Change In Trim Due To Change In Density 4t714e

Change In Trim Due To Change In Density Prepared by Capt. B. Ceylan When a ship es from water of one density to water of another the hydrostatic draft changes. Furthermore, the change in the position of the center of the buoyancy may cause the trim to change.

Figure 1 Let the ship in Figure 1 float in salt water at the waterline WL. B represents the position of the center of buoyancy and the G the center of the gravity. For equilibrium, B and G must be in the same vertical line. If the ship now es into the fresh water, the mean draft will increase. Let W1L1 represent the new waterline and bthe center of gravity of the extra volume of water displaced. The center of buoyancy of the ship will move from B to B1in the direction directly towards b. The force of buoyancy now acts vertically upwards through B1 and the ship’s weight acts vertically downward through G. The ship will then change trim to bring the centers of gravity and buoyancy back in to the same vertical line. ∆: displacement G: center of gravity B: center of buoyancy B1: shifted center of buoyancy F: center of floatation q: density q1: reduced density d: difference between LCB and LCF V: underwater volume of the vessel v: increment of volume Change of Trim = {W x BB1} / MCTC * The horizontal shift of the center of buoyancy BB1 can be found from BB1 = v x d / V + v * Moment changing trim = ∆ x BB1 * The increment of volume can be found from

v = Sinkage x area of waterplane v = s x[(TPC x 100)/q] Sinkage can be found from (D + s) / D = q / q1 equals s = D x [(q/q1) - 1]

Example: A vessel is floating at drafts: Fwd: 11.914 m, Mid: 11.900 m and Aft: 12.024 m in 1.000 ton/m3 water density. She is to enter water density 0.9954 ton/m3. Find her drafts Fwd, Mid and Aft in dock water. ( Figures are taken from the hydrostatic table of M/V North Princess) MCTC = 974.57 ton m/cm. TPC = 64.4 ton/cm LCF = 0.62 (A) LCB = (-) 5.90 (F) Length = 216.9 m Displacement = 70650.50 tons Initial mean draft: dml = da – ( l/L ) x t = 12.024– (107.865 / 216.9) x 0.11 = 11.969 m (Initial Mean Draft also can be found from (11.914 + 12.024) / 2 = 23.938 / 2 = 11.969 m) Final mean draft: dmF = dml x q1/qF = 11.969 x 1.000/0.9954 = 12.024 m Sinkage: s = dmF – dml = 12.024 – 11.969 = 0.055 m V=∆/q = 70650.50 / 1.000 m3 = 70650.50 m3 v = V ( q1 / qF – 1 ) = 70650.50 x ((1.000 / 0.9954) – 1) = 326.49 m3 BB1 = v x d / V + v

= (326.49 x (5.90+0.62)) / (70650.50 + 326.49) = 2128.71/ 70976.99 m = 0.03 m Trim = moment changing trim / MCTC = ∆ x BB1 / MCTC = 70650.50 x 0.03 / 974.57 = 2.2 cm = 0.022 m by Fwd Change of trim aft = l/L x change of trim = 107.865 / 216.9 x 2.2 = 1.1 cm = (-) 0.011 m (aft) Change of trim forward = 2.2 – 1.1 = 1.1 cm = (+) 0.011 m (fwd)

Initial Draft Sinkage Trim Final Draft

Fwd (m)

Mid (m)

Aft (m)

11.914

11.900

12.024

0.055

0.055

0.055

11.969

11.955

12.079

(+) 0.011

(-) 0.011

11.98

11.955

12.068

2nd Method: Change of Trim= [∆ x {D2 – D1} x {LCF – LCB}] / D1 x MTC2 D1= 11.969 m D2= 12.024 m LCF= 0.62 LCB = (-) 5.90 MTC2 = 976.22 Trim = [70650.50 x {12.024 – 11.969} x {0.62+5.90}] / 11.969 x 976.22 = 0.022 m by Fwd 3rd Method: Change of Trim = [{LCB1 - LCB2} x W] / MTC x 100 LCB1 @ 11.969 = (-) 5.90 LCB2 @ 12.024 = (-) 5.87 Trim = [{5.90 – 5.87} x 70650.50] / 974.57 x 100 = 0.022 m

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