AIR NAVIGATION. Key Revision. Press F5 to start.

AIR NAVIGATION Key Revision Press F5 to start.

This presentation may be used either as a revision aid or as a self-test program. Revision Self-test Instructions

This presentation may be used either as a revision aid or as a self-test program. To revise, just use the down arrow or left mouse button to progress. The correct answer(s) will be highlighted. Press down arrow again to go on to the next question. To test yourself, use the mouse to left click directly on the highlighted a) b) c) or d) alongside to the correct answer. If you are correct your answer will be highlighted. If your answer was incorrect you will be invited to try again. Click OK to go back and try again. If you don t want to try again and just want the answer, left click or down arrow will highlight the correct one.

AIR NAVIGATION For revision, press the down arrow or left mouse button to advance. Press up arrow to go back.

AIR NAVIGATION To test your knowledge, click directly on the highlighted a) b) c) or d) alongside the correct answer.

Contents List. Click on a chapter. AIR NAVIGATION Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Distance, speed and time. The triangle. The 1 in 60 rule. Compasses. Weather. exit

AIR NAVIGATION Chapter 1 Return to contents list Distance, speed and time. exit

Distance, Speed & Time Pilots must make regular checks of their Estimated Time of Arrival (ETA) at destination as well as estimated times for passing waypoints enroute. These are necessary for ATC reports, and vital for ensuring sufficient fuel remains to reach the destination.

Regular checks of Estimated Time of Arrival (ETA) are important. These calculations help the crew to determine that: a) The aircraft has sufficient fuel to reach the destination. c) They are flying the shortest route. b) The wind velocity will not change. d) The drift is correct.

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Aircrew are always aware of their Estimated Time of Arrival (ETA). Why is this? a) Fuel flow rate depends on ETA. c) It is important for fuel calculations and air traffic control purposes. b) It is the easiest calculation to do. d) A revised ETA tells them the wind has changed.

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Distance on the Earth Distance on the earth s surface is measured in Nautical Miles. One Nautical Mile (nm) is the equivalent to one minute of latitude, (one sixtieth of a degree).

Distance on the Earth Degrees of latitude and longitude are marked with the symbol. Minutes of latitude and longitude are marked with the symbol.

Distance on the earth's surface is measured in nautical miles (nm). Which of the following is true? a) One nm is equal to one minute of latitude. b) One nm equals 1/10,000th of the distance from the North Pole to the Equator. c) One nm is equal to 5280 feet. d) One nm is equal to one minute of longitude.

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One degree of latitude is equal to: a) 360 nms b) 60 nms c) 60 kms d) 1 nm

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One degree of latitude is equal to: a) 360 nms b) 60 nms c) 60 kms d) 1 nm

One minute of latitude on the earth's surface is equal to: a) 1 nautical mile. b) 60 nautical miles. c) 1 knot. d) 1 km.

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One minute of latitude on the earth's surface is equal to: a) 1 nautical mile. b) 60 nautical miles. c) 1 knot. d) 1 km.

Measuring Distance Nautical maps do not have scales on the borders. We use the scale shown along each meridian. If dividers are used to measure distance, the degrees and minutes scale on the nearest meridian should be used to convert that distance into nautical miles. The degrees and minutes on the parallels of latitude should not be used for measuring purposes because convergence towards the poles shrinks the scale.

Distances should not be measured using parallels as they converge towards the poles. Longitude 0º 15ºE 30ºE 45ºE Only at the equator does one degree of longitude equal 60 nm.

Distances measured using scales along the meridians will be accurate. 75º N 60º N Latitude 45º N 30º N 15º N 0º

The LATITUDE of a point is its distance measured in degrees and minutes: a) From the Greenwich (Prime) Meridian. c) North or South of the Equator. b) From the true North Pole. d) From the true South Pole.

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The distance between two points on a navigation chart can be measured with dividers. What scale will then be used to convert that distance to nautical miles? a) The minute scale along a meridian close to the area of interest on the chart. c) The minute scale along a parallel of latitude. b) 1:50,000 scale. d) Any meridian scale off any chart.

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Change of Latitude 54N 53N 52N 51N Keil 54 20 N If two places are on the same meridian, it is possible to determine how far apart they are by calculating the differences in Latitude. In this example Keil is due north of Wartzburg. 50N Wartzburg 49 48 N

Change of Latitude 54N 53N 52N 51N Keil 54 20 N Remembering that one minute of latitude is one nautical mile, we can see that Wartzburg is just 12 nautical miles south of the 50 line of latitude. 49 48 plus 12 = 50 00. 50N Wartzburg 49 48 N

Change of Latitude 54N 53N 52N 51N Keil 54 20 N Each degree of latitude between 50 North and 54 North is a further 60 nautical miles. 4 times 60 = 240 nautical miles. 50N Wartzburg 49 48 N

Change of Latitude 54N 53N Keil 54 20 N Finally, we can see that Keil is another 20 North of latitude 54 North. 52N 51N So another 20 nautical miles must be added to our total. 50N Wartzburg 49 48 N

Change of Latitude 20 54N 53N 52N 51N Keil 54 20 N 240 12 + 240 + 20 = 272 Keil is therefore 272 nautical miles due north of Wartzburg. 12 50N Wartzburg 49 48 N

In Germany, Kiel is due north of Wartzburg. If Kiel's latitude is 54 20N and Wartzburg's is 49 48N how far are they apart? a) 272 nm b) 2720 nm c) 27.2 nm d) 227 nm

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In Germany, Kiel is due north of Wartzburg. If Kiel's latitude is 54 20N and Wartzburg's is 49 48N how far are they apart? a) 272 nm b) 2720 nm c) 27.2 nm d) 227 nm Each degree is 60 nm each minute is 1 nm. Wartzburg is 12 minutes South of 50N, Kiel 20 minutes North of 54N. 50N to 54N is 4 degrees. Each degree is 60 nm. 4 x 60 + 12 + 20 = 272 nm.

