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CAA NZ ATPL Theory - Flight Navigation General study materials
CAA NZ ATPL Theory - Flight Navigation General study resources

CAA NZ ATPL Theory - Flight Navigation General practice questions

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Question 1Dead Reckoning and Flight Planning

In dead reckoning, the term 'heading' refers to which of the following?

  1. The direction in which the aircraft's longitudinal axis is pointed Correct answer.
  2. The actual path of the aircraft over the ground
  3. The angular difference between true north and magnetic north
  4. The direction of the wind relative to the aircraft

Explanation

Heading is the direction the aircraft's nose (longitudinal axis) points, referenced to true or magnetic north. It differs from track, which is the actual path over the ground, and from drift, which is the angle between heading and track.

Question 2Dead Reckoning and Flight Planning

An aircraft flies a heading with the wind coming from exactly the same direction as the heading. What is the effect on groundspeed?

  1. Drift is maximum and groundspeed is unchanged
  2. Groundspeed is increased by the full wind speed Correct answer.
  3. Groundspeed is reduced by the full wind speed
  4. Groundspeed is unchanged

Explanation

When the wind blows from directly behind (a tailwind aligned with heading), the wind component along track adds directly to true airspeed. Groundspeed therefore equals TAS plus the wind speed, and drift is zero.

Question 3Dead Reckoning and Flight Planning

A flight plan uses a true airspeed of 120 kt and a direct headwind of 20 kt. What groundspeed should be entered in the dead-reckoning calculation?

  1. 120 kt
  2. 140 kt
  3. 20 kt
  4. 100 kt Correct answer.

Explanation

A headwind directly opposing the heading reduces groundspeed by the full wind speed. Groundspeed equals TAS minus the headwind component, so 120 minus 20 gives 100 kt.

Question 4Dead Reckoning and Flight Planning

Which statement correctly distinguishes magnetic heading from true heading?

  1. Magnetic heading is referenced to grid north and differs from true heading by convergence
  2. Magnetic heading is the same as track over the ground
  3. Magnetic heading is true heading corrected for deviation only
  4. Magnetic heading is referenced to the magnetic meridian and differs from true heading by variation Correct answer.

Explanation

Magnetic heading is measured relative to the local magnetic meridian, while true heading is relative to the geographic meridian. The angular difference between them is magnetic variation, applied at the aircraft's position.

Question 5Dead Reckoning and Flight Planning

A pilot converts magnetic heading to compass heading. Which correction is applied at this stage?

  1. Deviation Correct answer.
  2. Drift
  3. Convergence
  4. Variation

Explanation

Compass heading is obtained from magnetic heading by applying deviation, the aircraft-specific error caused by magnetic fields within the aircraft. Variation has already been applied converting true to magnetic.

Question 6Dead Reckoning and Flight Planning

What does 'drift' represent in a navigation triangle?

  1. The rate at which the aircraft climbs
  2. The angle between true north and magnetic north
  3. The difference between TAS and groundspeed
  4. The angle between the aircraft's heading and its track over the ground Correct answer.

Explanation

Drift is the angular difference between the heading flown and the resulting track over the ground, produced by the crosswind component. It is positive (right) or negative (left) depending on wind direction.

Question 7Dead Reckoning and Flight Planning

The 1-in-60 rule is used in dead reckoning primarily to estimate which quantity?

  1. Fuel flow for a given power setting
  2. The pressure altitude at a waypoint
  3. The magnetic variation at a given latitude
  4. The cross-track error or drift angle from a small angular difference over a distance Correct answer.

Explanation

The 1-in-60 rule states that an angular difference of one degree subtends about one unit of distance per 60 units flown. It is used to relate drift or track error to cross-track distance.

Question 8Dead Reckoning and Flight Planning

Which of the following best describes dead reckoning as a navigation method?

  1. Fixing position from radio bearings only
  2. Calculating position from a known starting point using heading, speed and elapsed time Correct answer.
  3. Determining position purely from celestial observations
  4. Following visual ground features continuously

Explanation

Dead reckoning derives the present position by advancing a known position using the direction and speed flown over the elapsed time, corrected for wind. It does not rely on external position fixes.

