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

Question 2 of 5: Shift of KG and TCG on Adding a Weight, and Restoring Ballast

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 2: Shift of KG and TCG on Adding a Weight, and Restoring Ballast

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 weight is added off the centreline and above the keel, shifting the ship's centre of gravity vertically and transversely; a second, smaller ballast weight must then be placed to restore $G$ to its original position.

Given data
Quantity	Symbol	Value
Initial displacement	$\Delta_0$	9906 LT
Initial $KG$	$KG_0$	23.19 ft
Initial $TCG$ (on centreline)	$TCG_0$	0
Equipment added	$w_1$	25 LT, $kg_1=30$ ft, $8$ ft stbd of centreline
Ballast to be added	$w_2$	20 LT

Find. New $KG$; new $TCG$; and the vertical/transverse position of the 20 LT ballast that restores $G$ to $(KG_0,\,TCG_0)$.

Approach. Take moments of weight about the keel (vertical) and about the centreline (transverse), for the added equipment first, then again for the ballast, solving each time for the unknown lever that returns the required centre of gravity.

New displacement. $\Delta_1=\Delta_0+w_1=9906+25=9931\text{ LT}$.
New $KG$. Take moments about the keel: $$KG_1=\frac{\Delta_0\,KG_0+w_1\,kg_1}{\Delta_1}=\frac{9906(23.19)+25(30)}{9931}\approx23.207\text{ ft}$$
New $TCG$. Take moments about the centreline ($TCG_0=0$): $$TCG_1=\frac{\Delta_0(0)+w_1(8)}{\Delta_1}=\frac{25(8)}{9931}\approx0.0201\text{ ft (}0.24\text{ in) to starboard}$$
Ballast to restore $G$ — vertical position. With $\Delta_2=\Delta_1+w_2=9951\text{ LT}$, require the final $KG$ to return to $KG_0=23.19\text{ ft}$: $$\Delta_1 KG_1+w_2\,x=\Delta_2\,KG_0\ \Rightarrow\ x=\frac{\Delta_2 KG_0-\Delta_1 KG_1}{w_2}=\frac{9951(23.19)-9931(23.207)}{20}\approx14.68\text{ ft above the keel}$$
Ballast to restore $G$ — transverse position. Require the final $TCG$ to return to $0$: $$\Delta_1 TCG_1+w_2\,y=0\ \Rightarrow\ y=\frac{-w_1(8)}{w_2}=\frac{-25(8)}{20}=-10\text{ ft}$$ $$\boxed{KG_1\approx23.207\text{ ft},\ \ TCG_1\approx0.020\text{ ft stbd},\ \ \text{ballast at }14.68\text{ ft above keel},\ 10.0\text{ ft to port}}$$

Question 2 — final results
Quantity	Value
New $KG$	$\approx23.207$ ft
New $TCG$	$\approx0.020$ ft to starboard
Ballast vertical position	$\approx14.68$ ft above keel
Ballast transverse position	$10.0$ ft to port of centreline

Topic: Shift of the centre of gravity by moments of weight (vertical and transverse).

Key relations: $KG_{new}=\dfrac{\Sigma(\text{weight}\times kg)}{\Sigma\text{weight}}$; the same weighted-average rule applies independently, and simultaneously, to $TCG$ about the centreline.

Why this works: The centre of gravity of a system of weights is the weighted average of the individual centres, and because the vertical and transverse directions are orthogonal, $KG$ and $TCG$ can be solved as two completely independent moment balances about the keel and about the centreline respectively. Restoring ballast is simply the same equation run backwards, solving for the unknown lever that brings the resultant back to a target position.

Common pitfall: mixing the sign convention for port/starboard (here starboard is taken positive, so the restoring ballast lever comes out negative, i.e. to port); forgetting to use the displacement after the equipment addition, $\Delta_1$, not $\Delta_0$, when solving for the ballast.

Source: Tupper, Introduction to Naval Architecture, Ch.3 (Equilibrium and centres of gravity).

← Question 1 Next: Question 3 →