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Oslo airport (Norway) is due north of Braunschweig airfield near Hanover (Germany). If their latitudes are 59 53N and 52 20N respectively, how far are they apart? a) 453 nm b) 454 nm c) 554 nm d) 445 nm 52 20N to 59 53N is 7 degrees and 33 minutes. Each degree is 60 nm, each minute is 1 nm. 7 degrees x 60 nm = 420nm, plus 33nm = 453nm.

Your destination airfield is situated due south of your departure airfield. If the two latitudes are 63 25N and 57 58N, how far are they apart? a) 327 b) 317 c) 323 d) 333

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Your destination airfield is situated due south of your departure airfield. If the two latitudes are 63 25N and 57 58N, how far are they apart? a) 327 b) 317 c) 323 d) 333

Dundee is due north of Abergavenny. If their latitudes are 56 27N and 51 50N, how far are they apart? a) 277 kms. b) 323 kms. c) 323 nms. d) 277 nms.

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Dundee is due north of Abergavenny. If their latitudes are 56 27N and 51 50N, how far are they apart? a) 277 kms. b) 323 kms. c) 323 nms. d) 277 nms.

Aircraft Speed On land we measure distance in miles and speed in miles per hour (mph). In aviation we use nautical miles (nm) to measure distances and speed is measured in nautical miles per hour, known as knots and abbreviated kts.

Aircraft Speed An aircraft measures speed through the air using an instrument called an Air Speed Indicator (ASI). The ASI compares the pressure caused by the aircraft s forward motion through the air (the Pitot pressure) with the pressure of the air surrounding the aircraft (the Static pressure). The faster the aircraft flies, the greater is the difference between these two pressures.

In aviation, speed is measured in: a) kilometres per hour (km/hr). b) miles per hour (mph). c) knots (kts). d) metres per hour (m/hr).

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In aviation, speed is measured in: a) kilometres per hour (km/hr). b) miles per hour (mph). c) knots (kts). d) metres per hour (m/hr).

The Air Speed Indicator (ASI) calculates speed by: a) Measuring the pressure difference between pitot and static pressures. b) Measuring the pitot pressure. c) Measuring the static pressure. d) Multiplying pitot pressure by static pressure.

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Calibrated Airspeed The indicated airspeed (IAS) is corrected for Pressure Error and Instrument Error to give a more accurate airspeed Calibrated Airspeed (CAS). IAS + Pressure Error + Instrument Error = CAS Pressure Error is caused by the airflow around the aircraft. Carefully positioning the pitot and static tubes can minimise, but not eliminate completely, this error.

Calibrated Air Speed (CAS) is: a) Pitot pressure minus static pressure. b) IAS after correction for pressure error and instrument error. c) Always less than IAS. d) Always greater than IAS.

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Calibrated Air Speed (CAS) is: a) Pitot pressure minus static pressure. b) IAS after correction for pressure error and instrument error. c) Always less than IAS. d) Always greater than IAS.

Calibrated Air Speed (CAS) equals Indicated Air Speed (IAS) plus corrections for: a) Altitude error. b) Pressure error. c) Instrument error. d) Pressure and instrument error.

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Calibrated Air Speed (CAS) equals Indicated Air Speed (IAS) plus corrections for: a) Altitude error. b) Pressure error. c) Instrument error. d) Pressure and instrument error.

True Airspeed As an aircraft flies higher the air becomes less dense, so the aircraft flies faster through the thinner air to achieve the same force on the pitot tube. To find the True Airspeed (TAS) at altitude the Calibrated Airspeed must now be corrected for air density changes caused by temperature and altitude. CAS + Density Error (temperature & altitude) = TAS

True Airspeed To summarise: IAS + Pressure & Instrument Error = CAS CAS + Density Error (temperature & altitude) = TAS

True Airspeed To summarise: IAS + Pressure & Instrument Error = CAS CAS + Density Error (temperature & altitude) = TAS I P I C D T

When Calibrated Airspeed (CAS) is corrected for altitude and temperature, it becomes: a) True Air Speed (TAS). b) Indicated Airspeed (IAS). c) Mach Number. d) Indicated Groundspeed.

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When Calibrated Airspeed (CAS) is corrected for altitude and temperature, it becomes: a) True Air Speed (TAS). b) Indicated Airspeed (IAS). c) Mach Number. d) Indicated Groundspeed.

Calculation of Flight Time If a car travels 120 miles at 60 mph, it will take 2 hours to complete the journey. Distance (D) Speed (S) 120 miles = Time (T) 60 mph = 2 hours

Calculation of Flight Time Similarly, if we know the distance and time taken, we can calculate the speed. Distance (D) Time (T) 120 miles = Speed (S) 2 hours = 60 mph

Calculation of Flight Time If we know the speed of the vehicle and the time the journey has taken, then we can calculate the distance covered. Speed (S) x Time (T) = Distance (D) 60 mph x 2 hours = 120 miles

Calculation of Flight Time Aircraft normally fly at faster speeds and cover greater distances, but the principle and the mathematics remain the same. Speed (S) x Time (T) = Distance (D) 600 kts x 2 hours = 1200 nm

How fast must an aircraft fly to cover 1200 nm in 3 hours? a) 400 kts b) 800 kts c) 400 mph d) 3600 kts

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How fast must an aircraft fly to cover 1200 nm in 3 hours? a) 400 kts b) 800 kts c) 400 mph d) 3600 kts 1200 nm in 3 hours requires the aircraft to cover 400 nm each hour (1200 / 3 = 400). 400 nm per hour = 400 knots.

A Hercules is flying at a groundspeed of 210 kts. How far will it travel in 3 hours? a) 630 nms. b) 70 nms c) 630 km. d) 210 nms.