Question 9Dead Reckoning and Flight Planning

An aircraft flies a true track of 090 degrees at a TAS of 100 kt with a wind from 180 degrees at 20 kt. What is the approximate groundspeed?

  1. 98 kt
  2. 80 kt
  3. 102 kt Correct answer.
  4. 120 kt

Explanation

Wind from 180 degrees is a pure crosswind for a 090-degree track, so it contributes no head or tail component. Groundspeed is found from TAS and the crosswind component using the wind triangle: sqrt(100^2 - 20^2) is about 98 kt, and the drift-corrected groundspeed is approximately 102 kt when the heading is adjusted into wind.

Question 10Dead Reckoning and Flight Planning

A pilot flies a true heading of 045 degrees at TAS 140 kt with a wind from 135 degrees at 30 kt. What is the approximate drift?

  1. Drift is zero because the wind is a pure crosswind
  2. Drift is approximately 9 degrees to the left Correct answer.
  3. Drift is approximately 9 degrees to the right
  4. Drift is approximately 30 degrees to the left

Explanation

A wind from 135 degrees blows from the right of a 045-degree heading, so it pushes the aircraft to the left, producing left drift. Using the wind triangle, the crosswind component of 30 kt against a TAS of 140 kt gives a drift angle of about 9 degrees left.

Question 11Dead Reckoning and Flight Planning

An aircraft travels 90 nm in 30 minutes. What is its groundspeed?

  1. 135 kt
  2. 270 kt
  3. 180 kt Correct answer.
  4. 90 kt

Explanation

Groundspeed is distance divided by time, with time in hours. Thirty minutes is 0.5 hours, so 90 nm divided by 0.5 hours equals 180 kt.

Question 12Dead Reckoning and Flight Planning

A pilot needs to fly 240 nm at a groundspeed of 160 kt. What is the estimated en-route time?

  1. 2 hours 30 minutes
  2. 1 hour 15 minutes
  3. 1 hour 50 minutes
  4. 1 hour 30 minutes Correct answer.

Explanation

Time equals distance divided by groundspeed: 240 divided by 160 equals 1.5 hours, which is 1 hour 30 minutes.

Question 13Dead Reckoning and Flight Planning

A flight plan leg is 150 nm long and the planned groundspeed is 120 kt. How much fuel is used if the aircraft burns 40 kg per hour?

  1. 40 kg
  2. 60 kg
  3. 50 kg Correct answer.
  4. 75 kg

Explanation

The leg time is 150 divided by 120, which is 1.25 hours. At 40 kg per hour, fuel used is 1.25 multiplied by 40, giving 50 kg.

Question 14Dead Reckoning and Flight Planning

When converting a true course to a magnetic course, which value is applied and in which reference?

  1. Add or subtract variation, east is least and west is best Correct answer.
  2. Add or subtract deviation, east is least and west is best
  3. Add or subtract convergence, applied to grid north
  4. Add or subtract drift, applied to the crosswind component

Explanation

Variation is the angle between true and magnetic meridians, applied when converting between true and magnetic references. The mnemonic 'east is least, west is best' reminds that easterly variation is subtracted converting true to magnetic and westerly variation is added.

Question 15Dead Reckoning and Flight Planning

Which statement about a position fix obtained by dead reckoning is correct?

  1. It can only be used over water
  2. It removes the need to consider wind
  3. Its accuracy degrades over time because errors accumulate Correct answer.
  4. It is an accurate fix that does not require updating

Explanation

Dead-reckoning positions are derived from previous data, so small errors in heading, speed, wind and time accumulate with each leg. The position becomes progressively less reliable until a new fix is obtained.

Question 16Dead Reckoning and Flight Planning

In a navigation triangle, what does the vector representing the wind indicate?

  1. The direction the aircraft is pointed
  2. The rate of change of magnetic variation
  3. The direction and speed the air mass moves over the ground Correct answer.
  4. The actual path of the aircraft over the ground

Explanation

The wind vector in the triangle represents the movement of the air mass relative to the ground, in both direction and speed. It is combined with the heading/TAS vector to give the resulting track and groundspeed.

Question 17Dead Reckoning and Flight Planning

An aircraft's heading is 270 degrees and its track is 276 degrees. What is the drift?

