Scribd
Upload
621 views5 pages

Stability Calculation

The document provides examples for calculating ship stability by finding the angle of list from transverse cargo shifts. It includes 5 examples of ships where cargo is shifted transversely by various amounts, and calculations are shown to find the angle of list using the formula: Tanθ = Listing Moment / (Ship Weight x GM). Key information includes ship weights from 5000-12000 tons, GM values from 0.3-1.2 meters, and cargo shifts ranging from 20-200 tons moving distances of 1.5-20 meters. The examples demonstrate how to calculate the new GM and find the angle of list caused by off-center cargo shifts.

Uploaded by

Lyne Angel Alvarez
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
621 views5 pages

Stability Calculation

The document provides examples for calculating ship stability by finding the angle of list from transverse cargo shifts. It includes 5 examples of ships where cargo is shifted transversely by various amounts, and calculations are shown to find the angle of list using the formula: Tanθ = Listing Moment / (Ship Weight x GM). Key information includes ship weights from 5000-12000 tons, GM values from 0.3-1.2 meters, and cargo shifts ranging from 20-200 tons moving distances of 1.5-20 meters. The examples demonstrate how to calculate the new GM and find the angle of list caused by off-center cargo shifts.

Uploaded by

Lyne Angel Alvarez
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 5

Stability Calculation

Intro – ( first slide )


Learning objective

why calculatingthe angle of list and its correction important to statical stability?
-

( 2nd slide )

1. On a ship of W 5000t , GM 0.3m, 20t was shifted transversely by 5m. Find the list.

Solution:

W = 5000t, GM = 0.3m, Weight shifted (w) = 20t & d= 5m

Listing moment = (20 x 5 ) = 100tm Tanθ = ( Listing Moment) /(W x GM) = 100/(5000 x 0.3) = 3.8 degree.

( 3rd slide )
why calculating the angle of list and its correction important to statical stability?
- Hernandez daw explain

(4th slide)
2. On a ship of W 8000t, GM 2.0m, If the following transverse shifting were done , find the
list:

• 200cargo shifted 4m to stbd


• 100t cargo shifted 2 m to port
• 100t cargo shifted 4m to port
• 50t stores shifted 20m to stbd

Solution:-

Listing moment due to shifting of cargo LM(1) = ( weight x distance ) = (200 x 4) = 800 tm (S)

Again, LM (2) = (Weight x distance ) = (100 x 2) = 200 tm (P)

LM(3) = ( weight x distance ) = (100 x 4) = 400 tm (P)

LM(4) = ( weight x distance ) = (50 x 20) = 1000 tm (S)

Final listing moment (LM) = LM(1) + LM(2) +LM(3)+ LM(4) = 800(S) + 200(P) + 400(P) + 1000(S) =1800(S)
+ 600(P)

So, final listing moments = (1800 – 600) = 1200 tm (S)

Now, Tanθ = (Final LM )/(W x GM) = (1200/(8000 x 2 ) = 4.29 degree (S)

( 5th slide)
3. If 200t cargo was shifted downwards by 10m and to starboard by 5m on a ship of W 10000t
, KG 7.0m, KM 7.4m, find the list.

Solution :-

Given : Cargo shifted (w) = 200t, Distance = 10m, d = 5m from CL to stbd W = 10000 t , KG = 7.0m & KM =
7.4m

Final W =10000t Final VM= 68000 tm Final KG = (final VM)/ (Final W) Final KG = (68000/10000) = 6.8m

Final GM= (KM – KG) = (7.4 – 6.8) = 0.6m

Listing moment (LM) = (weight x distance ) = (200 x 5) = 1000tm

tanθ = LM/(W x GM) = 1000 /(10000 x 0.6) = 9.46 degree

( 6th slide )
4. A quantity of grain estimated to be 100t shifts transeversly by 12m and upwards by 1.5m,
on a ship of W 12000t , GM 1.2m. Find the list caused.

Solution:

Grain shifted (w) = 100t, Distance = 12m & transversely, d = 1.5( î) W = 12000 t, GM = 1.2m

LM caused = (weight x distance) = (100 x 12) = 1200 tm

GG1 (î) = (distance x weight) /W = (1.5 x 100) / 12000 = 0.0125 m

Final GM = (GM – GG1) = (1.2 – 0.0125)m = 1.1875m

Tanθ = LM /(W x GM ) = 1200/(12000 x 1.1875) = 4 degree 8’.

( 6th slide )
5. A ship displaces 4950t and has KG 4.85m, KM 5.79m. cargo weighing 50t is loaded 1.25m
above the keel and 4m port of the centre line . Find the list

Solution:

W = 4950 t , KG = 4.85m KM = 5.79m Weight loaded (w) = 50 t KG = 1.25m d = 4 m to CL to port

LM caused = (weight x distance ) = (50 x 4) = 200 tm to port .

Final W = 5000 t Final VM = 24070 tm Final KG = (Final VM)/( Final W) Final KG = (24070/5000) =
4.814m

Final GM = (KM –KG) = (5.79 – 4.814) m = 0.976m.

Tanθ = LM/(Weight x GM) = 200/ (5000 x 0.976) θ = 2.34 degree

( 7th slide )

SEATWORK

You might also like