Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Experimental procedure (GulfStream)
In the beginning of the experimental work we started by measuring the weight of
the empty ship model, using three scales and the weight of the mast, using only one
scale. We obtained that the weight of the empty model is 13121 (4792+3504+4825)
grams [g], and the weight of the mast is 64 grams [g]. Then we determined the density
of water which was 0.998g/cmᶟ. After that, we selected the fixed and the mobile weights
to put inboard and determined the centre of gravity of each one. After selected the
weights, we measured that the draft marks are located 24.1 centimeters [cm] fwd of the
aft perpendicular and 25 centimeters aft of the fwd perpendicular. We continued with
the experimental work by placing the model to float in the tank and waited until the
caused waves calm down. Then we put the fixed and the mobile weights inboard. The
distribution of the weights was according to the following order:
. the first disk of 500 grams [g] was placed 22.6 centimeters [cm] from the aft
perpendicular of the model.
. the mast (64 grams [g]) was placed 30.1 centimeters [cm] from the aft perpendicular of
the model.
. the 5 kilograms [kg] weight was placed 81.6 centimeters [cm] from the aft
perpendicular of the model.
. the second disk of 500 grams [g] was placed from 131.1 centimeters [cm] from the aft
perpendicular of the model.
. the 997 grams [g] (mobile weight) weight was placed 76.3 centimeters [cm] from the
aft perpendicular of the model.
Finally, we read and recorded the four drafts (two in the aft marks, and two in
the fore marks) to each experiment: amidships, fore, aft, port and starboard.
We read the following drafts:
Aft-Starboard
Aft-Port
Fore-Starboard
Fore-Port
Amidships
7.1 [cm]
7.7 [cm]
6.2 [cm]
7.7 [cm]
Starboard
8.2 [cm]
6.5 [cm]
7.3 [cm]
6.1 [cm]
Port
5.7 [cm]
8.6 [cm]
5.0 [cm]
8.0 [cm]
Aft
7.5 [cm]
8.1 [cm]
5.3 [cm]
6.2 [cm]
Fore
6.0 [cm]
6.4 [cm]
7.0 [cm]
7.8 [cm]
Table 1
We recorded also the inclinometer readings, wich were:
. φ = [𝑡𝑔−1(2.7/78)] = 1.98˚ (mobile weight placed amidships).
. φ = [𝑡𝑔−1(7.5/78)] = 5.49˚ (mobile weights placed starboard).
. φ = [𝑡𝑔−1(7.7/78)] = 5.64˚ (mobile weight placed port).
. φ = [𝑡𝑔−1(1.5/78)] = 1.10˚ (mobile weight placed aft).
. φ = [𝑡𝑔−1(3.7/78)] = 2.72˚ (mobile weight placed fore).
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Buoyancy experiment
Calculation of the immersion amidships, the immersion at the centre of
floatation and trim:
Figure 1 – immersion amidship and trim (Amidships and Starboard)
Figure 2 – immersion amidship and trim (Port, Fore, Aft)
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Figure 3 – Centre of floatation (initially with the weights inboard, then for the empty model)
Figure 4 – Results obtained for the five trims and immersions, and for the centre of
floatation (empty model and model with weights inboard)
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Description of the procedure used to find the centre of floatation:
1st – Interpolation of the hydrostatic tables to find the LCF (abscissa of centre of
floatation) of the loaded model.
2nd – Calculation of the difference between the amidships immersion and the
centre of floatation quotas, and after that, calculation of the centre of floatation quota for
the loaded ship:
𝑑𝑇 = (𝐿𝐶𝐹𝑃𝑃𝐴𝑅 − 𝑥𝑃𝑃𝐴𝑀 )𝑡𝑔(𝛼)  𝐾𝐹 = 𝐼𝑀𝑁 + 𝑑𝑇
3rd – To find the centre of floatation of the empty ship, we only need to subtract
the weights influence:
𝐿𝐶𝐹 =
𝐾𝐹 =
∑ 𝑝𝑖 𝑥𝑖 + 𝐿𝐶𝐹0 𝐷0
∑ 𝑝𝑖 +𝐷0
∑ 𝑝𝑖 𝑧𝑖 + 𝐾𝐹0 𝐷0
∑ 𝑝𝑖 +𝐷0
;
;
Looking to the results, we can verify that the ship model has a initial trim aft of
the ship model. We can also verify that when we placed the mobile weight in the aft
perpedicular, the model gains a trim aft, and when we placed it in the fwd
perpendicular, it gains a trim fwd, as was expected.
