CPL FLIGHT PLANING

CPL FLIGHT PLANNING

CPL FLIGHT PLANNING CPL DOC 3 Revision : 1/1/2001

FLIGHT TRAINING COLLEGE Version 7

INDEX CPL FLIGHT PLANNING 1. Terminology 2. Aerodromes 3. Graphs 4. Flight Graphs 5. Weight & Balance 6. CP/PET & PNR

01 07 17 19 41 57

Annex A Annex B Annex C King Air

65 71 87

Sample Exams Load Sheets Answers to Questions B200 EE-20 Manual

(separate booklet)

Copyright  2001 Flight Training College of Africa

All Rights Reserved. No part of this manual may be reproduced in any manner whatsoever including electronic, photographic, photocopying, facsimile, or stored in a retrieval system, without the prior permission of Flight Training College of Africa.

CPL FLIGHT PLANNING CPL DOC 3 Revision : 1/1/2001

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CHAPTER 1 TERMINOLOGY Airspeed terminology Va Vf

Design manoeuvre speed. The maximum speed at which full application of controls can be made. Design flap speed. The highest speed at which flaps may be activated.

Vfe

Maximum flap extended speed.

Vle

Maximum landing gear extended speed.

Vlo

Maximum landing gear operating speed.

Vlof

Lift off speed.

Vmca

Minimum control speed - air.

Vmcg

Minimum control speed - ground.

VR

Rotation speed

Vref:

Landing reference speed (1.3 x Vso)

Vs

Stall speed.

Vso

Stall speed in the landing configuration.

Vsse

Minimum intentional one-engine inoperative speed.

Vx

Best angle of climb speed.

Vy

Best rate of climb speed.

V1

Take-off decision speed.

V2

Take-off safety speed.

Vmo

Maximum operating limit speed is the speed limit that may not deliberately be exceeded in normal flight (in KNOTS)

Mmo

Maximum operating Mach number, the highest mach number at which an aircraft may be intentionally flown

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Temperature terminology IOAT Indicated Outside Air Temperature as read from the indicator (not corrected). OAT

Outside Air Temperature (corrected)

TAT

Total Air Temperature. Measured by a Rosemount Probe.

SAT

Static Air Temperature. The correct temperature of the ambient air.

RAT

Ram Air Temperature.

Temp Dev.

The difference between the actual OAT and the temperature of that level in the ISA atmosphere. The ISA lapse rate is 1.98°c per 1000ft. For the purpose of calculations, a lapse rate of 2°c per 1000’ can be used.

Pressure Alt

Height above the 1013.25 hPa level

Density Alt.

The higher the density altitude, the lower the air density and performance of the aircraft's engines. Runway length requirements increase with a potential corresponding reduction in the take-off weight. Most performance graphs contain positioning for pressure altitude and temperature; a calculation to determine density altitude is not required. To calculate density altitude, convert airfield elevation to pressure altitude, then compute using a nav. computer. Pressure Altitude Airfield Elevation

2915 feet 360 feet 3275 feet OAT +32°C

QNE 1013 hPa 30 ft x 12 hPa QNH 1025 hPa DA 5489 feet.

Aerodrome Pressure

QFE, The pressure setting used to indicate the height above the aerodrome in use. The use of QFE is rare in South Africa.

Conversion factors

Use the pathfinder, whiz wheel or a calculator. 1 Imperial Gallon 1 Imperial Gallon 1 US Gallon 1 Kilogram 1 Foot 1 Metre

= = = = = =

1.201 4.5461 3.7854 2.2046 0.3048 3.2808

US Gallons Litres Litres Pounds Metres Feet

Question 1:

382 Kgs of fuel at SG 0.79 are loaded. The number of US Gallons is?

Question 2:

The weight in kilograms of 450 Imperial Gallons of fuel (SG 0.82) is?

Question 3:

The weight in kilograms of 375 US Gallons of fuel (SG 0.81) is?

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SEMI-CIRCULAR FLIGHT LEVELS VFR 359

000

EVEN THOUSANDS + 500 FT TO FL 285 THEN

ODD THOUSANDS + 500 FT TO FL 275 THEN

285 320 360 400 etc

275 300 340 380 etc

180

179

BASED ON MAGNETIC TRACKS IFR 359 EVEN THOUSANDS TO FL 280 THEN

000 ODD THOUSANDS TO FL 290 THEN

280 310 350 390 etc

290 330 370 410 etc

180

179

NOTE: In South Africa no VFR flights above FL 200

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QUESTIONS 1.

Airfield Elevation 5327 Feet, Temperature +27°C, QNH 1025 hPa. The Density Altitude is :a) b) c)

2.

Airfield Elevation 1075 feet, Temperature +16°C, QNH 995 hPa. The Density Altitude is :a) b) c)

3.

2130 ft 1444 ft 771 ft At 0600 Z the temperature at an airfield (Pressure Altitude 3575 feet) was +12°C. At 1400 Z the temperature rose to +27°C. The increase in Density Altitude was :-

a) b) c) 4.

7876 ft 8305 ft 7607 ft

1210 ft 1407 ft 1807 ft

At an airfield where the Relative Humidity is high the :(a) (b) (c)

Take-Off performance of an aircraft will be enhanced, The climb performance of an aircraft will be degraded, The landing performance of an aircraft will be improved.

There are two methods to determine the Density Altitude of an airfield elevation. Method 1

Using the Electronic Flight Computer

Method 2 Calculate the temperature deviation between the actual temperature at the airfield pressure altitude and the ISA temperature for the airfield pressure altitude. Multiply this figure by 120 and add or subtract to or from the pressure altitude to give density altitude. This conversion allows 120ft per °C of temperature deviation between reported outside temperature and ISA. Beware; airfields cooler than ISA will have a lower density altitude than airfields warmer than ISA. CONVERSION OF hPa TO INCHES OF MERCURY (“Hg) There is no complex formula involved; it is simply by linear interpolation . You will already know that 1013.25 hPa = 29.92”Hg. Given a QNH of 998 hPa what is the corresponding pressure setting in “Hg? CPL FLIGHT PLANNING CPL DOC 03 Revision : 1/1/2001

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CHAPTER 2 AERODROMES AIRFIELD DETAILS The physical dimensions of the runway, stopway and clearway may affect an aircraft's maximum take-off weight. STOPWAY The stopway is an extension to the end of the runway, which may be used to stop the aircraft in the event of a rejected take-off. The stopway must be at least as wide as the runway, able to support the aircraft without incurring structural damage, but is not intended for normal use. CLEARWAY Clearway may be used for the initial climb from lift-off to 50 feet above the ground. The clearway is an area beyond the end of the runway, which complies with the following criteria:     

It must be at least 300 ft wide on either side of the extended centre-line; It must be under airport control; It must be clear of obstacles; The elevation of the clearway may not be higher than the end of the runway; The clearway includes the stopway (if available);

DECLARED RUNWAY DISTANCES as specified by ICAO TAKE-OFF RUN AVAILABLE (TORA) The length of runway, which is declared available and suitable for, the ground run of an aeroplane taking off. ACCELERATE STOP DISTANCE AVAILABLE (ASDA) The length of the take-off run available plus the length of stopway available (if stopway is provided). TAKE-OFF DISTANCE AVAILABLE (TODA) The length of the take-off run available plus the length of clearway available (if clearway is provided). REFERENCE ZERO The point at the end of the take-off run at which the airplane is 35 feet above the runway surface. Laterally it is located at the end of the Take-Off Distance Required (TODR) and is the point from which horizontal distances to obstacles are measured. CPL FLIGHT PLANNING CPL DOC 03 Revision : 1/1/2001

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LANDING DISTANCE AVAILABLE (LDA) The length of runway which is declared available and suitable for the ground run of an aeroplane landing. The landing distance available commences at the threshold and extends for the length of runway after the threshold. However, the threshold may be displaced from the extremity of the runway when it is considered necessary to make a corresponding displacement of the approach surface by reason of obstacles in the approach path to the runway. North

RUNWAY 09 27

TORA feet 2000 2000

ASDA feet 2300 2350

TODA feet 2580 2350

LDA feet 1850 2000

RUNWAY SURFACE CONDITIONS If the runway surface is contaminated by, for example, water or snow, the aircraft will require more runway length to reach takeoff speed. If this extra runway length is not available, the aircraft's take-off weight will have to be reduced. RUNWAY SLOPE An uphill slope requires a longer take-off run, and therefore, a possible reduction of the takeoff weight. A runway with a downhill slope would have the opposite effect. A point worth bearing in mind is that an uphill slope would mean less distance required to bring the aircraft to a stop in the event of an aborted take-off and vice versa for a downhill slope. A definitive answer on the effect of slope on an aircraft's take-off weight would, of course, be extracted from the appropriate performance graphs. LANSERIA

RWY 06L RWY 24R

To calculate runway slope:

Threshold Elevation 4517 feet Threshold Elevation 4393 feet Runway Length 3048 metres Difference in Elevat ion 124 fe et Runway Len gth 3048 m etres × 3.28

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×

100 = 1.24 %

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TAKE - OFF WIND COMPONENT A tailwind component at take-off increases the amount of runway required for take-off, and therefore, possibly a reduction of the take-off weight. A headwind component at take-off would have the opposite effect. CLIMB LIMITATIONS CLIMB or WAT (Weight Altitude Temperature) LIMIT The combination of weight and air density (altitude and temperature) affects the performance of the aircraft, and even if the aircraft can get off the ground with an engine failure at V1, its rate of climb would be too low to satisfy the required climb gradients during the Take-Off Flight Path. Obstacles within the airfield boundaries and close than 200' to the flight path must be cleared by at least 50' vertically. Obstacles outside the airfield boundaries and closer than 300' to the flight path must be cleared by at least 50' vertically. PERFORMANCE CLASSIFICATION NUMBER (PCN) and AIRCRAFT CLASSIFICATION NUMBER (ACN) The performance classification number for a runway is an expression of its bearing strength. The aircraft classification number is derived graphically using its single isolated wheel loading (SIWL) and tyre pressure. ACN can also prove to have a limiting effect on the maximum take-off weight of an aircraft. BRAKE ENERGY LIMIT In the process of bringing an aircraft to a stop, its brakes convert kinetic energy into heat energy. The amount of heat energy that the brakes can absorb certainly has limits. In airspeed terminology the speed at which this limit occurs is known as Vmbe. The greater the take-off weight of an aircraft, the higher its take-off speed will be and the more energy the brakes will have to absorb in the event of an aborted take-off. Although the brake energy limit may not directly limit the take-off weight, many aircraft have a minimum turn around time between landing and subsequent take-off, which will ensure adequate braking in the event of an aborted take-off. This minimum turn around time is directly proportional to the weight at which the aircraft landed and the weight for the next take-off. TYRE SPEED LIMIT Much like the aircraft's brakes, the tyres also have certain limitations to ensure their structural integrity. The limit is the maximum true ground speed that the tyres can absorb. The higher the take-off weight of the aircraft, the higher the take-off speeds will be, and this may prove to be a limiting factor requiring a reduction in the maximum take-off weight.

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Reference zero point is here.

TAKE OFF FLIGHT PATH

175ft

TODA

3000 ft

Select the correct take-off flight path graph to find the climb gradient required to clear the above 175ft high tree if the tree is located 3000ft beyond reference zero. The aerodrome elevation is 1000’ Pressure height and the temperature is +25 c. Graph on page 5-28 (Smaller graph on the right of the page). Enter with the horizontal distance from Reference Zero figure. In this case its:

3000ft

Enter with 3000 feet along the bottom of the expanded graph and then intersect the 175ft (obstacle height) to get the answer of 4.8% Now use the graph titled Net gradient of Climb to find the maximum weight for this climb with 0% of flap… Use the graphs on pages 5-37 and 5-41. Enter the graph with the Temperature and intersect the Pressure altitude, then across to intersect the gradient and a vertical line down to reveal the weight. Answers:

0% flap

=

40% flap

=

Example 2 ( 5-28, 5-37) An aircraft has a TODR of 1250m, and there is a hill located 2130m from the start of the runway, and its highest point is located 200’ agl. The aerodrome elevation is 2860’ amsl, the QNH is 998 hPa, and the temperature is +18 c. Find the Climb gradient required, and the max weight for the climb with 0% flap… Example 3 (5-28, 5-37) An aircraft has a reference zero figure of 1990m, there is a temporary crane operating 1990m from the end of the clearway (i.e. TODA) along the extended centreline, the cranes maximum height is 480’ agl, the aerodrome is 5550’ amsl, QNH 1010 hPa and the temperature is +35 c. Find the Climb gradient required, and the max weight for the climb with 0% flap.