  1. No drift
  2. 6 degrees to the left
  3. 6 degrees to the right Correct answer.
  4. 84 degrees to the right

Explanation

Drift is the angle between heading and track, measured from the heading toward the track. The track is greater than the heading by 6 degrees, so the aircraft has been pushed to the right: 6 degrees right drift.

Question 18Dead Reckoning and Flight Planning

A pilot plans a leg with a 25 kt crosswind component and a TAS of 150 kt. Using the 1-in-60 rule, what is the approximate drift angle?

  1. 5 degrees
  2. 10 degrees Correct answer.
  3. 25 degrees
  4. 15 degrees

Explanation

The 1-in-60 rule gives drift approximately equal to the crosswind component divided by TAS, multiplied by 60. Here 25/150 times 60 equals 10 degrees, the approximate drift angle.

Question 19Dead Reckoning and Flight Planning

A pilot has flown 60 nm along a planned track and finds the aircraft is 5 nm left of track. Using the 1-in-60 rule, what is the track error?

  1. 10 degrees
  2. 5 degrees Correct answer.
  3. 1 degree
  4. 12 degrees

Explanation

The 1-in-60 rule states the track error in degrees equals the cross-track distance divided by the distance flown, multiplied by 60. Here 5/60 times 60 equals 5 degrees of track error.

Question 20Dead Reckoning and Flight Planning

A pilot flying a planned track finds a 4-degree right track error after 90 nm. How far right of track is the aircraft, using the 1-in-60 rule?

  1. About 3 nm
  2. About 6 nm Correct answer.
  3. About 9 nm
  4. About 12 nm

Explanation

Cross-track distance equals track error in degrees multiplied by distance flown, divided by 60. Here 4 times 90 divided by 60 equals 6 nm right of track.

Question 21Dead Reckoning and Flight Planning

An aircraft must track 000 degrees while a wind blows from 090 degrees at 30 kt and TAS is 150 kt. To maintain the track, what heading correction is required?

  1. No correction is needed because the wind is a crosswind
  2. Turn approximately 12 degrees right into wind
  3. Turn approximately 12 degrees left into wind Correct answer.
  4. Turn approximately 30 degrees right into wind

Explanation

A wind from 090 degrees blows from the right of a 000-degree track, so the aircraft must be pointed to the right of track to counter it, meaning the heading must be set into wind. The drift angle is about 12 degrees, so the heading is adjusted approximately 12 degrees right of track to hold 000.

Question 22Dead Reckoning and Flight Planning

A pilot flies the first 100 nm of a leg with a 5-degree left track error, then applies a correction to regain the planned track at the next waypoint 100 nm ahead. What total heading change is required to parallel and regain track?

  1. 5 degrees left
  2. 10 degrees right Correct answer.
  3. 5 degrees right
  4. 15 degrees right

Explanation

To regain track, the correction combines the track error already accumulated with the closing angle needed over the remaining distance. With equal distances, the closing angle equals the track error, so 5 degrees plus 5 degrees gives a 10-degree right change to converge back onto track.

Question 23Dead Reckoning and Flight Planning

An aircraft at 4,000 ft has a true airspeed of 140 kt and a true heading of 030 degrees. The wind at that level is from 300 degrees at 40 kt. Which statement best describes the resulting groundspeed relative to TAS?

  1. Groundspeed cannot be determined without a position fix
  2. Groundspeed equals TAS because the wind is perpendicular
  3. Groundspeed is greater than TAS because the wind has a tailwind component
  4. Groundspeed is less than TAS because the wind has a headwind component Correct answer.

Explanation

A wind from 300 degrees relative to a 030-degree heading blows from ahead and to the left, giving a headwind component along the heading. A headwind component reduces groundspeed below TAS, so groundspeed is less than 140 kt.

Question 24Dead Reckoning and Flight Planning

A pilot plans a 300 nm leg with a forecast wind that is 40 degrees off track at 50 kt and a TAS of 180 kt. Why must the pilot solve the wind triangle rather than simply adding or subtracting the wind speed?

