25-Nav-A1 Fundamentals of Naval Architecture · May-98-Nav-A1 2016

Question 3 of 7: Freeboard, Floodable Length & Corrected Draft Marks After Loading

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations, May 2016 — 98-Nav-A1 Fundamentals of Naval Architecture, 3 hours, closed book (5 questions constitute a complete exam paper, first five as they appear in the answer book are marked, each of equal value; all 7 answered below for full study coverage).

Reference texts: Tupper, Introduction to Naval Architecture; Lewis (ed.), Principles of Naval Architecture (PNA); IMO, International Code on Intact Stability (IS Code).

Question 3: Freeboard, Floodable Length & Corrected Draft Marks After Loading

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Freeboard and floodable length

Freeboard is the vertical distance, measured amidships, from the waterline to the top of the freeboard deck at side (the deck line), i.e. the height of the reserve of buoyancy and watertight/weathertight structure above the waterline. Minimum freeboards are assigned by load-line rules (draft marks and Plimsoll disc) as a function of ship length, block coefficient, superstructure and sheer, to keep enough reserve buoyancy and freeboard-deck immersion margin for the intended service and season.

Floodable length at any point along the ship's length is the maximum length of hull, centred at that point, that can be flooded (to the assumed permeability) without submerging the margin line (a line drawn a standard distance below the bulkhead deck at side). Plotting floodable length along the ship's length gives the floodable-length curve, from which the required transverse bulkhead spacing (subdivision) is derived so that no single compartment, or required group of adjacent compartments, exceeds the permitted floodable length at its centre — the basis of SOLAS/IS Code subdivision and damage-stability requirements.

(b) New draft-mark readings after loading a weight

Given. The draft marks are offset from the perpendiculars, so the readings must first be corrected to true drafts at $FP$/$AP$, a weight is then added, and the new perpendicular drafts are converted back to mark readings.

Given data
Quantity	Symbol	Value
Length	$L$	150.00 m
Forward mark reading (before)	$d_{F,mark}$	7.70 m, 2.00 m aft of $FP$
Aft mark reading (before)	$d_{A,mark}$	8.25 m, 20.00 m fwd of $AP$
Weight added	$w$	250.00 tonnef, 20.00 m fwd of midship
Tonnes per cm immersion	$TPC$	26 t/cm
Moment to change trim 1 cm	$MCT1cm$	250 tonnef·m
Centre of flotation	$LCF$	1.80 m fwd of midship

Find. The new forward and aft draft-mark readings after the weight is placed.

Approach. (1) Extrapolate the mark readings to true drafts at $FP$/$AP$ using the trim rate between the marks. (2) Apply the standard sinkage + change-of-trim procedure at the perpendiculars. (3) Convert the new perpendicular drafts back to mark positions using the ship's new trim rate.

True drafts at the perpendiculars (before loading). The marks are $150-2.00-20.00=128.0\text{ m}$ apart, with observed trim $8.25-7.70=0.550\text{ m}$ over that span, so the trim rate is $0.550/128.0=0.0042969\text{ m/m}$. Extrapolating outward to the perpendiculars: $$d_{FP}=7.70-2.00(0.0042969)\approx7.6914\text{ m}, \qquad d_{AP}=8.25+20.00(0.0042969)\approx8.3359\text{ m}$$
Sinkage and trimming moment from the added weight. $\text{sinkage}=250.0/26=9.615\text{ cm}=0.0962\text{ m}$. Lever from $F$: $20.00-1.80=18.2\text{ m}$ forward of $F$ (weight forward of $F$ ⇒ trim by the head): $$\text{Trimming moment}=250.0(18.2)=4550\text{ tonnef}\cdot\text{m} \ \Rightarrow\ COT=\frac{4550}{250}=18.20\text{ cm}$$
Apportion and apply the trim. $l_F=L/2-1.80=73.2\text{ m}$, $l_A=L/2+1.80=76.8\text{ m}$: $$\Delta d_{FP}=18.20\left(\frac{73.2}{150}\right)\approx8.882\text{ cm (increase)}, \quad \Delta d_{AP}=18.20\left(\frac{76.8}{150}\right)\approx9.318\text{ cm (decrease)}$$ $$d_{FP}'=7.6914+0.0962+0.0888\approx7.8764\text{ m}, \qquad d_{AP}'=8.3359+0.0962-0.0932\approx8.3389\text{ m}$$
Convert back to mark readings. New trim at the perpendiculars is $8.3389-7.8764=0.4625\text{ m}$ over $L=150\text{ m}$, i.e. rate $0.003084\text{ m/m}$: $$d_{F,mark}'=d_{FP}'+2.00(0.003084)\approx7.8825\text{ m}, \qquad d_{A,mark}'=d_{AP}'-20.00(0.003084)\approx8.2772\text{ m}$$ $$\boxed{d_{F,mark}'\approx7.883\text{ m}, \qquad d_{A,mark}'\approx8.277\text{ m}}$$

Question 3 — final results
Quantity	Value
True draft at FP / AP (before)	$7.691$ m / $8.336$ m
Parallel sinkage	$\approx 9.62$ cm
Change of trim	$\approx 18.20$ cm (by the head)
New forward mark reading	$\approx 7.883$ m
New aft mark reading	$\approx 8.277$ m

Topic: Freeboard and subdivision (floodable length); correcting draft-mark readings to the perpendiculars.

Key relations: trim rate $=(\text{aft mark}-\text{fwd mark})/(\text{distance between marks})$; extrapolate outward by each mark's own offset to get true $d_{FP}$, $d_{AP}$; then the standard sinkage/$COT$ procedure; then interpolate back at the (new) trim rate to recover mark readings.

Why this works: Draft marks are painted at a convenient hull location, not necessarily at the perpendiculars, so any straight extrapolation of the observed waterline (constant along the length for a rigid trimmed hull) must be carried the extra distance to $FP$/$AP$ before the naval-architecture formulas (which are defined at the perpendiculars) can be applied — and carried back afterward if the question asks for mark readings rather than perpendicular drafts.

Common pitfall: applying the sinkage/trim formulas directly to the raw mark readings (skipping the perpendicular correction), or forgetting to re-correct the *new* trim back to mark positions using the *new* (not the original) trim rate.

Source: Tupper, Introduction to Naval Architecture, Ch.4 (Trim) & Ch.10 (Subdivision and damage stability); IMO IS Code, Ch.2.

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