🇮🇳 GATE Marine Engineering · flashcards
GATE Marine Engineering Naval Architecture and Ocean Engineering Flashcards
50 question-and-answer cards covering Naval Architecture and Ocean Engineering as it is examined in GATE Marine Engineering. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Naval Architecture and Ocean Engineering deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
In the inclining experiment, how is the angle of heel $\theta$ measured using a pendulum (plumb line)?
A pendulum of length $\ell$ is suspended; the horizontal deflection $a$ of its bob is measured. The heel angle is given by $$\tan\theta = \frac{a}{\ell}.$$
List key precautions for a valid inclining experiment.
Calm water and weather, ship freely floating (moorings slack), minimal free surfaces (tanks empty or pressed full), no loose weights, all known weights accounted for, only essential personnel aboard at fixed positions, and accurate draft/density readings.
State the formula for the vertical shift of the centre of gravity when a mass $w$ is added on board.
$$GG' = \frac{w \cdot d}{\Delta + w}$$ where $d$ is the vertical distance between the centre of the added mass and the original $G$, and $\Delta$ is the original displacement. $G$ moves toward the added mass.
State the formula for the shift of the centre of gravity when a mass $w$ is removed from a ship.
$$GG' = \frac{w \cdot d}{\Delta - w}$$ where $d$ is the distance from the removed mass to the original $G$. The centre of gravity moves directly away from the position of the removed weight.
In which direction does the ship's centre of gravity move when a mass is added, removed, or shifted?
Added mass: $G$ moves toward the added weight. Removed mass: $G$ moves away from the removed weight. Shifted mass: $G$ moves in the same direction as, and parallel to, the shift of the mass.
Give the formula for the transverse (horizontal) shift of $G$ due to moving a mass $w$ a distance $d$ across the ship.
$$GG_H = \frac{w \cdot d}{\Delta}$$ where $d$ is the transverse distance the mass is moved and $\Delta$ is the displacement. $G$ moves parallel to the shift, causing the ship to heel.
What heel angle results from a transverse shift of mass, expressed via $GM$?
The ship heels until $G$ is again vertically below $M$. The angle satisfies $$\tan\theta = \frac{GG_H}{GM} = \frac{w \cdot d}{\Delta \cdot GM}.$$
What is the free surface effect (FSE) and how does it affect stability?
A free surface of liquid in a partly filled tank shifts toward the low side as the ship heels, moving the liquid's centre of gravity and causing a virtual rise of the ship's $G$. This reduces $GM$ and the righting arm, decreasing stability.
Give the formula for the virtual rise in $G$ (loss of $GM$) due to free surface effect in one tank.
$$GG_v = \frac{i \cdot \rho_t}{\Delta} = \frac{i \cdot \rho_t}{\rho_s \nabla}$$ where $i$ is the second moment of the free surface area about its own centreline, $\rho_t$ the density of the tank liquid, $\rho_s$ the sea-water density, and $\nabla$ the displaced volume. This $GG_v$ is the free surface correction subtracted from $GM$.
For a rectangular tank of length $l$ and breadth $b$, what is the free surface moment of inertia, and how does subdividing the tank help?
$i = \dfrac{l b^{3}}{12}$. Since FSE depends on $b^{3}$, fitting one or two longitudinal divisions (subdivisions) drastically reduces it: with $n$ equal divisions the effect falls to $1/n^{2}$ of the undivided value.
What is the effect of a suspended mass on a ship's stability?
A suspended mass (e.g. a load on a crane hook) acts as though its weight is located at the point of suspension, not at its actual position. This raises the effective centre of gravity, causing a virtual rise of $G$ and a reduction in $GM$.
Give the expression for the virtual rise of $G$ when a mass $w$ is suspended a distance $h$ above its stowed position.
$$GG_v = \frac{w \cdot h}{\Delta}$$ where $h$ is the height of the point of suspension above the original position of the mass and $\Delta$ the displacement. $G$ effectively rises to the suspension point the instant the load is lifted.
Why does the small-angle stability formula $GZ = GM\sin\theta$ fail at large angles of heel?
At large angles the metacentre $M$ no longer stays fixed because the waterplane shape changes significantly, deck edge immersion and bilge emergence alter the buoyancy distribution. $GZ$ must instead be obtained from cross curves of stability or wall-sided/Atwood's formula.
