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BD Deck Officer Class 2 (Chief Mate) study materials
BD Deck Officer Class 2 (Chief Mate) study resources

BD Deck Officer Class 2 (Chief Mate) practice questions

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Question 1Navigation (terrestrial, celestial, electronic and passage planning)

What does magnetic variation represent at a vessel's position?

  1. The angle between magnetic north and compass north caused by the ship
  2. The angle between true north and the local magnetic meridian Correct answer.
  3. The angle between true north and the gyrocompass north
  4. The angle between the ship's heading and its wake

Explanation

Variation is the angular difference between geographic north and the local direction of the Earth's magnetic field, shown on charts as isogonic lines. It changes with position and slowly with time. Deviation is the separate error caused by the vessel's own magnetic field.

Question 2Navigation (terrestrial, celestial, electronic and passage planning)

A marine sextant is used principally to measure which quantity?

  1. The azimuth of the body relative to the observer's meridian
  2. The altitude of the body above the visible horizon Correct answer.
  3. The hour angle of the body measured westward from the meridian
  4. The declination of the body measured from the celestial equator

Explanation

The sextant measures the vertical angle between the visible horizon and the celestial body, giving the observed altitude after index-error and dip corrections. Azimuth is a bearing, declination is a coordinate on the celestial sphere, and hour angle is a time-related coordinate.

Question 3Navigation (terrestrial, celestial, electronic and passage planning)

In the four-stage passage planning model, what is the main purpose of appraisal?

  1. To gather and assess all information relevant to the passage Correct answer.
  2. To execute the plan and record the vessel's positions
  3. To monitor the vessel's progress against the planned track
  4. To calculate the final fuel consumption after arrival

Explanation

Appraisal is the first stage, in which the navigator collects and evaluates charts, publications, weather, traffic, and vessel constraints. Planning, execution, and monitoring follow. Appraisal identifies hazards and information gaps before a route is drawn.

Question 4Navigation (terrestrial, celestial, electronic and passage planning)

Which statement describes a rhumb line on the Earth's surface?

  1. It is the shortest track between two points on the globe
  2. It is a line that follows a single parallel of latitude
  3. It crosses every meridian at the same angle Correct answer.
  4. It is a great circle that passes through both poles

Explanation

A rhumb line, or loxodrome, maintains a constant true course and therefore crosses each meridian at the same angle. It appears as a straight line on a Mercator chart but is longer than the great-circle track except along a meridian or the equator. The great circle is the shortest surface route.

Question 5Navigation (terrestrial, celestial, electronic and passage planning)

A celestial observation gives Ho 34°26.0′ and Hc 34°20.0′ with azimuth 090°. What is the intercept and how is it plotted?

  1. 6.0′ away from the body along azimuth 270°
  2. 6.0′ away from the body along azimuth 090°
  3. 6.0′ towards the body along azimuth 270°
  4. 6.0′ towards the body along azimuth 090° Correct answer.

Explanation

Intercept equals Ho minus Hc, so here it is plus 6.0′, meaning the observed altitude is greater than the calculated altitude. A positive intercept is plotted from the dead-reckoning position toward the body's azimuth; a negative one is plotted away from it.

Question 6Navigation (terrestrial, celestial, electronic and passage planning)

On an ECDIS using a vector ENC, what is the function of the safety contour?

  1. It defines the outer limit of the radar display
  2. It divides safe from unsafe water for alarms Correct answer.
  3. It marks the boundary of the chart horizontal datum
  4. It sets the maximum draught permitted for the vessel

Explanation

The safety contour is a selected depth contour that ECDIS uses to distinguish navigable water from shoal or unsafe water. If the vessel's safety setting crosses it, the system can trigger an alarm during route checking or monitoring. It is a depth boundary, not the vessel's maximum draught.

Question 7Navigation (terrestrial, celestial, electronic and passage planning)

A vessel steams 60 nautical miles due north from latitude 10°00′N. What is the new latitude?

