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

Question 5 of 7: Waterplane Geometry by Simpson's Rule

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 5: Waterplane Geometry by Simpson's Rule

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.

Given. Eleven equally-spaced half-ordinates (10 stations FP to AP) of the design waterplane, and the ship's displacement, to be integrated by Simpson's First Rule.

Half-ordinates $y_i$ (m), Simpson multipliers $S_i$
Station	FP	2	3	4	5	6 (mid)	7	8	9	10	AP
$y_i$	0.00	6.40	9.75	11.25	12.35	13.10	13.35	13.30	13.10	12.20	11.28
$S_i$	1	4	2	4	2	4	2	4	2	4	1

Find. Waterplane area $A_W$; longitudinal centroid distance from midship; transverse inertia $I_T$; longitudinal inertia about the centroid $I_L$; $BM_T$; $BM_L$.

Half-breadth curve $y_i$ vs. station (waterplane outline, one side; multiply by 2 for full breadth).

Approach. Apply Simpson's First Rule with interval $h=L/10$ to the area, then to the first moment about midship (for the centroid), then to $\int y^3\,dx$ for $I_T$ and to $\int x^2\,y\,dx$ for $I_L$ about midship, shifting $I_L$ to the true centroid with the parallel-axis theorem.

Interval and area. $h=135.00/10=13.5\text{ m}$. $\Sigma(S_i y_i)=333.38$, so $$A_W=2\cdot\frac{h}{3}\Sigma(S_iy_i)=2\left(\frac{13.5}{3}\right)(333.38)\approx\boxed{3000.4\text{ m}^2}$$
Centroid from midship. With $x_i$ measured $+$forward from midship ($x=\pm67.5,\pm54.0,\ldots,0$), $\Sigma(S_iy_ix_i)=-2533.95$, so the first moment is $M=2(h/3)(-2533.95)\approx-22{,}805.6\text{ m}^3$, and $$\bar{x}=\frac{M}{A_W}=\frac{-22{,}805.6}{3000.4}\approx\boxed{-7.60\text{ m (i.e. 7.60 m aft of midship)}}$$
Transverse inertia $I_T$. A strip of half-breadth $y$ contributes $\tfrac{2}{3}y^3\,dx$ about the centreline. $\Sigma(S_iy_i^3)=48{,}721.2$, so $$I_T=\frac{2}{3}\cdot\frac{h}{3}\Sigma(S_iy_i^3)=\frac{2}{3}(4.5)(48{,}721.2)\approx\boxed{1.462\times10^5\text{ m}^4}$$
Longitudinal inertia $I_L$ about the centroid. $\Sigma(S_ix_i^2y_i)=424{,}259.8$, so about midship $I_{L,mid}=2(h/3)(424{,}259.8)\approx3.8183\times10^6\text{ m}^4$. Shifting to the true centroid ($\bar{x}=-7.60\text{ m}$) by the parallel-axis theorem: $$I_L=I_{L,mid}-A_W\bar{x}^2=3.8183\times10^6-3000.4(7.60)^2\approx\boxed{3.645\times10^6\text{ m}^4}$$
$BM_T$ and $BM_L$. $\nabla=\Delta/\rho=72{,}100.0/1.025\approx70{,}341.5\text{ m}^3$: $$BM_T=\frac{I_T}{\nabla}=\frac{1.462\times10^5}{70{,}341.5}\approx\boxed{2.078\text{ m}}, \qquad BM_L=\frac{I_L}{\nabla}=\frac{3.645\times10^6}{70{,}341.5}\approx\boxed{51.82\text{ m}}$$

Question 5 — final results
Quantity	Value
Waterplane area $A_W$	$\approx 3000.4\text{ m}^2$
Centroid from midship	$\approx 7.60$ m aft
Transverse inertia $I_T$	$\approx 1.462\times10^5\text{ m}^4$
Longitudinal inertia $I_L$ (about centroid)	$\approx 3.645\times10^6\text{ m}^4$
$BM_T$	$\approx 2.078$ m
$BM_L$	$\approx 51.82$ m

Topic: Numerical integration (Simpson's Rule) of waterplane hydrostatic properties.

Key relations: $A_W=2\frac{h}{3}\Sigma S_iy_i$; $M=2\frac{h}{3}\Sigma S_iy_ix_i$; $I_T=\frac{2}{3}\frac{h}{3}\Sigma S_iy_i^3$; $I_{L,mid}=2\frac{h}{3}\Sigma S_ix_i^2y_i$; parallel-axis shift $I_L=I_{L,mid}-A_W\bar{x}^2$; $BM=I/\nabla$.

Why this works: Simpson's Rule fits a parabola through each set of three consecutive ordinates, giving a good approximation to smoothly-curved hull waterlines from a handful of station offsets. Because a half-breadth strip's own transverse inertia about the centreline is $\tfrac{2}{3}y^3\,dx$ (not $\tfrac{1}{3}$, and not $y^3\,dx$), that factor must be carried through the Simpson summation; the longitudinal inertia is always easiest to sum about midship first and then shifted to the true centroid by the parallel-axis theorem, since the *tabulated* $x_i$ are naturally referenced to midship.

Common pitfall: using $\tfrac{1}{3}y^3$ instead of $\tfrac{2}{3}y^3$ for $I_T$ (forgetting the two-sided waterplane); reporting $I_L$ about midship instead of about the actual centroid, which the question explicitly asks for.

Source: Tupper, Introduction to Naval Architecture, Ch.2 (Numerical integration) & Ch.3 (Hydrostatics); PNA Vol.1, Ch.II.

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