Definitions Trim is the difference between the draft

Definitions Trim is the difference between the draft at the forward perpendicular (FP) and the draft at the aft perpendicular(AP). If there is no difference, ship is on even keel. Change of Trim is the difference between the original trim and the final trim. Center of Floatation (LCF) or Tipping center(TC) is the geometrical 2 -dimensional center of the waterplane. It is the point about which the ship trims. In effect it is the fulcrum of the waterplane. Trim ratio aft k Upper dec θ LCF θ C. o f T. θ P Length LB Trim ratio for’d FP AP

Trim ratio aft k Upper dec LCF θ θ C. o f T. Trim ratio for’d θ P Length LB AP tanθ= Change of trim/ LBP tanθ= Trim ratio for’d/LCF to FP tanθ= Trim ratio aft / LCF to AP Change of Trim x LCF to FP/LBP = trim ratio for’d in meters Change of Trim x LCF to AP/LBP = trim ratio aft in meters FP

Trimming moment If a weight is shifted on a vessel in a fore and aft direction, or is loaded or discharged forward or aft of the tipping center, a trimming moment is created. MTI( the moment to change trim one inch) Using the English system, suppose that a certain trimming moment is sufficient to change the trim of a vessel one inch. MTC (the moment to change trim one centimeter) Using the metric system, however, the trimming moment is expressed in meter-metric tons and the change of trim is in centimeters.

True mean draft = draft amidships + correction ZF= KY = KX + XY Then, SO: Correction XY= When center of flotation is in the same direction from amidships as the maximum draft, the correction to be added to the mean of the drafts

A ship’s minimum permissible freeboard is at a true mean draft of 8. 5 m. The ship’s length is 120 m, center of flotation being 3 m aft of amidships. TPC = 50 tons. The present drafts are 7. 36 m F and 9. 00 m A. Find how much more cargo can be loaded. Trim = Draft – draft forward= 9. 00 -7. 36 m = 1. 64 m by the stern Correction = t x FY/L = 1. 64 x 3 / 120 = +0. 04 m True mean draft = draft amidships + correction = (9. 00 + 7. 36)/2 + 0. 04 m = 8. 22 m Load mean draft = 8. 50 m, so increase in draft = 0. 28 m = 28 cm Cargo to load = increase in draft required x TPC = 28 x 50 = 1400 tons

Change of trim caused by loaded or discharged weights Example 1 Shifted Weight. The weight of 100 tons of seawater is pumped from fore peak to the after peak, a distance of 400 feet. MTI at the vessel’s mean draft of 20 feet (21 feet forward; 19 feet aft) is 1, 000 foot-tons. What is the change of trim and the new draft ? (assume the change in trim will be equal at both ends) Chang of Trim = Trimming moment /MT 1 = (100 tons)(400 feet)/1, 000 foot-tons/inch = 40 inches by stern Forward Initial drafts Mean Aft 21’ 00” 20’ 00” 19’ 00” Change of trim 1’ 08” (-) Final drafts 19’ 04” 1’ 08” (+) 20’ 00” 20’ 08”

Change of trim caused by loaded or discharged weights Example 2, loaded weight. 55 tons are loaded on a vessel 40 feet forward of the tipping center and 80 tons, 60 feet aft of the tipping center. What effect does this loading have on the vessel’s initial drafts of 16 feet forward and 15 feet 6 inches aft? TPI is 45 and MT 1 is 867 (picked off deadweight scale for the draft after loading). Change in mean draft = weight loaded/TPI = (135 tons)/(45 tons/inch)= 3 inches increase Trimming moment: 80 tons x 60 feet = 4, 800 foot- tons(aft) 55 tons x 40 feet = 2, 200 foot-tons(forward) Net trimming moment: 2600 foot-tons (AFT) Change of trim = (2600 foot-tons)/867 foot-tons/inch = 3 inches (aft)

Forward Mean Aft Initial drafts 16’ 00” 15’ 09” 15’ 06” Change of mean draft 0’ 03”(+) 16’ 03” 16’ 00” 15’ 09” Change of trim Final draft 0’ 01. 5”(-) 16’ 01. 5” 0’ 01. 5”(+) 16’ 00” 15’ 10. 5” The tipping center does not coincide with amidships except for one draft (on some vessels it never coincides)

Calculating Exact distribution of trim change Example 3 Distribution of Trim Change. Given a ship with a length between the forward and after draft marks of 400 feet, and its center of flotation 10 feet aft of amidships, if the change of trim is 8 inches by bow, how does the draft change forward and aft? Trim ratio for’d = Change of Trim x LCF to FP/LBP = (8 inches)(210 feet/ 400 feet) = +4. 2 inches Trim ratio aft = Change of Trim x LCF to AP/LBP = (8 inches)(190 feet/ 400 feet) = -3. 8 inches

Example 4 A vessel 120 m long MTC is 100 tons-meters, TPC is 25, draft forward is 6. 00 meter and draft is 6. 60 meter. A weight of 250 tons is loaded 12 meter forward of the center of flotation which is 2 meter abaft amidships. Calculate the new end drafts forward and aft. Change of draft = weight loaded/TPC = 250/25 = 10 cm = 0. 1 meter Change of trim = Trimming moment/MTC = 250 x 12/100 = 30 cm by the head Trim ratio for’d = Change of Trim x LCF to FP/LBP = (30 cm)(62 meter/ 120 meter) = +15. 5 cm Trim ratio aft = Change of Trim x LCF to AP/LBP = (30 cm)(58 meter/ 120 meter) = -14. 5 cm Forward Aft Initial drafts 6. 00 meter 6. 60 meter Change of mean draft 0. 1 meter(+) 6. 1 meter 6. 70 meter Trim Ratio +0. 155 meter -0. 145 meter Final draft 6. 255 meter 6. 56 meter