  1. 11°00′N Correct answer.
  2. 10°60′N
  3. 10°30′N
  4. 11°30′N

Explanation

Along a meridian, one minute of latitude corresponds to one nautical mile, so 60 nautical miles equals 1° of latitude. Steaming north increases the latitude from 10°00′N to 11°00′N, with longitude unchanged.

Question 8Navigation (terrestrial, celestial, electronic and passage planning)

Which IMO instrument provides guidelines for voyage planning under SOLAS regulation V/34?

  1. The International Regulations for Preventing Collisions at Sea
  2. The International Safety Management Code
  3. IMO Resolution A.893(21) Correct answer.
  4. The International Convention on Standards of Training, Certification and Watchkeeping

Explanation

SOLAS V/34 requires a passage plan, and IMO Resolution A.893(21) gives the associated guidelines, including appraisal, planning, execution, and monitoring. The COLREGs address collision avoidance, the ISM Code addresses safety management, and STCW addresses training and watchkeeping.

Question 9Navigation (terrestrial, celestial, electronic and passage planning)

At upper meridian transit, a body has observed altitude 62°10.0′ and declination 20°05.0′S. The observer is in the northern hemisphere and the body bears due south. What is the observer's latitude?

  1. 7°45′N Correct answer.
  2. 47°55′N
  3. 27°50′N
  4. 7°45′S

Explanation

For a body culminating due south of a northern-hemisphere observer, latitude equals zenith distance minus southerly declination. Zenith distance is 90° minus Ho, or 27°50′. Subtracting the 20°05′S declination gives 7°45′N.

Question 10Navigation (terrestrial, celestial, electronic and passage planning)

A vessel plans a route on ECDIS using raster chart display system (RCDS) mode. Which limitation applies compared with a vector ENC?

  1. Full route safety checking against chart objects is unavailable Correct answer.
  2. The raster chart cannot display the vessel's own ship symbol
  3. The raster chart cannot be corrected by notices to mariners
  4. The raster chart cannot be georeferenced to the vessel position

Explanation

Raster charts are scanned images without separate object attributes, so ECDIS cannot apply the same automatic route safety checks or anti-grounding alarms as it can with vector ENC data. RCDS is still georeferenced, displays own ship, and can be updated. The limitation concerns the lack of vector attribution.

Question 11Navigation (terrestrial, celestial, electronic and passage planning)

On a navigational chart, soundings and drying heights are referenced to which level?

  1. Chart datum, normally lowest astronomical tide Correct answer.
  2. Mean sea level at the port
  3. Mean high water springs
  4. Highest astronomical tide

Explanation

Charted soundings, drying heights and depth contours are referred to chart datum, which is set at or near lowest astronomical tide so that charted depths represent the least depth expected in normal conditions. Heights of lights and bridges are referenced to a high-water level instead. The distinction matters when calculating under-keel clearance.

Question 12Navigation (terrestrial, celestial, electronic and passage planning)

On an ECDIS route monitor, what does cross-track error represent?

  1. The distance from own ship to the next waypoint
  2. The perpendicular distance between own ship and the planned track Correct answer.
  3. The distance run since the last waypoint was passed
  4. The difference between the ordered course and the actual course over ground

Explanation

Cross-track error is measured at right angles to the planned leg, with a port or starboard sense, so it shows how far the ship has drifted off the intended track. The ECDIS uses it to trigger the off-track alarm and to display the course needed to regain track. It is a lateral distance, not a progress or heading value.

Question 13Navigation (terrestrial, celestial, electronic and passage planning)

Applying index correction and dip to a sextant altitude gives which altitude?

  1. True altitude
  2. Zenith distance
  3. Apparent altitude Correct answer.
  4. Observed altitude

Explanation

Index correction removes the instrument's own error and dip lowers the visible horizon to the observer's eye level, so the result is the apparent altitude. Refraction, semi-diameter and parallax are then applied to the apparent altitude to obtain the observed altitude used in sight reduction.