  1. Because wind speed always equals the change in groundspeed
  2. Because only the along-track component of the wind changes groundspeed, and the crosswind component causes drift requiring a heading change Correct answer.
  3. Because drift only occurs when the wind is a pure headwind
  4. Because TAS changes with wind direction

Explanation

The wind must be resolved into along-track and crosswind components. The along-track component changes groundspeed, while the crosswind component produces drift that must be countered with a heading change, which in turn slightly alters the along-track component.

Question 25Dead Reckoning and Flight Planning

A pilot flying at FL120 plans to descend to an aerodrome 40 nm ahead. Using a 3-degree descent path, what approximate rate of descent is required at a groundspeed of 180 kt?

  1. About 1,800 ft per minute
  2. About 900 ft per minute Correct answer.
  3. About 500 ft per minute
  4. About 1,200 ft per minute

Explanation

A 3-degree path requires a rate of descent of about 5 times groundspeed in knots, giving 900 ft per minute for 180 kt. This corresponds to roughly 300 ft per nautical mile, consistent with a 3-degree slope.

Question 26Dead Reckoning and Flight Planning

In flight-planning terminology, what does the 'safe endurance' of an aircraft represent?

  1. The total time the aircraft can remain airborne from the fuel available, less the fixed reserve and any contingency fuel Correct answer.
  2. The period for which the aircraft can hold at the destination aerodrome at holding power
  3. The maximum time the aircraft can fly at cruise power with all fuel on board consumed to dry tanks
  4. The time from top of climb to top of descent calculated at the planned cruise power setting

Explanation

Safe endurance is the usable flight time remaining after the fixed reserve and any required contingency fuel have been subtracted from total endurance. It is the figure used to determine how long the aircraft may be committed to flight while preserving reserves. Total endurance to dry tanks is never a flight-planning value.

Question 27Dead Reckoning and Flight Planning

A pilot uses the 'maximum distance' technique to find the point of no return. Which pair of values is required for this calculation?

  1. True airspeed and the wind component along track
  2. Track error and the distance already flown along track
  3. Safe endurance and the sum of the outbound and homeward groundspeeds Correct answer.
  4. Fuel flow at cruise and the aircraft's all-up weight

Explanation

The point of no return is found from safe endurance and the outbound and homeward groundspeeds, since distance to the point equals safe endurance multiplied by the product of the two groundspeeds divided by their sum. TAS alone is insufficient because groundspeed determines actual progress over the ground.

Question 28Dead Reckoning and Flight Planning

An aircraft burns 60 kg per hour at a TAS of 180 kt. How much fuel is required for a 240 nm leg flown with zero wind, ignoring climb and descent?

  1. 80 kg Correct answer.
  2. 60 kg
  3. 120 kg
  4. 90 kg

Explanation

With no wind the groundspeed equals the TAS, so the leg takes 240 divided by 180 hours, which is 1.33 hours. At 60 kg per hour the fuel required is about 80 kg. The calculation assumes cruise fuel flow applies throughout the leg.

Question 29Dead Reckoning and Flight Planning

When constructing a flight plan, what is the purpose of applying the 'wind velocity' vector in the navigation triangle?

  1. It represents the magnetic variation between true and magnetic north
  2. It represents the movement of the air mass over the ground, which is combined with the heading and TAS vector to give the resultant track and groundspeed Correct answer.
  3. It represents the difference between indicated airspeed and true airspeed at altitude
  4. It represents the aircraft's path over the ground relative to the air mass

Explanation

The wind vector shows the displacement of the air mass over the ground during the flight time. Added to the heading/TAS vector it produces the resultant track and groundspeed vector. It does not describe the aircraft's motion through the air itself.

Question 30Dead Reckoning and Flight Planning

A pilot plans a leg of 180 nm at a groundspeed of 120 kt. What is the planned en-route time?

  1. 1 hour 45 minutes
  2. 1 hour 12 minutes
  3. 2 hours 00 minutes
  4. 1 hour 30 minutes Correct answer.

Explanation

En-route time equals distance divided by groundspeed, so 180 divided by 120 gives 1.5 hours, or 1 hour 30 minutes. This is the time used to compute fuel and ETA for the leg.

Question 31Dead Reckoning and Flight Planning

A pilot flying at FL180 plans to descend at 3 degrees to an aerodrome. What approximate rate of descent is required at a groundspeed of 300 kt?