The immersions amidship are similar to all the five experiments, with a slight
difference of 4 millimeters [mm] from the highest value to the lowest one.
For the study of the centre of floatation we admitted that it is a good
approximation to consider the centre of floatation obtained for the experiment with the
mobile weight almost amidships and generalize it for all the others experiments. We can
do this because the centre of floatation has to be the same whenever we have an equal
weight inboard, independently of the positions of the weights. Therefore, studying the
obtained centre of floatation point for the experiments with the mobile weight almost
amidships we can check that its quota value is almost the same of the immersion
amidships, with a small diference, in order of three decimal places of the millimeter.
Then, calculating the centre of floatation of the empty model we verify that its quota
value is higher than with the loaded ship, due to the absence of extra weights. It is easy
to understand that the centre of floatation is located aft of the amidships perpendicular,
due to the initial trim aft of the ship model.
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Calculation of the hull volume and of the displacement using the table of
offsets:
Figure 5 – Hull volume and displacement
Figure 6 – Results obtained for the hull volume and fot the displacement
Description of the procedure used to find the hull volume:
1st – Interpolation of the hydrostatic tables to find the hull volume.
Description of the procedure to find the displacement:
1st – Determinate the displacement using the formula:
D = V ρ , where ρ is the density of the water in the tank
Looking for the obtained results we can check that the value for the hull volume
is an acceptable number, since, with an easy account, we can estimate the volume of a
retangular prism with the lenght, width, and amidships immersion of the model, where
the model would fit without many free space:
𝑉𝑝 = 1600 × 254 × 71.7 =
3
3
29138880 𝑚𝑚 . So, it is verifiable that 22773447 mm is lower than 29138880 mm3, it
is a little bit more than 78% of that retangular prism volume.
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Determination of the total weight of the model and of its light weight.
Figure 7 – Experimental light displacement and total weight
Figure 8 – Results obtained for the light displacement and for the total weight
We know from the experimental work that the light weight of the model is
13121g and the total weight is 20182g (light weight of the model and the weights that
we put inboard). What we verify by calculating the displacement is the difference
between the real weight of the vessel and the weight that we obtained with the
experimental values. We will approach this topic better when we study the Archimedes
Principle.
Comparison of the displacement and the weight to verify the Archimedes
Principle. Interpret quantitatively any discrepancy by making use of the concept of
tons per cm, TPC, analysing the error in terms of variation of draft, eventually
related to reading errors.
Figure 9 – Buoyancy centre (initially with the weights inboard, then for the empty model) and TPC
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Figure 10 – Results obtained for the buoyancy centre (empty model and model with weights inboard) and for the TPC
Archimedes Principle says: “Any object, wholly or partially immersed in a fluid,
is buoyed up by a force equal to the weight of the fluid displaced by the object.”
To verify the Archimedes Principle we will find the point where the force is applied
(B), and then the concept of tons per cm, TPC.
Description of the procedure used to find the centre of buoyancy:
1st – Interpolation of the hydrostatic tables to find the LCB (abscissa of centre of
buoyancy) of the loaded model.
2nd – Interpolation of the hydrostatic tables to find the KB (quota of centre of
buoyancy) of the loaded model.
3rd – To find the centre of buoyancy of the empty ship, we only need to subtract
the weights influence:
𝐿𝐶𝐵 =
𝐾𝐵 =
∑ 𝑝𝑖 𝑥𝑖 + 𝐿𝐶𝐵0 𝐷0
∑ 𝑝𝑖 +𝐷0
∑ 𝑝𝑖 𝑧𝑖 + 𝐾𝐵0 𝐷0
∑ 𝑝𝑖 +𝐷0
;
;
Description of the procedure used to study the Archimedes Principle, using the
concept of tons per cm, TPC:
1st – Determinate the difference between the real weight of the vessel and the
weight that we obtained with the experimental values.
𝑉
2nd – Determinate the floatation area: 𝐴𝑓 = 𝐼 ;
𝑓
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
3rd – Determinate the unit displacement (TPC): 𝐷 𝑢 = 𝜌 𝐴𝑓 ;
It is possible to verify that the draft variation is 0.8cm when we consider that the
mobile weight is placed almost amidships (3.7 cm of difference between the centre of
gravity of the mobile weight and the amidships perpendicular). We could improve this
result by considering that the floatation centre is the average of the five floatation
centres that we obtained by studying all the five events (different positions of the
mobile weight), instead of considering that the floatation centre is the same of the event
with the mobile weight almost amidships. If we would do that, we would obtain a draft
variation of 0.5cm, substantially minor than the previous value obtained. These values
are affected by some errors committed during the experimental work, such as: errors in
the drafts readings, the water in the tank wasn’t completely stopped, the draft marks can
present some errors in the scales, illumination of the tank wasn’t the best, interference
of water meniscus (1.5mm) and possible errors in the table of offsets.