Question 14Navigation (terrestrial, celestial, electronic and passage planning)

A passage plan prepared in accordance with SOLAS V/34 should cover the intended voyage from where to where?

  1. Pilot station to pilot station
  2. The departure breakwater to the arrival breakwater
  3. The departure berth to the arrival berth Correct answer.
  4. The first waypoint to the last waypoint of the planned track

Explanation

The IMO guidelines expect the plan to cover the whole voyage from berth to berth, including all pilotage waters, even where a pilot will be embarked. Planning only the sea passage leaves the most constrained and highest-risk legs without a documented plan. The plan is therefore continuous from the departure berth to the arrival berth.

Question 15Navigation (terrestrial, celestial, electronic and passage planning)

At latitude 60°N, one minute of longitude measured along the parallel is approximately what distance?

  1. 1.0 nautical miles
  2. 0.5 nautical miles Correct answer.
  3. 1.15 nautical miles
  4. 0.87 nautical miles

Explanation

One minute of longitude subtends one minute of arc along the parallel, and its length in nautical miles equals one minute of latitude multiplied by the cosine of the latitude. At 60°N the cosine is 0.5, so a minute of longitude measures about 0.5 M. A minute of latitude remains close to 1 M everywhere.

Question 16Navigation (terrestrial, celestial, electronic and passage planning)

The equation of time is applied to convert between which two quantities?

  1. Local apparent time and zone time
  2. Sidereal time and mean solar time
  3. Greenwich mean time and coordinated universal time
  4. Apparent solar time and mean solar time Correct answer.

Explanation

Apparent solar time, taken from the real Sun, runs unevenly through the year because of the Earth's orbital eccentricity and the tilt of its axis. The equation of time is the difference between apparent and mean solar time and is applied when converting a time derived from the Sun to the mean time kept by chronometers and UTC.

Question 17Navigation (terrestrial, celestial, electronic and passage planning)

On a type-approved ARPA, what is the function of a guard zone?

  1. Alarm when a tracked target enters the set area Correct answer.
  2. Reduce sea clutter returns close to own ship
  3. Define the minimum CPA applied during trial manoeuvres
  4. Record the radar picture automatically for later playback and analysis

Explanation

A guard zone is a sector or area set around own ship, commonly ahead, that raises an acoustic and visual alarm when an acquired target enters it. It gives early warning of approaching traffic without waiting for CPA and TCPA limits to be breached. CPA limits and picture recording are separate functions.

Question 18Navigation (terrestrial, celestial, electronic and passage planning)

A tidal range is 3.6 m. Using the rule of twelfths, what is the rise of tide four hours after low water?

  1. 1.8 m
  2. 0.3 m
  3. 1.2 m
  4. 3.0 m Correct answer.

Explanation

The rule of twelfths divides the range into twelve parts that accumulate as 1, 2, 3, 3, 2 and 1 over the six hours of the rise. After four hours the accumulated fraction is (1+2+3+4)/12, which is 10/12, so the rise is 3.6 × 10/12 = 3.0 m. The remaining 0.6 m is gained in the last two hours.

Question 19Navigation (terrestrial, celestial, electronic and passage planning)

The Sun's declination is 20°N and the observer's latitude is 30°N. Using sin amplitude = sin d / cos lat, what is the true amplitude at sunrise?

  1. E 43.2° N
  2. E 20.0° N
  3. E 23.3° S
  4. E 23.3° N Correct answer.

Explanation

Substituting the values gives sin A = sin 20° / cos 30° = 0.342 / 0.866 = 0.395, so the amplitude is 23.3°. Because the declination is northerly the body rises north of east, written E 23.3° N. Amplitude is measured from the prime vertical, not from the meridian.