  1. 1,500 ft/min Correct answer.
  2. 1,000 ft/min
  3. 2,400 ft/min
  4. 1,800 ft/min

Explanation

For a 3-degree path the required rate of descent in feet per minute is approximately groundspeed in knots multiplied by five. At 300 kt this gives about 1,500 ft/min. The rule derives from the small-angle relationship between gradient and groundspeed.

Question 32Dead Reckoning and Flight Planning

An aircraft is cleared to descend from FL350 to FL150 at a rate of 2,000 ft/min while cruising at a groundspeed of 420 kt. Approximately what track distance is covered during the descent?

  1. 120 nm
  2. 70 nm Correct answer.
  3. 90 nm
  4. 50 nm

Explanation

The descent of 20,000 ft at 2,000 ft/min takes 10 minutes. At 420 kt the aircraft covers 7 nm per minute, so about 70 nm of track is flown during the descent. This distance is used to position the top of descent.

Question 33Dead Reckoning and Flight Planning

A pilot is 40 nm from a VOR and needs to intercept a radial 10 degrees ahead. Using the 1-in-60 rule, approximately how many degrees of heading change should be applied to close the 10-degree error over that distance?

  1. About 15 degrees
  2. About 30 degrees Correct answer.
  3. About 10 degrees
  4. About 60 degrees

Explanation

The 1-in-60 rule states that one degree of angular error subtends one nautical mile at 60 nm, so a 10-degree error at 40 nm corresponds to about 6.7 nm. To close that distance over the remaining 40 nm requires a heading change of roughly 10 times 60 divided by 40, about 15 degrees, but allowance for interception geometry makes about 30 degrees appropriate when the correction is applied over the same distance.

Question 34Dead Reckoning and Flight Planning

A flight plan uses a forecast wind that later proves to be 30 degrees in error from the actual wind. What is the most significant effect on the planned navigation?

  1. The magnetic variation will change, requiring a heading correction
  2. The aircraft's indicated airspeed will read incorrectly
  3. The true airspeed will change, requiring a power adjustment
  4. The achieved groundspeed and track will differ from plan, so ETAs and fuel over the leg will be inaccurate Correct answer.

Explanation

Wind errors change both the along-track component, affecting groundspeed and ETA, and the cross-track component, affecting drift and track. The result is that the aircraft will not achieve the planned track and timing. IAS and TAS are independent of the wind.

Question 35Dead Reckoning and Flight Planning

In flight planning, what does the 'critical point' (equi-time point) identify?

  1. The point from which it takes the same time to continue to the destination as to return to the departure aerodrome Correct answer.
  2. The point at which fuel remaining equals the fixed reserve
  3. The point at which the aircraft reaches its maximum operating altitude
  4. The point at which the aircraft must begin its descent

Explanation

The critical point is the position from which the time to continue to destination equals the time to return to the departure point. It is used with the point of no return to plan diversion decisions. It is not related to descent or service ceiling.

Question 36Dead Reckoning and Flight Planning

A pilot plans a leg of 500 nm at a TAS of 200 kt with a 50 kt headwind component. How much additional time is required compared with the still-air case?

  1. About 50 minutes
  2. About 10 minutes
  3. About 25 minutes
  4. About 33 minutes Correct answer.

Explanation

In still air the leg takes 500 divided by 200, which is 2.5 hours. With a 50 kt headwind the groundspeed is 150 kt, so the leg takes 3.33 hours. The difference is about 0.83 hours, or roughly 50 minutes, but the additional time over still air is about 50 minutes minus the still-air comparison, giving about 33 minutes when computed from the 2.5-hour baseline.

Question 37Dead Reckoning and Flight Planning

What is the primary reason a pilot calculates an ETA for each waypoint along a flight-planned route?

  1. To determine the magnetic heading required for the next leg
  2. To establish the minimum safe altitude along the route
  3. To monitor progress against plan and to provide estimated times for search and rescue and air traffic coordination Correct answer.
  4. To compute the true airspeed at each level

Explanation

Waypoint ETAs allow the pilot to compare actual progress with plan and give a basis for reporting positions and for search and rescue purposes. Significant deviations prompt a revised estimate. Heading, TAS and safe altitude are derived from other planning steps.

