Change in Trim Due to Change in Density

8/13/2019 Change in Trim Due to Change in Density

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Change In Trim Due To Change In Density

Prepared by Capt. B. Ceylan

When a ship passes 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 1float in salt water at the waterline WL. Brepresents the position of thecenter of buoyancy and the Gthe center of the gravity. For equilibrium, Band Gmust be in thesame vertical line.

If the ship now passes into the fresh water, the mean draft will increase. Let W1L1represent thenew waterline and bthe center of gravity of the extra volume of water displaced. The center of

buoyancy of the ship will move from Bto B1in the direction directly towards b. The force ofbuoyancy now acts vertically upwards through B1and the ships weight acts vertically downwardthrough G. The ship will then change trim to bring the centers of gravity and buoyancy back in tothe same vertical line.

: displacementG: center of gravityB: center of buoyancyB1: shifted center of buoyancyF: center of floatation

q:density

q1: reduced densityd: difference between LCBand LCFV:underwater volume of the vesselv: 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


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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/m3water density. She is to enter water density 0.9954 ton/m3. Find her drafts Fwd, Mid and Aft indock water. (Figures are taken from the hydrostatic table of M/V North Princess)

MCTC = 974.57 ton m/cm.TPC = 64.4 ton/cmLCF = 0.62 (A)LCB = (-) 5.90 (F)Length = 216.9 mDisplacement = 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


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= (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/Lx 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)

2nd Method:

Change of Trim= [x {D2 D1} x {LCF LCB}] / D1 x MTC2

D1= 11.969 mD2= 12.024 mLCF= 0.62LCB = (-) 5.90MTC2 = 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 = (-) 5.90LCB2 @ 12.024 = (-) 5.87

Trim = [{5.90 5.87} x 70650.50] / 974.57 x 100

= 0.022 m

End...

Fwd (m) Mid (m) Aft (m)

Initial Draft 11.914 11.900 12.024

Sinkage 0.055 0.055 0.055

11.969 11.955 12.079

Trim (+) 0.011 (-) 0.011

Final Draft 11.98 11.955 12.068


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Change in Trim Due to Change in Density