Inclining experiment
Calculation of centre of gravity (KG and LCG) with the added weights
centred, and also in light condition.
Figure 11 – Centre of gravity (Study of its quota KG) for the model with weight inboard
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Figure 12 – Centre of gravity (study of its abscissa LCG) for the model with weight inboard, and
then study of the centre of gravity of the empt model
Figure 13 – Results obtained for the centre of gravity for the models with weght inboard and empty
Description of the procedure used to find the centre of gravity:
1st – Determinate GM (transverse metacentric height) for the experiments with
the mobile weight port and starboard, applying the formula: 𝐺𝑀 =
∑ 𝑝𝑖 ∆𝑦𝑖
𝐷 𝑡𝑔𝜃
;
2nd – Interpolation of the hydrostatic tables to find the 𝐵𝑀𝑇 (transverse
metacentric radius) of the loaded model.
3rd – Determinate KG (centre of gravity height) for the experiments with the
mobile weight port and starboard, applying the formula: 𝐾𝐺 = 𝐾𝑀 − 𝐺𝑀  𝐾𝐺 =
𝐾𝐵 + 𝐵𝑀𝑇 − 𝐺𝑀 ;
4th – Determinate the average value for KG (using the experiments with the
mobile weight port and starboard): 𝐾𝐺 =
𝐾𝐺𝐸𝑠𝑡. +𝐾𝐺𝐵𝑜𝑚𝑏.
2
;
5th – Determinate 𝐺𝑀𝐿 (longitudinal metacentric height) for the experiments
with the mobile weight aft and fore, applying the formula: 𝐺𝑀𝐿 =
∑ 𝑝𝑖 ∆𝑥𝑖
𝐷 𝑡𝑔𝜑
;
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
6th – Determinate the difference between the abscissas of centre of buoyancy and
centre of gravity : 𝑥 = 𝐺𝑀𝐿 𝑡𝑔𝜑 ;
7th – Determinate LCG (abscissa of centre of gravity) for the experiments with
the mobile weight aft and fore: 𝐿𝐶𝐺 = 𝐿𝐶𝐵 − 𝑥 ;
8th – Determinate the average value for LCG (using the experiments with the
mobile weight aft and fore): 𝐿𝐶𝐺 =
𝐿𝐶𝐺𝑅é +𝐿𝐶𝐺𝑉𝑎𝑛𝑡𝑒.
2
;
9th – To find the centre of gravity of the empty ship, we only need to subtract the
weights influence:
𝐿𝐶𝐺 =
𝐾𝐺 =
∑ 𝑝𝑖 𝑥𝑖 + 𝐿𝐶𝐺0 𝐷0
∑ 𝑝𝑖 +𝐷0
∑ 𝑝𝑖 𝑧𝑖 + 𝐾𝐺0 𝐷0
∑ 𝑝𝑖 +𝐷0
;
;
Calculation of the inclining moments due to displaced weights on board and
calculation of the righting moment acting on the model. Make use of the concept of
metacentric height, GM.
Figure 14 – Inclining moment (transverse and longitudinal moments)
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Figure 15 – Righting moment (transverse and longitudinal moments)
Figure 16 – Results obtained for the study of inclinig and righting moments
Description of the procedure used to find the incling and righting moments:
1st – Determinate 𝑀𝐼 (transverse inclining moment) for the experiments with the
mobile weight port and starboard, applying the formula: 𝑀𝐼 = 𝑝 ∆𝑦 𝑐𝑜𝑠𝜃 ;
2nd – Determinate the average value for 𝑀𝐼 (using the experiments with the
mobile weight port and starboard): 𝑀𝐼 =
𝑀𝐼 𝐸𝑠𝑡. +𝑀𝐼 𝐵𝑜𝑚𝑏.