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A Hercules is flying at a groundspeed of 210 kts. How far will it travel in 3 hours? a) 630 nms. b) 70 nms c) 630 km. d) 210 nms. 3 hours @ 210 kts (nautical miles per hour) = 630 nautical miles.

A Tornado flies from its base to a target in 30 minutes. If the distance is 250 nms, what speed is it flying at? a) 125 kts. b) 500 kts. c) 750 kts.. d) 800 kts.

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A Tornado flies from its base to a target in 30 minutes. If the distance is 250 nms, what speed is it flying at? a) 125 kts. b) 500 kts. c) 750 kts. d) 800 kts. 250 nms in 30 minutes means the aircraft would cover 500 nms in one hour, giving a speed of 500 kts.

A Nimrod flies on patrol for nine hours at a speed of 300 kts. How far does it travel in this time? a) 2400 nms. b) 2700 nms. c) 3000 nms. d) 3900 nms.

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A Nimrod flies on patrol for nine hours at a speed of 300 kts. How far does it travel in this time? a) 2400 nms. b) 2700 nms. c) 3000 nms. d) 3900 nms.

A Hercules flies from A to B, a distance of 1000 nms at a groundspeed of 250 kts. How long does the flight take? a) 3 hrs 20 mins. b) 4 hrs. c) 3 hrs 30 ins. d) 5 hrs.

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A Hercules flies from A to B, a distance of 1000 nms at a groundspeed of 250 kts. How long does the flight take? a) 3 hrs 20 mins. b) 4 hrs. c) 3 hrs 30 mins. d) 5 hrs.

Units of Time All military and commercial aviators use the same time. This is known as either Greenwich Mean Time (GMT) or Universal Time (UT).

Universal Time (UT) time is used as standard in military and commercial aviation. By what other name is it known? a) British Summer Time (BST). b) European Daylight Saving Time (EDST). c) Greenwich Mean Time (GMT). d) Local Time (LT) i.e. the time of the country over which the aircraft is flying.

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AIR NAVIGATION Chapter 2 Return to contents list The triangle. exit

Vectors and Velocity Whenever we talk about aircraft or wind movement, we must always give both direction and speed of that movement. Direction and speed together are called a velocity. A velocity can be represented on paper by a line called a vector.

Vectors and Velocity The bearing of the line represents the direction of the movement.

Vectors and Velocity The bearing of the line represents the direction of the movement. The length of the line represents speed.

Vectors and Velocity Imagine trying to sail a boat across a fast flowing river. A

Vectors and Velocity This line with one arrow represents the velocity of the boat (its speed and the direction it was pointed). A

Vectors and Velocity If a boat is pointed directly across a river, the flow of the river will push the boat downstream. A B

Vectors and Velocity The line with three arrows represents the speed and direction of the current. A B

Vectors and Velocity Although the boat started off pointing at A it finished up at B because of this current. A B

Vectors and Velocity Joining the ends of these two lines with a third line completes the vector triangle. A B

Vectors and Velocity This line, with two arrows, is the resultant and represents the actual course of the boat. A B

The Air Triangle A similar triangle can represent an aircraft s movement through air which is itself moving (wind). heading and true airspeed (HDG/TAS) track and groundspeed wind speed and direction Note that the wind vector describes the direction the wind is blowing from northerly in this example.

The Air Triangle The angle between the heading and the track, caused by the wind, is called drift. heading and true airspeed (HDG/TAS) drift track and groundspeed wind speed and direction

Velocity consists of: a) Speed only. b) Direction only. c) Speed and direction together. d) Several speed vectors together.

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Velocity consists of: a) Speed only. b) Direction only. c) Speed and direction together. d) Several speed vectors together.

A vector is a representation on paper of: a) Speed. b) Time. c) Direction. d) Direction and speed.

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A vector is a representation on paper of: a) Speed. b) Time. c) Direction. d) Direction and speed.

A vector is a line, drawn to represent a velocity. This is achieved by: a) The bearing represents knots at all times. c) The length represents mph at all times. b) The bearing represents speed and the length represents direction. d) The bearing represents direction and the length represents speed.

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In the diagram, vector 2 is added to vector 1. What is vector 3 (A-C) known as? a) The ready vector. b) Current. c) The resultant vector. d) Drift. B 1 2 A 3 C

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In the diagram, vector 2 is added to vector 1. What is vector 3 (A-C) known as? a) The ready vector. b) Current. c) The resultant vector. d) Drift B 1 2 A 3 C

In the triangle of velocities, DRIFT is: a) The bearing of the wind vector. c) The angle between heading and track vectors. b) The angle between the wind and track vectors. d) The angle between heading and wind vectors.

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In the air triangle, the heading vector includes 2 components. They are: a) Heading and wind velocity. b) Heading and groundspeed. c) Heading and drift. d) Heading and true air speed.

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In the air triangle, the heading vector includes 2 components. They are: a) Heading and wind velocity. b) Heading and groundspeed. c) Heading and drift. d) Heading and true air speed.

In the air triangle, the track vector includes 2 components. They are: a) Track and drift. b) Track and heading. c) Track and groundspeed. d) Track and true air speed.

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In the air triangle, the track vector includes 2 components. They are: a) Track and drift. b) Track and heading. c) Track and groundspeed. d) Track and true air speed.

In the air triangle, the wind vector includes 2 components. They are: a) Wind speed and drift. b) Wind speed and heading. c) Wind speed and the direction the wind is blowing from. d) Wind speed and track.

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In the Air Triangle shown below, name the components of the 3rd side, represented by a dotted line: a) Wind velocity. b) Heading and true airspeed. c) Drift and groundspeed. d) Track and groundspeed.

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In the Air Triangle shown below, name the components of the 3rd side, represented by a dotted line: a) Wind velocity. b) Heading and true airspeed. c) Drift and groundspeed. d) Track and groundspeed.