Question 20Navigation (terrestrial, celestial, electronic and passage planning)

An ENC cell is compiled on a local horizontal datum while the GNSS position is referenced to WGS 84. How does a type-approved ECDIS ensure the position plots correctly?

  1. It applies the vessel's gyro error to the position before plotting
  2. It transforms the position between chart datum and WGS 84 Correct answer.
  3. It requires the navigator to offset every waypoint manually
  4. It reduces the displayed scale of the chart to limit the error

Explanation

Where an ENC is compiled on a local horizontal datum, the ECDIS applies a defined datum transformation so that WGS 84 positions are converted to the chart datum before plotting. The parameters are held in the cell metadata and applied automatically; the shift is usually small but can reach several hundred metres. Manual offsets are not the intended method.

Question 21Navigation (terrestrial, celestial, electronic and passage planning)

A vessel's position is confirmed by keeping two charted landmarks in transit. What does this observation establish?

  1. The vessel is on the charted leading line through both objects Correct answer.
  2. The vessel's speed over the ground has been verified
  3. The depth of water in the channel has been confirmed
  4. The set and rate of the tidal stream have been determined

Explanation

Two objects in transit define a position line that runs through both objects and the observer, so the vessel lies somewhere on that line; the charted line joining them is the leading line. It is a line of position only, and a second observation is needed to fix the vessel's position along it. Transits are commonly used to keep a vessel in the dredged channel.

Question 22Navigation (terrestrial, celestial, electronic and passage planning)

What information does a careful observation of Polaris give the navigator?

  1. The direction of the magnetic meridian
  2. Longitude by chronometer
  3. Local apparent time at the observer
  4. Latitude, since its altitude approximates the observer's latitude Correct answer.

Explanation

Polaris lies close to the celestial pole, so its observed altitude, corrected for refraction and for its small polar distance, is very nearly equal to the observer's latitude. The Polaris tables provide the small correction for the star's offset from the pole. It cannot give longitude, which depends on accurate time and the body's hour angle.

Question 23Navigation (terrestrial, celestial, electronic and passage planning)

In a satellite positioning receiver, what does the horizontal dilution of precision (HDOP) describe?

  1. The number of satellites currently being tracked
  2. The age of the almanac data stored in the receiver
  3. The quality of the horizontal geometry of the satellites used Correct answer.
  4. The vertical error of the fix multiplied by a scaling factor

Explanation

Dilution of precision is a purely geometric factor: widely spread satellites give a low value and a strong fix, while satellites clustered in a small part of the sky give a high value and a weak fix. HDOP refers specifically to the horizontal component. Multiplied by the ranging error, it estimates the position error.

Question 24Navigation (terrestrial, celestial, electronic and passage planning)

During the execution stage of a voyage, how should the agreed passage plan be treated by the bridge team?

  1. It is cancelled whenever the ARPA shows no targets ahead
  2. It is fixed and may not be altered for any reason during the voyage
  3. It is continuously monitored, and any deviation requires a reassessment and a record Correct answer.
  4. It becomes advisory once the vessel is underway and may be amended at will

Explanation

Execution is the monitoring stage: positions are plotted, the vessel's progress along the planned track is checked, and any departure from the plan is a deliberate decision taken after reassessing the new circumstances. Changes must be recorded, and the master is kept informed. The plan is a working safety document, not a fixed line that can never move.

Question 25Navigation (terrestrial, celestial, electronic and passage planning)

Greenwich hour angle of Aries is 100°00.0′, a star's sidereal hour angle is 250°00.0′ and the observer's longitude is 40°00.0′W. What is the star's local hour angle?

  1. 250°00.0′
  2. 310°00.0′ Correct answer.
  3. 350°00.0′
  4. 030°00.0′

Explanation

First add the star's sidereal hour angle to the GHA of Aries to obtain the star's Greenwich hour angle: 100° + 250° = 350°. Then apply longitude, subtracting west longitude (or adding east longitude), which gives an LHA of 310°. A west longitude always reduces the Greenwich hour angle.

