ATP

FLIGHT PLANNING

ATP Flight Planning FLIGHT TRAINING COLLEGE

ATP DOC 7
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 INDEX

ATP FLIGHT PLANNING

1. Planning Revision 01
2. Cruise Performance 13
3. Flight Plans 19
4. Graphs 37
5. Weight & Balance 43
6. Critical point 47
7. Point of No Return 49

Annex A Sample Exams 53

Annex B Answers to Questions 63

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.

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 CHAPTER 1

PLANNING REVISION

THE 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.

The Maximum Payload is :-

Least of 3 method

Find the least of: Max Take off Weight

Max Landing Weight + Trip Fuel
Max Zero Fuel Weight +Total Fuel
The question is:
An aircraft has the following weight schedule, find the take off weight and the maximum
payload that can be carried..

Max Ramp Weight 89 700kg

Basic Weight 47 000kg
Max Brakes Release Weight 89 350kg
Max Landing Weight 72 600kg
Max Zero Fuel Weight 63 500kg
Trip Fuel 12 462kg
Reserve Fuel 4680kg

Step 1
Calculate fuel at Brakes Release….

Trip Fuel 12 462kg

Reserve Fuel 4680kg
FUEL ON BOARD 17 142kg

Step 2

Calculate the 3 limitations…..

MTOW 89 350kg
MLW+trip fuel 85 062kg (72 600+12 462)
MZFW+fob 80 642kg (63 500+17 142) 

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Step 3

In this case the payload would be:

MZFW 63 500kg
- Basic weight 47 000kg
PAYLOAD 16 500kg

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
-Taxi Fuel
-Reserve Fuel
-Basic Weight
PAYLOAD

WEIGHT LIMITATIONS

The following weight limitations apply to the B747:

 Operating Empty Weight

 Maximum Zero Fuel Weight
 Maximum Ramp Weight
 Maximum Structural Take-Off weight
 Maximum Landing Weight

What is the amount of start up and taxi fuel?

Using the above figures found in the manual, what is the maximum amount of start up and
taxi fuel that can be carried?

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RUNWAY CONSIDERATIONS

The physical dimensions of the Runway, Stopway and Clearway have a very definite effect
on an aircraft's Maximum Take-Off Weight.

THE 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.

THE CLEARWAY

The Clearway is an area beyond the end of the runway, which complies with the following
criteria:

- must be at least 250 ft wide on either side of the extended centre-line;

- must be under airport control;
- must be clear of obstacles;
- may have a maximum upward slope of 1.25 %;
- the Clearway includes the Stopway (if available);
- the maximum usable Clearway for the purpose of calculations is 50 % of the runway
- length.

HOW THESE DIMENSIONS AFFECT THE MAXIMUM TAKE-OFF WEIGHT

Stopway 300ft Displaced threshold Stopway 200ft

03 21
100ft
600ft
End of runway Runway length 1950ft Clearway

TAKE-OFF RUN AVAILABLE (TORA)

The length of runway available for the ground run of an aircraft taking off.

Runway 03 : 1950 ft
Runway 21 : 1950 ft

TAKE-OFF DISTANCE AVAILABLE (TODA)

The length of the TORA plus the Clearway (if available).

NOTE: The Clearway includes the Stopway.

Runway 03 : 2550 ft
Runway 21 : 2250 ft

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TAKE-OFF DISTANCE REQUIRED (TODR)

This is the greater distance of :

The total distance required to reach a height of 35 ft at a speed not less than V2, after failure
of the critical power unit at V1 (also known as accelerate go distance required).
Or
115 % of the total demonstrated distance required to reach V2 at 35 ft over the runway with
all engines operative.

ACCELERATE STOP DISTANCE AVAILABLE (ASDA)

The length of the TORA plus the Stopway (if available).

Runway 03 : 2150 ft
Runway 21 : 2250 ft

ACCELERATE STOP DISTANCE REQUIRED (ASDR)

The Accelerate Stop Distance for either the engine out or all engine case, is the sum of the
following:
1. Acceleration distance from brake release to speed V1
2. Distance from V1 to application of full braking
3. Distance with full braking applied to stop. (one braking aid not considered)

 A recognition time, i.e. between the actual engine failure and recognition by the pilot
is built into the engine out Accelerate Stop Distance .
 Minimum time intervals between V1 and each subsequent braking action are given.
 The Accelerate Stop Distance required is the greater of either the engine out or all
engine Accelerate Stop Distances.

REFERENCE ZERO

This is the reference to which the co-ordinates of the various points in the take-off flight path
are referred. It is defined as the end of the Take-off Distance Required and is 35 feet below
the flight path at this point.

LANDING DISTANCE AVAILABLE (LDA)

The length of the runway, which is declared to be available and suitable for a landing. The
landing distance available commences at the threshold and in most cases corresponds to the
physical length of the runway. HOWEVER, if the threshold is displaced, the landing distance
available is reduced and will normally be the distance between the threshold and the far end
of the runway.

Runway 03 : 1850 ft
Runway 21 : 1950 ft

BALANCED FIELD

The balance field occurs when the accelerate stop distance required equals the engine out
distance to 35ft.

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DECISION SPEED

The V1 speed is the speed which the pilot used as a reference in deciding whether to
continue or to abort the take off .

V1 is selected so that:
1. If the take-off is continued after an engine failure recognized at or above V1
speed it will be possible to attain a height of 35ft within the scheduled take off
distance, or
2. If the take-off is abandoned at or below the V1 speed it will be possible to
bring the aircraft to a stop within the scheduled accelerate stop distance.

The scheduled take off field lengths are based on stopping if an engine failure is recognized
before reaching V1 and on continuing the take off if the engine failure is recognized after V1.

An appropriate time interval is allowed between recognition of the failure and completion of
all actions required for rejection. It is also required that one of the available braking systems
must not be used during certification or if all braking devices are used that an additional
factor be applied.

RUNWAY SURFACE CONDITIONS

If the runway surface is contaminated by, for example, water or snow, the aircraft will require
more runway length to reach take-off 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, possibly a reduction of the take-
off 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 rejected 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.

Working out runway slope:

Munich Germany, runway 08R

08 end elev. 1486ft

26 end elev. 1498ft
Runway length 4000m.

What is the slope on Rwy 08R?

WIND COMPONENT AT TAKE-OFF

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.

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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.

THE TAKE-OFF FLIGHTPATH

This will be that portion of the flightpath, which begins at a height of 35ft above the runway
i.e. the end of the Take-Off Distance Required. It should be assumed that this height is
reached at the end of TODR, with one engine inoperative. The Take-Off Flightpath ends
when the aircraft is 1500ft AGL or clear of the last obstacle.

THE NET TAKE-OFF FLIGHTPATH

This represents the Gross Take-Off Flightpath, reduced by the amount considered necessary
to allow for various contingencies, i.e. the need to manoeuvre, unavoidable variations in
piloting technique, temporary below average performance and take-off on wet or slushy
runways, etc.

This amount considered necessary:

0.8 % for aeroplanes with 2 power units; (reduction on climb gradient)

0.9 % for aeroplanes with 3 power units;
1.0 % for aeroplanes with 4 power units;

The Net Take-Off Flightpath should clear en-route obstacles by at least 35 ft with one engine
inoperative.
V2

First lets look at brakes release to 35ft…

Vmcg Vef V1 V1 (MBE) VR Vmu Vlof

V1 Decision Speed Vr Rotation speed V2 Initial climb out

speed

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 1st Segment 2nd Segment 3rd Segment 4th Segment

Gross
Gear Up
400ft
Net

1500ft

FIRST SEGMENT

Critical engine inoperative.

Take-off thrust.
Landing gear extended.
Flaps in take-off position.
Take-off speed V2.
Minimum climb gradient 0.5 %.

SECOND SEGMENT

Critical engine inoperative.

Take-off thrust.
Landing gear retracted.
Flaps in take-off position.
Take-off speed V2.
Minimum climb gradient 3.0 %.
Extends to a height of at least 400 ft.

THIRD SEGMENT

Critical engine inoperative.

Take-off thrust.
Landing gear retracted.
Flaps transitioning to retracted.
Climb gradient required, nil.

FOURTH SEGMENT

Critical engine inoperative.

Take-off thrust.
Landing gear retracted.
Flaps retracted.
Minimum climb gradient 1.7 %.
This sector leads up to a height of 1500 ft above the runway, or until the final
obstacle is cleared.

NOTE:
The gradients quoted are for a four engined jet aeroplane with one engine inoperative.

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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. 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 a rejected take-off. Although the brake
energy limit (known as Vmbe) 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 a rejected 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.

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 definite 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 that will determine the aircraft's actual take-off weight.

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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 = pr²

Where r = the radius of the circular object.

To perform the actual floor load calculation, use the formula:

MAX WEIGHT = AREA x MAX FLOOR LOAD

EXAMPLE:

The maximum floor load of an aircraft's cargo hold is 700 Kg/m². What is the
maximum weight that can be loaded onto a pallet that measures 2 m by 3 m?

MAX WEIGHT = AREA x MAX FLOOR LOAD

= (2 m x 3 m) x

= 4200 Kg

Care should always be taken to perform calculations using common units of

measurement.

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DEFINITIONS

The following is a list of definitions relevant to the ATP syllabus:

Va : Design manoeuvre speed. The maximum speed at which full

application of controls can be made.

Vf : 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.