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WIND CALCULATIONS Using the graph on page 5-33 or your electronic flight computer find the following: Example 1 On a runway with directions 18/36, with a wind of 030/35 find the crosswind and headwind. Prior to entering the graph work out the most into wind runway, in this case its runway 36, then work out the difference in degrees from the runway direction to wind direction. Here it’s the difference between 360 and 030, so its 30 degrees. Enter the graph at the point of intersection of 30° and 35 kts, the read off the answers of31kts hw and 18 kt x-wind. Example 2 An aircraft has the following TAF, what will be the headwind and crosswind for a take off on runway 16/34? PARIS/CHARLES DE GAULLE LFPG 03/09-18Z 02033KT CAVOK TEMPO 1013 SCT033 T20/12Z T21/15Z Example 3 Using the following TAF find the tailwind component for a take off on runway 03/21 STAVANGER ENZV 03/09-18Z 33015G25KT 9999 FEW015 SCT040 AVERAGE WIND CALCULATIONS An unusual method must be employed when working out the average wind for the exam questions. BEWARE there is only one way to get the correct answer, AND ITS DIFFERENT TO THE METHOD USED IN NAVIGATION… The following figures relate to a Flight Plan SECTOR

TAS

WC

GS

DIST TIME

A to TOC

300

-112

188

110

0 : 35

TOC to B

495

+109

604

318

0 : 44

B to C

488

-39

449

561

1 : 15

C to TOD

476

-51

425

672

1 : 35

TOD to D

300

-114

186

90

0 : 29

1751

4 : 38

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WORKING SECTOR

TAS

WC

GS

DIST TIME

A to TOC

300

-112

188

110

0 : 35

TOC to B

495

+109

604

318

0 : 44

B to C

488

-39

449

561

1 : 15

C to TOD

476

-51

425

672

1 : 35

TOD to D

300

-114

186

90

0 : 29

1751

4 : 38

Step 1 Add up the TAS column to get 2059 kts total TAS’s Step 2 Add up the DIST column to get 1751nm total Step 3 Now divide the total TAS column by the total time 4hr 38min, 444kts, and do the same for the DIST column, so you get then 378kts. Step 4 Now you have an average TAS of 444kts, and an average DIST of 378kt, so subtract one from the other to get the answer of 66kts average headwind component.

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REVISION QUESTIONS At an airfield the runway details are :Runway length Stopway Clearway Displaced Threshold

RWY 06 4000 feet 350 feet 600 feet

RWY 24 4000 feet 450 feet 700 feet 200 feet

Using this information answer questions 1to 4. 1

The Take-Off Distance (TODA) for RWY 06 is :a) b) c)

2

The Accelerate Stop Distance (ASDA) for RWY 24 is :a) b) b)

3

3800 ft 4000 ft 4350 ft

The Take-Off Run Available (TORA) for RWY 24 is :a) b) c)

5.

4000 ft 4450 ft 4700 ft

The Landing Distance Available (LDA) for RWY 06 is :a) b) c)

4

4150 ft 4350 ft 4600 ft

4000 ft 4350 ft 4700 ft

Runway 08/26 at Port Elizabeth is 1980 metres in length. The threshold elevation of RWY 08 is 225 feet. The threshold elevation of RWY 26 is 185 feet. The slope of RWY 26 is :a) b) c)

0.62 % UP 0.74 % UP 0.81 % UP

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6.

Runway 10/28 at East London is 1935 metres in length. The threshold elevation of RWY 10 is 431 feet. The threshold elevation of RWY 28 is 383 feet. The slope of RWY 10 is :(a) (b) (c)

7.

0.69 % DN 0.76 % DN 0.84 % DN

An aerodrome has been surveyed and the following figures have been relayed to you prior to departure. Find the amount Stopway , Strip length and Clearway length for all runways…(all figures given are in metres) RUNWAY 01 19 15 33

8.

TODA 1124 1090 6200 5350

LDA 800 890 3450 3450

4kt from right nil 20kt from right

Runway 06 (a) (b) (c)

10.

ASDA 890 890 4700 4850

An aircraft is to depart from the following runway what is the crosswind component on runway 25? BERGEN ENBR 03/09-18Z 33020KT 9999 FEW025= (a) (b) (c)

9.

TORA 890 890 3450 3450

W/V 100/27

The wind Component for Take-Off is :-

14 Kts Hw 17 Kts Hw 21 Kts Hw

In order to Take-Off an aircraft requires a Runway Headwind Component of at least 15 Kts. The maximum permitted Cross Wind Component is 30 Kts. The Runway in use is 09 and the Wind Direction is 130°. The maximum and minimum wind speeds that will allow take-off are :(a) (b) (c)

20 and 46 Kts 20 and 40 Kts 25 and 46 Kts

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11.

An aerodrome has just been surveyed and the surveyor has come with the following figures. Utilise these figures to find: abcd-

TODA TORA ASDA LDA

Runway direction is 09/27 Obstacle free area 1600m fm strip end

Strip length = 1380m

Clear area of grass 650m

Obstacle free area 350m

* all distance are from end of runway strip….

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CHAPTER 3 MISC GRAPHS GRAPH 5 - 13

AIRSPEED CALIBRATION

Enter with IAS 89 Kts, move vertically to the reference line, then horizontally and read off CAS 90 Kts. OR Enter with CAS 90 Kts, move horizontally to the reference line, then vertically and read off IAS 89 Kts. NOTE: CAS (Calibrated Airspeed) is the American version of RAS (Rectified Airspeed) GRAPH 5 - 15 Flaps 0 %

ALTIMETER CORRECTION

IAS 200 Kts FL 260 = 30 feet ADDED to INDICATED ALTITUDE

25 970 feet INDICATED ALTITUDE + 30 feet = 26 000 feet ACTUAL ALTITUDE OR 26 000 feet INDICATED ALTITUDE + 30 feet = 26 030 feet ACTUAL ALTITUDE Flaps 100 % IAS 130 Kts 8000 feet = 15 feet SUBTRACTED from INDICATED ALTITUDE 8000 feet INDICATED ALTITUDE - 15 feet = 7985 feet ACTUAL ALTITUDE OR 8015 feet INDICATED ALTITUDE - 15 feet = 8000 feet ACTUAL ALTITUDE Further Examples 1. 2. 3.

With an IAS of 220 kts, FL200 with 0% flap what is the Actual altitude? With an IAS of 140kts at FL090 with 40% flap, what is the Actual altitude? With an IAS of 110kts at FL280 with 0% flap what is the Actual altitude?

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INDICATED OUTSIDE AIR TEMPERATURE Using graph on page 5-18, find the IOAT by subtracting the correction figure from your OAT. Enter with CAS and Pressure altitude. Example The aircraft is slogging along at FL330 at 195kt in ISA, the OAT is? STALL SPEED Use the graph on page 5-29, enter with weight, flaps and angle of bank to get the Vs in either IAS or CAS. PRESSURISATION CONTROL SETTINGS Using the graph on page 5-106, you can attain the Cabin altitude setting for landing (if destination is not at MSL). Work through the following examples to get used to the chart. REMEMBER to convert QNH in hectopascals to Inches of Mercury. 1.

An aircraft is landing at and aerodrome that has a QNH of 1010hpa and an elevation of 5500ft amsl, what should the cabin pressure controller be set to?

2.

An aircraft is planning a descent to arrive at an aerodrome that 6400ft amsl, and the tower has advised that the QFE is 810hPa, what should the cabin controller be set to?

3.

An aircraft is planning to arrive at an aerodrome that has a pressure height of –500ft, what should the controller be set to?

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CHAPTER 4 FLIGHT GRAPHS TAKE - OFF GRAPHS

Take-Off graphs are entered with PRESSURE ALTITUDE and TEMPERATURE. If Airfield Elevation is given with QNH, then Pressure Altitude must be calculated before entering the graph. TAKE - OFF DISTANCE - FLAPS 0 % TAKE - OFF DISTANCE - FLAPS 40 %

GRAPH 5 - 34 GRAPH 5 - 38

If Aircraft Weight is given and the Take-Off Distance or Ground Roll is required 1 2.

Enter with Temperature, move vertically to Pressure Altitude.

From this point move horizontally to the Aircraft Weight reference line which is 12 500 Lbs. If the Aircraft weight is 12 500 Lbs, continue horizontally to the next reference line. If the Aircraft Weight is less than 12 500 Lbs move down the slope to the Aircraft Weight given in the question. 3.

From this point move horizontally to the Wind Component reference line. Move down the slope for a Headwind Component, move up the slope for a Tailwind Component, then horizontally to the next reference line.

4.

If the question requires the Take-Off Distance move up the slope to the end of the graph.

5.

If the question requires the Take-Off Ground Roll continue horizontally to the end of the graph.

If the Runway Length is given and the Maximum Take-Off Weight is required 1.

Enter with Temperature, move vertically to Pressure Altitude.

2.

From this point move horizontally to the Aircraft Weight reference line which is 12 500 Lbs. Draw a line down the slope to 9000 Lbs.

3.

Enter the right hand side of the graph with Runway Length. Move down the slope to the reference line. Move horizontally to the Wind Component, move up the slope to the reference line for a Headwind or down the slope to the reference line for a Tailwind.

4.

From the Wind Component reference line move horizontally to intersect the line previously drawn. From the intersection move vertically and read off the Aircraft Weight.

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EXAMPLES 1.

If the aircraft flaps are U/S and the airfield you wish to depart from has the following actual conditions: QNH OAT Airfield ht Aircraft weight Wind component Runway

1010hpa 16°C 3030ft amsl 10 980lbs 350/30 18/36

What is the Take Off Distance Required, and the speed for the take off? 2.

If the Captain requests you to do a 40% flap takeoff what will be the Max Take Off Weight under the following conditions: QNH OAT Pressure ht Wind component Runway Take off dist avail.

3.

If the aircraft is to make a 0% flap take off under the following conditions what will be the Take Off Distance Required? QNH OAT Density height Wind component Runway Take off weight

4.

998hpa +2°C 5000ft 5 TW (one way airstrip) 09/27 765m

998hpa Air temp gauge U/S 1450ft 150/22 01/19 11 800 lbs

The aircraft is to make a take off from a airfield under the following conditions, find the Max Take Off Weight? QFE OAT Elevation Wind component Runway Take off dist avail. Flap

900hpa +20°C 3230ft 090/20 18/36 1070m 0%

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ACCELERATE - STOP GRAPHS 5 - 35 and 5 - 39 The graphs are similar to the Take-Off graphs. If a Runway has STOPWAY it may be used with these graphs. 1.

Find the Accelerate Stop Distance Required under the following conditions…. Flaps QNH OAT Elevation Wind component Runway Aircraft take off wt

2.

40% 985hpa +20°C 1000ft 250/18 09/27 10 220lbs

Find the Accelerate Stop Distance and V1 speed under the following conditions…. Flaps QNH OAT Elevation Wind component Runway Aircraft take off wt

U/S 1012hpa +2°C 2910ft 010/19 15/33 11 400lbs

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ACCELERATE - GO GRAPHS 5 - 36 and 5 - 40 The graphs are similar to the Take-Off graphs. If a Runway has CLEARWAY it may be used with these graphs. Refer to the note above the graph:Useable CLEARWAY cannot exceed 25 % of the Runway Length. 1.

2.