Example 5. A vessel 150 m in length, 18 m in breadth, MTC 150 tons-meters, TPC 25 tons/cm is drawing 6. 35 m F and 6. 65 m A and loads are following: 230 tons in No. 1 hold 50 m forward of the LCF 800 tons in N 0. 3 hold 20 m forward of the LCF 500 tons in NO. 4 hold 21 m abaft the LCF She discharges 200 tons from No. 2 hold 36 m forward of the LCF She discharges 105 tons from F. P. 60 m forward of the LCF The center of flotation is 5 m abaft amidships. Calculate the new end drafts. Change in mean draft = weight loaded/TPI = (1225 tons)/(25 tons/cm)= 49 cm increase Trimming moment: 230 tons x 50 m = 1, 1500 m- tons(forward) 800 tons x 20 m = 16, 000 m-tons(forward) Change of trim 500 tons x (-21 )m = -10, 500 m-tons (aft) =3500/150 -200 tons x 36 m= - 7, 200 m- tons (aft) =23. 3 by the head -105 tons x 60 m = - 6, 300 m- tons(aft) Net trimming moment: 3, 500 m-tons (forward) Trim ratio forward= 80/150 x 23. 33= 12. 44 cm increase Trim ratio aft= 70/150 x 23. 33 = 10. 89 cm decrease

Calculating MTC and MT 1 ML Change of trim LBP G’ θ B’ AP G LCF θ B θ FP K w x d is the moment which has changed the trim one Tanθ=GG’/GML= Change of trim/length of the vessel GG’ = GML/100 L (if change of trim is 1 cm) MTC = W x GML/ 100 L tons-meters centimeter. GML/100 L = w x d / W

Approximation of MTC for a box-shaped vessel M L θ LBP G ’ LC B F ’ AP G B FP K MTC = W x GML/100 L now GML ≈BML and BML = L 2/12 d Hence, Where, T= TPC= 1. 025 x L x B /100 For oil tankers, ; For General Cargo ships,

Another formula for MT 1 is : b k 0. 65 28 0. 70 29 0. 75 30 0. 80 31 MTI = k x (TPI)2/B Where B = breadth k = constant depending upon the value of block coefficient, b Example. A vessel in a certain draft condition has a block coefficient of 0. 75, breadth of 60 feet, and TPI of 50. Calculate MT 1 = 30 x (50)2/60 = 1, 250 foot-tons/inch

Example 5 A vessel of 6600 tons displacement 120 m in length, GM L 140 m is drawing 4. 8 m F 4. 5 m A. the center of flotation is 2 m abaft amidships. How much cargo should be discharged from No. 2 Hold which is 16 m forward of amidships, so that the vessel would be trimmed 15 cm by the stern. 2 m LCF 4. 5 m Length LBP=120 m w 16 m 4. 8 m AP FP

LCF 4. 5 m 2 m 16 m w Length LBP=120 m 4. 8 m AP Present draft 4. 80 m F, 4. 50 m A; Present trim = 0. 30 by the head Required trim = 0. 15 by the stern; then change of trim required = 0. 45 m by the stern MTC = W x GML/100 L = 6600 x 140/100 x 120 = 77 tons-meters The moment caused = the moment required 18 w = 45 x 77; w = 192. 50 tons FP

Change of draft at one end only A vessel floating at a draft of 21 feet forward and aft is to be loaded in such a way as to have final drafts of 21 feet forward and 22 feet aft. Where must the weight be loaded? How many tons? TPI is 48. MT 1 is 1, 035. Change of trim is 12 inches Change of mean draft is 6 inches 6 x 48 is 288 tons to be loaded ( change in mean draft x TPI) Change in trim = trimming moment/MTI so, 12 = 288 d/1, 035 d = 12 x 1, 035/288 = 43. 1 feet aft of center flotation

Example 6 A vessel drawing 6. 75 m forward, 7. 75 m aft, MTC 140 tons-meters, TPC 15 has cargo space available in No. 2 and 4 holds, 50 m forward and 40 meters abaft the center of flotation which is at amidships. How much cargo should be loaded in each hold if the ship is to complete loading with a mean draft of 8. 0 and trimmed 15 cm by stern? Present draft : 6. 75 m F 7. 75 m A Trim now : 1. 00 m by stern Trim required : 0. 15 m by stern Change of trim required: 0. 85 by head Present Mean draft 7. 25 m Mean draft required 8. 00 m Change of mean draft : 75 cm Change of trim = (w x d) / MTC then( w x d )= change of trim required x MTC = 85 x 140 = 11900 Cargo to be loaded= change of mean draft x TPC= 75 x 15= 1125 tons Let w tons to be loaded in No. 2 hold, then (1125 -w) tons will be loaded in No. 4 hold 50 w- 40(1125 -w)=11900, then w = 632. 22 tons in No. 2 hold and 492. 78 tons in No. 4 hold