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QUESTIONS

1. Maximum Take-Off Weight : 377 000 kg

Maximum Landing Weight : 285 000 kg
Maximum Zero Fuel Weight : 263 000 kg
Operating Empty Weight : 174 000 kg
Total distance : 3200 nm
Average TAS : 500 kts
Average fuel flow : 12000 kg/hr
Reserve fuel : 22 000 kg

a) Determine the maximum payload in still air conditions.

a. 89 000kg
b. 88 000kg
c. 86 000kg

b) Determine the maximum payload in an average W/C of -50 kts for the flight.
a. 89 000kg
b. 88 000kg
c. 86 000kg

2. Using the same data as in (A), beyond which value of headwind would the payload
have to be reduced?
a. 84kts
b. 82kts
c. 83kts

3. A flat-bottomed 44 imperial Gallon drum with a diameter of 56 cm is to be loaded into

an aircraft. The SG of the fuel is 0.75. The mass of the drum is 20 kg. The
maximum floor load of the aircraft is 700 Kg/m². Can the drum be loaded?
a. Yes
b. No

4. FL 260, M 0.76, TAS 450 kts. What is the temperature deviation?

a. -48c
b. -5c
c. -37c

5. FAGC elevation 5327 ft, QNH 1007.4 hPa. Temperature + 24°C. What is the density
altitude?
a. 2400ft
b. 7903ft
c. 5503ft

6. When does third segment climb end?

a. Flaps up
b. V2
c. Landing gear up

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7. The following details apply to a certain runway 15/33. Runway length 9200ft.
Runway 15 has a displaced threshold of 400ft. Runway 33 has Stopway 250ft and
clearway 600ft.

What is the landing distance available on runway 15?

a. 9200ft
b. 8800ft
c. 9600ft

8. With reference to question 7 above, what is the take-off distance available on runway
33?
a. 8600ft
b. 9200ft
c. 9800ft

9. Plan a flight from A-B. Fuel is prohibitively expensive at B, so plan not to uplift before
returning to A. Plan to carry minimum fuel on departure at A, because fuel isn’t cheap
there either

OEW 174 000 kg

MZFW 263 000 kg
MTOW (Structural) 377 000 kg
MLW 285 000 kg
Distance A-B 1200 NM. F/F constant 10 000 kg/hr
W/C A-B is -50 KTS along track. TAS 500 kts.
Carry 1000 KG for an approach at each point (B and A).
Contingency Fuel 4 % for each leg.

What is the maximum payload that can be uplifted from A?

a. 87 309kg
b. 51 358kg
c. 52 425kg

10. The following dimensions apply to runway 09/27.

Runway length 2000 m.

09 : Displaced threshold 200 m Stopway 300 m. Clearway 580 m.

27 : Stopway 250 m. Clearway 350 m.

What is the take-off distance available on runway 09?

a. 2000m
b. 1420m
c. 2580m

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 CHAPTER 2

CRUISE PERFORMANCE

The aim of this chapter is to illustrate the various methods of expressing and calculating an
aircraft's performance. In doing so, students should refresh their memories concerning the
use of the 1/x function on their calculators. The use of this function can greatly reduce the
workload during performance problems.

NOTE: You are reminded to be aware of the difference between Specific Gravity (SG) which
relates to 1 imperial gallon or 1 litre of water and Specific Weight (SW) which relates to the
US gallon only.

BASIC PERFORMANCE

The three most common expressions of an aircraft's performance are:

TAS - which should be written as

G/S - which should be written as

FUEL FLOW - which can be expressed as either :

volume of fuel per hour and written as

or weight of fuel and written as

RANGE PERFORMANCE

An aircraft's range is defined as:

The amount of distance that can be flown for a given amount of fuel.

Distance can be expressed in either ANM or GNM.

Fuel can be expressed in either volume - LITRES/US GAL/IMP GAL

or in weight - KG/LBS

The combination of two factors, distance and fuel, gives rise to numerous expression
of range performance.

E.g. :

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PERFORMANCE CALCULATIONS
Performance calculations involve the following format:

Given a series of performance expressions, select two, to mathematically create a new

performance expression.

EXAMPLE:

TAS 180 kts, G/S 160 kts, F/F 400 Lbs/hr.

Express the aircraft's performance in :

i) ANM/LBS

ii) LBS/GNM

SOLUTION

i) Write down the series of performance expressions as follows:

Write down the new performance expression that is required:

From the series of performance expressions, select two which will satisfy the
requirements in terms of units:

Mathematically combine the two selected expressions by division or multiplication to

derive the new performance expression:

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ii)

IN CLASS EXAMPLES

1. Given: TAS 240 kts WC +30 kts TW GS 270 kts Fuel Flow 750 Lbs/Hour

GS 270 Kts
Then aircraft performance is ? = 0.36 GNM / LB
Fuel Flow 750 Lbs / Hour
OR
Fuel Flow 750 Lbs / Hour
= 2.7778 LB / GNM
GS 270 Kts

2. 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 :-

3 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 :-

4 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 GS

48.36 anm/1000kg x 46.06gnm/1000kgs = 461kts GS

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QUESTIONS

1. Track 058° (T), W/V 240/20, TAS 240 kts, Fuel flow 1500 LBS/Hr

Calculate performance expressions for:

a) GNM/LB
b) LBS/ANM

2. TAS 220 kts, Performance 0.2 GNM/LB, Performance 6.0 LBS/ANM.

Calculate:

a) The fuel flow

b) The ground speed

3. TAS 210 kts, Performance 6.5 LBS/GNM, Fuel flow 166.5 US GAL/HR, SG 0.8.

Calculate:

a) G/S
b) The wind component

4. Cruising at FL 160, TAS 220 kts, F/F 1500 LBS/HR, Track 238° (T), W/V 040/20.
At FL 180, the TAS will be 230 kts and the fuel flow only 1400 LBS/HR, but the W/V
360/05. Will it be worthwhile to climb to FL 180?

5. From the following data, calculate:

a) The best FL to cruise at.

b) The maximum range with a total of 7400 LBS fuel if a 1 hour reserve is
required at destination.

FL TAS W/C F/F (LBS/HR)

150 210 +8 1500
170 230 -2 1400
190 240 -26 1300

6. On a flight from A - B, distance 800 nm, the aircraft has 900 USG of fuel (SG 0.79)
excluding reserves. What is the maximum TAS at which the aircraft can fly the route
if there is a constant wind component of -10 kts for the route?

The following actuals apply:

TAS F/F

220 1500 LBS/HR

240 1800 LBS/HR
280 2200 LBS/HR

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7. TAS 500 kts, W/C -20 kts, F/F 12 000 kg/hr. What is the maximum range that the
aircraft can achieve with 92 000 kg fuel in the tanks and a minimum reserve
overhead destination of 22 000kg?

8. TAS 500 kts, F/F 4743 USG/hr, SG 0.67, performance 0.037 GNM/kg. What is the
wind component?

9. FL 310, TAS 501kts, W/C -41 Kts, F/F 9956 kg/hr

FL 350, TAS 505 kts, F/F 9050 kg/hr.

Beyond what value of wind component would it not be economical to climb to FL

350?

10. FL 290 TAS 501kts, performance 94.8 ANM/1 000 kg, Zero wind.

FL 350 TAS 490kts, performance 108.4 ANM/1 000 kg.

Beyond what value of headwind would it not be economical to climb to FL 350?

11. A-B B-C

TAS 492 TAS 492
W/C -12 W/C -18
DIST 506 NM DIST 481 NM
FF 12 200 kg/hr FF 12 800 kg/hr

Determine the aircraft’s performance in ANM/1000 KG from A-C.

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 CHAPTER 3

FLIGHT PLAN

The CAA requirement for this section is to perform some, or all of, the following calculations,
using the B747 manual:

 A climb.
 A cruise sector (normally two).
 A descent.
 A diversion to alternate.
 A hold.

In essence, this is almost the same as completing an entire flight plan except -

 no blank pro-forma is provided;

 the questions are isolated and do not relate to one another.

One of the most important points to bear in mind when answering these questions, is the
sequence of fuel used throughout a flight:

RAMP WEIGHT
- TAXI FUEL (600 kg)
TOW
- TRIP FUEL
- 4 % (Standard contingency or reserve)
- APPROACH FUEL (Standard 1 100 kg)
Landing Weight at destination or overshoot
- ALTERNATE FUEL
- 4 % of ALTERNATE FUEL (Standard contingency or reserve)
Landing Weight at Alternate or Enter Hold
- HOLDING FUEL
- MINIMUM IN TANKS (Standard 1 600 kg)
ZFW

OPTIMUM ALTITUDE CALCULATION

Refer graph:

Using the graph, find the Optimum altitude if the TOM is 340 000kg, and the aircraft will
cruise at Long Range Cruise (LRC) 30 800ft.

Find the optimum altitude for a flight from Luxemburg to Munich, distance is 236nm, the TOC
weight is estimated to be 245 600kg, and the temperature deviation is expected to be ISA
+15c. Find the optimum altitude under these conditions for this flight… 27 500ft

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CLIMB CALCULATIONS

Reference page 2-4-1 to page 2-4-8.

In order to calculate a climb the following data must be given:

 The aircraft TOW.

 The FL to which you wish to climb.
 The temperature at that FL.
 The elevation of the departure airport.

The tables will give you:

- Time to climb.
- Fuel to climb.
- TAS in the climb.

All from SEA LEVEL to the chosen FL. At the bottom of each page is a fuel correction table
for departure airport elevation.

EXAMPLE:

Departure airport PA 6000ft. Climb to FL 330 TOW 340 000 kg.

Temperature at FL 330 is ISA + 10C.

Refer to page 2-4-3 for ISA + 10°C.

Against TOW of 340 000 kg and FL 330, read off:

- 22 MINS to climb.
- 8 950 kg of fuel to climb (from sea level)
- 142 NAM to climb
- 448 Kts TAS in the climb

From the bottom of the page adjust the climb fuel for the departure airport elevation.

6 000ft = -850 kg of fuel

8 950 kg - 850 kg = 8 100 kg of fuel to climb from 6 000ft to FL 330.

Further data that can be calculated for a climb sector: If the CAA offers the true track
and wind for the climb.

Refer to page 1-2-2 for drift angles and wind components. The left hand table is for climb,
descent and diversion and the right hand table is for the cruise. The vertical axis of the table
is for the angular difference between the wind and the track. The horizontal axis is for the
wind speed.

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Relating to the previous example, if the true track for the climb is 360° and the wind is
060/40.

The angular difference between the wind and the track is 60°. Look this up on the left-hand
table. The 60 is listed with 300 because a wind 60° off track is the same as a wind 300° off
track, except the one will drift you left and the other will drift you right. Now look up 60°
against 40 Kts wind speed.

