity 67 m forward of amidships). Find the minimum amount of water to load and also
give the final draft forward.
Ex 16-23, P.170 Continue……
B.S = !
"#$
= 4'2.5
1&
= 7.2 (cm) = 0.072 (m)
T. moment = w ´ d
= 215.8 ´ 70
= 15106 t-m (distance from CF) (by the Head)
C. O. T = ".)*)+,-
.$"$
= '2'&6
34&
= 36 (cm) = 0.360 (m)
C. O. D. A = /
0
´ C.O.T
= 6%
'3&
´ 0.360 = 0.172 m
C. O. D. F = C. O. T – C. O. D. A
Present DF 5.70 m DA 6.30 m
B.S ± + 0.072 m + 0.072 m
C. O. D. F + 0.188 m C. O. D. A - 0.172 m
Final DF 5.960 m DA 6.200 m
= 0.360 – 0.172 = 0.188 m
Example 2, P. 140
Moment of statical stability = W x GZ 
= 3198 x 0.560
= 1792.36 t-m (1790.9) t-m
GZ = (GM + .
-
BM tan2 𝛉) sin 𝛉
= (1.0 + .
-
x 3 x tan2 25) sin 25
= 1.3262 x sin 25
= 0.560 mKM = KB + BM
= 2.0 + 3.0
= 5.0 m
BM = '!
%& × )
= %&!
%& × *
= 3.0 m
A box-shaped vessel 65 m x 12 m x 8 m has KG 4 m, and is floating in salt water upright on an
even keel at 4 m draft F and A. Calculate the moments of statical stability at (a) 5 degrees and (b)
25 degrees heel.
(OOW) BOE
W = L x B x draft x density of water
= 65 x 12 x 4 x 1.025
= 3198 ts
KB = ½ draft
= 2.0 m
At 5 degree heel angle,
Moment of statical stability = W x GM x sin 𝛉
= 3198 x 1.0 x sin 5
= 278.72 t - m
GM = KM - KG
= 5.0 - 4.0
= 1.0 m
At 25 degree heel angle,
Ex 15 - 5, P. 142
Moment of statical stability = W x GZ 
= 1691.25 x 0.2392
= 404.59 t-m 
GZ = (GM + .
-
BM tan2 𝛉) sin 𝛉
= (0.472 + .
-
x 1.172 x tan2 24) sin 24
= 0.5882 x sin 24
= 0.2392 m
KM = KB + BM
= 2.0 + 1.172
= 3.172 m
BM = '!
%& × )
= +.-!
%& × *
= 1.172 m
(OOW) BOE
W = L x B x draft x density of water
= 55 x 7.5 x 4 x 1.025
= 1691.25 ts
KB = ½ draft
= 2.0 m
At 6 degree heel angle,
Moment of statical stability = W x GM x sin 𝛉
= 1691.25 x 0.472 x sin 6
= 83.72 t - m
GM = KM - KG
= 3.172 – 2.7
= 0.472 m
The image part with 
relationship ID rId3 was 
not found in the file.
At 24 degree heel angle,
A box-shaped vessel 55 m x 7.5 m x 6 m has KG 2.7 m, and floats in salt water on an even keel at 4
m draft F and A. Calculate the moments of statical stability at (a) 6 degrees heel and (b) 24 degrees
heel.
Example 3, Pg. 128 OOW BOE
A ship of 8000 tonnes displacement has a GM = 0.5 m. A quantity of grain in the hold, estimated at
80 tonnes, shifts and, as a result, the centre of gravity of this grain moves 6.1 m horizontally and
1.5 m vertically. Find the resultant list.