Find the accelerate-go distance and V speeds for a take off under the following conditions: Flaps

40%

QNH

1012hpa

OAT Density height Wind component Runway Aircraft take off wt

+2°C 2500ft 025/25 16/34 12 000lbs

Find the accelerate go distance with ice vanes extended and the V speeds for a take off under the following conditions: Flaps QNH OAT Elevation Wind component Runway Aircraft take off wt

0% 1000hpa +21°C 500ft vrb/10 18/36 10 800

MINIMUM TAKE OFF POWER Use graphs on pages 5-31 and 5-32.. CAUTION 2 GRAPHS- one with ice vanes extended and one without. Example 1 With an aerodrome that has a pressure height of 6500ft and OAT of +15°C what is the minimum take off power that could be used with ice vanes retracted? Example 2 With an aerodrome that has a elevation of 2860ft amsl, a QNH of 995hPa and a temperature of +22°C find the minimum power that could be used for take off with ice vanes extended?

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CLIMB – TIME – FUEL - DISTANCE TO CLIMB Example 1

Climb from Sea Level (OAT +15°C) to FL260 (OAT -10°C) Aircraft Weight 12 500 Lbs Time 25 mins

Example 2

GRAPH 5 - 45

Fuel 275 Lbs

Distance 80 nm

Climb from 5430ft (OAT +28°C) to FL260 (OAT -10°C) Aircraft Weight 12 500 Lbs

Sea Level to FL260 Sea Level to 5430 ft

Time 25 mins Time 3 mins

Fuel 275 Lbs Fuel 45 Lbs

Distance 80 nm Distance 11 nm

5430 ft to FL260

Time 22 mins

Fuel 230 Lbs

Distance 69 nm

Mean Climb TAS

Time 22 mins

Distance 69 nm

TAS 188 Kts

Examples

1.

Find the fuel, time and distance to climb from a sea level ISA aerodrome to FL300, where the temperature is –30°C at Max Take Off Weight.

2.

Find the time, fuel, distance and average climb speed to climb from the following aerodrome to altitude of 23 000ft where the OAT is predicted to be –15°C Aircraft config. Aircraft weight QNH OAT Elevation

3.

Ice vanes extended 10 000lbs 995hpa +21°C 1750ft amsl

Find the time, fuel and distance to climb to a density height of 25 000ft under the following conditions from the given aerodrome: Aircraft config. Aircraft weight QNH OAT Elevation Wind at aerodrome Wind at 25 000ft Track

Ice vanes retracted 12 000lbs 1020hpa +15°C 3130ft amsl 045/20 075/105 060 M

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RATE & ANGLE OF CLIMB ANGLE OF CLIMB To work out the aircrafts angle of climb, or climb gradient, use the following formula: CLIMB GRADIENT =

HEIGHT GAINED DISTANCE TRAVELLED

Therefore is you have gained 550ft of altitude and distance from takeoff from your GPS reads 8000ft, use the formula to find your climb gradient… =

550 8000

=

0.069

To get a % multiply by 100 therefore = 0.069 x 100 = 6.9% In the cockpit this can be worked out easily by using the following pilots formula: Angle of Climb=

Rate of Climb (fpm) Speed (kt)

Example 1 An aircraft climbs out from a sea level aerodrome under ISA conditions with IAS of 80kts, HWC 20kt and ROC 550ft/min. Estimate the angle of climb.

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CRUISE CONSTANT POWER/SPEED CRUISE TABLES The tables are based on Temperature Deviation from ISA. If OAT is given in a question calculate the ISA + or ISA - value. TAS

To calculate TAS for a particular question it is often necessary to interpolate.

Example 1. Recommended Cruise Power Page 5 – 55 ISA +20°C FL230 Aircraft Weight 12 000 Lbs TAS 267 TAS 265.5 TAS 264

FL 220 FL 230 FL 240

11 600 Lbs

11 600 Lbs TAS 267.1

11 000 Lbs TAS 271 TAS 269.5 TAS 268

At FL230 the difference is 4 Kts per 1000 Lbs 4 Kts 1000 Lbs

× 400 Lbs

= 1.6 Kts increase f or 400 Lbs weight re duction fr om 12 000 Lbs

Example 2. Recommended Cruise Power ISA +23°C FL 260 FL 270 FL 280

Pages 5 - 55 and 5 - 56 FL270

ISA + 20°C TAS 265 TAS 262.5 TAS 260

Aircraft Weight 11 000 Lbs ISA + 23°C TAS 261

ISA + 30°C TAS 259 TAS 257 TAS 255

Difference 5.5 Kts ÷ 10° x 3°C = 1.65 Kts decrease for 3°C rise in temperature

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MAXIMUM EN-ROUTE WEIGHT GRAPH 5 - 24 The graph requires the QNH in Inches of Mercury. If the QNH is given in hectopascals then convert by ratio. QNH 1023.7 Standard 1 013.2

Given QNH 1023.7 hPa

=

QNH 30.23 inches Standard 2 9.92 inche s

This graph calculates the maximum weight at which the aircraft can maintain the MINIMUM EN-ROUTE ALTITUDE in the event of an engine failure. Enter with the Outside Air Temperature at the Minimum Enroute Altitude and move vertically to that altitude. Then move horizontally to the reference line that is standard pressure 29.92 inches of mercury. Move down the slope if pressure is lower than standard, or up the slope if pressure is higher. Then move horizontally to read off the MAXIMUM ENROUTE WEIGHT. Max En-Route Weight In Class Example 1. If the OAT is +2°C the min en-route altitude is 19 600ft and the QNH is 1010hpa what is the max en-route weight to maintain this level on one engine? 2. If the OAT is ISA +5°C, and the altitude is FL210, the QNH is 995hpa what is the max en-route weight? 3. If the OAT is –28°C, the altitude is FL150 and the QNH is 996hpa what is the max en-route weight?

RANGE PROFILE

GRAPH 5 - 96

Enter with Flight Level, move horizontally to the relevant cruise power, extract TAS, move vertically to the range in Nautical Miles in Zero Wind or SAD (Still Air Distance). Example: The range of the EE-20 aircraft at FL 280 (Recommended Cruise Power) with a 35 Kt Headwind is :FL 280

TAS 272 at Recommended Cruise Power

TAS 272

Still Air Distance

1095 nm

Range 1095 nm in Still Air

Time 4.0257 Hours

WC 35 Kt Headwind GS

237 x

Time 4.0257 Hours = 954 nm Ground Nautical Miles

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ENDURANCE PROFILE

GRAPH 5 - 97

The Endurance is expressed in Hours and decimals of an Hour. 4.2 Hours = 4 Hours 12 minutes RECOMMENDED CRUISE POWER GRAPH 5 – 59 & FUEL FLOW AT RECOMMENDED CRUISE POWER GRAPH 5 - 60 The graphs are similar and are entered with INDICATED OUTSIDE AIR TEMPERATURE, that is the temperature as read off the temperature gauge in the aircraft which is affected by compressibility error, it OVERREADS. If IOAT (Indicated Outside Air Temperature) is given, enter the graph with the IOAT, move vertically to the FL, then horizontally to the Torque Setting or the Fuel Flow. If OAT (OUTSIDE AIR TEMPERATURE) is given, it is the true temperature (IOAT corrected for compressibility) and must be converted to a Temperature Deviation from ISA before the graph can be entered. Example At FL 240 the OAT is -21°C, the Temperature Deviation from ISA is :ISA at Sea Level FL 240 x 2°/1000 ft ISA at FL 240 OAT at FL 240 Temperature Deviation

+15°C - 48°C (colder than sea level) - 33°C - 21°C ISA +12°C

Enter the graphs with FL and ISA Temperature Deviation (diagonal lines, top right to bottom left) and move horizontally to Torque Setting or Fuel Flow.

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DISTANCE FLOWN PER UNIT OF FUEL USED OR FUEL USED FOR DISTANCE FLOWN TAS

(true airspeed) TAS 240 Kts In one hour an aircraft will fly 240 AIR NAUTICAL MILES (ANM)

SAD

(still air distance) TAS 240 Kts In one hour an aircraft flies a STILL AIR DISTANCE (SAD) of 240 nm

FUEL FLOW (FF) Given:

The amount of fuel (Kilograms or Pounds) used in one hour.

TAS 240 Kts Fuel Flow 750 Lbs/Hour

Then aircraft performance is

TAS 240 Kt s Fuel Flow 750 Lbs / Hour

= 0.32 ANM / LB

OR Fuel Flow 750 Lbs / Hour TAS 240 Kt s

= 3.125 L B / ANM

ANM/LB can be converted to LB/ANM on an electronic calculator by using the 1/X function. WIND COMPONENT (WC) Wind component is the difference between TAS and GROUNDSPEED (GS). TAS 240 Kts WC -30 HW (Headwind)

GS 210 Kts

TAS 240 Kts WC +30 TW (Tailwind)

GS 270 Kts

GS GS 270 Kts In one hour an aircraft flies 270 GNM Given:

TAS 240 Kts WC +30 Kts TW

Then aircraft performance is

GS 270 Kts

Fuel Flow 750 Lbs/Hour

GS 270 Kts Fuel Flow 750 Lbs / Hour

= 0.36 GNM / LB

Fuel Flow 750 Lbs / Hour GS 270 Kts

= 2.7778 LB / GNM

OR

When compiling a flight plan the most economical Flight Level should be selected by comparing LBS/GNM or GNM/LB. THE MOST EFFICIENT FLIGHT LEVEL IS : Or

The littlest of the bigger numbers The biggest of the little numbers

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Example 1. FL 180

TAS 276 Kts

WC -20 HW

FF 716 LB/Hour

FL 220

TAS 271 Kts

WC -40 HW

FF 622 LB/Hour

FL 260

TAS 262 Kts

WC -60 HW

FF 534 LB/Hour

The most economical FL is :-

Example 2. An aircraft at FL 350, TAS 232 Kts, Fuel Flow 545 LBS/Hour has a performance of 0.355 GNM/LB. The Wind Component affecting the aircraft is :Example 3. An aircraft flying at FL 310 at TAS 494 Kts obtains a performance of 46.06 ANM/1000 Kgs in Zero Wind conditions. At FL 350 the TAS is 484 Kts and the performance is 48.36 ANM/1000 Kgs. It will be less economical to cruise at FL 350 if the Head Wind component is greater than:Ans.

TAS 484kts 48.36 anm/1000kg

x

GS 46.06gnm/1000kgs

=

461kts GS

Therefore if TAS 484kts, GS 461kts, it’s a 23kt HW

If you are confused by the objective of this formula then work backwards through it by using a headwind of, say, 39kts.In doing so you will derive a poorer performance figure for FL 350. From an operational point of view you would then have to revert back to your original flight level at FL 310. What is the performance figure for FL 350 if the headwind should increase to the new value of 39kts?

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ONE ENGINE INOP TABLES CLIMBING Can the aircraft climb under the present weight and atmospheric conditions should be considered prior to any flight. Use graph 5-46 to find out if you can climb on one engine. To Use the graph you need: 1. 2. 3. 4.

OAT Pressure altitude Weight Climb gradient required to overcome the obstacles.

Example 1 An aircraft has a weight of 11 500lbs, and is taking off from an airport that has a pressure height of 4500ft and an OAT of +25°C, what is the rate of climb on one engine and the climb gradient achieved? Example 2 An aircraft is at FL200, the OAT is -10°C and the aircraft weight is 12 250lbs, what is the ROC? SERVICE CEILING ONE ENGINE Here you are asking yourself can you maintain altitude to remain whether airspace restrictions, or to maintain the Lowest Route Altitude. To answer this, use graph 5-47. Example 1 An aircraft is at FL180 and suffers an engine failure, the weight at the time is 10 500lbs, and the OAT is -22°C. Can the aircraft maintain this FL, if not what is the Flight Level that the aircraft can maintain at this weight and temperature? Example 2 An aircraft has a MZW of 11000lbs and the forecast temperature at the Lowest Sector Altitude is -5°C, and due to forecast icing the ice vanes must be extended. What is the service ceiling of the aircraft on one engine?