In the Air Triangle shown below, name the components of the 3rd side, represented by a dotted line: a) Wind direction and speed. b) Heading and true airspeed. c) Drift and groundspeed. d) Drift.

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In the Air Triangle shown below, name the components of the 3rd side, represented by a dotted line: a) Wind direction and speed. b) Heading and true airspeed. c) Drift and groundspeed. d) Drift.

Solving the Vector Triangle As we have demonstrated, the vector triangle consists of 6 elements: heading and true airspeed windspeed and direction track and groundspeed Providing we know four of the elements of the vector triangle (any four) it is possible to calculate the other two.

The Air Triangle of velocities can be used to calculate flight data. There are 6 elements in total. How many elements are needed to calculate those missing? a) 2 b) 4 c) 5 d) 6

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The Air Triangle of velocities can be used to calculate flight data. There are 6 elements in total. How many elements are needed to calculate those missing? a) 2 b) 4 c) 5 d) 6 Any of the four elements will enable the remaining two to be found.

Mental Calculations Despite modern equipment, pilots must still make quick, accurate calculations in their heads. Every pilot should know the distance his aircraft will cover in one minute for any given groundspeed. For instance, a Tornado flying at 420 kts groundspeed will cover 7 nm per minute. If the next turning point is 35 nm away, dividing 35 by 7 tells him he will be there in 5 minutes.

Mental Calculations The table on the next page shows how many nautical miles per minute are covered at various groundspeeds. Ensure you know the examples in italics.

Mental Calculations Groundspeed (kts) nm/minute 60 1 120 2 180 3 240 4 300 5 360 6 420 7 480 8 540 9

You are flying at 120 kts groundspeed. How long will it take to fly 20 nms? a) 60 minutes. b) 10 minutes. c) 6 minutes. d) 2 minutes.

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You are flying at 120 kts groundspeed. How long will it take to fly 20 nms? a) 60 minutes. b) 10 minutes. c) 6 minutes. d) 2 minutes. 120 kts is 120 nms per hour or 2 nms per minute. To fly 20 nms at 2 nms per minute would take 10 minutes.

You are flying a Tornado at 420 kts groundspeed. How many miles do you travel each minute? a) 42 nm b) 8 nm c) 7 nm d) 6 nm

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You are flying a Tornado at 420 kts groundspeed. How many miles do you travel each minute? a) 42 nm b) 8 nm c) 7 nm d) 6 nm There are 60 minutes in each hour. 420 divided by 60 is 7 nm per minute.

You fly between 2 features on the ground and you notice it takes 3 minutes. If the features are 18 nm apart, what is your groundspeed? a) 54 kts b) 180 kts c) 280 kts d) 360 kts

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You fly between 2 features on the ground and you notice it takes 3 minutes. If the features are 18 nm apart, what is your groundspeed? a) 54 kts b) 180 kts c) 280 kts d) 360 kts 3 minutes to cover 18 nm is 6 nm per minute. Each hour is 60 minutes, you will cover 6 x 60 = 360 nm per hour.

Estimated Time of Arrival (ETA) Pilots are constantly calculating and updating their ETA s, both for turning points and final destination. ETA s are important for both fuel and Air Traffic Control purposes. If an aircraft fails to arrive at its destination on time, then ATC will initiate overdue action.

An aircraft departs from base, but does not arrive at the destination on its Estimated Time of Arrival (ETA). What action will Air Traffic Control take? a) No immediate action is required. b) Close down and go home. c) Contact the departure base. d) Initiate overdue action.

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AIR NAVIGATION Chapter 3 Return to contents list The 1 in 60 rule. exit

The 1 in 60 rule A Track Required B A line drawn on the map between departure and destination (or from one turning point to another) is known as the Track Required.

The 1 in 60 rule Track Made Good pinpoint A Track Required B If the aircraft drifts off track, then the line from our departure airfield to our present position is known as Track Made Good (TMG).

An aircraft is flying from point A to point B. Halfway a pinpoint fix shows it to be off track. A line between point A and the fix would be known as: a) Drift. b) Revised track. c) Track made good. d) Track required.

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The 1 in 60 rule Track Made Good pinpoint A Track Required Track Error B The angle between the Track Required and the Track Made Good (the actual track of the aircraft) is called the Track Error (TE).

The 1 in 60 rule Track Made Good pinpoint A Track Required Track Error B The 1 in 60 rule states if an aircraft flies a Track Made Good (TMG) one degree off the Track Required, after 60 miles of flying the aircraft will be one mile off track.

The 1 in 60 rule 60 miles A Track Made Good T E = 10 degrees Track Required pinpoint 10 miles B Similarly, if an aircraft flies a Track Made Good (TMG) ten degrees off the Track Required, after 60 miles of flying the aircraft will be ten miles off track.

The 1 in 60 rule 60 miles A Track Made Good T E = 10 degrees Track Required pinpoint 10 miles B The pilot now has the information he requires to get the aircraft back on track.

The 1 in 60 rule 60 miles A Track Made Good T E = 10 degrees Track Required pinpoint 10 miles B The rule works for track errors up to 23 degrees.

The 1 in 60 rule 30 miles A Track Made Good T E Track Required pinpoint 4 miles B Where the aircraft has flown less than 60 miles the triangle has to be extended to determine how far the aircraft would be off track after 60 miles.

The 1 in 60 rule 30 miles A Track Made Good T E Track Required pinpoint 4 miles B In this example the aircraft is 4 miles off track after just 30 miles. After 60 miles it would be 8 miles off track giving a Track Error (TE) of 8 degrees.

The 1 in 60 rule 40 miles A Track Made Good T E Track Required pinpoint 6 miles B Here the aircraft is 6 miles off track after 40 miles. After 60 miles therefore, it would be 9 miles off track giving a Track Error (TE) of 9 degrees.