Question 26Navigation (terrestrial, celestial, electronic and passage planning)

In a tidal stream atlas, the tabulated set and rate of the stream are normally referred to which event?

  1. The time of mean low water springs at the standard port
  2. High water at the designated standard port Correct answer.
  3. The time of sunrise at the standard port
  4. The time of local high water at the vessel's position

Explanation

Tidal stream atlases give the stream for each hour relative to high water at the standard port named on the atlas, usually with a spring and neap rate. The navigator finds the standard port's high water time from the tide tables and counts the hours from it. A correction is then applied for the difference in time of high water between the standard port and the vessel's position.

Question 27Navigation (terrestrial, celestial, electronic and passage planning)

What is the purpose of the trial manoeuvre facility on a type-approved ARPA?

  1. To check the radar range calibration against a fixed charted object
  2. To record the vessel's position automatically in the electronic logbook
  3. To align the heading marker with the ship's fore-and-aft line
  4. To predict the effect of a proposed course or speed change on tracked targets Correct answer.

Explanation

The trial manoeuvre function lets the operator enter a proposed course and speed and see, after a short delay representing the vessel's response to the helm and engine, where the tracked targets would be relative to the vessel. It is used for planning an avoiding action before committing to it. The prediction is only as good as the target data and does not replace the OOW's judgement.

Question 28Navigation (terrestrial, celestial, electronic and passage planning)

On a gnomonic chart, how are great circles represented?

  1. As straight lines only in the polar regions
  2. As curves concave to the equator
  3. As rhumb lines crossing each meridian at the same angle
  4. As straight lines between any two points Correct answer.

Explanation

A gnomonic projection is made from the centre of the Earth onto a plane tangent to it, so every great circle projects as a straight line. This makes the gnomonic chart the standard tool for plotting a great circle track by joining departure and destination with a straight line and reading off intermediate waypoints. Those waypoints are then transferred to a Mercator chart for day-to-day navigation.

Question 29Navigation (terrestrial, celestial, electronic and passage planning)

A circumpolar star is observed at upper meridian transit at a sextant-corrected altitude of 70°00′ and at lower meridian transit at 20°00′, both culminations being on the meridian to the north. What is the observer's latitude?

  1. 90°00′N
  2. 65°00′N
  3. 25°00′N
  4. 45°00′N Correct answer.

Explanation

For a circumpolar star the sum of its upper and lower meridian altitudes equals twice the observer's latitude, so the latitude is (70° + 20°)/2 = 45°N. The declination can then be found from the lower transit as 90° + 20° − 45° = 65°N, which confirms the geometry. Using both culminations avoids any dependence on the star's declination being tabulated.

Question 30Navigation (terrestrial, celestial, electronic and passage planning)

A lighthouse charted 40 metres above sea level subtends a vertical sextant angle of 20.0′. Using distance in nautical miles ≈ 1.856 × height in metres ÷ angle in minutes, what is the approximate distance off?

  1. 2.3 nautical miles
  2. 1.9 nautical miles
  3. 9.3 nautical miles
  4. 3.7 nautical miles Correct answer.

Explanation

Substituting gives 1.856 × 40 ÷ 20.0 = 3.71, so the distance off is about 3.7 nautical miles. The formula converts the small-angle relationship into nautical miles, where one minute of arc subtends roughly 1.856 metres at one nautical mile. For precise work, the observer's height of eye and dip should be allowed for.

Question 31Navigation (terrestrial, celestial, electronic and passage planning)

The magnetic variation at the vessel's position is 7°W and the deviation for the heading being steered is 4°E. What is the compass error?

  1. 3°W Correct answer.
  2. 3°E
  3. 11°W
  4. 11°E

Explanation

Compass error is the algebraic sum of variation and deviation, taking west as negative and east as positive. Here −7° + 4° = −3°, so the error is 3° west. An easterly deviation therefore partly cancels a larger westerly variation.