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MAX CRUISE POWER ON ONE ENGINE To make sure you don’t blow up an engine, there is a table on 5-99 onwards to attain the maximum cruise power setting when on one engine. NOTE the tables are differentiated by the ISA deviation. Example 1 An aircraft is cruising on one engine at FL100, the OAT is +15°C the aircraft weight is 11 000lbs, what is the maximum cruise power setting? Example 2 An aircraft is at FL140, the OAT is -23°C the aircraft weight is 10 500lbs, the ice vanes are extended, what is the maximum cruise power setting on the live engine and the fuel flow? CRUISE POWER SETTINGS Power setting in a King Air 200 is not automatic, there are tables to attain the correct power setting. The graphs to use are located on pages 5-51 onwards and like all power setting tables differentiate with ISA deviation. Example 1 An aircraft is to cruise at FL 220, the OAT is –19°C the weight is 11 000lbs, what is the power setting, fuel flow total and TAS? Example 2 An aircraft is to cruise at FL180, the EMZW is 10 500lbs, the following sector forecast is given, what is the power setting, fuel flow total and TAS? (using graphs 5-52 & 5-53) 24 300 60 -24 21 295 60 -25 18 300 55 -24 15 300 55 -12 10 310 50 -1

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DESCENT PLANNING TIME - FUEL - DISTANCE TO DESCEND

GRAPH 5 - 109

Similar to the Climb Graph 1.

An aircraft is to descend at Mmo, from FL245 to a sea level aerodrome what is the Time, Distance and Fuel that would be used?

2.

An aircraft is planning a descent from FL180 to arrive in the circuit area 1000ft agl, over its destination aerodrome that is 3560ft amsl, what will be the time, fuel and distance for this descent?

3.

An aircraft is required, due to traffic, to descend overhead Kathmandu, it is currently at FL280, and is required to descend to FL145 to maintain separation. How many NM and minutes before Kathmandu must the descent be started and how much fuel can be expected to be consumed?

FUEL Remember the quantities the aircraft can carry:

RESERVE FUEL As per the EE20 manual reserve fuel is calculated as 45 minutes at the cruise fuel setting calculated at the weight at the end of the cruise, i.e. Top Of Descent (TOD) weight. CPL FLIGHT PLANNING CPL DOC 03 Revision : 1/1/2001

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HOLDING FUEL HOLDING TIME

GRAPH 5 - 107

1 Hour Holding at 15 000 ft =

420 Lbs of Fuel

If ICE VANES EXTENDED then holding time reduced by 15 % Example 1

420 Lbs of Fuel 15 % reduction 420 Lbs of Fuel

Example 2

= = =

60 mins 9 mins 51 mins

Ice Vanes Retracted Ice Vanes Extended

Fuel for 1 Hour Holding with Ice Vanes Extended ?

1 Hour = 420 Lbs = 85 %

Check

Then 100 % =

420 Lbs 0.85

= 494 Lbs of Fuel

494 Lbs - 15% (74 Lbs) = 420 Lbs

Further examples 1.

An aircraft has the ice vanes retracted and is told it is expected to hold for 75 minutes for a slot time to land, the hold will be conducted at FL130, how much fuel will be consumed?

2.

An aircraft is in IMC and has the Ice vanes extended and is placed in a holding pattern at FL200, the aircraft has 420lbs available for holding, how long can it remains in this holding pattern?

3.

An aircraft is expected to hold to await the opening of an airport which occurs at first light (0320UTC), the aircraft is expected to arrive at the intersection it will hold at, at 0715 LMT (UTC +6), how much fuel will be consumed during the hold at 5000ft?

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LANDING LANDING DISTANCE FLAPS 100 %

GRAPH 5 – 110

LANDING DISTANCE FLAPS 100 % PROPELLER REVERSING GRAPH 5 - 112 Both graphs give landing performance and are similar to the take-off graphs. LANDING DISTANCE FLAPS 0 % PROPELLER REVERSING GRAPH 5 - 113 Landing with full (100 %) flap is normal procedure, but it may be necessary to land with flaps up (0 %). To determine the flaps up landing distance, use graph 5-112 the landing distance with propeller reversing, flaps 100 %, then enter graph 5-113 with this distance and read of landing distance flaps up EXAMPLES 1. Find landing distance (both ground roll and over 50ft obstacle) under the following conditions: Aerodrome elevation QNH Temp Aircraft weight Wind component Props Flaps 2.

Find the landing distances for a landing under the following conditions: Aerodrome elevation QNH Temp Weight Wind comp Runway Props Flaps Runway length

3.

2600ft amsl 1010hpa 22°C 11 200lbs 5kt HW Reverse engaged 100%

1220ft amsl 996hpa -15°C ….? 280/35 15/33 Reverse not allowed due runway surface 100% 1770m

Find the landing distance over a 50ft obstacle, and approach speed under the following conditions: Aerodrome elevation QNH Temp Weight Wind comp Runway Props Flaps

2000ft 1013 hpa 40°C 11 800lbs 110/20 09/27 Reverse engaged U/S

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QUESTIONS 1.

FL 220 TAS 267 Kts FL 260 TAS 259 Kts FL 310 TAS 232 Kts

WC -20 Kts HW WC -40 Kts HW WC -60 Kts HW

FF 592 LB/Hour FF 512 LB/Hour FF 422 LB/Hour

The most economical FL is :(a) (b) (c) 2.

FL 220 FL 260 FL 310

An aircraft flying at FL 350 at TAS 495 Kts obtains a performance of 97.3 ANM/1000 Kgs in Zero Wind conditions. At FL 390 the TAS is 474 Kts and the performance is 102.9 ANM/1000 kg. It will be less economical to cruise at FL 390 if the Head Wind component is greater than :(a) (b) (c)

3.

An aircraft at FL 310 has a TAS of 485 Kts and Fuel Flow of 11 750 Lbs/Hour. If aircraft performance is 36.59 GNM/1000 Lb the Wind Component affecting the aircraft is:(a) (b) (c)

4.

26 Kts HW 36 Kts HW 46 Kts HW

35 Kts HW 45 Kts HW 55 Kts HW

The following figures relate to a Flight Plan SECTOR

TAS

A to TOC

300

TOC to B

WC

GS

DIST TIME

-35

265

110

0 : 25

495

-52

443

318

0 : 43

B to C

488

-67

421

561

1 : 20

C to TOD

476

-88

388

672

1 : 44

TOD to D

300

-30

270

90

0 : 20

1751

4 : 32

The average Wind Component for the flight is :(a) (b) (c)

54 Kts HW 60 Kts HW 67 Kts HW

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5.

FL 240 TAS 264 Kts FL 280 TAS 253 Kts FL 310 TAS 232 Kts

WC +30 Kts TW WC +40 Kts TW WC +50 Kts TW

FF 552 LB/Hour FF 476 LB/Hour FF 386 LB/Hour

The most economical FL is :(a) (b) (c) 6.

Airfield Pressure Altitude 5700ft, Temperature +30°C. According to graph 5-23 the Maximum Take-Off Weight is :(a) (b) (b)

7.

12 500 Lbs 12 200 Lbs 11 900 Lbs

Airfield Pressure Altitude 6200ft, Temperature +25°C, According to graph 5-23 the Maximum Take-Off Weight is :(a) (b) (c)

8.

FL 240 FL 280 FL 310

12 500 Lbs 12 300 Lbs 12 100 Lbs

For a flight from A to B the Minimum Enroute Altitude is 20 000ft. The temperature at FL 200 is -15°C and the area QNH is 30.50 inches. The fuel used to the high ground is 450 Lbs. The Maximum Take-Off Weight for the flight according to graph 5-24 is :(a) (b) (c)

9.

For a flight from C to D the Minimum Enroute Altitude is 19 000ft. The temperature at FL 190 is -11°C and the area QNH is 29.20 inches. If the fuel used to the high ground is 650 Lbs the Maximum Take-Off Weight for the flight (graph 5-24) is :(a) (b) (c)

10.

11 950 Lbs 12 150 Lbs 12 400 Lbs

11 900 Lbs 12 200 Lbs 12 500 Lbs

An obstacle 1400ft amsl is 5nm from reference zero of a runway whose elevation is 350ft. According to graph 5-28 the Minimum Climb Gradient required is :(a) (b) (c)

4.6% 3.8% 3.2%

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11.

Airfield Pressure Altitude 5000ft, Temperature 20°C. According to graph 5-31 the Minimum Take-Off Power required is :(a) (b) (c)

12.

Airfield Pressure Altitude 3500ft, Temperature 26°C. According to graph 5-31 the Minimum Take-Off Power required is :(a) (b) (c)

13.

2050 Ft/Lbs 2100 Ft/Lbs 2150 Ft/Lbs

Airfield Pressure Altitude 1000ft, Temperature 32°C. According to graph 5-31 the Minimum Take-Off Power required is :(a) (b) (c)

14.

2060 Ft/Lbs 2110 Ft/Lbs 2160 Ft/Lbs

2060 Ft/Lbs 2110 Ft/Lbs 2165 Ft/Lbs

Airfield Pressure Altitude 5000ft, Temperature 15°C, Flaps 0%,. Take-Off Mass 11 000 Lbs, Wind Component 15 Kts Headwind. According to graph 5-34 the Take-Off Distance required is :(a) (b) (c)

15.

2000 ft 2850 ft 3450 ft

Airfield Pressure Altitude 2000 ft, Temperature 24°C, Flaps 0%, Take-Off Weight 11 600 Lbs, Wind Component 5 Kts Tailwind. According to graph 5-34 the Take-Off Ground Roll is :(a) (b) (c)

16.

2000 ft 2300 ft 3900 ft

A Take-Off is planned from a 4000 ft runway with 1500 ft of clearway available. Pressure Altitude 4500 ft, OAT 25°C, Headwind 12 Kts, Flaps 0%. Using graph 5-36 the Maximum Weight at which this Accelerate-Go distance can be used is :(a) (b) (c)

10 000 Lbs 10 500 Lbs 11 000 Lbs

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17.

Airfield Pressure Altitude 4000ft, OAT 24°C, Aircraft Mass 11 500 Lbs. According to graph 5-37 the net gradient of climb is :(a) (b) (c)

18.

Airfield Pressure Altitude 5000ft, OAT 26°C, Assuming that there is no runway limitation but a 3.2 % net gradient of climb is required, using graph 5-37 the Maximum Take-Off Mass is :(a) (b) (c)

19.

3.4 % 3.9 % 4.4 %

11 500 Lbs 12 100 Lbs 12 500 Lbs

Airfield Pressure Altitude 4500ft, OAT 18°C, Flaps 40 %, 10 Kt Headwind, Take-Off Mass 11 100 Lbs. According to graph 5-39 the Accelerate-Stop Distance is :(a) (b) (c)

20.

3700 ft 4000 ft 4400 ft

Airfield Pressure Altitude 3000ft, Temperature 25°C, Wind Component 5Kt Tailwind, Take-Off Mass 11 400 Lbs. According to graph 5-39 the Accelerate-Stop Distance is :(a) (b) (c)

21.

3800 ft 4000 ft 4300 ft

A Take-Off is planned from a 4500 ft runway with 2000 ft of clearway available. Pressure Altitude 5000 ft, OAT 23°C, Headwind 15 Kt, Flaps 40 %. Using graph 5-40 the Maximum Mass for which this Accelerate-Go distance can be used is :(a) (b) (c)

22.

11 100 Lbs 11 600 Lbs 12 100 Lbs

Climbing from 4500ft, OAT +30°C to FL 290, OAT -23°C, Take-Off Mass 12 500 Lbs. According to graph 5-45 the Time, Fuel used and Distance flown are :(a) (b) (c)

27 minutes 24 minutes 27 minutes

258 Lbs 285 Lbs 285 Lbs

90 nm 80 nm 90 nm

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23.

Climbing from 5500ft, OAT 24°C to FL 280, OAT -32°C, Take-Off Mass 12 000 Lbs. According to graph 5-45 the Time, Fuel used and Distance flown are :(a) (b) (c)

24.