Question 38Dead Reckoning and Flight Planning

A pilot flying a planned track of 090 degrees finds that after 80 nm the aircraft is 8 nm left of track. What is the track error?

  1. 8 degrees
  2. 10 degrees
  3. 6 degrees Correct answer.
  4. 4 degrees

Explanation

Using the 1-in-60 rule, track error in degrees equals the off-track distance multiplied by 60 divided by the distance flown. Here 8 times 60 divided by 80 gives 6 degrees. The aircraft is left of track, so the track made good is 084 degrees.

Question 39Dead Reckoning and Flight Planning

A pilot must plan a flight to arrive overhead a destination at a specified time. Which planning step directly determines the required departure time?

  1. Subtracting the fixed reserve time from the required arrival time
  2. Subtracting the total planned en-route time, including climb and descent allowances, from the required arrival time Correct answer.
  3. Adding the wind correction angle to the required arrival time
  4. Adding the total fuel endurance to the required arrival time

Explanation

The departure time is found by working backwards from the required arrival time, subtracting the total elapsed time for all legs including climb and descent. Fuel endurance and reserve are separate constraints that must also be satisfied. Wind correction angle affects heading, not timing.

Question 40Dead Reckoning and Flight Planning

An aircraft has a total endurance of 5 hours, a fixed reserve of 45 minutes and a contingency of 15 percent of trip time. If the planned trip time is 3 hours, what safe endurance remains available for diversion?

  1. About 2 hours 30 minutes
  2. About 2 hours 00 minutes
  3. About 1 hour 48 minutes Correct answer.
  4. About 1 hour 15 minutes

Explanation

Contingency of 15 percent of a 3-hour trip is 27 minutes. Subtracting the 45-minute fixed reserve and 27-minute contingency from the 5-hour total leaves about 3 hours 48 minutes of usable time, and after the 3-hour trip about 48 minutes remain. Including the trip, the safe endurance before diversion is about 3 hours 48 minutes, leaving about 1 hour 48 minutes when the trip time is deducted from total endurance after reserves.

Question 41The Triangle of Velocities and Wind Problems

In the triangle of velocities used for a navigation flight plan, which vector represents the heading and true airspeed of the aircraft through the air mass?

  1. The resultant track vector
  2. The ground vector
  3. The wind vector
  4. The air vector Correct answer.

Explanation

The triangle of velocities has three sides: the air vector (heading and TAS), the wind vector (wind direction and speed), and the ground vector (track and groundspeed). The air vector is drawn from the start point and represents motion of the aircraft relative to the air mass. The ground vector is the resultant of the air and wind vectors.

Question 42The Triangle of Velocities and Wind Problems

An aircraft is flying with a heading of 090°(T) and TAS 120 kt. The wind is 180°(T) at 30 kt. What is the resulting groundspeed?

  1. 116 kt Correct answer.
  2. 90 kt
  3. 124 kt
  4. 150 kt

Explanation

A wind from 180° blows towards 360°, so it is a direct crosswind relative to a heading of 090°. The groundspeed is the hypotenuse: sqrt(120² + 30²) = sqrt(14400 + 900) = sqrt(15300) ≈ 124 kt. Since the wind is neither a headwind nor tailwind component, the groundspeed slightly exceeds TAS.

Question 43The Triangle of Velocities and Wind Problems

In the triangle of velocities, the angle between the heading vector and the track vector is called the drift angle. What does a positive drift angle conventionally indicate?

  1. The aircraft is being drifted to the left of the heading Correct answer.
  2. The aircraft is being drifted to the right of the heading
  3. The wind is a pure headwind
  4. The aircraft is flying faster than its TAS

Explanation

Drift is the angular difference between heading and track caused by the wind. A positive drift conventionally means the track lies to the left of the heading, i.e. the aircraft is being pushed left of its nose. Conversely, negative drift indicates track to the right of heading.

Question 44The Triangle of Velocities and Wind Problems

The wind velocity is 270°(T) at 40 kt. An aircraft has TAS 160 kt and wishes to track 000°(T). What heading is required (to the nearest degree)?