Therefore the drift is

The wind component is

- 5° is the drift (to the left)

- 22 Kts is the wind component (head-wind)

So by applying the above:

The groundspeed is kts

The GNM for the climb is kts

The true heading is 

This completes the data that can be calculated for a climb. The most popular CAA question
for the climb is only the fuel calculation, but with a correction for departure airport elevation.
Remember that even though it may be offered, the wind will have no effect on either the fuel
or the time to climb.

EXAMPLE:

TOW 336 500 kg. Climb to FL 310. Departure airport elevation 5 500ft
ISA + 15°C. Track 165° (T). Wind 070/60.

Because of the TOW, this example requires interpolation to derive the answers.

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CRUISE CALCULATIONS

Reference page 3-2-1 to page 3-2-6(a).

The first set of tables is for a M0.84 cruise and the second set of tables is for a Long Range
Cruise (LRC). The vertical axis of the table is for the gross weight of the aircraft and the
horizontal axis is for the flight level at which the aircraft is cruising.

The graphs will give you: - the TAS;

- the total fuel flow.

This means that to use the aircraft weight at the beginning of a cruise sector would be to
incur an unfair penalty in terms of fuel flow. Likewise, to use the aircraft weight at the end of
a cruise sector would be too generous in terms of fuel flow. Instead, the weight halfway
through the cruise sector should be used as being most representative of the aircraft weight
throughout the cruise sector. This weight is called the mid-zone weight (MZW).

An MZW should always be derived twice in order to refine its accuracy. The first MZW is
called the approximate MZW and the second MZW is called the actual MZW.

In order to calculate a cruise sector, the following data must be given:

- cruise speed (MO.84 or LRC);

- flight level;
- temperature;
- true track and distance for the sector;
- wind;
- aircraft gross weight at the commencement of the cruise sector.

EXAMPLE:

LRC ISA + 10°C FL 310. Aircraft weight at the commencement of the cruise sector 290
000 kg. Sector distance 620 NM. True track 020°. Wind 160/70.

Refer to page . 3-2-5(a). Against FL 310 and 290 000 kg, read off.- TAS 501kts.
- FF 11 408 kg.

Now determine the wind component and drift angle from table 1-2-2. (Right-hand table for
the cruise.) The angle between the track is 140° (vertical axis) and the wind speed is 70 Kts
(horizontal axis).

- Drift is 5°.
- Wind component + 52 Kts (tailwind).

Now do the computations for the cruise leg:

TAS 501 Kts + 52 Kts = 553 Kts G/S

DISTANCE 620 NM  553 Kts G/S = 1 HR 07 MINS
1 HR 07 MINS x FF 11 408 KG/HR = 12 790 KG FUEL BURN
12 790 KG  2 = 6 395 KG = 283 605 KG A/C WEIGHT AT HALFWAY OR
APPROXIMATE MZW.

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Now using the approximate MZW, do the computations for the cruise sector again to
determine the actual MZW.
Against FL 310 and 283 605 kg

283 605 KG - 280 000 KG

= 0,3605 INTERPOLATION FACTOR
290 000 KG - 280 000 KG

TAS : 501 Kts - 497 Kts = 4 Kts

4 Kts  0.3605 = 1.442 Kts
497 Kts + 1.442 Kts = 498.442 Kts

FF : 11 408 KG/HR - 11 004 KG/HR = 404 KG/HR

404 KG/HR x 0.3605 = 146 KG/HR
11 004 KG/HR + 146 KG/HR = 11 150 KG/HR

TAS 498.442 Kts + 52 Kts = 550.442 Kts G/S

Distance 620 NM  550.442 Kts = 1 HR 08 MINS
1 HR 08 MINS x 11 150 KG/HR = 12 559 KG FUEL BURN
12 559 KG  2 = 6 279 KG FUEL TO HALFWAY
290 000 KG - 6 279 KG = 283 721 KG A/C weight at halfway or actual MZW

Now using this actual MZW, do the computations for the cruise sector one last time to
derive the CAA answer.
Against FL 310 and 283 721 kg.

283 721 KG - 280 000 KG

= 0,3721 INTERPOLATION FACTOR
290 000 KG - 280 000 KG

TAS : 501 KTS - 497 KTS = 4 KTS

4 KTS  0.3721 = 1.4884 KTS
497 KTS + 1.4884 KTS = 498.4884 KTS

FF : 11 408 KG/HR - 11 004 KG/HR = 404 KG/HR

4 KTS  0.3721 = 150 KG/HR
11 004 KG/HR + 150 KG/HR = 11 154 KG/HR

TAS 498.4884 KTS + 52 KTS = 550.4884 KTS G/S

DISTANCE 620 NM  550.4884 KTS = 1 HR 08 MINS
1 HR 08 MINS  11 154 KG/HR = 12562.4 KG FUEL BURN

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As the aircraft continues on its flight, it becomes progressively lighter as fuel is burned off. At
a certain point, the aircraft will be light enough to make a climb to the next higher flight level
feasible.

Refer to page 2-3 (Altitude Capability).

The first column gives the flight level, the second column gives the cruise speed and the
fourth column gives the weight at which a climb to that flight level is optimum.

With reference to the previous example, the aircraft, at LRC, could climb from FL 310 to FL
350 when the weight is 281 000 kg or less.

A very simple method is used for calculating this en-route climb: assume 100 KG FUEL
BURN PER 1 000ft CLIMB. Subtract the climb fuel from the weight at the commencement
of the sector to derive a new weight at the commencement of the sector i.e. assume the
aircraft went vertically up at 100 kg/1 000ft but with no distance or time used up on that
cruise sector at the new flight level.

The typical example of this type of question would be:

EXAMPLE:

M0.84. FL 330 climbing to FL 370. Cruise sector 496 NM. Temperature at FL 370 ISA
+15C. Wind 260/60. True track 140°. Weight at commencement of the cruise sector
260 000 kg.

FL 370 - FL 330 = 4 000ft

4 000ft  100 KG/1 000ft = 400 KG
260 000 KG - 400 KG = 259 600 KG weight at commencement of cruise sector.

Refer to page 3-2-3.

259 600 KG - 250 000 KG

= 0,96 INTERPOLATION FACTOR
260 000 KG - 250 000 KG

Refer to page 1-2-2 to determine the W/C and drift angle:

260° - 140° = 120° at 60 KTS

6° left drift
+ 27 KTS W/C

TAS : 498 KTS

F/F : 10 048 KG/HR - 9 704 KG/HR = 344 KG/HR
344 KG/HR  0.96 = 330 KG/HR
9 704 KG/HR + 330 KG/HR = 10 034 KG/HR

TAS 498 KTS + 27 KTS = 525 KTS/GS

DISTANCE 496 NM  525 KTS = 57 MINS
57 MINS  10 034 KG/HR = 9 480 KG FUEL BURN
9 480 KG  2 = 4 740 KG FUEL BURN TO HALFWAY
259 600 KG - 4 740 KG = 254 860 KG APPROXIMATE MZW

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 254 860 KG - 250 000 KG
= 0,486 INTERPOLATION FACTOR
260 000 KG - 250 000 KG

TAS 498 KTS and G/S 525 KTS remain unchanged, therefore, time remains 57 mins.
F/F : 10 048 KG/HR - 9 704 KG/HR = 344 KG/HR
344 KG/HR  0.486 = 167 KG/HR
9 704 KG/HR + 167 KG/HR = 9 871 KG/HR
9 871 KG/HR  57 MINS = 9 326 FUEL BURN
9 326 KG  2 = 4 663 KG FUEL TO HALFWAY

259 600 KG - 4 663 KG = 254 937 KG ACTUAL MZW

254 937 KG - 250 000 KG

= 0,4937 INTERPOLATION FACTOR
260 000 KG - 250 000 KG

TAS, G/S and time remain unchanged.

F/F : 10 048 KG/HR - 9 704 KG/HR = 344 KG/HR

344 KG/HR  0.4937 = 170 KG/HR
9 704 KG/HR + 170 KG/HR = 9 874 KG/HR F/F
9 784 KG/HR  57 MINS = 9 328 KG FUEL BURN PLUS
400 KG IN THE CLIMB = 9 728 KG FUEL BURN FOR THE SECTOR

Although not a common occurrence it may happen that the aircraft has to descend to a new
flight level before continuing in the cruise (perhaps due to changing track requiring a change
from odds to evens or vice versa or perhaps due to conflicting traffic joining the same
airway).

Should the CAA pose such a question in the exam, the solution is simple - pretend it never
happened, just continue the cruise at the new flight level with the descent having no effect
whatsoever on time, speed or fuel for that sector.

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TIME TO DESCEND….
Reference page 4-1.

In order to calculate a descent, the following data must be given to you:

- Flight Level from which you are descending;

- The speed at which the aircraft will descend. There are 4 different types of
descent that the table provides for:

M0.84/290 KTS/250 KTS or M0.84/320 KTS/250 KTS or

M0.86/340 KTS/250 KTS or M0.88/340 KTS/250 KTS.

The table will give you:

- Time to descend;
- Fuel to descend;
- Distance (NAM) to descend.

The effect of destination airport elevation on time, fuel and distance to descend may be
ignored.

EXAMPLE: Descend from FL 370 at M0.84/290/250.

Time to descend : 25 minutes

Fuel to descend : 1 550 kg
Distance to descend : 140 NAM
Average TAS in the descent : 140 NAM  25 mins = 336 Kts.

If the CAA offers you the true track and average wind in the descent then:

- Derive the wind component from page 1-2-2;

- TAS  W/C = G/S in the descent.
- G/S  TIME = GNM to descend.

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DIVERSION TO AN ALTERNATE…
Reference page 2-2 and page 3-5-5.

In order to calculate a flight to alternate, the first step must be to select a suitable flight level.
For this purpose, use the bottom graph on page 2-2.

The following data must be given to you in order to use this graph:

- The trip distance to alternate.