State the wall-sided formula for the righting arm $GZ$ at moderately large angles.
For a wall-sided ship: $$GZ = \left(GM + \tfrac{1}{2} BM \tan^{2}\theta\right)\sin\theta.$$ The $\tfrac{1}{2}BM\tan^{2}\theta$ term accounts for the rise of the metacentre and is valid until the deck edge immerses.
What is the angle of loll and what condition causes it?
The angle of loll is the angle at which a ship with a small initial negative $GM$ (unstable upright, $GM < 0$) comes to rest, heeled to one side but in stable equilibrium. It arises because as the ship heels, $BM$ increases until $GZ$ becomes zero with positive slope.
Give the formula for the approximate angle of loll for a wall-sided vessel with negative $GM$.
Setting $GZ = 0$ in the wall-sided formula (with $GM$ negative): $$\tan\theta_{loll} = \sqrt{\frac{-2\,GM}{BM}}.$$ The vessel lolls to this angle on either side.
How does a ship at an angle of loll behave, and how should it be corrected?
It floats heeled and can flop suddenly to the same angle on the opposite side, which is dangerous. Correction: lower the effective $G$ (restore positive $GM$) by removing free surfaces, then add low weight or ballast on the centreline / low side first, never the high side.
What is a curve of statical stability ($GZ$ curve)?
A graph of the righting arm $GZ$ (vertical axis) against the angle of heel $\theta$ (horizontal axis) for a given displacement and $KG$. It characterizes a ship's stability over the full range of heel.
From a $GZ$ curve, how do you read the initial metacentric height $GM$?
$GM$ equals the slope of the $GZ$ curve at the origin. Practically, erect a vertical at $\theta = 1$ radian ($57.3^\circ$); the height of the tangent at the origin intercepted there equals $GM$ (since $GZ \approx GM\theta$ for small $\theta$).
List the key features read from a curve of statical stability.
Initial slope (= $GM$), maximum $GZ$ and the angle at which it occurs, the range of stability (heel range where $GZ > 0$), the angle of vanishing stability (where $GZ$ returns to zero), the deck-edge immersion point of inflection, and the area under the curve (dynamical stability).
Define the range of stability and the angle of vanishing stability.
The range of stability is the span of heel angles over which $GZ$ is positive (the ship still rights itself). The angle of vanishing stability is the heel angle at which $GZ$ falls back to zero; beyond it the ship will capsize.
What are cross curves of stability (KN curves) and how are they used?
Cross curves plot the righting lever (KN or GZ assuming an assumed $KG$) against displacement for several fixed heel angles. They are used to derive the $GZ$ curve for any actual loading using $GZ = KN - KG\sin\theta$.
Define dynamical stability and state how it is obtained from the $GZ$ curve.
Dynamical stability is the work done in heeling the ship to a given angle against the righting moment, i.e. the energy needed to incline it. It equals the displacement times the area under the $GZ$ curve up to that angle: $$\text{Dynamical stability} = \Delta \int_{0}^{\theta} GZ \, d\theta.$$
How is the area under the $GZ$ curve commonly computed, and what stability-criteria areas are specified by IMO?
The area is computed numerically using Simpson's Rule. IMO intact stability criteria specify minimum areas: at least $0.055\ \text{m·rad}$ up to $30^\circ$, $0.090\ \text{m·rad}$ up to $40^\circ$ (or downflooding), and $0.030\ \text{m·rad}$ between $30^\circ$ and $40^\circ$, with $GZ \geq 0.20\ \text{m}$ at $30^\circ$ and minimum $GM \geq 0.15\ \text{m}$.
What this deck covers
The Naval Architecture and Ocean Engineering deck follows the GATE Marine Engineering Naval Architecture and Ocean Engineering syllabus — 9 chapters and 71 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 5.6 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 233 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Naval Architecture and Ocean Engineering flashcards FAQ
How many Naval Architecture and Ocean Engineering flashcards are in this GATE Marine Engineering deck?
50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these GATE Marine Engineering flashcards free?
Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.
What do the Naval Architecture and Ocean Engineering cards cover?
They follow the GATE Marine Engineering Naval Architecture and Ocean Engineering syllabus — 9 chapters and 71 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.