Stability NAU 205 Lesson 2

Calculation of the Ship’s Vertical Center of Gravity, KG NAU 205 Ship Stability Steven D. Browne, MT

Center of Gravity (G)

Light Ship KG Inclining Experiment

Moment The product of a quantity and its distance from a reference point. Example: A 40 lb child is sitting 4 ft from the center of a seesaw. What is the moment? Ans: 160 ft-lbs

Moment To balance, both moments must be equal. Example: A 40 lb child is sitting 4 ft from the center of a seesaw. Where must an 80 lb. child sit to balance the seesaw? Ans: 160 ft-lbs = 80 lbs. x distance Distance = 2 ft.

Using moments to find KG

Wt. (in lbs)Dist. (in ft.)Moments (in ft.-lbs) 100 X10=1, X30=4, lbs5500 5,500 ft.-lbs / 250 lbs = 22 ft. G’ is 22 ft. from the left end

Using moments to find KG

KG = total moments / total displacement

Example 1: A vessel floating at her light draft displaces 5,000 tons. The light ship KG is 20’. 200 tons are loaded 10’ above the keel and 300 tons 5’ above the light ship KG. What is the new KG? Using moments to find KG

Example 1: A vessel floating at her light draft displaces 5,000 tons. The light ship KG is 20’. 200 tons are loaded 10’ above the keel and 300 tons 5’ above the light ship KG. What is the new KG? Using moments to find KG Wt. (in tons)KG (in ft.)Moments (in ft.-tons) 5000 X20=100, X10= 2, X25= 7, tons109,500 ft-tons 109,500 ft.-tons / 5,500 tons = 19.9 ft. KG = 19.9 ft.

Example 2: The vessel in example 1 has the following weights removed: 700 tons 5’ above the keel, 300 tons 2’ above the keel and 150 tons 15’ above the keel. What is the new KG? Using moments to find KG

Wt. (in tons)KG (in ft.)Moments (in ft.-tons) 700X 5=3, X 2= X15=2,250 1,150 tons6,350 ft-tons Original5,500 tons 109,500 foot-tons Discharged-1,150 tons- 6,350 foot-tons Final 4,350 tons 103,150 foot-tons New KG = 103,150 ft-tons / 4,350 tons = 23.7 feet Example 2: The vessel in example 1 has the following weights removed: 700 tons 5’ above the keel, 300 tons 2’ above the keel and 150 tons 15’ above the keel. What is the new KG? Using moments to find KG

Wt. (in tons)KG (in ft.)Moments (in ft.-tons) 5,500X19.9=109, X 5= -3, X 2= X15= -2,250 4,350 tons103,150 ft-tons New KG = 103,150 ft-tons / 4,350 tons = 23.7 feet Example 2 (Second Method): The vessel in example 1 has the following weights removed: 700 tons 5’ above the keel, 300 tons 2’ above the keel and 150 tons 15’ above the keel. What is the new KG? Using moments to find KG

Calculating the shift of G (GG’) Loading, discharging or shifting 1 load: Method 1: KG’ = total moments / total displacement GG’ = [KG – KG’] Method 2: GG’ = (w x d) / Δ w = weight of load d = distance of shift or distance from G (for loading and discharging) Δ = new displacement

Calculating the shift of G (GG’) Rules about the movement of G: 1.When a weight, g, is loaded, the vessel’s center of gravity, G, will move toward the loaded weight. 2.When a weight, g, is discharged, G will move away from the discharged weight. 3.When a weight, g, is shifted, G will move parallel to the shifted weight. 4.When a weight, g, is suspended, G will move toward the point of suspension.

Calculating the shift of G (GG’) Example 3: Suppose that on a 10,000 ton vessel with a KG of 25 ft, 200 tons are shifted vertically upwards 20 ft. What is GG’? Method 1 Wt. (tons)KG (ft.)Moments (ft.-tons) Initial Cond.10,000X25 = 250,000 Shift200X20+ 4,000 Final10, ,000 KG = 254,000 ft.-tons / 10,000 ft.-tons = 25.4 ft. GG’ = 25 ft. – 25.4 ft. = 0.4 ft. upwards

Calculating the shift of G (GG’) Example 3: Suppose that on a 10,000 ton vessel with a KG of 25 ft, 200 tons are shifted vertically upwards 20 ft. What is GG’? Method 2: GG’ = (200 tons x 20 feet) / 10,000 tons GG’ = 0.4 ft upwards

Example 4: 300 tons of salt water are loaded in a stbd. deep tank with the center 10 ft. above the keel. Displacement before loading was 9,700 tons, KG of 25 ft. Find the vertical shift of G. Calculating the shift of G (GG’) GG’ = (300 tons x 15 ft) / 10,000 tons GG’ = 0.45 ft downwards

Example 5: A tank weighing 60 tons is discharged from the portside of an upper deck with a KG of 45 ft. Displacement of the ship before discharging was 6,060 tons, KG 20 ft. Find the vertical shift of G. Calculating the shift of G (GG’) GG’ = (60 tons x 25 ft) / 6,000 tons GG’ = 0.25 ft downwards

GG’ with a suspended weight

Example 6: You need to lift a 30 ton container. The head of your jumbo boom is 80 ft. above the container. The ship’s displacement 10,000 tons. What is GG’ the moment the container is off the deck? GG’ with a suspended weight GG’ = (30 tons x 80 ft) / 10,000 tons GG’ = 0.24 ft upwards

QUESTIONS? Stability