- The approximate temperature at the higher levels and whether the flight is
ODDS or EVENS.
- The take off weight. This weight may not be given to you directly and require
a small calculation

EXAMPLE:

Distance to alternate 325 NM. Forecast temperature for the upper levels ISA + 15°C.
TOW 315 000 kg. Trip fuel 106 000 kg. ODDS sector.

First calculate the overshoot weight TOW 315 000 kg.

- 106 000 kg (trip fuel)

- 4 240 kg (4 % contingency or reserve)
- 1 100 kg (standard approach at destination)
203 660 kg

Now, enter the graph with the distance, temperature and weight:

- Answer 37 600ft.
- Therefore, choose FL 370.
If this had been an evens sector, choose FL 350 (the next one lower), not FL
390, as it would be above optimum.

Now, in order to calculate the time and fuel to alternate, use the graph on page 3-5-5.

EXAMPLE:

As above, but also given a wind component of -50 Kts (headwind).

First calculate the landing weight at alternate - a catch 22 question, because you are trying
to calculate the fuel to alternate. .

TOW 315 000 kg

- 106 000 kg (trip fuel)
- 4 240 kg (4 %)
- 1 100 kg (approach)
203 660 kg (overshoot weight)
- XXXX kg (fuel to alternate)
- XXXX kg (4 % of alternate fuel)
XXXX kg (landing weight at alternate

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In order to perform this calculation, a standard must be used. The standard is:

TOW 315 000 kg

- 106 000 kg (trip fuel)
- 13 000 kg (standard figure)
196 000 kg (landing weight at alternate)

Enter to graph with this weight:

- 8 700 kg fuel to alternate.

Continue up to the all weights line:

- 0.865 HRS (52 MINS) time to alternate.

As can be seen, the standard figure of 13 000 kg is a sensible one, because now with the
benefit of having completed the question:

TOW 315 000 kg

- 106 000 kg (trip fuel)
- 4 240 kg (4 %)
- 1 100 kg (approach)
203 660 kg (overshoot weight)
- 8 700 kg (alternate fuel)
- 348 kg (4 %)
194 712 kg (landing weight at alternate)

Using the standard 13 000 kg, we derived a landing weight of 196 000 kg which is
certainly close enough.

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HOLDING AT ALTERNATE
Reference page 4-3. Always use the top table (flaps up) unless otherwise stated. In order to
use this table, the following data must be give:

- The flight level for the hold.

- The time for which you wish to hold (normally a standard 30 mins).
- The aircraft weight in the hold.

As can be seen from the table, as the aircraft weight reduces in the hold, so the fuel flow
decreases, in other words, a holding MZW must be calculated in order to derive the fuel flow
in the hold. Once again, a standard figure is used. This standard figure assumes that a
standard 30 minutes hold would correspond to a fuel burn of 4 400 kg, therefore, halfway
through the hold, the aircraft would have burned off 2 200 kg.

EXAMPLE: TOW 315 000 kg. Trip fuel 106 000 kg. Fuel to alternate 8 700 kg. Hold
for 30 mins at FL 150.

First calculate the holding MZW

TOW 315 000 kg

- 106 000 kg (trip fuel)
- 4 240 kg (4 %)
- 1 100 kg (standard approach)
203 660 kg (land at destination or overshoot)
- 8 700 kg (fuel to alternate)
- 348 kg (4 %)
194 712 kg (land at alternate or enter hold)
- 2 200 kg (standard fuel for ½ hold)
192 512 kg (holding MZW)

Enter the table with this weight and FL 150:

192 412 KG - 190 000 KG

 1 870 KG / HR - 1 780 KG / HR
200 000 KG - 190 000 KG

= 21.7 KG/HR + 1 780 KG/HR = 1801.7 KG/HR

1801.7 KG/HR PER ENGINE  4 ENGINES = 7 206.8 KG/HR
7206.8 KG/HR  ½ HOUR (30 MINS) = 3603 KG

The CAA could also have given the same question except requiring the holding MZW to be
calculated in reverse.

EXAMPLE: As above except given ZFW 188 612 kg.

First calculate the holding MZW.

ZFW 188 612 KG

+ 1 600 KG (minimum in tanks)
+ 2 200 KG (standard fuel for ½ hold)
192 412 KG (holding MZW)

Clearly the same answer would be derived as before.

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DESTINATION HOLD
Although not part of the flight planning process at all, it is a very popular CAA question.

Reference: page 4-3.

EXAMPLE:

The B747 arrives overhead destination with 23 500 kg fuel in tanks. ZFW 222 000 kg.
Fuel to alternate 8 000 kg. Contingency 320 kg. Fuel to hold at alternate 4 000 kg. How
long can the aircraft legally hold overhead destination at FL 250, before it has to divert to
alternate?

First determine how much of the 23 500 kg the aircraft has to keep in tanks to satisfy the
Law.

1 100 kg must remain for an approach at destination

+ 8 000 kg must remain to reach the alternate
+ 320 kg must remain for legality (4 % contingency)
+ 4 000 kg must remain in case the aircraft has to hold at alternate as well
+ 1 600 kg will remain as minimum in tanks
15 020 kg must not be used

Now determine how much fuel is available for the hold overhead destination.

23 500 kg - 15 020 kg = 8 480 kg

Now determine the holding MZW.

ZFW 222 000 kg

+15 020 kg
237 020 kg weight just before diverting to
alternate i.e. at the end of the
hold over destination.

If 8 480 kg is available for the hold, then ½ of this plus the weight at the end of the hold will
give the holding MZW.

237 020 kg + 4 240 kg = 241 260 kg MZW

Enter the flaps up table with this weight at FL 250:

241 260 KG - 240 000 KG

 2 170 KG - 2 080 KG = 11,34 KG
250 000 KG - 240 000 KG

2 080 KG + 11.34 KG = 2 091.34 KG PER ENGINE PER HOUR

2 091.34 KG  4 = 8 365.36 KG TOTAL F/F PER HOUR
8 480 KG (AVAILABLE FUEL)  8 365.36 KG/HR = 1 HR 01 MIN

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THE SIMPLIFIED FLIGHT PLAN

The function of a simplified flight plan is the same as that of a full flight plan, albeit
abbreviated.

The climb plus all the cruise sectors plus the descent, are all found on one graph.
The diversion to alternate is found on the same graph as the full flight plan.
The holding fuel need not be calculated because it is an assumed standard for
simplified flight planning i.e. 30 MINS and 4 400 kg.

If the holding fuel need not be calculated and the alternate time and fuel are calculated in the
same way as per the full flight plan, there is only one question that can be asked:
Determine the trip time and fuel for a particular flight.

Reference: pages 3-5-1 to 3-5-4.

Graphs 3-5-1 and 3-5-1(a) are for a M0.84 cruise at a constant flight level.
Graph 3-5-2 is for a 350 Kts IAS short distance cruise.
Graphs 3-5-3 and 3-5-3(a) are for LRC at a constant flight level.
Graph 3-5-4 is for a cruise with a step climb to optimum altitude.

EXAMPLE:

Trip distance 3500 nm. Mean W/C for the trip -50 Kts. Brake release weight 310 000 kg.
Mean temperature deviation for the trip ISA + 10°C. Determine the trip fuel and time for a
step climb to optimum altitude, graph 3-5-4.

Trip fuel = 84 000 kg.

Using the all weights line, the trip time = 8 hrs

EXAMPLE:

Trip distance 3500 nm. Mean W/C for the trip -50 Kts. Brake release weight 310 000 kg.
Mean temperature deviation for the trip ISA + 10°C. M0.84. FL 310. Determine the trip fuel
and time using graph 3-5-1(a).

Part of the graph requires the landing weight (at destination),a catch 22 situation because the
landing weight is dependant on the trip fuel which we are trying to determine. This is the
solution:

Assume a standard TAS of 500 Kts.

Assume a standard F/F of 12 000 kg/hr.

Now:
TAS 500 KTS - W/C 50 KTS = G/S 450 KTS
DISTANCE 3 500 NM  G/S 450 KTS = TIME 7 HRS 47 MINS
7 HRS 47 MINS  F/F 12 000 KG/HR = 93 333 KG TRIP FUEL
TOW 315 000 KG - 93 333 KG = 221 667 KG LANDING WEIGHT

Use this weight on the graph to determine the trip fuel and time.
- Trip fuel = 90 000kg
- Trip time = 8hrs

The same procedure would apply for a LRC and a 350 KTS IAS cruise.

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RE-DESPATCH FLIGHT PLAN

Re-despatch flight planning is a method of increasing an aircraft's payload by eliminating a

portion of the 4 % contingency fuel.

METHOD

On the flight from A - B - C, with alternate at D, the following fuel must be carried.

A - B - C Burn-off
A-B-C 4%
Approach fuel (1100 kg)
C - D Burn off
C-D 4%
30 minutes holding fuel
Taxi fuel (600 kg)
Minimum in tanks (1600 kg)

On this flight the fuel requirement is so great that the payload would be significantly limited,
perhaps to the extent of the flight being unprofitable. The alternative would be to carry less
fuel and to make and en-route stop for refuelling.

This en-route stop is called point E, with alternate F.

On the flight from A - B - E, with alternate F, the following fuel must be carried :

A - B - E Burn-off
A-B-E 4%
Approach fuel (1100 kg)
E - F Burn-off
E-F 4%
30 minutes holding fuel
Taxi fuel (600 kg) Minimum in tanks (1600 Kg)

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THE RE-DESPATCH FLIGHT PLAN

Point C is called the Scheduled Destination.

Point E is called the Inflight Diversion.

Point B is called the Re-despatch point.

Calculate flight A - B - E - F in the normal fashion.

Calculate flight A - B - C in the normal fashion, except : eliminate the 4 % fuel from A-B.

Eliminating the 4 % fuel for leg A - B means that more payload can be carried.

If the flight now reaches point B, the re-despatch point, and the fuel remaining in the tanks
indicates that only the B/O has been used (as in fact should be the case), the flight may
legally continue to point C with not only B/O, but also 4 % in the tanks.

If, however, the flight reaches point B, and the fuel remaining in the tanks indicates that the
aircraft has used more than just the B/O (perhaps due to inaccurate forecast winds), it will
have to set course for point E and again, not only have B/O, but also 4 % in the tanks.