Question 32Navigation (terrestrial, celestial, electronic and passage planning)

On a Mercator chart, how is a rhumb line represented?

  1. A straight line that always stays parallel to the equator
  2. A curved line that converges towards the nearer pole Correct answer.
  3. A series of separate arcs joined at every meridian
  4. A straight line cutting meridians at a constant angle

Explanation

A rhumb line cuts every meridian at the same angle, so it plots as a straight line on a Mercator chart. This follows directly from the projection's construction, which stretches latitude to match the east–west stretching of longitude.

Question 33Navigation (terrestrial, celestial, electronic and passage planning)

Civil twilight is the period during which the centre of the Sun lies between the horizon and how many degrees below it?

  1. 3°
  2. 12°
  3. 6° Correct answer.
  4. 18°

Explanation

Civil twilight runs from sunset until the Sun's centre is 6° below the horizon, and from that depth until sunrise. Nautical twilight uses 12° and astronomical twilight uses 18°, so the 6° figure is the civil definition.

Question 34Navigation (terrestrial, celestial, electronic and passage planning)

At meridian passage the Sun bears due South of the observer. Its true altitude is 60°00′ and its declination is 10°00′N. What is the observer's latitude?

  1. 20°N
  2. 40°N Correct answer.
  3. 30°N
  4. 50°N

Explanation

When the Sun bears due south, the observer is north of the sub-solar point, so latitude equals declination plus zenith distance. Zenith distance is 90° − 60° = 30°, and adding the 10°N declination gives 40°N.

Question 35Navigation (terrestrial, celestial, electronic and passage planning)

In the intercept method of sight reduction, what does the intercept represent?

  1. The correction applied to allow for atmospheric refraction
  2. The angular error built into the sextant arc
  3. The time difference between predicted and observed meridian passage
  4. The difference between observed and computed altitudes along the azimuth Correct answer.

Explanation

The intercept is the difference between the observed altitude and the computed altitude for the assumed position, plotted from that position along the azimuth. A positive intercept is plotted towards the geographical position of the body and a negative one away from it.

Question 36Navigation (terrestrial, celestial, electronic and passage planning)

On an ECDIS, what does the safety contour define?

  1. The charted depth at which the vessel would touch bottom
  2. The maximum draught line marked on the vessel's hull
  3. The boundary between navigable and unsafe water selected by the navigator Correct answer.
  4. The outer limit of the route corridor used for cross-track alarms

Explanation

The safety contour is chosen by the navigator, normally at or deeper than the safety depth, to separate water the vessel can safely navigate from water it must avoid. Crossing it triggers an alarm, so it acts as an electronic no-go boundary rather than a fixed charted depth.

Question 37Navigation (terrestrial, celestial, electronic and passage planning)

On an ARPA display, what does the time to closest point of approach (TCPA) indicate?

  1. The time taken for a target to cross the heading marker
  2. The time remaining before a target reaches its minimum predicted range Correct answer.
  3. The interval between successive rotations of the radar antenna
  4. The delay before a newly detected target is acquired by the ARPA

Explanation

TCPA is the predicted time until the range to a target reaches its minimum, that is, the closest point of approach. A negative TCPA indicates that the target is already opening, having passed its closest point.

Question 38Navigation (terrestrial, celestial, electronic and passage planning)

A vessel steams at 12 knots through the water on a course of 000°T. A current sets 090°T at 3 knots. What are the course and speed made good?

  1. 090°T at 12.4 knots
  2. 346°T at 12.4 knots
  3. 014°T at 12.4 knots Correct answer.
  4. 000°T at 15.0 knots

Explanation

The current is a vector at right angles to the heading, so the made-good vector is the hypotenuse of a right triangle with sides 12 and 3 knots. The speed is √(12² + 3²) ≈ 12.4 knots, and the course is arctan(3/12) ≈ 14° east of north, or 014°T.

