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

Question 5 of 5: Correcting the GZ Curve for the Actual Loading Condition

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 5: Correcting the GZ Curve for the Actual Loading Condition

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. A static-stability ($GZ$) curve computed for an assumed $KG=6.5\text{ m}$, and the ship's actual weight distribution before sailing, from which the true $KG$ differs from the assumed value.

Given data
Quantity	Symbol	Value
Assumed $KG$ (curve as given)	$KG_{assumed}$	6.5 m
Total displacement	$\Delta$	$4200+9100+1500+200=15{,}000$ t
Actual $KG$	$KG_{actual}$	to be computed
$GZ(\theta)$, assumed curve	—	tabulated above, $\theta=0$–$90^\circ$

Find. The corrected $GZ$ curve, $GM$ read from it, the angle of vanishing stability, the angle of maximum stability, and the maximum righting moment.

Static stability curves: assumed $KG=6.5\text{ m}$ (grey, dashed) and corrected for the actual loading, $KG\approx6.14\text{ m}$ (blue, solid). The dashed red line is the tangent at the origin, whose height at $1$ radian gives $GM$.

Approach. Find the actual $KG$ by taking moments of the loading condition about the keel, apply the standard small-angle $KG$-correction to every point of the given $GZ$ curve, then read $GM$, the vanishing-stability angle, and the maximum-$GZ$ angle off the corrected curve.

Actual $KG$ for the loading condition. $$KG_{actual}=\frac{4200(6.0)+9100(7.0)+1500(1.1)+200(7.5)}{15{,}000}=\frac{92{,}050}{15{,}000}\approx6.137\text{ m}$$
Correct the GZ curve for the KG difference. Since $KG_{actual}
$GM$ from the corrected curve. For small angles $GZ\approx GM\sin\theta$; using the $15^\circ$ point: $$GM\approx\frac{GZ(15^\circ)}{\sin15^\circ}=\frac{0.204}{0.2588}\approx0.788\text{ m}$$
Angle of vanishing stability. $GZ_{actual}$ is still positive at $75^\circ$ ($0.301\text{ m}$) but negative at $90^\circ$ ($-0.267\text{ m}$); interpolating linearly between these two tabulated points for the zero crossing: $$\theta_{vanish}=75+15\left(\frac{0.301}{0.301+0.267}\right)\approx83.0^\circ$$
Angle of maximum stability and maximum righting moment. The corrected curve peaks at the tabulated $45^\circ$ point ($GZ_{max}\approx0.837\text{ m}$), rising from $0.542\text{ m}$ at $30^\circ$ and falling sharply to $0.435\text{ m}$ at $60^\circ$: $$\boxed{GM\approx0.79\text{ m},\quad \theta_{vanish}\approx83.0^\circ,\quad \theta_{max}\approx45^\circ,\quad M_{max}=\Delta\cdot GZ_{max}\approx15{,}000(0.837)\approx12{,}554\text{ t}\cdot\text{m}}$$

Question 5 — final results
Quantity	Value
Actual $KG$	$\approx6.137$ m
$GM$ (from corrected curve)	$\approx0.79$ m
Angle of vanishing stability	$\approx83.0^\circ$
Angle of maximum stability	$\approx45^\circ$
Maximum $GZ$	$\approx0.837$ m
Maximum righting moment	$\approx12{,}554$ t·m

Topic: Correcting a $GZ$ (static stability) curve for a different centre of gravity height.

Key relations: $GZ_{actual}(\theta)=GZ_{assumed}(\theta)+(KG_{assumed}-KG_{actual})\sin\theta$; small-angle $GZ\approx GM\sin\theta$; righting moment $=\Delta\cdot GZ$.

Why this works: A $GZ$ curve is computed hydrostatically for one assumed $KG$, but a righting arm is measured from $G$, so raising or lowering $G$ by $\delta$ simply adds or subtracts $\delta\sin\theta$ at every angle — this is the standard "correction to the curve of statical stability" used whenever the as-loaded $KG$ differs from the value the curve was drawn for, avoiding a full recomputation from the hull form. Because $\sin\theta\to\theta$ (radians) for small $\theta$, the same relation gives the initial $GM$ directly from the low-angle end of the corrected curve, without needing $KM$ or $BM$ at all.

Common pitfall: adding the correction with the wrong sign (a lower actual $KG$ than assumed makes the ship more stable, i.e. $GZ$ increases); reading the vanishing-stability angle from the uncorrected (assumed-$KG$) curve instead of the corrected one; picking the maximum $GZ$ from a coarse table without checking whether the curve is still rising or has already turned over between tabulated points.

Source: Tupper, Introduction to Naval Architecture, Ch.3 (Statical stability & the $GZ$ curve); IMO, International Code on Intact Stability (IS Code) §2.2 (stability criteria referencing $GM$, vanishing-stability angle, and $GZ_{max}$).

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