Using the 1 in 60 rule, calculate how many miles off track an aircraft will be if it flies 60 nm with a track error of 2 degrees. a) 60 nms b) 6 nms c) 4 nms d) 2 nms

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Using the 1 in 60 rule, calculate how many miles off track an aircraft will be if it flies 60 nm with a track error of 2 degrees. a) 60 nms b) 6 nms c) 4 nms d) 2 nms

An aircraft flies a track made good, 3 degrees in error from the required track. Using the 1 in 60 rule, how many miles off track will the aircraft be after 60 miles of flying? a) 2 nms b) 6 nms c) 1 nm d) 3 nms

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An aircraft is flying from A to B, after 20 nms it is found to be 3 nms off track. What is the track error? a) 6 degrees. b) 2 degrees. c) 9 degrees. d) 4 degrees.

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An aircraft is flying from A to B, after 20 nms it is found to be 3 nms off track. What is the track error? a) 6 degrees. b) 2 degrees. c) 9 degrees. d) 4 degrees. If the aircraft was 3 nms off track after 20 nms, projecting ahead it would be 9 nms off track after 60 nms, therefore 9 degrees.

The 1 in 60 rule 120 miles A Track Made Good T E Track Required pinpoint 6 miles B Where the aircraft has flown more than 60 miles before obtaining a pinpoint, you must determine how far the aircraft was off track back at the 60 mile point.

The 1 in 60 rule 120 miles A Track Made Good T E Track Required pinpoint 6 miles B In this example the aircraft is 6 miles off track after 120 miles, so would have been 3 miles off track after 60 miles, a 3 degree Track Error.

The 1 in 60 rule 90 miles A Track Made Good T E Track Required pinpoint 6 miles B Here the aircraft is 6 miles off track after 90 miles, so would have been 4 miles off track after 60 miles, that s a 4 degree Track Error.

The 1 in 60 rule A 60 miles 60 miles pinpoint Track Made Good TE Track Required B Once a pilot has determined his position and therefore his Track Error, then he can adjust his heading.

The 1 in 60 rule A 60 miles 60 miles pinpoint Track Made Good TE Track Required B The aircraft is to the left of the Track Required, so must turn to the right. If the Track Error is 8 degrees, by how many degrees should he change heading?

The 1 in 60 rule A 60 miles 60 miles pinpoint Track Made Good TE Track Required B If he turns by only eight degrees (the Track Error) the pilot will only parallel the Track Required. A further eight degrees (16 degrees in all) will put him back on track after another 60 miles.

The 1 in 60 rule 60 miles 60 miles A pinpoint Track Made Good TE Closing Angle Track Required Revised Track B This new track is known as the Revised Track. The angle between the Revised Track and the original Track Required is the Closing Angle (CA).

An aircraft is flying from point A to point B. A pinpoint fix shows it to be off track. A line from the pinpoint fix to point B would be known as: a) Track made good. b) Track required. c) Revised track. d) Heading required.

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An aircraft is flying from A to B, a distance of 120 nm. Halfway, a fix shows the aircraft to be 4 nm right of track. What heading change does the pilot require to reach point B? a) 8 degrees right b) 8 degrees left c) 4 degrees right d) 4 degrees left

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20 nm after takeoff for a pre-planned destination, a pilot finds that he is 3 nm off track. By how much does the pilot need to turn to regain the intended track after flying a further 20 nm? a) 18 degrees b) 9 degrees c) 6 degrees d) 3 degrees

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The 1 in 60 rule A 30 miles 60 miles TMG pinpoint Revised Track CA Track Required B When the pinpoint is not exactly halfway along the track, a little more thought is required. Each triangle must be evaluated separately.

The 1 in 60 rule A 30 miles 60 miles TMG pinpoint Revised Track 4nm CA Track Required B After 30 miles the aircraft is 4 miles off track. The Track Error is therefore 8 degrees. Turning eight degrees to the right will parallel the Track Required.

The 1 in 60 rule A 30 miles 60 miles TMG pinpoint Revised Track CA Track Required B After 30 miles the aircraft is 4 miles off track. The Track Error is therefore 8 degrees. Turning eight degrees to the right will parallel the Track Required.

The 1 in 60 rule A 30 miles 60 miles TMG pinpoint Revised Track 4 nm CA Track Required The pilot must turn further right to get onto the Revised Track but by how much? B He is paralleling the Track Required, so must continue to turn through an angle equivalent to the Closing Angle (CA).

The 1 in 60 rule A 30 miles 60 miles TMG pinpoint Revised Track 4 nm CA Track Required B As the aircraft was 4 miles off track, using the 1 in 60 rule gives a Closing Angle (CA) of 4 degrees.

The 1 in 60 rule A 30 miles 60 miles TMG pinpoint Revised Track 4 nm CA Track Required B As the aircraft was 4 miles off track, using the 1 in 60 rule gives a Closing Angle (CA) of 4 degrees. The pilot must turn 12 degrees in all, 8 degrees to parallel the Track Required and 4 degrees to get onto the Revised Track.

The 1 in 60 rule A 30 miles 60 miles TMG pinpoint Revised Track CA Track Required B As the aircraft was 4 miles off track, using the 1 in 60 rule gives a Closing Angle (CA) of 4 degrees. The pilot must turn 12 degrees in all, 8 degrees to parallel the Track Required and 4 degrees to get onto the Revised Track.