2
;
3rd – Determinate 𝑀𝐼 (longitudinal inclining moment) for the experiments with
the mobile weight aft and fore, applying the formula: 𝑀𝐼 = 𝑝 ∆𝑥 𝑐𝑜𝑠𝜑 ;
4th – Determinate the average value for 𝑀𝐼 (using the experiments with the
mobile weight aft and fore): 𝑀𝐼 =
𝑀𝐼 𝑅é +𝑀𝐼 𝑉𝑎𝑛𝑡𝑒
2
;
5th – Determinate 𝑀𝐸 (transverse righting moment) for the experiments with the
mobile weight port and starboard, applying the formula: 𝑀𝐸 = 𝐷 𝐺𝑀 𝑠𝑖𝑛𝜃 ;
6th – Determinate the average value for 𝑀𝐸 (using the experiments with the
mobile weight port and starboard): 𝑀𝐸 =
𝑀𝐸 𝐸𝑠𝑡. +𝑀𝐸 𝐵𝑜𝑚𝑏.
2
;
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
7th – Determinate 𝑀𝐸 (longitudinal righting moment) for the experiments with
the mobile weight aft and fore, applying the formula: 𝑀𝐸 = 𝐷 𝐺𝑀𝐿 𝑠𝑖𝑛𝜑 ;
8th – Determinate the average value for 𝑀𝐸 (using the experiments with the
mobile weight aft and fore): 𝑀𝐸 =
𝑀𝐸 𝑅é +𝑀𝐸 𝑉𝑎𝑛𝑡𝑒
2
;
Considering the obtained values from the transverse moments (inclining and
righting), it is possible to verify that they are coherent, because the absolute value of
both are equal, like we were expecting, since the transverse inclining moment must be
opposite of the transverse righting moment to have the ship model in a stable mode. The
same succeeded with the longitudinal moments (inclining and transverse). We can also
verify that the value for the longitudinal moments is bigger than the value obtained for
the transverse moment, because 𝐺𝑀𝐿 (longitudinal metacentric height) is bigger than
GM (transverse metacentric height).
Comparison of the position of the centre of buoyancy B with the centre of
gravity G, determining the resulting moment. Verification and quantitative
interpretation of the discrepancy BG, relating it to the trim of the model, which is
a crucial aspect of the report. Use the concept of moment to trim one cm, MTC.
Figure 17 – Resulting moment and MTC
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Figure 18 – Results obtained for the resulting moment and MTC
Description of the procedure used to find the resulting moment:
1st – Determinate x, distance between the abscissa of buoyancy centre and
gravity centre: 𝑥 = 𝐿𝐶𝐵 − 𝐿𝐶𝐺;
2nd – Determinate the impulsion and the force applied in centre of gravity:
𝐼=𝐷𝑔;
𝐹𝐺 = 𝑃 𝑔 ;
3rd – Determinate 𝑀𝑅 (resulting moment): 𝑀𝑅 = (𝐼 − 𝑃) 𝑥 ;
Description of the procedure used to find MTC:
1st – Interpolation of the hydrostatic tables to find the 𝐵𝑀𝐿 (longitudinal
metacentric radius) of the loaded model.
2nd – Determinate 𝐼0𝑦𝑦 (moment of inertia): 𝐼0𝑦𝑦 = 𝐵𝑀𝐿 𝑉 ;
3rd – Determinate 𝑀𝑢 (moment to trim one cm): 𝑀𝑢 =
𝜌 𝐼0𝑦𝑦
𝐿𝑝𝑝
;
The distance between the centre of buoyancy and the centre of gravity of the
model with the weights inboard is 35.1 millimeters, and for the empty ship this distance
is 54 millimeters. We obtained a resulting moment of 876468.4 N*mm for the model
with weights inboard and a resulting moment of 1348135.4 N*mm for the empty model.
We can verify that the centre of gravity is aft than centre of buoyancy due to the initial
trim aft of the model.
The moment to trim one cm (MTC) is the required inclining moment to trim one
centimeter. In our case, MTC is equal to 35223.5 grams (0.345 kN). Apllying a
longitudinal inclining moment equal to the obtained in our study we get a trim of 22.6
millimeters.
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Arquitectura Naval
Diogo Silva, nº 72834, LEAN
2013/2014
Comparison and interpretation of these moments and the discrepancies
found. Analyse the possible causes of error. Determine LCG and KG for the light
ship and estimate the absolute and relative errors of these values.
The comparison of the results obtained for the inclining and righting moments
was done two topics ago. We verified that the absolute value of both (inclining and
righting transverse moment/ inclining and righting longitudinal moment) are equal, so it
is possible to affirm that the values obtained for the LCG and KG are acceptable.
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