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- 1. TrimPengantar Teknologi Kelautan Adi Wirawan Husodo
- 2. • Trim is also known as ‘longitudinal stability’• Trim is measured as the difference between the draft forward and aft. – If difference is zero ship is on even keel – If forward draft is greater than aft draft trimming by bow – If aft draft is greater than forward draft trimming by stern
- 3. • Consider a ship to be floating at rest still water and on an even keel (fig 15.1)• Now let a weight w, already on board, be shifted aft through a distance d. The center of gravity of the ship will shift from G to G1, parallel to the shift of center of gravity of the weight shifted. GG1= (w x d)/W W x GG1=w x d…….trimming moment
- 4. • When trimmed (fig 15.2), the wedge of buoyancy LFL1 emerges and wedge WFW1 is immersed• The volume of immersed wedge = the volume of emerged wedges• F , the point which the ship trim, is the center of gravity of water plane area (center of flotation or tipping centre)• ML (longitudinal metacentre) is the point of intersection between verticals through the longitudinal positions of centre of buoyancy. The vertical distance between the center of gravity and the longitudinal metacentre is called longitudinal metacenrric height (GML)
- 5. Moment to change Trim one cm (MCT 1 cm)• Moment required to change trim by 1 cm• MCT 1 cm = (W x GML)/100L W : vessel’s displacement, ton GML : long. Metacentric height, m L : Vessel’s length, m• Fig 15.3 (a) ship in equilibrium• Fig 15.3 (b) weight w is shifted through d distance. The center of gravity will shift from G to G1• Causing the trimming moment of W x GG1
- 6. Moment to change Trim one cm (MCT 1 cm)• See fig 15.3 (c)• The ship will trim to bring the centre of gravity and buoyancy into the same vertical line the ship is again in equilibrium• The tipping centre (F) is l metre from aft GG1 = (w x d)/W = GML tan θ tan θ =(w x d)/(W x GML) tan θ = t/L…….fig 15.4.(b)
- 7. Moment to change Trim one cm (MCT 1 cm)• Let the change of trim due to shifting is 1 cm• Then w x d is the moment to change trim 1 cm• tan θ= 1/100 LBut• tan θ=(w x d)/(W x GML)• tan θ= MCT 1 cm/(W x GML)• MCT 1 cm/(W x GML) = 1/100 L MCT 1 cm = (W x GML)/100 L
- 8. To find the change of draft due to change of trim• Trim will cause a change in the draft forward and draft aft. One will be increased and one will decreased.• A is a new draft aft, F is a new draft forward. Trim = A-F• x is the change of draft aft, y is the change draft forward• See the triangles WW1F1 and WW1C
- 9. To find the change of draft due to change of trim x cm t cm lm Lm l m x t cm x cm Lm• Change of draft aft in cm = (l/L) x Change of trim in cm• l = distance F from aft, m• L = ship’s length, m• x+y=t• Change of draft F in cm = Change of trim – Change of draft A
- 10. The effect of shifting weight already on boardA ship 126 m long is floating at draft of 5.5 m F and 6.5 m A. The centre of flotation is 3 m aft of amidships. MCT 1 cm = 240 ton. m, displacement 6000 tons. Find the new drafts if a weight 120 ton already on board is shifted forward a distance of 45 m.Trimming moment =wxd = 120 x 45 = 5400 ton m by the headChange of trim = trimming moment/MCT 1 cm = 5400/240 = 22.5 cm by the headChange of draft aft = (l/L) x Change of trim = (60/126) x 22.5 = 10.7 cm
- 11. The effect of shifting weight already on boardChange of draft Forward = Change of trim – Change of draft Aft = 22.5 – 10.7 = 11.8 cmOriginal draft 6.500 m A 5.500 m FChange due trim -0.107 m +0.118 mNew draft 6.393 m 5.618 m
- 12. The effect of loading and/or discharging weight• When a weight is loaded at the centre of flotation it will produce no trimming moment, but the ship’s draft will increase uniformly so that the ship displaces an extra weight of water equal to the weight loaded. If the weight is now shifted forward or aft away from the center of flotation, it will cause a change of trim. cause both the bodily sinkage and a change of trim• When a weight is being discharged, if the weight is first shifted to the centre of flotation it will produce a change of trim, and if it is then discharged from the centre of flotation, the ship will rise bodily. TPC = the mass which must be loaded or discharged to change w Bodily sinkage or rise cm the ship’s mean draft by 1 cm TPC (ton) WPA TPC ton 97.56 WPA = water plane area, m2
- 13. The effect of loading and/or discharging weightA ship 90 m long is floating at drafts 4.5 m F and 5.0 m A. The centre of flotation is 1.5 m aft of amidships. TPC 10 tons. MCT 1 cm is 120 ton m. Find the new drafts if a total weight of 450 tons is loaded in a position 14 m forward of amidshipsBodily sinkage = w/TPC Change of trim = trim moment/MCT 1 cm = 450/10 = (450 x 15.5)/120 = 45 cm = 58.12 cm by the headChange of draft aft = (l/L) x Change of trim = (43.5/90) x 58.12 = 28.09 cmChange of draft Forward = Change of trim – Change of draft Aft = 58.12 – 28.09 = 30.03 cm
- 14. The effect of loading and/or discharging weightOriginal draft 5.000 m A 4.500 m FBodily sinkage + 0.450 m + 0.450 m 5.450 m 4.950 mChange due trim - 0.281 m + 0.300 mNEW DRAFT 5.169 m A 5.250 m F

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