W = 8000 t, GM = 0.50 m, w = 80 t, d(H) = 6.1 m, d(V) = 1.5 m
Horizontal shift of G,
GGH (GG1 ) = . × )"
/
= 0$ × #.%
0$$$
= 0.061 m
Vertical shift of G,
GGV (G1G2 ) = . × )#
/
Tan q = 11$
12
= 0$ × %.-
0$$$
Vert. shif of G = 0.015 m
Initial GM = 0.500 m
Final GM = 0.485 m
= $.$#%
$.*0-
= 0.1258
q (List) = 7.17° (or) 7° 10.1’
Ex 14 - 6, Pg. 132 OOW BOE
W = 9000 t, GM = 0.50 m, w = 100 t, d(H) = 10.0 m, d(V) = 1.5 m
Horizontal shift of G,
GGH (GG1 ) = . × )"
/
= %$$ × %$.$
3$$$
= 0.111 m
Vertical shift of G,
GGV (G1G2 ) = . × )#
/
Tan q = 11$
12
= %$$ × %.-
3$$$
Vert. shif of G = 0.017 m
Initial GM = 0.500 m
Final GM = 0.483 m
= $.%%%
$.*04
= 0.2298
q (List) = 12.94° (or) 12° 56.5’
A quantity of grain, estimated at 100 tonnes, shifts 10 m horizontally and 1.5 m vertically in a ship 
of 9000 tonnes displacement. If the ship’s original GM was 0.5 m, find the resulting list. 
Ex 14 -9, Pg. 132
Moment about Keel
Weight x KG = Moment about the keel
Final KG = 5678! 29:;7?!8@;:;7?!8@;:;7?!8@;:;7?!8@;:;7?!8@;:;7?!8@;:;7'4 × 6
= 2.00 m
KM = KB + BM
= 3.00 + 2.00
= 5.00 m
Ex 9-3, Pg. 68
A ship is 150 m long, has 20 m beam, load draft 8 m, light draft 3 m. The block coefficient at the 
load draft is 0.766, and at the light draft is 0.668. Find the ship’s deadweight. 
Deadweight = (Load-Light) displacement
Load disp: = L x B x load draft x load Cb x ⍴sw
= 150 x 20 x 8 x 0.766 x 1.025
Light disp: = L x B x light draft x light Cb x ⍴sw
= 150 x 20 x 3 x 0.668 x 1.025
Deadweight = (Load-Light) displacement
= 18843.6 ts
= 6162.3 ts
= 18843.6 – 6162.3 ts
= 12681.3 ts
OOW BOE
Example , Pg. 58
(a) Construct a displacement curve from the following data:
Draft (m) 3 3.5 4 4.5 5.0 5.5 
Displacement (tonnes) 2700 3260 3800 4450 5180 6060 
(b) If this ship’s light draft is 3 m, and the load draft is 5.5 m, find the deadweight. 
(c) Find the ship’s draft when there are 500 tonnes of bunkers, and 50 tonnes of fresh water and 
stores on board. 
(d) When at 5.13 m mean draft the ship discharges 2100 tonnes of cargo and loads 250 tonnes of 
bunkers. Find the new mean draft. 
(e) Find the approximate TPC at 4.4 m mean draft. 
(f) If the ship is floating at a mean draft of 5.2 m, and the load mean draft is 5.5 m, find how 
much more cargo may be loaded. 
(b) Load draft 5.50 m
Light draft 3.00 m
Deadweight 3360
displacement 6060
displacement 2700
(c) Light displacement
Bunkers
Fresh water and stores
2700 tonnes
+500 tonnes
+ 50 tonnes
New displacement 3250 tonnes
OOW BOE
3000
(a) (c) 3250 ts 3.48 m
Displacement in tonnes
3500 4000 4500 5000 5500 6000
3.00
3.50
4.00
4.50
5.00
5.50
6.00
Dr
af
t 
in
 m
et
re
s
Draft Displacement 
3.00 2700
3.50 3260
4.00 3800
4.50 4450
5.00 5180
5.50 6060
(d) At 5.13m 5380ts
Cargo dish; -2100ts
Bunker load;+ 250ts
New Disp; 3530ts
New draft 3.745 m
(3.775 m) 
(e) 4.50 m 4450 ts
4.30 m 4175 ts
0.20 m 275 ts
For 1cm; (275 ÷20) 
= 13.75 ts
(f) 5.50 m 6060 ts
5.20 m - 5525 ts
535 ts
Ex 8 - 7 , Pg. 61
(a) Construct a displacement curve from the following data:
Draft(m) 1 2 3 4 5 6 
Displacement (tonnes) 335 767 1270 1800 2400 3100 
(b) The ship commenced loading at 3 m mean draft and, when work ceased for the day, the mean draft was 4.2 
m. During the day 85 tonnes of salt water ballast had been pumped out. Find how much cargo had been loaded. 