Question 39Navigation (terrestrial, celestial, electronic and passage planning)

A great circle track starts at 40°N with an initial course of 060°T. What is the approximate latitude of the vertex?

  1. 48°N Correct answer.
  2. 56°N
  3. 40°N
  4. 32°N

Explanation

For a great circle, the vertex latitude satisfies cos(latitude of vertex) = sin(initial course) × cos(start latitude). Thus sin 60° × cos 40° ≈ 0.663, whose arc cosine is about 48.4°N, so 48°N is the closest option.

Question 40Navigation (terrestrial, celestial, electronic and passage planning)

A navigator measures a horizontal sextant angle between two charted landmarks to fix the vessel's position. What does this angle define?

  1. The vessel's distance from the nearer of the two landmarks
  2. A circle of position passing through the observer and both landmarks Correct answer.
  3. The bearing of the vessel from the midpoint between the two landmarks
  4. A straight line of position perpendicular to the line joining the landmarks

Explanation

A horizontal angle measured between two charted objects places the observer on a circular arc through both objects, because every point on that arc subtends the same angle. Two such arcs, or one arc plus another position line, give the fix.

Question 41Cargo handling, stowage and care of cargoes

In cargo stowage, what does the stowage factor of a cargo describe?

  1. The mass of cargo in tonnes that one cubic metre of hold space will accept
  2. The volume in cubic metres occupied by one tonne of the cargo Correct answer.
  3. The maximum load density that the tank top can accept, in tonnes per square metre
  4. The ratio of the cargo's mass in air to its mass in vacuum, as a percentage

Explanation

The stowage factor is the volume, in cubic metres, that one tonne of a cargo occupies when stowed, and it normally includes an allowance for broken stowage. Dividing the available hold volume by the stowage factor gives the cargo intake in tonnes.

Question 42Cargo handling, stowage and care of cargoes

What is the main purpose of laying dunnage on a tank top before loading bagged cargo?

  1. To keep the cargo clear of the deck, allow drainage and spread its weight Correct answer.
  2. To increase the load density that the tank top plating is permitted to carry
  3. To lower the ship's centre of gravity and increase the metacentric height
  4. To mark the cargo's stowage position for the cargo plan and the charterer

Explanation

Dunnage lifts the cargo off steel surfaces, keeps it clear of water and dirt, allows air to circulate and spreads concentrated loads over a larger area of the tank top so that the permitted load density is not exceeded.

Question 43Cargo handling, stowage and care of cargoes

In cargo securing calculations under the CSS Code, what does MSL mean?

  1. The load a securing device can be expected to withstand in service, in kN Correct answer.
  2. The maximum speed at which a ship may proceed with deck cargo stowed
  3. The maximum load a single deck fitting may carry in still water
  4. The maximum stack height permitted for a container stowed on deck

Explanation

MSL stands for Maximum Securing Load, the load that a securing device such as a lashing rod, wire or chain can be expected to withstand in service. It is expressed in kN and is used with the CSS Code when calculating the number of lashings required.

Question 44Cargo handling, stowage and care of cargoes

Under the IMDG Code segregation table, two packages of incompatible dangerous goods are stowed 'away from' each other. What does this require?

  1. They are stowed on deck only, on opposite sides of the ship
  2. They are stowed at opposite ends of the same hold, at least 24 metres apart
  3. They are effectively segregated with a minimum horizontal separation of 3 metres Correct answer.
  4. They are stowed in different holds separated by a complete deck

Explanation

In the IMDG segregation table, 'away from' is the least stringent of the four terms. It requires effective segregation so that incompatible goods cannot interact dangerously, with a minimum horizontal separation of 3 metres.

Question 45Cargo handling, stowage and care of cargoes

A bulk carrier loaded with grain has a computed angle of heel due to the assumed grain shift of 14 degrees; the angle at which the deck edge would immerse is 16 degrees. How does this compare with the Grain Code criterion?