An aircraft flying from A to B finds itself 6 nms off track. It has a further 60 nms to travel. What is the required closing angle? a) 10 degrees b) 6 degrees c) 3 degrees d) 2 degrees

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An aircraft flying from A to B finds itself 6 nms off track. It has a further 60 nms to travel. What is the required closing angle? a) 10 degrees b) 6 degrees c) 3 degrees d) 2 degrees

An aircraft flying from A to B finds itself 4 nms off track. It has a further 60 nms to travel. What is the required closing angle? a) 6 degrees b) 4 degrees c) 3 degrees d) 2 degrees

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An aircraft flying from A to B finds itself 4 nms off track. It has a further 60 nms to travel. What is the required closing angle? a) 6 degrees b) 4 degrees c) 3 degrees d) 2 degrees

An aircraft flying from A to B is found to be off track at the pinpoint shown below. The pilot calculates the track error as 6 degrees and the closing angle of 3 degrees. By how much does the pilot need to turn to reach point B? a) 9 degrees to the left b) 9 degrees to the right c) 2 degrees to the left d) 2 degrees to the right fix TE CA A B

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An aircraft flying from A to B is found to be off track at the pinpoint shown below. The pilot calculates the track error as 12 degrees and the closing angle of 8 degrees. By how much does the pilot need to turn to reach point B? a) 20 degrees to the right b) 12 degrees to the right c) 8 degrees to the right d) 4 degrees to the right fix TE CA A B

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The 1 in 60 rule 40 miles 30 miles pinpoint A TMG TE 6 m Track Required Revised Track CA B Here is an example of the most complex problem you will face in the examination. After flying for 40 miles the aircraft is 6 miles off track. It has a further 30 miles to travel. By how much does the pilot need to turn to regain track at B?

The 1 in 60 rule 40 miles 20 miles pinpoint A TMG TE 6 m Track Required 9 miles B To solve this problem you must consider the two triangles individually. Extend the first triangle the full 60 miles to find out how far you would be off track there that s the track error in degrees.

The 1 in 60 rule 40 miles 20 miles pinpoint A TMG TE 6 m Track Required 9 miles B After 60 miles you would be 9 miles off track. The Track Error is therefore 9 degrees. This is the angle you must turn right to parallel the Track Required. To regain track, let s look at the second triangle.

The 1 in 60 rule 30 miles 30 miles 12 miles A pinpoint 6 m Track Required Revised Track CA B As you only have 30 miles to travel the Closing Angle can be found by projecting the triangle back to 60 miles. This gives a Closing Angle of 12 degrees and this must be added to the 9 degrees already turned to regain track at B.

The 1 in 60 rule 40 miles 30 miles pinpoint A TMG TE 6 m Track Required Revised Track CA B The Track Error of 9 degrees plus the Closing Angle of 12 degrees requires a turn of 21 degrees to regain track at B. In the examination always draw the triangles on scrap paper to check the sums.

The 1 in 60 rule 40 miles 30 miles pinpoint A TMG TE 6 m Track Required Revised Track CA B For simplification, in all of these examples the aircraft has been off track to the left. Examination questions have Track Errors left and right.

An aircraft flying from A to B finds that after 40 nms it is 4 nms off track. It has a further 60 nms to travel. By how much does the pilot need to turn to regain the intended track at B? a) 12 degrees b) 10 degrees c) 6 degrees d) 4 degrees

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An aircraft flying from A to B finds that after 30 nms it is 4 nms left of track. It has a further 40 nms to travel. By how much does the pilot need to turn to regain the intended track at B? a) 16 degrees to the right b) 14 degrees to the right c) 14 degrees to the left d) 12 degrees to the left

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An aircraft flying from A to B finds that after 20 nms it is 2 nms right of track. It has a further 40 nms to travel. By how much does the pilot need to turn to regain the intended track at B? a) 12 degrees to the left b) 9 degrees to the left c) 6 degrees to the left d) 6 degrees to the right

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An aircraft flying from A to B finds that after 40 nms it is 6 nms right of track. It has a further 30 nms to travel. By how much does the pilot need to turn to regain the intended track at B? a) 24 degrees right b) 21 degrees left c) 18 degrees left d) 12 degrees left

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AIR NAVIGATION Chapter 4 Return to contents list Compasses. exit

Magnetic Variation The earth s molten core creates a weak magnetic field which resembles the field around a bar magnet. This magnet inside the earth is inclined slightly to the earth s axis, so true north and magnetic north are not in the same place.

Magnetic North Pole True North Pole

Magnetic Variation The difference in direction between true and magnetic north (variation) is at its greatest in polar areas. Also, just as the lines of magnetic force go into the ends of a bar magnet, they angle steeply into the earth in polar areas causing compass needles to dip as they try and align themselves. For these reasons, compasses are very inaccurate in polar areas.

Where are variation values at their greatest? a) In the Northern hemisphere. b) In polar regions. c) At the equator. d) In the Southern hemisphere.

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Where are variation values at their greatest? a) In the Northern hemisphere. b) In polar regions. c) At the equator. d) In the Southern hemisphere.

As a compass nears the Magnetic North Pole, the compass detector will try and point at the magnetic material inside the Earth. This tilting is called: a) Drip. b) Drop. c) Dip. d) Variation.

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As a compass nears the Magnetic North Pole, the compass detector will try and point at the magnetic material inside the Earth. This tilting is called: a) Drip. b) Drop. c) Dip. d) Variation.

The Direct Indicating Compass The Direct Indicating Compass (DIC) is a very simple aircraft compass. Because of its limitations it is only used as a standby on most aircraft. Its limitations are: It only reads correctly in unaccelerated straight and level flight. It only reads magnetic heading. It is unreliable at high magnetic latitudes.

The Direct Indicating Compass The Direct Indicating Compass (DIC) does, however, have three significant advantages. These advantages are: It is simple and reliable. It is cheap and lightweight. It does not require any form of power.

When would a Direct Indicating Compass (DIC) be most accurate? a) In unaccelerated flight. b) In a turn. c) In a steady climb. d) In a steady descent.

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When would a Direct Indicating Compass (DIC) be most accurate? a) In unaccelerated flight. b) In a turn. c) In a steady climb. d) In a steady descent.