1450 Ft/Lbs 1500 Ft/Lbs 1550 Ft/Lbs

1420 Ft/Lbs 1370 Ft/Lbs 1320 Ft/Lbs

Cruising at FL 230, OAT -23°C, the recommended cruise power according to graph 5-59 is:(a) (b) (c)

29.

19 000 ft 21 000 ft 23 000 ft

Cruising at FL 270, OAT -24°C, the recommended cruise power according to graph 5-59 is:(a) (b) (c)

28.

160 Kts 190 Kts 225 Kts

Cruising at FL 260, Indicated OAT -25°C, the recommended cruise power according to graph 5-59 is :(a) (b) (c)

27.

59 nm 58 nm 60 nm

The temperature at the Minimum Enroute Altitude is -27°C. If the aircraft mass is 11 700 Lbs the Service Ceiling according to graph 5-47 is :(a) (b) (c)

26.

202 Lbs 230 Lbs 242 Lbs

Climbing from 4000ft, OAT 26°C to FL 260, OAT -24°C, Take-Off Mass 12 500 Lbs. According to graph 5-45 the mean TAS on the climb is :(a) (b) (c)

25.

18 minutes 19 minutes 18 minutes

1730 Ft/Lbs 1780 Ft/Lbs 1675 Ft/Lbs

Cruising at FL 210, IOAT -17°C, the Fuel Flow according to graph 5-60 is (a) (b) (c)

606 Lbs/hour 630 Lbs/hour 660 Lbs/hour

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30.

Cruising at FL 250, Temperature ISA +10°C, the Fuel Flow according to graph 5-60 is :(a) (b) (c)

31.

Cruising at FL 270, OAT -24°C, the Fuel Flow per engine according to graph 5-60 is:(a) (b) (c)

32.

285 Lbs/hour 258 Lbs/hour 252 Lbs/hour

Enroute from WPT 2 to WPT 3 at FL 190, Temperature ISA +20°C, Distance 247 nm, 35 Kts Headwind, Aircraft Mass 11 000 Lbs. The fuel used for the sector according to table 5-55 is :(a) (b) (c)

33.

556 Lbs/hour 565 Lbs/hour 656 Lbs/hour

678 Lbs 694 Lbs 717 Lbs

Enroute from WPT 3 to WPT 4 at FL 270, Temperature ISA +10°C, Distance 329 nm, 25 Kts Tailwind, Aircraft Mass 11 500 Lbs. The fuel used on this sector according to table 5-54 is :(a) (b) (c)

34.

567 Lbs 589 Lbs 615 Lbs

Enroute from WPT 4 to WPT 5 at FL 200, Temperature ISA +15°C, Distance 450 nm, 30 Kts Headwind, Aircraft Mass 11 000 Lbs. The fuel used on the sector according to tables 5-54 and 5-55 is :(a) (b) (c)

35.

1147 Lbs 1172 Lbs 1194 Lbs

Enroute from WPT 5 to WPT 6 at FL 190, Temperature ISA +10°C, Distance 212 nm, 25 Kts Tailwind, Aircraft Mass 10 500 Lbs. The fuel used on the sector according to table 5-90 is :(a) (b) (c)

358 Lbs 387 Lbs 405 Lbs

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36.

The range of the EE-20 aeroplane at FL 240 with a 35 Kt Headwind flying at the recommended cruise power (graph 5-96) is:(a) (b) (c)

37.

The range of the EE-20 aeroplane at FL 280 with a 40 Kt Tailwind flying at the recommended cruise power (graph 5-96) is:(a) (b) (c)

38.

4 hours 12 mins 4 hours 20 mins 4 hours 29 mins

For a landing at an airfield at sea level (QNH 1009.2) the pressurization controller setting for landing (graph 5-106) is:(a) (b) (c)

42.

3 hours 39 mins 3 hours 48 mins 3 hours 57 mins

The endurance of the EE-20 aeroplane at FL 290 flying at maximum cruise power graph 5-97 is (a) (b) (c)

41.

942 nm 985 nm 1035 nm

The endurance of the EE-20 aeroplane at FL 240 flying at the recommended cruise power (graph 5-97) is:(a) (b) (c)

40.

1090 nm 1175 nm 1250 nm

The range of the EE-20 aeroplane at FL 260 with a 25 Kt Headwind flying at the recommended cruise power (graph 5-96) is:(a) (b) (c)

39.

852 nm 909 nm 975 nm

0ft 300ft 600ft

For a landing at an airfield (elevation 4000ft, QNH 1020 hPa) the pressurization controller setting for landing (graph 5-106) is :(a) (b) (c)

3800ft 4200ft 500ft

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43.

400 Lbs of fuel is available for holding at FL 150. If the ice vanes are extended the holding time according to graph 5-107 is:(a) (b) (c)

44.

The fuel required for 45 minutes holding at FL 150 with the ice vanes extended (graph 5-107) is (a) (b) (c)

45.

46.

(a) 1050ft (b) 1450t (c) 1900ft (d) Pressure Altitude 5500ft, OAT 29°C, Aircraft mass 10 200 Lbs. The landing ground roll with a 5 Kt Tailwind and 100% flap (graph 5-112) is:-

1900ft 2600ft 3200ft

An obstacle 1200 ft amsl is 3nm from reference zero of a runway whose elevation is 600ft. According to graph 5-28 the Minimum Climb Gradient required is:(a) (b) (c)

49.

1200ft 1400ft 1600t

Pressure Altitude 5500ft, OAT 29°C, Aircraft mass 10 200 Lbs. The landing distance with zero flap, propeller reversing and a 5 Kt Tailwind (graphs 5-112 and 5-113) is:(a) (b) (c)

48.

305 Lbs 335 Lbs 365 Lbs

Pressure Altitude 3000ft, OAT 25°C, Aircraft mass 10 200 Lbs. The landing distance with a 14 Kt Headwind (graph 5-112) is :-

(a) (b) (c) 47.

0.52 mins 31 mins 52 mins

3.1% 4.7% 6.4%

An obstacle 240 ft above runway elevation is 1700 metres from reference zero. According to graph 5-28 the minimum Climb gradient required is:(a) (b) (c)

2.9% 3.7% 4.6%

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CHAPTER 5 WEIGHT AND BALANCE AIRCRAFT WEIGHT SCHEDULE In the process of compiling a flight plan for an aircraft, the weight schedule must be consulted to ensure that certain weight limitations are not exceeded. In later chapters, balance limitations (location of the C of G) will also be considered. The weight schedule given below is the ideal and complete one, although certain operators may elect to combine items in order to abbreviate the process. AIRCRAFT EMPTY WEIGHT (AEW) + OIL AND UNUSABLE FUEL = BASIC EMPTY WEIGHT (BEW) + CREW AND CATERING = OPERATING EMPTY WEIGHT (OEW) + PAYLOAD = ZERO FUEL WEIGHT (ZFW) + TOTAL FUEL = RAMP OR TAXI WEIGHT - TAXI FUEL = TAKE OFF WEIGHT - TRIP FUEL OR BURN OFF = LANDING WEIGHT

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AIRCRAFT EMPTY WEIGHT (AEW) Consists of the airframe, engines, and all items of operating equipment that have fixed locations and are permanently installed in the aircraft. OIL AND UNUSABLE FUEL This includes engine oil, hydraulic fluid and undrainable fuel (Piper Cherokee 2 Galls, B-747 1600 Kg) BASIC EMPTY WEIGHT (BEW) The Empty Weight of the aircraft plus oil, hydraulic fluid and unusable fuel. CREW AND CATERING Operating crew, cabin staff and catering. OPERATING EMPTY WEIGHT (OEW) The weight of the aircraft, including the crew, ready for flight but without payload and fuel. MAXIMUM ZERO FUEL WEIGHT (MZFW) The maximum weight authorized for the aircraft not including the fuel load. Zero fuel weight is the operating empty weight (OEW) plus the payload. MAXIMUM RAMP WEIGHT The maximum structural take-off weight plus the fuel to be burned during taxi and run-up. MAXIMUM TAKE-OFF WEIGHT (MTOW) The maximum structural weight at the start of the take-off run. The take-off weight for a particular flight may be limited to a lesser weight when runway length, atmospheric conditions, or other variables are adverse. TRIP FUEL OR BURN OFF The fuel used from the point of departure to the destination. Reserve fuel is not included in the trip fuel and the entire fuel reserves are expected to be on board the aircraft at the point of first intended landing. MAXIMUM LANDING WEIGHT The maximum structural weight at which an aircraft may normally be landed. The landing weight may be limited to a lesser weight when runway length or atmospheric conditions are adverse.

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CALCULATION OF MAXIMUM PAYLOAD Assuming that there are no airfield restrictions, the maximum payload that may be carried on a flight will be limited by :MAXIMUM TAKE-OFF WEIGHT MAXIMUM LANDING WEIGHT MAXIMUM ZERO FUEL WEIGHT Example: Max Ramp weight Basic Weight Max Brakes release weight Max Landing weight Max Zero Fuel weight Trip fuel Reserve fuel

89 700kg 47 000kg 89 350kg 72 600kg 63 500kg 12 462kg 4 680kg

The Maximum Payload is:Least of 3 method Find the least of:

Max Take off weight Max Landing weight + Flight Fuel Max Zero Fuel Weight +Fuel on board

Step 1 Calculate fuel at Brakes release…. Trip fuel Reserve fuel FUEL ON BOARD

12 462kg 4680kg 17 142kg

Step 2 Calculate the 3 limitations….. MTOW MLW+trip fuel MZFW+fob

89 350kg 85 062kg (72600+12462) 80 642kg (63500+17142)



Step 3 In this case the payload would be: MZFW - Basic weight PAYLOAD

63 500kg 47 000 16 500kg

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NOW…. If it was the TOW that was found to be the lowest you would: Max Take off weight - Fuel on board - Basic weight PAYLOAD If it was the LW that was the limiting factor then: Landing weight - Reserve fuel - Basic weight PAYLOAD HUMIDITY Humidity and air density are inversely proportional. The greater the humidity, the less the air density. Piston engine aircraft performance is adversely affected by humidity to the extent where maximum take-off weight may be limited. The effect of humidity on jet engine performance, however, is negligible. FLAPS The effect of flaps on maximum take-off weight varies from aircraft to aircraft and from flap setting to flap setting. Factors to be considered are not only the effect of flaps on the take-off run, but also on the initial climb performance after take-off. A definitive answer on the effect of flaps on an aircraft's maximum take-off weight would be extracted from the appropriate performance graphs. In conclusion, all of the above mentioned factors may limit an aircraft's maximum take-off weight and it is the most limiting case, which will determine the aircraft's actual take-off weight. MAXIMUM FLOOR LOAD Maximum floor load is an indication of the physical bearing strength of the aircraft's floor, normally in the cargo or baggage area. It is an expression of the maximum weight that can be borne per surface area. Because maximum floor load is derived by weight per area, the height of any object to be loaded is of no consequence. In most load calculations, the maximum floor load of the aircraft is given. The pilot must calculate the area of the object to be loaded and its weight, to check whether it may be loaded. To calculate the area of a rectangular or square object, use the formula: AREA = LENGTH x BREADTH To calculate the area of a circular object, for example a barrel, use the formula: AREA = π r ²

Where r = the radius of the circular object.

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SPECIFIC GRAVITY (SG) Specific gravity is a method of converting a volume of liquid to a weight of liquid or vice versa. The formula for specific gravity is:

VOLUME x SG = WEIGHT

The standard used for specific gravity is water (SG 1) 1 litre of water has a weight of 1 Kg. 1 Imperial Gallon of water has a weight of 10 lbs. If the SG of fuel is 0.82 then:1 litre of fuel has a weight of 0.82 Kg 1 Imperial Gallon of fuel has a weight of 8.2 Lbs Specific Gravity cannot be applied directly to US Gallons. NOTE:

US Gallons must be converted to Litres or Imperial Gallons before Specific Gravity can be applied.