  1. It is acceptable, provided the metacentric height stays positive throughout
  2. It is not acceptable, because the limit for grain shift is 5 degrees
  3. It is not acceptable, because 12 degrees is the maximum heel permitted Correct answer.
  4. It is acceptable, because the deck edge does not immerse until 16 degrees

Explanation

The Grain Code requires the angle of heel from the assumed grain shift not to exceed 12 degrees or the angle of deck edge immersion, whichever is smaller, and that a residual area remains under the righting lever curve. Here 12 degrees governs, so a heel of 14 degrees fails.

Question 46Cargo handling, stowage and care of cargoes

Under the Code of Safe Practice for Ships Carrying Timber Deck Cargoes, what is the maximum permitted height of the deck cargo?

  1. Two thirds of the ship's extreme breadth
  2. One third of the ship's extreme breadth Correct answer.
  3. Three quarters of the ship's extreme breadth
  4. One half of the ship's extreme breadth

Explanation

The Code of Safe Practice for Ships Carrying Timber Deck Cargoes limits the height of the deck cargo to one third of the ship's extreme breadth. This keeps the vertical centre of gravity and the windage within reasonable bounds while preserving intact stability.

Question 47Cargo handling, stowage and care of cargoes

A hold containing a hygroscopic cargo has a hold-air dew point of 18 degrees Celsius. The outside air dew point is 12 degrees Celsius. What is the correct ventilation decision?

  1. Ventilate only if the outside dew point is higher than the hold dew point
  2. Keep the hatches closed, because hygroscopic cargoes must never be ventilated
  3. Keep the hatches closed, because outside air is always colder than hold air
  4. Ventilate, because the outside dew point is lower than that of the hold air Correct answer.

Explanation

Ventilation is decided by comparing dew points: outside air may be admitted when its dew point is lower than that of the hold air, because it then reduces rather than adds to the risk of condensation on the cargo or the ship's structure.

Question 48Cargo handling, stowage and care of cargoes

A hold has a bale capacity of 9,500 cubic metres. Cargo with a stowage factor of 1.90 cubic metres per tonne is to be loaded, and broken stowage is estimated at 8 per cent of the hold volume. What quantity of cargo can be loaded?

  1. 4,200 tonnes
  2. 4,600 tonnes Correct answer.
  3. 4,000 tonnes
  4. 5,000 tonnes

Explanation

Broken stowage reduces the usable volume: 9,500 cubic metres less 8 per cent leaves 8,740 cubic metres. Dividing this by the stowage factor of 1.90 cubic metres per tonne gives 4,600 tonnes.

Question 49Cargo handling, stowage and care of cargoes

What preparation of the bilges is required before loading a dry bulk cargo that is easily damaged by water?

  1. Leave the bilges as found, since dry cargo does not produce water
  2. Clean and dry the bilges and bilge wells, test the suctions and cover the wells with burlap Correct answer.
  3. Fill the bilges with fresh water to keep the double bottom cool
  4. Paint the bilge wells with cement wash and leave the suctions open

Explanation

Before loading dry cargo that is easily damaged by water, the bilges and bilge wells must be cleaned, dried and tested, and the wells covered with burlap so that the cargo cannot choke the suctions. The sounding pipes should also be checked.

Question 50Cargo handling, stowage and care of cargoes

In a draft survey, what does the constant represent?

  1. An allowance for unaccounted weights such as mud, unpumped water and stores Correct answer.
  2. The difference between the displacements calculated at the forward and after perpendiculars
  3. The density correction applied to the displacement for the water at the survey berth
  4. The thickness correction between moulded and extreme drafts

Explanation

The constant is found by comparing the lightship displacement in the stability booklet with the displacement calculated from the survey when the ship's condition is known. It absorbs unaccounted weights such as mud, unpumped water and stores, together with booklet inaccuracies.