25-Nav-A1 Fundamentals of Naval Architecture · May-16-Nav-A1 2017
Question 1 of 5: Parallel Sinkage and Change of Trim from an Added Weight
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Tupper, Introduction to Naval Architecture; Lewis (ed.), Principles of Naval Architecture (PNA); IMO, International Code on Intact Stability (IS Code).
Question 1: Parallel Sinkage and Change of Trim from an Added Weight
Given. A ship at level trim takes on a single weight aft of both amidships and the centre of flotation, so it both sinks bodily and trims by the stern.
Given data
Quantity
Symbol
Value
Length
$L$
529 ft
Initial draft (level trim)
$d_0$
21.25 ft
Displacement
$\Delta$
8880 LT
Tons per inch immersion
$TPI$
52 LT/in
Moment to trim 1 inch
$MT1$
1601 ft·LT/in
Centre of flotation
$LCF$
45 ft aft of amidships
Weight added
$w$
120 LT, 122 ft aft of amidships
Find. The final forward draft $d_F'$, aft draft $d_A'$, and the mean draft.
Profile schematic (not to scale): the weight sits aft of the centre of flotation $F$, so the ship sinks bodily and trims by the stern — $WL_0$ is the initial waterline, $WL_1$ the final trimmed waterline.
Approach. Compute the bodily (parallel) sinkage from $TPI$, the total change of trim from the trimming moment about $F$ and $MT1$, then apportion the change of trim to the forward and aft perpendiculars by their distances from $F$.
Parallel sinkage. The weight acts uniformly on the whole waterplane through $TPI$:
$$\text{sinkage}=\frac{w}{TPI}=\frac{120}{52}=2.308\text{ in}=0.192\text{ ft (at both ends)}$$
Trimming moment and total change of trim. The weight's lever from $F$ is $122-45=77\text{ ft}$ aft of $F$, so the ship trims by the stern:
$$\text{Trimming moment}=w(77)=120(77)=9240\text{ ft}\cdot\text{LT}\ \Rightarrow\ COT=\frac{9240}{MT1}=\frac{9240}{1601}\approx5.771\text{ in}$$
Apportion the trim about $F$. Distances from $F$ to the perpendiculars: $l_F=L/2+45=309.5\text{ ft}$, $l_A=L/2-45=219.5\text{ ft}$ (check: $l_F+l_A=529\text{ ft}=L$):
$$\Delta d_F=COT\cdot\frac{l_F}{L}=5.771\left(\frac{309.5}{529}\right)\approx3.377\text{ in (decrease)},\qquad \Delta d_A=COT\cdot\frac{l_A}{L}=5.771\left(\frac{219.5}{529}\right)\approx2.395\text{ in (increase)}$$
New drafts. Forward draft gets the parallel sinkage minus its share of trim; aft draft gets the parallel sinkage plus its share:
$$d_F'=21.25+0.192-\frac{3.377}{12}\approx21.161\text{ ft},\qquad d_A'=21.25+0.192+\frac{2.395}{12}\approx21.642\text{ ft}$$
$$\boxed{d_F'\approx21.161\text{ ft},\quad d_A'\approx21.642\text{ ft},\quad d_{mean}=\frac{d_F'+d_A'}{2}\approx21.401\text{ ft}}$$