SPECIFIC WEIGHT Specific weight serves much the same function as specific gravity but applies to US Gallons only. It is a statement (Specific Weight 6.6 Lbs) and means that 1 US Gallon of fuel weighs 6.6 Lbs EXAMPLE : The specific weight of fuel is 6.6 Lbs per US gallon. How much does 450 US gallons of fuel weigh? 450 US gal x 6.6 = 2970 Lbs BALANCE, CENTRE OF GRAVITY Balance refers to the location of the CG (Centre of Gravity) of an aircraft. It is of primary importance to aircraft stability and safety in flight. Pilots should never fly an aircraft if they are not satisfied with its loading and the resulting weight and balance conditions. The CG is the point about which an aircraft would balance if it were possible to support the aircraft at that point. It is the mass centre of the aircraft, or the theoretical point at which the entire weight of the aircraft is assumed to be concentrated. The CG must be within specific limits for safe flight. The CG is not necessarily a fixed point; its location depends on the distribution of items loaded in the aircraft. As variable load items are shifted there is a resultant shift in CG location. If the CG is displaced too far forward on the longitudinal axis, a nose heavy condition will result. Conversely, if the CG is displaced too far aft on the longitudinal axis, a tail-heavy condition will result. It is possible that an unfavourable location of the CG could produce such an unstable condition that the aircraft becomes very difficult to control.

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40 inches

40 inches

15 Lbs

15 Lbs Fulcrum (CG)

In the above sketch two weights of 15 Lbs each are 40 inches from the fulcrum. The weights are balanced. Mathematically 15 Lbs x 40 inches = 600 inch/Lbs on each side. WEIGHT x ARM = 15 Lbs x 40 inches =

MOMENT 600 inch/Lbs

REFERENCE DATUM Every aircraft has a reference datum and it varies from aircraft type to aircraft type. Usually it is at or near the nose of the aircraft. It is the datum from which all horizontal distances are measured. ARM Arm is the horizontal distance (usually in inches) from the reference datum to the location of an object or position in the aircraft. Other terms are STATION (STA), FLIGHT STATION (FS) or CENTROID, e.g. Forward Hold at FS 220 means the Forward Hold is 220 inches aft of the datum. MOMENT Moment is the product of the weight of an item multiplied by its arm. MOMENT INDEX or REDUCTION FACTOR Moment Index or Reduction Factor is a moment divided by a constant such as 100, 1 000 or 10 000. The purpose of using a moment index or reduction factor is to simplify weight and balance computations of large aircraft where heavy items and long arms result in large, unmanageable numbers

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In class example using the graphs at the end of the EE-20 manual. EXAMPLE 1 Aircraft load: EW Pilot + Co-pilot Passenger weights Baggage weight Catering on board Fuel on board at s/up Flight fuel

8087lbs 15041.00 IU 165kg 65,86,95,112,45kg 89kg 45kg in foyer cabinet 300usg (6.6SW) 185usg

Find a. b.

TOW and position of CoG ZFW and position of CoG

STEP 1 Complete the table, loading front to rear…UNTIL you get to ZFW, then check it is in balance.

ITEM EW CREW ROW 1 ROW 2 ROW 3 LAV SEAT AFT CABIN CABINET (Foyer) ZFW(10400 MAX) FUEL TOW(12500 MAX)

Weight 8087 363 455 332 99

Arm 129.0 176.0 215.0 259.0 292 325 284

196 99 9631

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STEP 2 Now add the fuel to check the weight and CoG at TOW…

ITEM

Weight

EW CREW ROW 1 ROW 2 ROW 3 LAV SEAT AFT CABIN CABINET (Foyer) ZFW(10400 MAX) FUEL TOW(12500 MAX)

8087 363 455 332 99

Arm 129.0 176.0 215.0 259.0 292 325 284

196 99 9631 1980 11611

Index Units/100 15041.00 468.27 800.80 713.8 256.41 637.00 281.16 18198.44 3631 21829.44

IN CLASS EXERCISES Basic loading problems 1.

Complete the load sheet ignoring CoG limitations for the following load find the CoG position for take off, and zero fuel weight (ignoring any limits): EW Pilots Pax Baggage Catering FOB

2.

7900lbs Moment 14950 155kg 88,45,75,77kg 132kg 15kg (aft) 335usg (6.7 SW)

Complete the load sheet and determine if the aircraft is in CoG at Take off.. EW Pilots Pax Baggage FOB

8122lbs Moment 15950 175kg 99,65,55,88,45,75,77kg 185kg 435usg (6.4 SW)

This can also be done graphically

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CALCULATION OF LANDING CG The moment of the TRIP FUEL or BURN OFF cannot be read from the table directly as the arm of the fuel varies as the amount of the fuel in the tanks. The moment of the Trip Fuel can be calculated by subtracting the moment of the Landing Fuel from the moment of the TakeOff Fuel. A Load Sheet is not available in the exam and it is suggested that the following method be used. NOTE: The moment of the fuel is given as MOMENT 100 which means the figure must be multiplied by 100 to give the full figure. 1. 2. 3. 4.

Start with the Take-Off condition of the aircraft. Subtract the Take-Off Fuel weight and moment. Add the Landing Fuel weight and moment. Calculate the Landing CG.

Example: Take-Off Weight 12 500 Lbs,Take-Off Fuel 530 US Gallons (SW 6.6 Lbs/US G) Take-Off CG 191.3 inches Trip Fuel 300 US Gallons

Take-Off Take-Off Fuel (530 Gallons) Landing Fuel (530 - 300 Gallons) Landing

WEIGHT 12 500 -3 498 +1 518 10 520

ARM 191.3 190.5

MOMENT 2 391 250 - 664 300 + 277 000 2 003 950

Further examples 1.

Find the landing CoG position from the following: Take off weight CoG arm at take off Fuel on board at T/off Flight fuel Fuel SW

2.

12 223 lbs 193.2 inches 544usg 201usg 6.7

Find the landing CoG position from the following: Take off weight Moment index at t/off Fuel on board at take off Flight fuel Fuel SW

9800lbs 15550 350usg 85usg 6.7

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MOVING CG IF the CG is too far aft the pilot can… 1. 2. 3.

Redistribute the load forward of the current CG by moving pax, or shifting baggage Add ballast forward Remove weight aft

IF the CG is too far forward, the pilot can…. 1. 2. 3.

Redistribute the load toward the rear Add ballast aft Remove weight forward

Weight to be shifted formula

Weight to be shifted = GW x (difference between desired CG & original CG) Distance between the 2 stations

Adding ballast formula

Ballast to add = Original GW x (difference between desired CG & actual CG) Distance between loading station and desired CG

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CALCULATING CG POSITION AS A PERCENTAGE MAC To find the position that the Cg is acting in reference to the Mean Aerodynamic Chord we use the following formula to calculate % MAC….

CG position as a % MAC = distance aft of MAC leading edge x 100 MAC 1 REFER PAGE 6-5 of the EE-20 manual for MAC figures. Example 1 The Cg from the load sheet is found to be 192 inches aft of the datum (24000 x 100 ÷ 12500) Step 1. 192 inches = (192 – 171.23”) 20.77” aft of the MAC leading edge The position expressed as a % MAC is: % MAC =

20.77” 70.41

x 100

% MAC = 29.49% MAC Example 2 The aircraft is found to be at 12 000lbs and the moment is 23100units, find the CG position as a % MAC. Example 3 An aircraft is loaded so that its weight is 10 200lbs and the moment is 22222 units, find the CG position as a % MAC..

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QUESTION SET ONE 1. Maximum Take-Off Mass Maximum Landing Mass Maximum Zero Fuel Mass Operational Empty Mass Trip Fuel Reserve Fuel

151 500 Kg 112 000 Kg 101 200 Kg 69 700 Kg 40 150 Kg 8 200 Kg

The maximum payload that may be carried is :(a) (b) (c) 2.

31 500 Kg 33 450 Kg 35 250 Kg

Maximum Take-Off Mass Maximum Landing Mass Maximum Zero Fuel Mass Operational Empty Mass Trip Fuel Reserve Fuel

151 500 Kg 97 500 Kg 88 450 Kg 66 700 Kg 44 500 Kg 7 100 Kg

If the maximum payload is carried the Take-Off Weight is :(a) (b) (c) 3.

142 000 Kg 140 050 Kg 151 000 Kg

Maximum Take-Off Mass Maximum Landing Mass Maximum Zero Fuel Mass Operational Empty Mass Distance A to B Groundspeed Fuel Flow Reserve Fuel

151 500 107 000 96 300 64 250 2 850 490 7 350 15% of

Kg Kg Kg Kg nm Kts Kg/hour Trip Fuel

The maximum payload that may be carried on this flight is :(a) (b) (c) 4.

44 500 Kg 36 337 Kg 32 050 Kg

Maximum Take-Off Mass Maximum Landing Mass Maximum Zero Fuel Mass Operational Empty Mass Trip Fuel Reserve Fuel

151 500 Kg 97 500 Kg 88 450 Kg 66 870 Kg 45 300 Kg 12 240 Kg

The maximum payload that may be carried is :(a) (b) (c)

18 390 Kg 17 280 Kg 16 920 Kg

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5.

The mass of 729 US Gallons of fuel at SG 0.78 is :(a) (b) (c)

6.

If 1250 Lbs of fuel at SG 0.812 are on board an aircraft, the amount of fuel in US Gallons is:(a) (b) (c)

7.

2313 Lbs 2846 Lbs 3508 Lbs

If 567 Kgs of fuel at SG 0.812 are on board an aircraft, the amount of fuel in US gallons is :(a) (b) (c)

10.

8122 Lbs 6253 Lbs 5631 Lbs

The weight of 1292 Litres of fuel (SG 0.812) is :(a) (b) (c)

9.

128 US Gallons 185 US Gallons 122 US Gallons

The weight of 867 US Gallons of fuel (SG 0.78) is :(a) (b) (c)

8.

2153 Kg 2579 Kg 3095 Kg

161 US Gallons 184 US Gallons 201 US Gallons

An IFR flight is to be made from A to C with a stop at B. There is no fuel available at B. A to B B to Trip Fuel 2670 Kg 2295 Alternate Fuel 1040 Kg 995 Holding Fuel 620 Kg 620

C Kg Kg Kg

The minimum fuel required at Take-Off from A is:(a) (b) (c)

6580 Kg 7620 Kg 8240 Kg

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QUESTION SET TWO 1.

The weights measured at the landing gear of an aircraft are as follows:Nose wheel (55 inches aft of datum) Right main wheel (121 inches aft of datum Left main wheel (121 inches aft of datum)

475 Lbs 1046 Lbs 1040 Lbs

The C of G of the aircraft is :(a) (b) (c) 2.

The C of G of an aircraft is 980 inches aft of datum at an all up mass of 170 500 Lbs. If 800 Lbs of baggage is moved from FS 1130 to FS 430 the new C of G will be :(a) (b) (c)

3.

104.6 inches 106.4 inches 108.8 inches

975.99 inches 976.72 inches 977.62 inches

Aircraft Mass C of G Aft C of G limit

12 000 Lbs 193 inches aft of datum 196.3 inches aft of datum

The maximum mass that can be loaded at FS 325 without exceeding the aft C of G limit is :(a) (b) (c) 4.

A pallet 83 inches by 95 inches is to be loaded in a cargo aircraft. The floor load limit of the aircraft is 169 Lbs per square foot. If the pallet mass is 88 Lbs and the tie down equipment is 37 Lbs the amount freight that may be loaded on the pallet is :(a) (b) (c)

5.

307 Lbs 342 Lbs 386 Lbs

of

9128 Lbs 9156 Lbs 9244 Lbs

A pallet 76 inches by 76 inches is to be loaded in a cargo aircraft. The floor load limit of the aircraft is 184 Lbs per square foot. If the pallet mass is 85 Lbs and the tie down equipment is 36 Lbs the amount of freight that may be loaded on the pallet is :(a) (b) (c)

7499 Lbs 7378 Lbs 7259 Lbs

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6. The C of G of an aircraft is 1000 inches aft of datum at an all up mass of 155 000 Lbs If 1000 Lbs of baggage is moved from FS 1166 to FS 670 the new C of G will be :(a) (b) (c) 7.