Should it occur that the fuel for flight A - B - C - D, even with a portion of the 4 % contingency
eliminated, is greater than the fuel for the flight A - B - E - F, this amount of fuel must
obviously be carried out would be carried on flight plan A - B - E - F as extra fuel.

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QUESTIONS

FLIGHT PLANNING

1. TOW 312 000 kg. Departure airport elevation 5 558ft. ISA + 12°C. Calculate the fuel
required to climb to FL 330.

2. TOW 307 600 kg. Departure airport elevation 3 700ft. ISA + 8°C. Calculate the fuel
required to climb to FL 310.

3. TOW 296 000 kg. Departure airport elevation 2 777ft. ISA + 4C. Climb track 340T.
Wind 270/70. What is the ground distance to TOC FL 330?

4. Aircraft weight 287 500 kg. ISA + 14C. Sector distance 521 nm. Track 328T. Wind
270/70. Calculate the fuel required for this sector at FL 310 and M0.84.

5. Sector distance 612 nm. LRC. FL 310 climbing to FL 350. Track 272T. Wind
300/40. Aircraft weight at the beginning of the sector 271 100 kg. ISA + 10°C.
Calculate the fuel required for this sector.

6. FL 310. ISA + 12°C. Track 250T. Wind 230/60. M0.84. Sector distance 447 nm.
Aircraft weight at the beginning of this sector 313 370 kg. What is the fuel flow on this
sector?

7. FL 330. ISA + 12°C. Track 115°T. Wind 250/55. Sector distance 405 nm. Weight
at the beginning of this sector 310 070 kg. What is the fuel required for this sector?

8. FL 310. ISA + 13°C. Track 275°T. Wind 210/40. LRC. Sector distance 284 nm.
Weight at the beginning of the sector 336 050 kg. What is the fuel flow for this sector.

9. Descend from FL 350. Track 260°T. Wind 205/35. What is the total ground distance
to descend at speeds MO.84/320/250?

10. Distance to alternate 270 nm. Track 280°T. VAR 11°W. TOW 345 000 kg.
Trip fuel 91 470 kg to destination. What is the correct FL to alternate at ISA + 15°C?

11. Distance to alternate 180 nm. Track 140°T. VAR 5°E. TOW 335 000 kg.
Trip fuel 76 130 kg to destination. What is the correct FL to alternate at ISA + 15°C?

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12. Distance to alternate 310 nm. W/C -10 Kts. TOW 344 600 kg. Trip fuel 100 900 kg.
Calculate the fuel and time to alternate.

13. Distance to alternate 410 nm. W/C + 28 Kts. TOW 351 000 kg. Trip fuel 85 250 kg.
Calculate the fuel and time to alternate.

14. TOW 345 000 kg. Trip fuel 91 470 kg. Fuel to alternate 8 300 kg. Calculate the fuel
required to hold for 30 mins. at FL 150.

15. ZFW 226 470 kg. Calculate the fuel required to hold at alternate at FL 200 for 30
mins.

16. An aircraft arrives overhead destination at FL 200 with 19 700 kg fuel in tanks. The
scheduled fuel to alternate is 7 200 kg, the contingency fuel is 288 kg and the fuel to
hold at alternate is 4 000 kg. How long can the aircraft safely hold at this level if the
ZFW is
218 000 kg?

17. An aircraft arrives overhead destination at FL 140 with 21 700 kg fuel in tanks. The
scheduled fuel to alternate is 6 000 kg, the contingency is 240 kg and the fuel to hold
at alternate is 4 400 kg. How long can the aircraft safely hold at this level if the ZFW
is
221 400 kg?

SIMPLIFIED PLANNING

1. Trip distance 3 200 nm. Average W/C for trip -25 Kts. TOW 315 000 kg average
upper level temperatures ISA + 10 with reference to Simplified Flight Planning Graph
3-5-4 “Step Climb to Optimum Altitude”, what is the trip fuel and time?

2. Trip distance 3 700 nm. Average W/C + 30 Kts. Constant FL 330. TOW 316 000
kg. Mean temperature deviation ISA + 15°C LRC. With reference to Simplified Flight
Planning Graph 3-5-3(a), calculate the fuel required.

3. Trip distance 4 000 nm. Average W/C - 45 Kts. Constant FL 310. TOW 325 000 kg.
M0.85. With reference to Simplified Flight Planning Graph 3-5-1(a), calculate the
trip fuel.

4. Distance from destination to alternate 375 nm. Mean W/C + 25 Kts. TOW 305 000
kg. Trip fuel 92 000 kg. Calculate the fuel to alternate with reference to graph 3-5-5.

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ATP FLIGHT PLANNING FLIGHT TRAINING COLLEGE
ATP DOC 7
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 CHAPTER 4

GRAPHS

The aircraft take-off weight can be restricted by 3 conditions. Remember these:

a) Maximum structural weight (377 842 kg)

b) Field Length Limited Take-off Weight

(Page 1-4-1 Flaps 10)
(Page 1-4-3 Flaps 20)

c) Climb limited Take-off Weight

(Page 1-4-2 Flaps 10)
(Page 1-4-4 Flaps 20)

If an actual take-off weight is given, it will automatically do away with the necessity to write
down the above three conditions.

The lowest of these 3 is the actual take-off weight.

Enter the take-off speeds table (Page 1-7) with this weight and extract V1, VR, V2.

Make V1 adjustments for slope and wind.

CASE 1 : If the speeds lie in the shaded area this means that the V1 speed may be
affected by Vmcg.

a) Enter the speeds table with the field length limited take-off weight.

b) If the speeds are out of the shaded area use the speeds determined in
step 2 - the take-off is legal.

c) If the speeds still fall in the shaded area, the take-off is not legal.

CASE 2 : If the problem states that either the anti-skid is inoperative or two brakes
are de-activated:

a) Enter the Two Brakes De-activated Table (page 5-5), or Anti-skid

inoperative Table (page 5-6). Extract a weight reduction and a V1
reduction.

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 b) Re-evaluate the three weights:

1) Maximum structural weight.

2) Field length limited take-off weight minus reduction.
3) Climb limited take-off weight.

c) Enter the speeds table with the lowest of these three weights to extract
VR and V2.

d) Enter the speeds table with the field length limited take-off weight
BEFORE REDUCTION to extract V1.

e) Adjust the V1 for :

1) Wind.
2) Slope.
3) Step vi (a) reduction.

f) The correct V1 will be the lower of this V1 and the V1 for the actual
TOW (corrected for slope and wind) as per para. A (ii). Always cross-
check to ensure that the chosen V1 is above Vmcg.

g) Should it be required to extract a V1 speed at weights greater than

390 000 kg, enter the speeds table at 390 000 kg, then increase the
V1 by 1 knot for every 3000 Kg above 390 000 kg.

IN CLASS EXAMPLES

1. Find the take off speeds under the following conditions:

TOM 350 000KG FLAPS 10 PH 2170ft COAT +21C
NIL WIND, NIL SLOPE…

2. Determine V1, Vr & V2 for a take off under the following conditions:
TOM 377 842KG FLAP 20 PH 4600FT COAT +25C
RWY 11 800FT NIL WIND NIL SLOPE PACKS ON
2 BRAKES U/S

3. If the TOM 274 500KG, FLAPS 20, ELEVATION 3700FT, QNH

1003.2MB, OAT +24C W/C –18KT, RWY LENGTH 10 200FT, RWY 18
ELEV 3659FT, RWY 36 2741FT.
What are the V speeds for takeoff?

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INTEGRATED RANGE

The integrated range tables (B747 manual, pages 3-3-1 to 3-3-15) are used to calculate a
simplified figure for the anticipated fuel burn-off for a particular trip distance.

SPECIAL NOTES

i) There are integrated range tables for LONG RANGE CRUISE at various flight
levels, and integrated range tables for MACH 0.84 cruise at various flight
levels.

ii) The fuel figures derived from these tables excludes climb and descent fuel,
thus in their pure form would be used in the event of a dog-leg or unplanned
routing.

iii) The tables are valid for ISA conditions. If conditions deviate from ISA, there
are adjustments for fuel required and TAS below the tables.

iv) The tables are valid for still air conditions. If there is a prevailing wind, the fuel
required will have to be adjusted.

IN CLASS EXAMPLES

1. LRC. FL270. CRUISE DISTANCE 4273 nm. Still air conditions.

ISA conditions. Weight at the commencement of the trip, 333 000 kg.
What is the fuel required?

i) Against 330 000 kg (left) and 3000 kg (top) read 7044.

ii) Subtract the trip distance 4273 from 7044 = 2771.

iii) Against 2771 in the table, read 230 000 Kg (left) and 4000 kg (top).

iv) The weight at the end of the trip is thus 234 000 kg.

v) The fuel required for the trip is thus:

333 000kg weight at commencement;

234 000kg weight at end;
99 000kg fuel required

vi) No fuel adjustment is required due to ISA conditions and still air prevailing.

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 2. What is the trip fuel using the integrated range tables, if:
TOM 338 280KG
DISTANCE 5162NM
FL310 MACH 0.84
W/C 0KTS TEMP DEVIATION +14C

QUESTIONS
TAKE OFF GRAPHS

1. TOW 345 000 kg, flaps 20, PA 3000ft, temperature + 27°C, headwind 15 Kts. 1 %
uphill slope. Determine the take-off speeds.

2. TOW 270 000 kg, flaps 10, PA 2800ft, temperature + 27°C, field length 9000ft, wind
calm, nil slope. Determine whether the take-off is legal and if so, what are the take-
off speeds?

3. Determine the take-off speeds at maximum take-off weight. PA 2000ft, temperature +

5°C, flaps 20, runway length 10 000ft, runway 09 elevation 1940ft, runway 27
elevation 2050ft, headwind 13 Kts, packs on, two brakes de-activated. Take-off on
runway 09.

4. TOW 304 000 kg, runway 05 elevation 3762 ft, runway 23 elevation 3562 ft. Runway
05 headwind 14 Kts, runway length 10 000 ft, flaps 10°, PA 4000 ft, temperature +
30° C, anti-skid inoperative. Determine the take-off speeds.