Which of the following statements is true, concerning the Direct Indicating Compass (DIC)? a) The DIC gives a reading of aircraft true heading. c) The DIC is not affected by turns and accelerations. b) The DIC needs only a small power supply. d) The DIC only reads magnetic headings.

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All RAF aircraft are equipped with a Direct Indicating Compass (DIC). Why is this? a) The DIC is the most accurate compass available. c) The DIC gives a reading of true heading. b) The DIC is not affected by turns or accelerations. d) The DIC is reliable and needs no power supply.

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The Gyro-Magnetic Compass The gyro-magnetic compass (GMC) was invented to overcome the limitations of the DIC. It has three main components: A magnetic detector (or flux valve). A turn / acceleration cut-out switch. A gyroscope. It can also power various compass repeaters around the aircraft (for navigators, WSOps etc).

The Gyro-Magnetic Compass The magnetic detector or flux valve is solid state equipment (i.e. no moving parts) which electrically senses the magnetic field. It is located in the wingtip in order to keep deviation to a minimum. It is, however, still subject to the same turning and acceleration errors as the DIC. A turn / acceleration cut-out switch prevents the gyroscope from receiving erroneous updates during aircraft manoeuvres.

The Gyroscope The gyroscope continues to point to the same direction, no matter what manoeuvres the aircraft may make. It is constantly being updated by signals from the magnetic detector, but only when the aircraft is in straight and level flight. The turn / acceleration cut-out switch ensures that only valid magnetic signals are used to update the gyroscope.

The Gyroscope Despite the fine tolerances employed in gyroscope production, no gyroscope is completely error free and will suffer real errors as a result. Gyroscopes will also tend to point to a position in space, although the earth rotates 360 each 24 hours, creating apparent errors. These real and apparent errors increase over time and are known as gyro wander.

Which of the following is not a component within a Gyro-magnetic compass system? a) A turn / acceleration cut-out switch. b) A gyroscope. c) A suspended magnet. d) A flux valve magnetic detector.

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Which of the following is not a component within a Gyro-magnetic compass system? a) A turn / acceleration cut-out switch. b) A gyroscope. c) A suspended magnet. d) A flux valve magnetic detector.

Which of the following statements about the Gyro-magnetic compass is true? a) When the aircraft climbs or descends, the flux valve takes over from the gyroscope. c) The Gyro-magnetic compass is less accurate than the Direct Indicating Compass. b) The gyroscope takes over from the flux valve whenever the aircraft turns. d) The flux valve controls the speed of the gyroscope.

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Which of the following is one advantage of a gyro-magnetic compass over a Direct Indicating Compass? a) A gyro-magnetic compass requires no electricity. c) A gyro-magnetic compass is cheaper. b) A gyro-magnetic compass can feed repeaters around the aircraft. d) A gyro-magnetic compass does not work during turns or accelerations.

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A gyroscope cannot be perfect, and so over a period of time it becomes inaccurate, this is called: a) Gyro wander. b) Variation. c) Gyro rigidity. d) Turn / acceleration error.

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A gyroscope cannot be perfect, and so over a period of time it becomes inaccurate, this is called: a) Gyro wander. b) Variation. c) Gyro rigidity. d) Turn / acceleration error.

Inertial Navigation Inertial navigation systems (INS) use accelerometers (highly accurate gyroscopes) to detect rate of change of position along three axes. Providing your start point is accurately entered you can obtain an instant read out of your position at any time. Combining laser gyroscope INS with satellite navigation systems has significantly enhanced accuracy in modern aircraft.

What principle does an Inertial Navigation System (INS) use to calculate the position of the aircraft? a) A gyroscope feeds position to the computer. c) It uses compass heading and doppler values to compute aircraft position. b) The navigator must update the system all the time. d) It is set accurately on the ground, then measures acceleration in the fore, aft and lateral.

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Within an Inertial Navigation System the movement of the aircraft is measured by sensors called: a) Axis. b) Accelerators. c) Accelerometers. d) Inertials.

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Within an Inertial Navigation System the movement of the aircraft is measured by sensors called: a) Axis. b) Accelerators. c) Accelerometers. d) Inertials.

AIR NAVIGATION Chapter 5 Return to contents list Weather exit

Meteorological Conditions Student pilots do not have the experience, and basic training aircraft do not have the instruments to safely fly in cloud and fog. Beginners may only fly in Visual Meteorological Conditions (VMC), when visibility is good and aircraft can remain clear of cloud. When this is not the case, Instrument Meteorological Conditions (IMC) apply and only pilots with instrument ratings may fly aircraft equipped with suitable instrumentation.

The Visual Circuit At flying training schools and training units, special conditions apply. Trainee pilots will not be allowed to take off and fly circuits unless the cloudbase and visibility meet the aerodrome controller s requirements.

Beginners may only fly in good weather conditions. These conditions are called: a) Instrument Meteorological Conditions (IMC). b) Runway Visual Range (RVR). c) Visual Circuits (VC). d) Visual Meteorological Conditions (VMC).

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In order to fly in Instrument Met Conditions (IMC), which of the following are required: a) A clear windscreen canopy. b) No cloud in the local area. c) An instrument rating only. d) The correct instrumentation and a suitable pilot instrument rating.

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In order to fly a visual circuit, a trainee pilot requires: a) No wind. b) Good visibility and no cloud in the sky. c) Good visibility and no wind. d) Visibility and cloudbase conditions to meet the aerodrome controller's requirements.

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Wind The take off runway will normally be the one which allows the aircraft take off to be made into wind. Into a strong wind an aircraft reaches flying speed quickly and needs a shorter take off run.