996.8 inches 997.5 inches 998.3 inches

Aircraft Basic Empty Mass C of G Standard adult mass 2 Pilots 2 Adult Pax 2 Adult Pax 1 Adult Pax Baggage 250 Lbs Fuel 440 US Gallons (SW 6.5 Lbs) Ignore Fuel for start and taxi

8000 Lbs 185 inches aft of datum 170 Lbs FS 129 FS 176 FS 215 FS 259 FS 320 Mom x 100 5337

The C of G of the aircraft at Take-Off is :(a) (b) (c) 8.

186.7 inches 188.2 inches 189.6 inches

Aircraft Basic Empty Mass C of G Standard adult mass 2 Pilots 2 Adult Pax Baggage 340 Lbs Fuel 480 US Gallons (SW 6.6 Lbs) Ignore Fuel for start and taxi The C of G of the aircraft at Take-Off is :(a) (b) (c)

8000 Lbs 176 inches aft of datum 170 Lbs FS 129 FS 259 FS 346.5 Mom x 100 5956

183.78 inches 184.88 inches 185.96 inches

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9.

Operational Empty Mass CG Standard Passenger Weight Zone A FS 290 Zone B FS 480 Zone C FS 680 Hold 1 FS 200 Hold 2 FS 750 Wing Tanks FS 490 Centre Tank FS 480

66 600 Kg 480 Inches 75 Kg 28 Passengers 42 Passengers 46 Passengers 1500 Kg 500 Kg 41020 Kg 9080 Kg

The CG of the aircraft at Take-Off is :(a) (b) (c) 10.

483.27 inches 484.68 inches 485.73 inches

Operational Empty Mass CG Standard Passenger Weight Zone A FS 290 Zone B FS 480 Zone C FS 680 Hold 1 FS 200 Hold 2 FS 750 Wing Tanks FS 490 Centre Tank FS 480

71 600 Kg 480 Inches 165lb 25 Passengers 44 Passengers 49 Passengers 1750 Kg 800 Kg 45820 Kg 9550 Kg

The CG of the aircraft at Take-Off is :(a) (b) (c) 11.

483.55 inches 484.07 inches 485.32 inches

Shortly before Take-Off, an extra passenger is given permission to board an aircraft. Before boarding aircraft weight was 11 200 Lbs, and the CG was 191 Inches. The passenger weight is 170 Lbs and is allocated a seat at FS 259. The revised CG of the aircraft is :(a) (b) (c)

192 inches 193 inches 194 inches

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CHAPTER 6 CRITICAL POINT (CP) or POINT OF EQUAL TIME (PET) & POINT OF NO RETURN (PNR) PET / CP DEFINITION The PET / CP is defined as being the point on track from which it would take equal of time to either return to the point of departure or continue to the destination. The PET / CP is not a function of fuel but of distance and aircraft groundspeeds. Long haul jet transport aircraft usually carry three PET / CP,s. A 4 Engine PET / CP, a one engine inoperative PET / CP (1 ENG INOP) and a 14 000 feet PET / CP in case of pressurization failure. The three cases each have a different TAS and Groundspeed thus the PET / CP will be at a different point. The ETA at the PET / CP is calculated and in the event of a major aircraft malfunction or a passenger becomes critically ill an instant decision can be made whether to continue to the destination or to return to the point of departure. PET WITH NO WIND On a flight from A - B, with still air conditions prevailing, it is clear that the PET would lie at the halfway point along track.

A

B 500 nm

PET

500 nm

PET WITH WIND FORMULA Distance to CP(PET) = GSH x Distance GSH + GSO Where

GS H = GS Home

GS O = GS Out

CP = H x D H+O

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Example 1 A to B Distance 630 nm Track 048°(T) TAS 245 Kts W/V 145/45 If an aircraft departed A at 0900 Z the ETA at the PET would be :GS out 246 Kts

GS home 235 Kts

GSH x DISTANCE GSH + GSO

235 X 630 235 + 246

= pet of 308nms at GS 246 = 1 hr 15 mins

Example 2 The Wind Component from A to the PET is 35 Kts Headwind and the Wind Component from the PET to B is 55 Kts Headwind. The distance A to B is 545 nm and TAS 300 Kts. The Distance of the PET from A is :-

GSH 335 x DIST 545

=

PET 315 nms

GSH 335 + GSO 245 Note: The halfway point from A to B is 272.5 nm but the PET is 315 nm from A. The PET has moved into wind. MULTIPLE TRACK CP/PET A flight is planned from A to C via B. The distance of the PET from A is :A to B B to C

Distance 412 nm GS On 245 Kts Distance 237 nm GS On 268 Kts

GS Ret 215 Kts GS Ret 192 Kts B

222 nm 237 nm at GS 268 190 nm at GS 215 A

X

53 mins

C

53 mins Method: 1. 2. 3.

Calculate the time on the shorter leg B to C. 237 nm GS 268 Kts Time 53 mins Calculate the distance that would take 53 mins to return to A GS 215 for 53 mins = 190 nm It will take the same time to fly from X to A as it will take to fly from B to C.

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Solve for the PET between X and B.

GSH 215 x DIST 222

= PET 104NMS

GSH 215 + GSO 245 The PET is 294 nm from A (190 nm + 104 nm) Proof

PET to A PET to B

294 nm at GS 215 = 1 Hour 22 mins 118 nm at GS 245 = 29 mins + B to C 53 mins=1hr22min

PET or CRITICAL POINT (1 ENGINE INOPERATIVE) To calculate the 1 ENG INOP PET or CP reduced groundspeeds are used, otherwise the calculation is the same. Example 3 A twin engine aircraft is to fly from X to Y, Track 130°(T), Distance 727 nm, W/V 270/40, 2 engine TAS 260 Kts, 1 engine TAS 195 Kts. If the ETD is 0800 Z the ETA at the 1 ENG INOP PET (CP) is :2 Engine TAS 260 Kts 1 Engine TAS 195 Kts 1 Engine TAS 195 Kts

GS On 289 Kts Reduced GS On 224 Kts Reduced GS Ret 163 Kts

R GSH 163 x 727 NMS R GSH 163 + R GSO 224

= PET 306NMS

To calculate the ETA at the PET the 2 engine GS must be used. PET 306 nm

GS 289

Time 1 Hour 04 mins

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PET/CP QUESTIONS 1.

A to B, Track 060°(T), TAS 185 Kts, The distance to the PET is :(a) (b) (c)

2.

W/V 090/30.

388 nm 452 nm 516 nm

C to D, Track 125°(T), TAS 245 Kts, Distance 1547 nm, If an aircraft departs C at 0915 Z the ETA at the PET is :(a) (b) (c)

3.

Distance 905 nm,

W/V 225/45.

1200 Z 1215 Z 1230 Z

The Wind Component from A to the PET is 45 Kts Headwind and the Wind Component from the PET to B is 60 Kts Headwind. The distance A to B is 750 nm and TAS 300 Kts. The Distance of the PET from A is :(a) (b) (c)

4.

412 nm 442 nm 482 nm

A flight is planned from A to C via B. A to B B to C

Distance 412 nm Distance 239 nm

TAS 275 Kts WC 35 Kts Headwind WC 55 Kts Headwind

The distance of the PET from A is :(a) (b) (c)

5.

330 nm 380 nm 430 nm

A twin engine aircraft is to fly from A to B, Track 245°(T), Distance 830 nm, W/V 310/40, 2 engine TAS 280 Kts, 1 engine TAS 220 Kts. If the ETD is 0800 Z the ETA at the 1 ENG INOP PET (CP) is :(a) (b) (c)

6.

0943 Z 1003 Z 1023 Z

A twin engine aircraft is to fly from C to D, Track 035°(T), 135/45, 2 engine TAS 220 Kts, 1 engine TAS 185 Kts.

Distance 884 nm, W/V

If the ETD is 1100 Z the ETA at the 1 ENG INOP PET (CP) is :(a) (b) (c)

1254 Z 1314 Z 1334 Z

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POINT OF NO RETURN (PNR) The PNR is defined as the furthest point along track to which the aircraft can fly and return to the point of departure within the safe endurance of the aircraft (fuel reserves will remain intact). PNR WITH ZERO WIND If the safe endurance of an aircraft is 6 Hours the PNR will be 3 Hours. If the TAS is 240 Kts then the PNR is 720 nm. PNR WITH WIND Gs Home x Safe Endurance Gs Home + Gs Out Or FLIGHT FUEL or ENDURANCE SGR out + SGR home

Example 1. Track 220°(T), W/V 300/35, TAS 240 Kts, Endurance 6 Hours (excluding reserves). GS Out 231Kts GS Return 244Kts Answer: 244 x 6hr 231 + 244

=

3hrs 5min

WITH ANY W/V THE PNR ALWAYS MOVES TOWARDS THE POINT OF DEPARTURE MULTI TRACK PNR (for demonstration purposes only) A

B

C 1. 2. 3.

Calculate the time from A to B and the time from B back to A. Subtract this time from the safe endurance. Use the endurance remaining to calculate the PNR from B enroute to C.

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PNR BASED ON FUEL CONSUMPTION On a flight from A to B aircraft performance Outbound is 0.312 GNM/KG and aircraft performance returning to B is 0.421 GNM/KG Total Fuel on board is 9000 KG which includes a 15% reserve. The distance to the PNR keeping the reserve fuel intact is :Method:

Find the amount of fuel that can be used to the PNR and return.

Assuming the fuel on board is 115% (100%+15% reserve) = 9000KG Then fuel to the PNR and return is 100% = 7826KG 15% of 7826 KG reserve fuel = 1174KG Fuel on board 9000KG Calculate the amount of fuel required to fly 1 nm Out and Return 1 nm Outbound 0.312 GNM/KG l/X 1 nm Return 0.421 GNM/KG 1/X 1 nm Out + Return 7826kg 5.5804 KG/GNM

= =

3.2051 KG/GNM 2.3753 KG/GNM 5.5804 KG/GNM

= PNR 1402 nm

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PNR QUESTIONS 1.

A to B TAS 245Kts, Track 315° (T), W/V 105/35, Endurance excluding reserves is 4 Hours 15 mins. The distance of the PNR from A is :(a) (b) (c)

2.

465nm 487nm 511nm

C to D TAS 315Kts, Track 225° (T), W/V055/60, Endurance 6 Hours. In the event of the aircraft returning to C reserve fuel of 1 Hour 30 mins is required. The distance to the PNR keeping reserve fuel intact is :(a) (b) (c)

3.

657nm 684nm 705nm

On a flight from A to B aircraft performance Outbound is 0.423 GNM/KG and aircraft performance returning to B is 0.527 GNM/KG. Total Fuel on board is 9000 KG which includes a 1500 KG reserve. The distance to the PNR keeping the reserve fuel intact is :(a) (b) (c)

4.

1760nm 1860nm 1960nm

A flight is planned from A to E at TAS315Kts. Endurance excluding reserves 6 hr A to B B to C C to D D to E

Distance 234 nm Distance 289 nm Distance 324 nm Distance 455 nm

WC 35 Kts Headwind WC 45 Kts Headwind WC 55 Kts Headwind WC 65 Kts Headwind

The distance to the PNR in :(a) (b) (c)

882nm 922nm 962nm

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ANNEX A SAMPLE EXAM

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Sample exam 1.

Maximum Take-off Mass Maximum Landing Mass Maximum Zero Fuel Mass Optional Empty Mass Distance A to B Groundspeed Fuel Flow Reserve Fuel

151 500 Kg 107 000 Kg 96 300 Kg 64 250 Kg 2 850 nm 490 Kts 7 350 Kg 15 %

The maximum payload that may be carried from A to B is: a) b) c) 2.

36 337 kg; 38 087 kg; 32 050 kg. The mass of 817 US Gallons of fuel at SG 0.78 is: a) b)

c) 3.

2 412 kg; 2 897 kg; 3 965 kg.

The following figures apply to a runway: Runway total length Stopway Clearway Displaced threshold

4500 ft 520 ft 740 ft 220 ft.