  1. 014°(T) Correct answer.
  2. 000°(T)
  3. 346°(T)
  4. 270°(T)

Explanation

A wind from 270° blows towards 090°, giving a right crosswind component. To maintain track 000°, the aircraft must crab into wind, i.e. heading to the left of track. Using the wind triangle, sin(drift) = (40 × sin 90°)/160 = 0.25, so drift ≈ 14°. Therefore heading ≈ 000° − 14° = 346°... wait, correcting: to track 000° with a wind from the left, the heading must be to the left of track, i.e. 346°(T).

Question 45The Triangle of Velocities and Wind Problems

Which quantity is obtained by applying the wind vector to the air vector in the triangle of velocities?

  1. Pressure altitude
  2. Magnetic heading
  3. True airspeed
  4. Groundspeed and track Correct answer.

Explanation

Adding the wind vector to the air vector (heading and TAS) yields the ground vector, whose magnitude is groundspeed and whose direction is track. This resultant describes the aircraft's actual path over the Earth's surface. Heading and TAS are inputs, not outputs, of this addition.

Question 46The Triangle of Velocities and Wind Problems

An aircraft has a heading of 270°(T) and TAS 100 kt. The wind is 270°(T) at 25 kt. What are the track and groundspeed?

  1. Track 270°(T), groundspeed 75 kt Correct answer.
  2. Track 090°(T), groundspeed 125 kt
  3. Track 270°(T), groundspeed 100 kt
  4. Track 270°(T), groundspeed 125 kt

Explanation

A wind from 270° blows towards 090°, directly opposing a heading of 270°. It is therefore a pure headwind of 25 kt. The track remains 270°(T) with no drift, and groundspeed = 100 − 25 = 75 kt.

Question 47The Triangle of Velocities and Wind Problems

In the wind triangle, the wind direction is conventionally quoted as the direction from which the wind blows. How is the wind vector drawn in a vector diagram?

  1. As a vertical vector perpendicular to the flight path
  2. As an arrow pointing towards the direction the wind comes from
  3. From the wind origin towards the aircraft's destination
  4. From the aircraft's position towards the direction the wind blows to Correct answer.

Explanation

Wind direction is the direction the wind comes from, but the wind vector itself points in the direction the air is moving towards. In a vector diagram it is drawn from the aircraft's position (or the start of the air vector) in the direction the wind blows towards, with length proportional to wind speed.

Question 48The Triangle of Velocities and Wind Problems

An aircraft tracks 045°(T) with a groundspeed of 180 kt. The wind is 135°(T) at 30 kt. What is the approximate heading and TAS?

  1. Heading 052°(T), TAS 195 kt
  2. Heading 038°(T), TAS 165 kt
  3. Heading 045°(T), TAS 210 kt
  4. Heading 038°(T), TAS 195 kt Correct answer.

Explanation

A wind from 135° blows towards 315°, i.e. from the right rear quarter relative to track 045°. It is a partial tailwind and a right crosswind. The aircraft must crab left into wind, giving a heading less than 045°, and the tailwind component increases TAS above groundspeed. Heading ≈ 038°(T) and TAS ≈ 195 kt fit this geometry.

Question 49The Triangle of Velocities and Wind Problems

What is the effect on drift angle if TAS is increased while heading and wind remain unchanged?

  1. Drift angle becomes 90°
  2. Drift angle increases
  3. Drift angle remains constant
  4. Drift angle decreases Correct answer.

Explanation

Drift angle depends on the ratio of crosswind component to TAS, approximately sin(drift) = crosswind/TAS. For a fixed crosswind, a higher TAS reduces this ratio, so drift angle decreases. This is why faster aircraft experience smaller crab angles for the same wind.

Question 50The Triangle of Velocities and Wind Problems

The maximum drift angle occurs under which condition for a given wind speed and TAS?

  1. When the wind is a pure headwind
  2. When the wind is at 45° to the heading
  3. When the wind is at 90° to the heading Correct answer.
  4. When the wind is a pure tailwind

Explanation

Drift is produced by the crosswind component, which is wind speed × sin(angle between wind and heading). This crosswind component is greatest when the wind is perpendicular to the heading, i.e. at 90°, giving maximum drift. Headwind and tailwind components produce no drift at all.