25-Nav-A1 Fundamentals of Naval Architecture · May-16-Nav-A1 2017

Question 3 of 5: KG from an Inclining Experiment (Least-Squares Fit)

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 3: KG from an Inclining Experiment (Least-Squares Fit)

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.

Check: the source table repeats the moment "528 (port)" for its last two rows instead of the expected mirrored 880 (port); this is transcribed exactly as printed on the exam paper. Real inclining-experiment readings show exactly this kind of scatter (friction, pendulum settling, hysteresis between successive weight shifts), which is precisely why the instructions say to plot the results and read a best-fit line, rather than compute $GM$ from any single reading.

Given. Five (moment, list-angle) pairs from moving a known weight across the deck, plus the ship's $KM$ and initial displacement (which includes the 28 LT inclining gear sitting on deck at $43$ ft above the keel).

Given data
Quantity	Symbol	Value
Displacement (with gear aboard)	$\Delta$	3400 LT
Transverse metacentric height	$KM$	26.5 ft
Inclining gear weight	$w_g$	28 LT
Inclining gear $KG$	$kg_g$	43 ft
Readings	$(M,\theta)$	see table above (5 pairs)

Find. $GM$ from the plotted data, then the ship's $KG$ corrected for removal of the inclining gear after the test.

Inclining-moment vs. list-angle readings (blue points) and the least-squares best-fit line through the corrected origin (red).

Approach. Correct every reading for the small permanent list observed at zero applied moment, fit a straight line of $\tan\theta$ against moment through the corrected origin, take $GM$ from its slope, recover $KG$ from $KM-GM$, then strip out the temporary inclining gear's contribution to get the ship's true $KG$.

Correct for the initial list. At zero applied moment the ship already lists $0.2^\circ$ to port — a small pre-existing asymmetry, not part of the experiment. Subtract this offset (add $0.2^\circ$, taking starboard positive) from every reading: $$\begin{aligned} (880,\,2.3)&\to(880,\,2.5^\circ)\qquad(528,\,1.2)\to(528,\,1.4^\circ)\qquad(0,\,-0.2)\to(0,\,0^\circ)\\ (-528,\,-1.5)&\to(-528,\,-1.3^\circ)\qquad(-528,\,-2.3)\to(-528,\,-2.1^\circ) \end{aligned}$$
Least-squares slope through the origin. For small angles $\tan\theta=M/(\Delta\,GM)$, so plotting $\tan\theta$ against $M$ gives a line of slope $1/(\Delta\,GM)$ through the origin. The best-fit slope (minimising the sum of squared residuals) is $$\text{slope}=\frac{\sum M_i\tan\theta_i}{\sum M_i^2}\approx5.132\times10^{-5}\ \text{deg}^{-1}\!\cdot\text{ft}^{-1}\cdot\text{LT}^{-1}$$ (using the five corrected pairs, including $(0,0)$, which contributes nothing to either sum).
Experimental $GM$. $$GM=\frac{1}{\Delta\cdot\text{slope}}=\frac{1}{3400(5.132\times10^{-5})}\approx5.731\text{ ft}$$
$KG$ as inclined (gear still aboard). $$KG_{inclined}=KM-GM=26.5-5.731\approx20.769\text{ ft}$$
Correct for removal of the inclining gear. The 28 LT of temporary gear (part of the 3400 LT tested) is removed after the test; take moments about the keel to strip it out: $$KG_{ship}=\frac{\Delta\,KG_{inclined}-w_g\,kg_g}{\Delta-w_g}=\frac{3400(20.769)-28(43)}{3372}\approx20.585\text{ ft}$$ $$\boxed{GM\approx5.731\text{ ft},\qquad KG_{ship}\approx20.585\text{ ft (gear removed)}}$$

Question 3 — final results
Quantity	Value
Best-fit slope, $\tan\theta/M$	$\approx5.132\times10^{-5}\text{ deg}^{-1}\text{ft}^{-1}\text{LT}^{-1}$
Experimental $GM$	$\approx5.731$ ft
$KG$ as inclined (gear aboard)	$\approx20.769$ ft
$KG$ of ship (gear removed)	$\approx20.585$ ft

Topic: Inclining experiment — determination of $GM$ and $KG$ from a moment/heel graph.

Key relations: $\tan\theta=\dfrac{w\,d}{\Delta\,GM}$ (small-angle inclining formula); $KG=KM-GM$; a weight temporarily aboard for the test is removed afterward by the usual moment-of-weight correction.

Why this works: A small transverse shift of weight produces a heel that is, for small angles, linearly proportional to the applied moment, with the constant of proportionality fixed by $\Delta\,GM$. Real pendulum/plumb-bob readings scatter about that line (friction in the pendulum, small movements of loose gear, imperfect zeroing), so the standard practice is to plot several moment/angle pairs and take the slope of the best-fit line through the corrected origin, rather than trust any single reading. Because the readings were taken with the inclining weights (and rigging) still aboard, the resulting $KG$ describes the ship as tested; removing that temporary weight afterward, by the same moment-of-weight method as Question 2, gives the ship's actual $KG$.

Common pitfall: using a single (moment, angle) pair instead of the fitted slope (the two "528 (port)" readings alone would give $GM$ values of 6.84 ft and 4.24 ft — both wrong); forgetting to zero the readings against the pre-existing $0.2^\circ$ list before fitting; forgetting the inclining-gear correction, which lowers the reported $KG$.

Source: Tupper, Introduction to Naval Architecture, Ch.3 (The Inclining Experiment); PNA Vol.1, Ch.II.

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