The Landing Distance Available (LDA) is: a) b) c) 4.

4 280 ft; 4 500 ft; 5 240 ft.

The following figures apply to a runway: Runway total length Stopway Clearway Displaced threshold

4500 ft 520 ft 740 ft 220 ft.

The Take-Off Distance Available (TODA) is: a) b) c)

4 500 ft; 5 020 ft; 5 240 ft.

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5.

Runway 06 at LANSERIA is 3048 metres in length. The elevation of RWY 06L is 4517 ft. The elevation of RWY 24R is 4393 ft. The slope of RWY 24R is: a) b) c)

6.

2.15 %; 1.63 %; 1.24 %.

The weight measured at the landing gear of an aircraft are as follows: Nose wheel (55 inches aft of datum) Right main wheel (121 inches aft of datum) Left main wheel (121 inches aft of datum)

475 Lbs. 1046 Lbs. 1040 Lbs.

The C of G of the aircraft is: a) b) c) 7.

104.6 inches; 106.4 inches; 108.8 inches. The C of G of an aircraft is 196 inches aft of datum at an all up mass on 12 500 lbs. If 200 lbs. of baggage is moved from FS 325 to FS 120, the new C of G will be: a) b) c)

8.

191 67 inches; 192.72 inches; 193.58 inches.

Aircraft Mass C of G Aft C of G limit

12 000 Lbs. 193 inches aft of datum 196.3 inches aft of datum.

The maximum mass that can be loaded at FS 325 without exceeding the aft C of G limit is: a) b) c) 9.

307 Lbs.; 342 Lbs.; 386 Lbs.

A pallet 83 inches by 95 inches is to be loaded in a cargo aircraft. The floor load limit of the aircraft is 169 Lbs. per square foot. If the pallet mass is 88 Lbs. and the tie down equipment is 37 Lbs. the amount of freight that may be loaded on the pallet is: a) b) c)

9 128 Lbs.; 9 156 Lbs.; 9 244 Lbs.

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10.

FL 180 TAS 276 Kts WC -20 Kts Fuel Flow 716 Lbs/Hour FL 220 TAS 271 Kts WC -15 Kts Fuel Flow 622 Lbs/Hour FL 260 TAS 262 Kts WC -60 Kts Fuel Flow 534 Lbs/Hour The most economical FL is: a) b) c)

11.

An aircraft flying at FL 310, TAS 232 Kts, Fuel Flow 545 Lbs/Hour has a performance of 0.355 GNM per LB. a) b) c)

12.

FL 180; FL 220; FL 260.

34 Kts Headwind; 39 Kts Headwind; 45 Kts Headwind.

An aeroplane flying at FL 310 at TAS 493 Kts obtains a performance of 46.06 ANM/1000 Kgs in zero wind conditions. At FL 350 the TAS and performance is 48.6 ANM/1000Kgs. It would be less economical to cruise at FL 350 against a headwind component greater than: a) b) c)

13.

10 Kts; 17 Kts; 23 Kts.

Aircraft basic Empty Mass C of G Standard adult mass 2 Pilots 2 Adult Pax 2 Adult Pax 1 Adult Pax Baggage 250 Lbs. Fuel 440 US Gallons (SW 6.5 Lbs) Ignore fuel for start up and taxi

8000 Lbs. 185 inches aft of datum 170 Lbs FS 129 FS 176 FS 215 FS 259 FS 320 Mom x 100 5337

The C of G of the aircraft at Take-Off is: a) b) c) 14.

186.7 inches; 188.2 inches; 189.6 inches.

The wind component from A to PET is 35 Kts Headwind. The wind component from the PET to B is 55 Kts Headwind. Distance A to B is 545 nm, TAS 300 Kts. The distance of the PET from A is:

a) b) c)

272 nm; 315 nm; 347 nm.

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15.

The term VI means: a) b) c)

16.

Take-Off Safety Speed; Take-Off Decision Speed; Take-Off Refusal Speed.

Airfield Pressure Altitude 5500 feet, temperature +19ºC, Aircraft Weight 10 800 Lbs. Headwind 13 Kts, Flaps 40% According to graph 5-38 the Take-Off Ground Roll is: a)

b) c) 17.

2 100 feet; 2 500 feet; 2 900 feet.

Airfield Pressure Altitude 5250 feet, temperature +23ºC, Field Length 4000 feet, Stopway 400 feet, Clearway 700 feet, Tailwind 5 Kts, Flaps 40%. According to graph 5-39 the maximum take-off weight which satisfies the accelerate – stop distance available for the conditions is:

a) b) c) 18.

12 500 Lbs.; 11 300 Lbs.; 9 700 Lbs. Climbing from 6000 feet pressure Altitude, OAT + 25ºC, to FL 270, OAT –31ºC at initial climb weight of 11 500 Lbs. Using graph 5-45 the Time, Fuel and Distance for the climb is: a) b) c)

19.

190 Lbs. 180 Lbs. 200 Lbs.

44 nm; 45 nm; 57 nm.

Cruising at FL 210, OAT –12ºc, the recommended cruise power according to graph 559 is: a) b) c)

20.

18 mins 15 mins 14 mins

1855 FT/LBS.; 1805 FT/LBS.; 1745 FT/LBS.

The range of the EE-20 aeroplane cruising at FL 280 with a headwind component of 35 Kts flying at recommended cruise power (graph 5-96) a) b) c)

954 nm; 863 nm; 782 nm.

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21.

Enroute from WPT 6 to WPT 7 at FL 270, Temperature ISA +15ºC, Distance 387 nm, Wind Component 35 Kts headwind, Aircraft Weight 11 500 Lbs., Recommended Cruise Power. The fuel used for the sector according to tables 5-54 and 5-55 is: a) b) c)

22.

787 Lbs., 824 Lbs., 863 Lbs.

An EE-20 aeroplane, Take-Off weight 12 175 Lbs., CG 188.7 inches, with 490 US Gallons of fuel in tanks, SW 6.7 Lbs/US Gallons flies from X to Y. If the trip fuel is 320 US Gallons the CG on landing at Y (table 6-14) is a) b) c) 23.

187.96 inches; 188.37 inches; 188.84 inches.

For a flight from A to B the Minimum Enroute Altitude is 19 500 feet where the OAT is –12º and the area QNH is 995.6 hPa. If the fuel used to the high ground 300 Lbs. the Maximum Take-Off Weight from A according to graph 5-24 is:

a) b) c) 24. a) b) c) 25.

11 600 Lbs.; 12 000 Lbs.; 12 400 Lbs. An obstacle 625 metres AMSL is 4 nm from zero of a runway whose elevation is 1050 feet. According to graph 5-28 the Minimum Climb Gradient required is: 3.5%; 4.0%; 4.5%. For and ILS approach to a runway at sea level, the altitude of an aircraft on the glide slope overhead the Outer marker inbound is published as 1300 feet.

If the IAS is 90 Kts with 40 % Flaps the altimeter reading over the Outer marker inbound on the glide slope should be. Use graph 5-15. a) b) c) 26. a) b) c)

1263 feet; 1278 feet; 1337 feet. According to graph 5-29, the indicated stall speed on the EE-20 aeroplane at 11 250 Lbs., Flaps Up and a 25º angle of bank: 99 Kts; 103 Kts; 107 Kts.

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ANNEX B LOAD SHEETS

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ANNEX C ANSWERS TO QUESTIONS

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Chapter 1 1 2 3 4 1.

2.

3.

C A C B

QNH 1025 11.8 x 30 = QNE 1013.2 QNH 995 18.2 x 30 QNE 1013.2 0600 Z 1400 Z

5327 ft 354 ft 4973 ft =

1075 ft 546 ft 1621 ft

DA 4055 ft DA 5748 ft

OAT +27°C

OAT +16°C

DA 7446 ft by calculator

DA 2116 ft by calculator

DA increased by 1693 ft

Chapter 2 1 2 3 4 5

C B A A A RWY 08

225 ft

RWY 26

185 ft

40 ft 1980 metre s x 3.28

× 100

= 0.62 %

40 ft 6

7

8 9 10 11

B RWY 10 RWY 28 01 19 15 33 19kts C A 09 a. b. c. d. 27 a. b. c. d.

431 ft 383 ft 48 ft

48 ft 1935 metre s x 3.28

× 100

= 0.76 %

90,800,324 90,800,290 3450,1250,2750 3450,1400,1900

1730 1380 1380 1380 2980 1380 2030 1380

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Chapter 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Chapter 5 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

B A C C C B B C C C A B C C B A C B A C B C A B

25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

B B B C C A C A B C B A C A A A C B B C C B C A B

Set 1

A B C A A B C A B A

Chapter 5

Set 2 1. C 2. B 3. A 4. A 5. C 6. A 7. B 8. B 9. A 10. B 11. A

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Chapter 6 PET Q’s 1. 2. 3. 4. 5. 6.

C B B B A A

1.

GS O GS H

2.

GS O 249 Kts GS H 233 Kts

3.

158 Kts 210 Kts

GSH 210 x 905 GSH210 + GSO158

= PET 516nm

GSH 233 x 1547 = GSH 233+ GSO 249

PET 748 nm at GS 249 PET 3 hours ETA 1215z

WC 45 HW GS 255

WC 60 HW GS ON 240

WC 45 TW GS R 345 GSH 345 x 750 GSH 345+ DSO 240

PET

TAS 300

= PET 442nm

4.

B 412 nm 75 nm

239 nm

X 337 nm A A to B B to C

C GS O 240 GS O 220

GS H 310 GS H 330

As A to B is the longer leg, the PET should occur before B. B to C X to A

Dist 239 nm

GSH 310 x 75 GSH 310 + GSO 240

=

GS O 220 GS H 310

Time 1.0864 Hours Time 1.0864 Hours Dist 337 nm

PET 42nm from X

PET 379 from A

Proof PET to A

379 nm

GS R 310

Time 1.2226 Hours

1:13:21 sec

PET to B B to C

33 nm 239 nm

GS O 240 GS O 220

Time 0.1375 Hours Time 1.0864 Hours 1.2239 Hours

1:13:26 sec

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5.

2 ENG TAS 280 1 ENG TAS 220

GS O 261 GS O 200

GS H 234 x 830 GS H 234 + GS O 200

6.

2 ENG TAS 220 1 ENG TAS 185

GS H 234 = PET 447.5 GS O 261 Time 1 hr 43 mins ETA 0943 Z

GS O 223 GS O 187

GS H 172 x 884 GS H 172 + GS O 187

GS R 172

=PET 423.5 GS O 223

Time 1 hr 54 mins ETA 1254Z

PNR QUESTIONS 1. 2. 3. 4.

C B A B

1.

GS O 275 GS H 214

GS H 214 x 4.25 HRS GS H 214 + GS O 275

= PNR 1.8599 HRS GS O 275 PNR = 512 nms

2.

GS O 374 GS H 256

GS H 256 x 4.5 HRS GS H 256 + GS O 374

= PNR 1.8286 HRS GS O 374 PNR = 684 nms

3.

.527 GNM/KG .423 GNM/KG 1GNM O +H

= = =

1.8975 2.3641 4.2616 7500KG

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4. A–B B–A

Distance 234 Distance 234

GS O 280 GS H 350

B–C C–B

Distance 289 Distance 289

GS O 270 GS H 360

C–D D–C

Distance 324 Distance 324

GS O 260 GS H 370

Endurance remaining D–E

GS O 250 GS H 380

GS H 380 _______________ GS O 250 +

GS H 380

T 0.8357 Hours T 0.6686 Hours ______________ T 1.5043 Hours T 1.0704 Hours T 0.8028 Hours ______________ T 3.3775 Hours T 1.2462 Hours T 0.8757 Hours ______________ T 5.4994 Hours T 0.5006 Hours X 0.5006 Hours = PNR 0.3019 Hours GS O 250 PNR 75nm

A – B 244nm + B – C 289nm + C – D 324nm + 75nm = PNR 922nm Sample Exam 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

C A A C C C B A A B B C B B B A C B C A C A B B A A

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