(c) If the ship’s light draft was 2 m find the mean draft after she had taken in 870 tonnes of water ballast and 
500 tonnes of bunkers. 
(d) Find the TPC at 3 m mean draft.
Draft(m) 
Displacement (tonnes) 
1.0
2.0
3.0
4.0
5.0
6.0
300 800 1300 1800 2300 2800 3300 
Draft(m) Disp;
1.0 335
2.0 767
3.0 1270
4.0 1800
5.0 2400
6.0 3100 
(b) Draft 4.2 m 1910 ts
Draft 3.0 m 1270 ts
Diff; 640 ts
BW P/O 85 ts
Cargo loaded 725 ts
(c) Draft 2.0 m 767 ts
BW P/In 870 ts
Bunker 500 ts
2137 ts Draft 4.57 m 
(d) Draft 3.10 m 
Draft 2.90 m 
1330 ts
1230 ts
Diff 20 cm 100 ts
TPC at 3.0 m = .,,
-,
= 5 ts
OOW BOE
Example 3, Pg. 39
A ship is loading in dock water of density 1010 kg per cu. m. FWA = 150 mm. Find the change in 
draft on entering salt water.
DWA (mm) = FWA x %$&- F G%&
&-
= 150 x %$&- F %$%$
&-
= 90 mm
Increase draft = 90 mm = 9 cm
OOW BOE
Example 4, Pg. 40 OOW BOE
A ship is loading in a Summer Zone in dock water of density 1005 kg per cu. m. FWA = 62.5 mm,
TPC = 15 tonnes. The lower edge of the Summer load line is in the waterline to port and is 5 cm
above the waterline to starboard. Find how much more cargo may be loaded if the ship is to be at
the correct load draft in salt water. Port Starboard
DWAIncrease draft
Thickness of Load line = 25 mm 
= 2.5 cm
Mean clearance = H9I6666 ?@
A+,=B-C *> ?@
= '4&
'.&
= 120 cu. m
Volume of Tank = Volume of oil + expansion
= 98 + 2% volume of oil
120 cu. m = 100 cu.m
100 cu.m is show actual volume 120 cu.m
98 cu.m …? = '4& × D5
'&&
= 117.6 cu.m (volume of Oil loaded)
Mass of Oil loaded = Volume of Oil x Density of Oil
= 117.6 x 0.88 x 1.0
= 103.5 tonnes
A tank holds 120 tonnes when full of fresh water. Find how many tonnes of diesel
oil of relative density 0.880 it will hold, allowing 2% of the volume of the tank for
expansion in the oil.
OOW BOE
Ex 3-1, Pg. 23 
Mass of fresh water = 120 tonnes
Volume of fresh water = . ?@
A+,=B-C *> ?@
= '4&
'.&
= 120 cu. m
Volume of Tank = Volume of oil + expansion
= 98 + 2% volume of oil
120 cu. m = 100 cu.m
100 cu.m is show actual volume 120 cu.m
98 cu.m …? = '4& × D5
'&&
= 117.6 cu.m (volume of Oil loaded)
Mass of Oil loaded = Volume of Oil x Density of Oil
= 117.6 x 0.84 x 1.0
= 98.78 tonnes
A tank holds 120 tonnes when full of fresh water. Find how many tonnes of diesel
oil of relative density 0.840 it will hold, allowing 2% of the volume of the tank for
expansion in the oil.
OOW BOE