5. Determine the take-off speeds at MTOW, Runway length 9000 ft, slope 1 % up, wind
calm, flaps 20°, 2 brakes de-activated, PA 5000 ft, temperature + 35° C.

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INTEGRATED RANGE

1. Mach 0.84 cruise, FL330, trip distance 3624 nm, weight at the commencement of the
trip 341 000 Kg. What is the fuel required?

2. Mach 0.84 cruise FL330, trip distance 3624 nm, weight at the commencement of the
trip 341 000 Kg, ISA + 15°C. What is the fuel required?

3. Mach 0.84 cruise, FL330, trip distance 3624 nm, weight at the commencement of the
trip 341 000 kg, average wind component for the trip -50 Kts. What is the fuel
required?

4. Mach 0.84 cruise, FL330, Trip distance 3624 nm, weight at the commencement of the
trip 341 000 kg, average wind component for the trip -50 Kts, ISA + 15°C. What is the
fuel required?

5. Trip distance 4170 nm. Weight at beginning of the trip 316 000 kg, WC - 40 Kts,
temperature ISA + 14° C. M0.84 cruise FL 280. Determine the trip fuel in kg.

6. Trip distance 3099 nm. Weight at beginning of the trip 281 500 kg, WC + 35 Kts,
temperature ISA + 8° C, M0.84, FL 290. Determine the trip fuel in kg.

7. Trip distance 3200 nm. Weight at the beginning of the trip 297 000 kg W/C - 22 Kts.
Temperature ISA + 13C. LRC. FL 310. Determine the trip fuel in kg.

8. Trip distance 1200 nm. Weight at the beginning of the trip 331 500 kg. LRC. FL
310. ISA + 12. W/C + 55 Kts. Determine the trip fuel in kg.

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 CHAPTER 5

WEIGHT AND BALANCE

When considering the subject of weight and balance, it is important to understand that there
are two very distinct matters under consideration:

i) Weight limitations as set out in the weight schedule.

ii) Balance (C of G) limitations as depicted on the aircraft's C of G envelope.

GENERAL TERMINOLOGY AND DEFINITIONS

CENTRE OF GRAVITY (C of G)

The centre of gravity is defined as that point on the aircraft through which the
total weight may be considered to act.

Centre of Gravity

REFERENCE POINT/DATUM

In order to express the position of the centre of gravity accurately, it is necessary to have a
reference point. This reference point is called the datum line. The choice of datum line
position varies from aircraft to aircraft and is of arbitrary importance only.

Once a datum line has been chosen, the location of the C of G can be accurately expressed;
e.g. 300ft aft of datum. 300 ft aft of datum

DATUM

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STATION/FLIGHT STATION/CENTROID/ARM

Clearly, if weight is added to the aircraft, for example, in the nose section, the C of G must
move forward to keep the aircraft in balance.

In order to accurately determine the location of the new C of G, it is necessary to determine

exactly where in the nose section the weight was added. Station/Flight
Station/Centroid/Arm are all terms which can be used to indicate a particular loading
position in inches aft of datum. For example, FS 50 means that the weight was added 50" aft
of datum.

Arm

MOMENT, REDUCTION FACTOR (RF) AND INDEX

In the course of determining the C of G of an aircraft, moments must be calculated.

A moment is the product of a weight and an arm; e.g. the weight of 10 000 kg at FS50=
10 000 x 50inches = 500 000kg/inches moment.

Certainly, with large aircraft, the moments calculated are cumbersome and difficult to work
with.

To solve this, a suitable reduction factor (RF) can be chosen and divided into the moment to
make it more workable. For example a RF of 10 000 could be used. When a moment is
divided by a RF, it becomes known as an index.

500 000 ¸ 10 000 = 50, a far more workable figure.

MEAN AERODYNAMIC CHORD (MAC) AND

LEADING EDGE MEAN AERODYNAMIC CHORD (LEMAC)

Another popular method of expressing the position of the C of G is as a percentage of the

mean aerodynamic chord.

The chord line is the line joining the leading and trailing edges of an aerofoil. On aerofoils
with taper and/or sweepback, a mean or average aerodynamic chord can be calculated.
LEMAC is then the leading edge of this mean aerodynamic chord.

E.G.: C of G 300ft aft of datum.

LEMAC 250ft aft of datum.
MAC 100 ft.

To express the C of G position as a percentage of the MAC:

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QUESTIONS

1. Shortly before take-off, an extra passenger is given permission to board the aircraft.
Before boarding, the weight of the aircraft was 11 200 Lbs and the C of G at
191inches aft. The passenger weighs 170 Lbs and is allocated a seat at FS 259.
What is the new C of G?

2. An aircraft has a MAC of 162 inches LEMAC is at FS 324. The C of G is 412 inches
aft of datum. What is the C of G position expressed as % MAC?

3. From the following data, calculate the position of the C of G in inches aft of datum and
as a percentage MAC.

OEW : 66 600 kg
C of G : 480inches
ZONE A FS 290 LOAD 28 STANDARD PASSENGERS
ZONE B FS 480 LOAD 42 STANDARD PASSENGERS
ZONE C FS 680 LOAD 46 STANDARD PASSENGERS
HOLD 1 FS 200 LOAD 1500 kg
HOLD 4 FS 750 LOAD 500 kg
WING TANK FS 490 LOAD 41 020 kg
CENTRE TANK FS 480 LOAD 9080 kg

STANDARD PASSENGER WEIGHT 75 kg

LEMAC 447 inches
MAC 144 inches

4. Aircraft weight 10 400 lbs. C of G 100". What is the furthermost aft FS that freight
weighing 500 lbs can be loaded without exceeding the rearmost C of G limit of 120"?

5. Shortly before take-off, a passenger at FS 259 moves to a seat at FS 315. The

passenger weights 170 lbs and the aircraft weight at the time of transfer was 12 200
lbs. By how many inches aft does the C of G move?

6. Before take-off, the aircraft weighs 11 000 lbs and the C of G is 191" aft of the datum.
Where does the C of G lie after take-off when the undercarriage is retracted? The
main wheels weigh 380 lbs. When extended they are at FS 85 and when retracted
they are at FS 91. The nosewheel has a negligible effect.

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 CHAPTER 6

THE CRITICAL POINT (CP)

The critical point is defined as : that point on track where, should one engine fail, it would
take the same amount of time to continue to destination with one engine inoperative, as it
would to return to departure with one engine inoperative. Thus, the CP is in essence, a one
engine inoperative Point of Equal Time.

METHOD FOR CALCULATING THE CP

All of the previously mentioned principles for the PET also hold true for the CP. For CP
calculations, a reduced (one engine inoperative) TAS will be given.

i) Use this reduced TAS to determine a reduced GSO and a reduced GSH.
Use this GSO and GSH in the CP formula to determine the distance to the
CP.

ii) Use the full TAS to determine a full GSO. Divide the distance to the CP by
the full GSO to determine the time to the CP.

THE CP FORMULA

CP DIST =

Multi Leg CP

Find the position of the Critical Point if:

Route Speed Distance W/C F.Flow

P-Q 420 360 +30 2100
Q-R 425 640 +55 2000
R-S 430 375 +20 1950
S-R 395 375 -20 1900
R-Q 380 640 -60 1850
Q-P 425 360 -25 1800

1.Draw the route:

2.Find the CP

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QUESTIONS

1. A - B track 084° (T) distance 607 nm, W/V 210/15, TAS 240 Kts, reduced TAS 180
kts.

Solve:

a) The distance to the CP

b) The time to the CP

2. A - B track 145° (T), distance 860 nm, W/V A - CP 060/20, CP - B 120/30, TAS 240
kts, reduced TAS 180 kts.

Solve:

a) The distance to the CP

b) The time to the CP

3. A - B track 316° (T), distance 1040 nm, W/V 090/30, TAS 240 Kts, reduced TAS 180
kts, fuel flow 1500 Lbs/Hr. Fuel flow with one engine inoperative 1250 Lbs/Hr. The
aircraft experiences an engine failure at the CP and elects to return to departure due
to better maintenance facilities. What was the fuel used on return at A?.

4. A - B distance 1072 nm, track 210° (T), TAS 490 Kts, reduced TAS 460 kts, W/V A -
CP 020/47, W/V CP - B 065/65. What is the distance and time to the CP?

5. Distance A - B 3140 nm, track 137° (T), W.V. 040/52, full TAS 480 kts, full F/F 9750
kg/hr. Reduced TAS 460 kts, reduced F/F 8240 kg/hr. If the aircraft experiences an
engine failure at the CP and elects to continue to destination, what will the fuel burn
off be on arrival at B?

6. A - B 634 nm W/C + 15 kts

B - C 380 nm W/C + 25 kts
C - D 642 nm W/C + 40 kts
4 engine TAS 490 kts F/F 6000 kg/hr.
3 engine TAS 430 kts F/F 5450 kg/hr.

All winds along track.

Fuel board 27 200 kg.

If the aircraft reaches the CP, and then returns, what will the fuel on board be on
arrival at A?

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 CHAPTER 7

THE POINT OF NO RETURN (PNR)

The PNR is defined as : that point on track which represents the furthest distance to which
the aircraft can go out and return within the safe endurance.

THE PNR WITH NO WIND

With still air conditions prevailing, the PNR will be at the half-way mark in terms of safe
endurance time.

A B

Safe endurance 4 hours, TAS 250 kts.

To determine the distance to the PNR, multiply the time out with the speed out;

i.e. 2 hrs x 250 kts = 500 nm.

THE PNR WITH WIND

Where there is a prevailing wind, the PNR will move into wind in terms of time. To fly
a given distance will take a longer time flying into a headwind and a shorter time
flying with a tailwind. The sum of the two times will equal the safe endurance.

A B

Safe endurance 4 hours, TAS 250 kts.

To determine the distance to the PNR, multiply the time out with the speed out;

i.e. 2 hrs 24 x 200 kts = 480 nm.