Crosswind In this example the aircraft is about to take off on runway 27 (runway direction 270 ). 27

Crosswind The wind is 20 kts from the direction 300 (300/20). 27

Crosswind This wind can be split into two parts the crosswind component 27

Crosswind This wind can be split into two parts the crosswind component and the headwind component. 27

Crosswind With a wind angle 30 off the runway direction the crosswind component will be 50% of the total wind speed. 27

Crosswind 10 10 10 With a wind angle 30 off the runway direction the crosswind component will be 50% of the total wind speed (i.e. 10 kts crosswind for 20 kts windspeed). 27

Crosswind With a wind angle 30 off the runway direction the headwind component will be 90% of the total wind speed. 27

Crosswind 2 18 With a wind angle 30 off the runway direction the headwind component will be 90% of the total wind speed (i.e. 18 kts headwind for 20 kts windspeed). 18 27

Crosswind 20 20 If a 20 kt wind is blowing at an angle of 90 to the runway direction, then the whole of that 20 kts of wind will be crosswind with no headwind component. 27

Crosswind Similarly, if a wind is blowing straight down the length of a runway then the whole of that wind will be headwind with zero crosswind component. 27

Why does an aircraft take off into wind? a) To increase the groundspeed at take off. c) To use the full length of the runway. b) To take off at a lower airspeed. d) To decrease the length of take off run.

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A wind is blowing at 90 degrees angle off the runway direction. If the wind speed is 20 kts, what is the crosswind component? a) 2 kts b) 10 kts c) 12 kts d) 20 kts

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A wind is blowing at 90 degrees angle off the runway direction. If the wind speed is 20 kts, what is the crosswind component? a) 2 kts b) 10 kts c) 12 kts d) 20 kts

The wind is blowing directly down the length of a runway. What is the crosswind component? a) Equal to the wind's speed. b) Equal to 3/4 of wind speed. c) Equal to half the wind speed. d) Zero.

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The wind is blowing directly down the length of a runway. What is the crosswind component? a) Equal to the wind's speed. b) Equal to 3/4 of wind speed. c) Equal to half the wind speed. d) Zero.

Shallow Fog Fog presents a unique problem for the pilot, particularly at night when runway lights can be clearly seen from overhead the airfield.

Shallow Fog Once on the glideslope, the pilot is looking at the fog at a much shallower angle and the reduced visibility may prevent the aircraft from landing.

Shallow Fog The slant visibility can be measured using equipment placed close to the runway touchdown zone.

Shallow Fog These yellow installations in the foreground measure the visibility in the direct vicinity of the runway.

Shallow Fog This slant visibility measurement, known as the Runway Visual Range (RVR) is passed to the pilot by ATC.

The airfield has a covering of shallow fog. A pilot circling directly overhead sees the runway lights clearly. However, on the approach to land he may have great difficulty seeing any lights. Why is this? a) Runway lights are designed to be seen from high level only. c) Fog will appear thicker when on the glide path because the pilot is looking at a shallower angle. b) Fog is more dense closer to the ground. d) The thickest fog always settles at the end of the runway.

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During periods of poor visibility due to fog, ATC will advise the pilot of the slant visibility along the runway. This visibility is measured accurately and is known as: a) Runway Range. b) Runway Visual Range. c) Runway Radar Range. d) Glide Slope Visibility.

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Precipitation Precipitation is rain, sleet, snow or hail. Heavy rain might restrict the visibility or even flood the runway. If the precipitation is frozen, or the temperature at ground level is below zero, the precipitation may stick to the airframe causing icing and create serious problems during take off.

The collective noun for rain, sleet, snow and hail is: a) Participation. b) VMC. c) IMC. d) Precipation.

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The collective noun for rain, sleet, snow and hail is: a) Participation. b) VMC. c) IMC. d) Precipitation.

What problems can be caused by heavy rain? a) Heavy snow. b) Runway Visual Range. c) Thunderstorms. d) Restricted visibility and runway flooding.

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What problems can be caused by heavy rain? a) Heavy snow. b) Runway Visual Range. c) Thunderstorms. d) Restricted visibility and runway flooding.

Engine Icing Both jet and piston engines are affected by icing. Whilst large and commercial aircraft have effective de-icing systems for engines and airframes, it is best to avoid icing conditions if possible. In light aircraft and high performance military aircraft with little or no anti-icing protection it is essential to avoid icing conditions completely.

What problems can be caused by precipitation at freezing temperatures? a) Crosswinds. b) Icing. c) Fog. d) Thunderstorms.

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What problems can be caused by precipitation at freezing temperatures? a) Crosswinds. b) Icing. c) Fog. d) Thunderstorms.

What can be the effects of heavy icing on an aircraft's performance? a) Loss of aerodynamics only. b) Loss of aerodynamics and reduced engine performance. c) It will fly much slower. d) There is no adverse affect on an aircraft's performance.

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What effect can icing have on the aerodynamics of an aircraft? a) The windscreen may freeze over. c) Ice forming on the leading edge of the wing will increase lift. b) Lift will decrease and weight will increase. d) There will be no adverse effect on the aerodynamics.

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A flight briefing indicates icing en-route. The aircraft has no ice protection. What advice would you give a novice pilot? a) Fly above the cloud. b) Fly slowly, the icing will have less effect. c) Fly quickly, the icing will have less effect. d) Plan a route avoiding icing conditions or cancel the flight.

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Conclusion An extra question has found its way into this examination which is not covered in the manual: The earth revolves from West to East on its axis. The sun appears to rise in the East, but as the Sun is stationary it is the earth rotating West to East.

Which way does the Earth revolve on its axis? a) East to West. b) West to East. c) North to South. d) South to North.

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Which way does the Earth revolve on its axis? a) East to West. b) West to East. c) North to South. d) South to North. The sun appears to rise in the East and set in the West. As the sun is actually stationary, it is the Earth which is revolving West to East.

AIR NAVIGATION The End Return to contents list exit

Q a) b) c) d)

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AIR NAVIGATION This has been a production