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THE PNR FORMULAE

PNR TIME =

Where : GSR = Ground speed return

GSO = Ground speed out
SAFE ENDURANCE = Total endurance less reserves,
and expressed in time.

The second formula for the PNR and the second formula for the PET are closely linked

PET (DIST) =

therefore

therefore PNR TIME =

AND ALSO:

PNR TIME =

therefore

therefore PET DIST =

.OR. PNR DISTANCE =

Finding a Multi Leg PNR:

Route Speed Distance W/C F.Flow

P-Q 420 360 +30 2100
Q-R 425 640 +55 2000
R-S 430 375 +20 1950
S-R 395 375 -20 1900
R-Q 380 640 -60 1850
Q-P 425 360 -25 1800

Total Fuel onboard is 7500kg + reserves

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Questions

1. A - B 941 nm
Track 084° (T)
TAS 240 kts
W/V 160/25
Safe endurance 6 hrs 16 mins

Solve a) The time to the PNR

b) The distance to the PNR

2. The total endurance of 9000 kg includes a reserve of 15 %

Performance out 0.312 Gnm/kg
Performance return 0.421 Gnm/kg

Solve : The distance to the PNR

3. A-B
Distance 762 nm
Track 040° (T)
W/V 090/20

B-C
Distance 502 nm
Track 090° (T)
W/V 210/10

C-D
Distance 801 nm
Track 135° (T)
W/V 120/25

Safe endurance 8 hours

TAS for all 3 legs is 240 Kts

Solve a) The time to the PNR

b) The distance to the PNR

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4. An aircraft is on a flight from FAJS to FACT via the KMV. Although the aircraft is able
to depart from FAJS, the RVR is below landing minima. In the event of having to turn
back, the aircraft will have to route KMV to FABL.

FAJS - KMV
Distance 240 nm
Track 248° (T)
W/V 180/20

KMV - FACT
Distance 390 nm
Track 240° (T)
W/V 270/30

KMV - FABL
Distance 80 nm
Track 125° (T)
W/V 180/20

TAS for all 3 legs is 240 kts

Safe endurance 4 hrs 08 mins

Solve a) The time to the PNR

b) The distance to the PNR.

5. On a flight from A - B, the safe endurance equals the flight time plus 25 %. Track
075° (T), W/V 220/45, TAS 470 kts distance 4875 nm. What is the time and distance
from the PET to the PNR?

6. Total endurance of 11 750 Kg includes a reserve of 20 %. Performance out 0.3

GNM/KG. Performance return 0.38 GNM/KG. What is the distance to the PNR?

7. An aircraft enroute from A to B, experiences an engine failure at point X, and returns

to A. Fuel on board before departure was 26 000 USG. Fuel on board when landing
back at A was 24 000 LBS. Fuel SG = 0.8. Track A-B 237T. Reduced TAS 460
KTS. Reduced FF 8240 KG.HR. Full TAS 480 KTS. Full FF 9750 KG/HR. W/V
040/52.

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 ANNEX A

SAMPLE EXAM

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1. Climb from JNB INTL. (elevation 5558ft) to FL 330. Take-off weight 342 700 kg.
Temperature ISA +10ºC. What is the fuel required?

a) 8 800 kg;
b) 7 940 kg;
c) 8 300 kg. (4)

2. Cruise at FL 350. Temperature ISA +15ºC. Long Range Cruise. Track 086º T.
Wind Velocity 310/40. Leg distance 378 nms. The aircraft weight at the
commencement of the leg is 308 600 kg. What is the fuel required?

a) 8 485 kg;
b) 7 940 kg;
c) 9 060 kg. (4)

3. Cruise at FL 290. Temperature ISA +12ºC. MACH 0.84. Track 186º T. Wind
velocity 270/40. leg distance 512 nms. the weight at the commencement of the leg
is 316 200 kg. What is the fuel flow on this leg?

a) 12 060 kg/hr;
b) 12 530 kg/hr;
c) 12 980 kg/hr. (4)

4. Take off weight 361 250 kg. Trip fuel A-B 98 700 kg. In the event of a go-around,
what will the flight level to alternate be, if the alternate distance is 386 nms and the
magnetic track to alternate 268º M, ISA +15ºC?

a) FL 350;
b) FL 310;
c) FL 390. (3)

5. Integrated range. FL 350. MACH 0.84. Trip distance 3600 nms. The weight at
the commencement of the trip 316 500 kg. The wind component of the trip is + 25
KTS. Temperature ISA +12ºC. What is the trip fuel?

a) 61 778 kg;
b) 63 842 kg;
c) 64 920 kg.

6. Determine the VI speed at the maximum take-off weight. Runway length 11 000ft.
Nil slope. Wind component – 10 KTS. Temperature +20ºC. Pressure altitude 5
500ft. Two brakes are de-activated. Flap 20.

a) 144 KTS;
b) 150 KTS;
c) 136 KTS. (4)

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7. Distance from destination to alternate is 410 nms. Wind component –20 KTS.
Overshoot weight is 246 000 kg. Determine the fuel to alternate:

a) 10 100 kg;
b) 11 400 kg;
c) 10 900 kg. (3)

8. Determine the maximum landing weight with anti-skid operative. Field length 8 500ft.
Two brakes inoperative. Dry runway. Pressure altitude 5 500ft. Headwind 5 KTS.
Flap 30

a) 286 500 kg;

b) 280 000 kg;
c) 276 000 kg. (3)

9. A–B 367 nms W/C -20 KTS

B–C 408 nms W/C -15 KTS
C–D 502 nms W/C Nil
D–E 319 nms W/C -20 KTS

The TAS throughout the flight is 512 KTS. Determine the distance to the PET.

a) 783 nms;
b) 821 nms;
c) 880 nms. (4)

10. The wind velocity from A to the CP is 260/26. The wind velocity from the CP to B is
210/40. TAS 260 KTS. Reduced TAS 210 KTS. The track from A to B is 170º T.
The distance from A to B is 962 nms. Determine the distance to the CP.

a) 600 nms;
b) 491 nms;
c) 518. (3)

11. FL 310 TAS 465 50 ANM/1 000 kg.

FL 370 TAS 420 56 ANM/1 000 kg.

The wind component at FL 310 which gives the same performance of GNM/1 000 kg
as for FL 370 with a 60 KTS headwind is:

a) 19 KTS;
b) 27 KTS;
c) 38 KTS. (4)

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12. A–B 475 NMS W/C –35
B–C 538 NMS W/C –47
C–D 398 NMS W/C –53
D–E 457 NMS W/C –38

TAS A – E is 260 KTS. Safe endurance excluding reserves is 8 HRS 45 MIN. If the
aircraft departed from A at 0900Z, the ETA for the PNR is:

a) 1405 Z;
b) 1417 Z;
c) 1429 Z. (4)
13. Basic operating empty weight 105 500 lbs. Standard passenger weight 170 lbs. C
of G 880”. Mac 180.9”. Lemac 860.5”. The loading details are as follows:

FORWARD COMPARTMENT 16 Passengers FS 582

AFT COMPARTMENT 59 Passengers FS 1028
FORWARD CARGO HOLD 21 50 lbs. FS 680
AFT CARGO COMPT 17 00 lbs. FS 1166
FUEL IN WING TANKS 23 000 lbs. FS 997.5
FUEL IN CENTRE TANK 26 500 lbs. FS 914.6

What is the C of G expressed in % Mac?

a) 22.56%
b) 24.79%
c) 26.82% (4)

14. Aircraft mass 298 500 kg. FL 330. Mach 0.84 cruise. Temperature ISA +5ºC.
Headwind 65 KTS. Sector distance 855 nms. According to the integrated range
tables, the sector fuel required is:

a) 19 000 kg;
b) 20 500 kg;
c) 21 600 kg. (3)

15. Airport pressure altitude 3 000ft Temperature 34ºC. Slope 1% uphill. 15 KTS
headwind. Runway length 9 000ft. Flap 20. PAX on. What is the maximum take-
off weight?

a) 300 000 kg.

b) 307 000 kg.
c) 314 000 kg.

16. Airport pressure altitude 5 000ft. Flap 30. Wet runway. Two brakes inoperative.
Anti-skid operative. Runway length 8 800ft. Headwind 10 KTS. According to the
Landing Performance Graph (5-3) the maximum landing weight?

a) 252 000 kg;

b) 267 000 kg;
c) 275 000 kg.

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17. Distance from X to Y 250 nms. Mean wind component + 30 KTS.
Take-off mass 341 000 kg. The trip fuel according to simplified flight planning graph
“
Step Climb to Altitude 3-5-4’ is:

a) 90 000 kg;
b) 94 500 kg;
c) 99 000 kg. (3)

18. Take-off with anti-skid inoperative. TOW 327 000 kg. Pressure altitude 5 500ft.
Temperature +25ºC. Runway length 12 000ft. Slope 1% downhill. Runway
headwind component 15 KTS. Packs off. Flap 10.

The VI speed is:

a) 123
b) 130
c) 136 (4)

19. Departure elevation 3 500ft. Estimated take-off mass 347 200 kg. With reference to
the B747 climb tables, the fuel required to climb for FL 330 with a temperature of ISA
+12 is:

a) 9 200 kg;
b) 9 670 kg;
c) 8 650 kg.

20. On a flight from Delta to Kilo with Lima as alternate:

TOW 343 000 kg. ZFW 233 700. Trip fuel 90 140 kg. Contingency fuel 3 605 kg.
Holding fuel 30 mins. Approach fuel 1 100 kg. Minimum in tanks 1 600 kg.
Distance Kilo to Lima 350 nms. Wind component + 21 KTS.

According to the B747 Alternate Planning Graph, 3-5-5, the fuel required from Kilo to
Lima is:

a) 7 900 kg;
b) 8 600 kg;
c) 9 300 kg. (3)

21. A B747 arrives overhead its alternate airport after diverting, at a mass of 247 450 kg.
According to table 4-3, the fuel required for a 30 minute hold at FL 150 with flap up is:

a) 4 405 kg;
b) 4 445 kg;
c) 4 485 kg. (3)

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22. Overhead waypoint 4 FL 290. Aircraft mass 325 365 kg. Temperature ISA +18ºC.
Distance to waypoint 5 is 570 nms. Wind component +59 KTS. According to the
B747 LRC tables, the fuel required from WPT 4 to WPT 5 is:

a) 12 230 kg;
b) 12 680 kg;
c) 13 130 kg. (4)

23. On a flight from Golf to Hotel with India as alternate.

AFW 233 000kg

FUEL TO ALTERNATE 8 500 kg.
4% CONTINGENCY 340 kg.
30 MINS HOLDING 4 337 kg.
APPROACH FUEL 1 100 kg.
MINIMUM IN TANKS 1 600 kg.

If the aircraft arrives overhead Hotel with 19 500kg of fuel in tanks, it could hold at FL
150 before diverting to India with reserves intact for:

a) 24 mins.;
b) 29 mins.;
c) 34 mins. (4)

24. On a flight from Cape Town on a track of 347ºT, an aircraft performs as follows:

4 engine TAS 480 KTS and 73.44 ANM/1000 kg fuel

3 engine TAS 420 KTS and 71.9 ANM/1000 kg fuel.
Wind velocity 290/43. Fuel available 39 600 kg.

The distance from Cape Town beyond which the aircraft will be unable to return to
Cape Town is:

a) 1534 nm;
b) 1439 nm;
c) 1485 nm. (4)

25. The following figures relate to an aircraft’s performance at various flight levels:

FL 250 G/S 300 KTS F/F 607 LBS/HR

FL 270 G/S 270 KTS F/F 560 LBS/HR
FL 290 G/S 250 KTS F/F 514 LBS/HR

The most economical FL is:

a) FL 250;
b) FL 270;
c) FL 280. (3)

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26. A 4 engine jet transport aircraft is cruising at FL 350. F/F 240 kg/hr per engine.
TAS 500 KTS. The aircraft’s performance is:

a) 46.87 GNM / 1000 kg;

b) 52.08 GNM / 1000 kg;
c) 57.29 GNM / 1000 kg. (3)

27. Floor load limit 173 LBS / sq. ft. pallet mass 79 LBS. tie downs mass 35 LBS.
Pallet dimensions 70ft X 70”. What is the maximum cargo that can be carried on this
pallet.

a) 5772 lbs.;
b) 5807 lbs.;
c) 5886 lbs. (3)

28. Aircraft mass 4800 LBS. C of G 82”. AFT C of G limit 85”. The maximum mass
that could be carried at station 155 without exceeding the AFT C of G limit is:

a) 227.3 LBS;
b) 214.9 LBS;
c) 205.9 LBS. (4)

29. The following distances app to runway 09/27:

RUNWAY TOTAL LENGTH 2000ft

RNWY 09 DISPLACED THRESHOLD 150ft RNWY 27
RNWY 09 STOPWAY 300ft STOPWAY 350ft
CLEARWAY 580ft CLEARWAY 350ft

The A.S.D.A. for Runway 27 is:

a) 2000ft;
b) 2350ft;
c) 2650ft. (2)

30. Aerodrome elevation 2480ft. QNH 1018.7. Runway 01 slope 1.2% up. Surface
wind 060/15. Temperature + 33ºC. Flap 20. At a break release weight of 353 250
kg, the minimum runway length required for take-off is;

a) 11 200 FT;
b) 12 300 FT;
c) 14 000 FT. (3)

31. Aircraft weight 352 500 kg, FL 290. M0.84. Cruise temperature ISA + 15ºC.
Tailwind 55 KTS. Sector distance 1500 nm. According to the integrated range
tables, the sector fuel is:

a) 35 200 kg;
b) 36 500 kg;
c) 38 000 kg. (4)

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32. Aerodrome pressure altitude 1000 ft. Runway length 7000ft. headwind 10 KTS.
Flaps 30. Wet runway. Two brakes inoperative. Anti-skid operative. According to
the landing performance Graph (5-3), the maximum landing mass is:

a) 223 000 kg;

b) 233 000 kg;
c) 247 000 kg. (4)

33. From XRAY to WHISKEY 3906 NMS. Mean W/C –25 KTS . Mean temperature
deviation + 12ºC. Take-off mass 368 800 kg. The trip time and fuel according to
simplified flight planning “Step Climb to Optimum Altitude’ (3-5-4) is:

a) 8 hrs 12 & 99 000 kg;

b) 8 hrs 36 & 108 000 kg;
c) 8 hrs 34 & 104 000 kg.

34. RUNWAY 20 12 300 FT ELEVATION 5143 FT

RUNWAY 02 12 300 FT ELEVATION 5266 FT
W/V 240/20. QNH 1028. TEMPERATURE +27ºC. Take-off mass 327 000 kg.
FLAP 20. V1 for take-off on Runway 20 is:

a) 146 KTS;
b) 149 KTS;
c) 153 KTS. (3)

35. Departure aerodrome elevation 4 250 ft. Estimated take-off mass 336 500 kg. With
reference to B747 Climb Tables, the fuel required to climb to FL 350, ISA +8ºC is:

a) 8 530 kg;
b) 8 950 kg;
c) 9 440 kg. (4)

36. A flight plan is to be prepared from JNB to Athens with Brindisi as the alternative
airport. According to the B747 Alternative Planning Graph (3-5-5), the fuel required
from Athens to Brindisi is:

Take-off mass 341 000 kg. ZFW 223 500 kg. Trip fuel 98 110 kg. Contingency
fuel 4% 3925 kg. Holding fuel 30 mins. Approach fuel 1100 kg. Minimum in tanks
1600 kg. Distance Athens to Brindisi 315 NMS. W/C –35 KTS.

a) 10 600 kg;
b) 9 950 kg;
c) 9 300 kg. (4)

37. A B747 has arrived over its alternate after a diversion. Mass 238 300 kg. According
to the B747 manual page 4-3, the fuel required for 30 minutes holding at FL 150 with
flaps up is:

a) 4 338 kg;
b) 4 392 kg;
c) 4 439 kg.

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38. Overhead WPT 3 at FL 310. Aircraft mass 312 770 kg. ISA +12ºC. Distance to
WPT 4 is 489 NMS. W/C –30 KTS. According to the B747 LRC tables, the fuel
required from WPT3 to WPT4 is:

a) 12 330 kg;
b) 12 780 kg;
c) 13 230 kg. (4)

39. The following figures relate to a B747 flight from JULIET to KILO with LIMA as the
alternate airport.
ZFW 233 700 kg.
Fuel to alternate 12 500 kg.
4 % contingency (ALT) 500 kg.
30 mins holding at alternate 4315 kg.
Approach 1100 kg
Minimum in tanks 1600 kg.

If the aircraft arrives overhead KILO 26 000 kg of fuel remaining, it could hold
overhead Kilo at FL 150 before diverting to LIMA with reserves intact for:

a) 26 mins;
b) 32 mins;
c) 38 mins. (4)

40. A B747 is flying at FL 330. Long range cruise. ISA +10. Aircraft mass 277 500 kg,
time 1315 Z. The time at which the aircraft mass has reduced to that at which FL
370 is the optimum altitude is:

a) 1514 Z;
b) 1519 Z;
c) 1524 Z. (4)

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ATP FLIGHT PLANNING FLIGHT TRAINING COLLEGE
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 ANNEX B

ANSWERS

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Chapter 1

. a) OEW : 174 000 KG

+ PAY : 89 000 KG
ZFW : 263 000 KG
+ FUEL : 98 800 KG (22 000 + 76800)
MTOW: 361 800 KG (landing weight restricted)

b) OEW : 174 000 KG

+ PAY : 89 000 KG
ZFW : 263 000 KG
+ FUEL : 107 333 KG (22 000 + 85 333)
MTOW: 370 333 KG (landing weight restricted)

2. OEW : 174 000 KG

+ PAY : 89 000 KG
ZFW : 263 000 KG
+ FUEL : 114 000 KG (22 000 + 92 000)
MTOW: 377 000 KG

92 000 ÷ 12 000 Kg/HR = 7.6 hrs

3200 nm ÷ 7.6 hrs = 417.4 kts
500 kts TAS - 417.4 kts G/S = 82.6 kts h/w

3. Calculate the mass of the drum:

V x SG = W
44 Imp Gal x 7.5 = 330 lbs

330 ÷ 2.205 = 149.7 Kg + 20 Kg = 169.7 Kg

Calculate the area of the drum:

AREA = pr²
= p x 0.28 m x 0.28 m
= 0.2463 m²

Calculate the maximum weight that the floor can sustain on an area this size.

W = A x FL

= 0.2463 m² x

= 172.4 Kg and the weight of the drum is only 169.7 Kg, therefore it can be
carried.

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4. Mno =

LSS =

= 592 kts

LSS = 38.945

592 = 38.945

15.2 =

15.2² = ºC + 273
231 = ºC + 273
ºC = 231 - 273
ºC = - 42ºC

At FL 260, the ISA temperature should be - 37ºC. The temperature is - 42ºC.

ISA deviation = ISA - 5ºC

5. Convert elevation to PA

1013.25 - 1007.4 = 5.85 hPa

5.85 hPa x 30 ft = 175.5 ft
PA = 5327 ft + 175.5 ft

= 5502.5 ft

Convert PA to DA

At 5502.5 ft, the ISA temperature should be + 4ºC. The temperature is + 4ºC. It is
therefore 20ºC too hot.

20ºC x 120 ft = 2400 ft

5502.5 + 2400 ft = 7902.5 ft

6. Third segment climb ends when the flaps are up.

7. You may not plan on landing before A displaced threshold or use the stopway at the
other end, therefore:

9200ft - 400ft - 8800ft.

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8. Remember, the clearway includes the stopway and the two must not be added
therefore:

9200ft + 600ft - 9800ft.

9. First calculate the fuel requirement.

A-B 1 200 nm  G/S 450 KTS = 2.67 hrs

2.67 hrs x 10 000 kg/hr = 26 667 kg
A-B approach = 1 000 kg
A-B contingency 26 667 x 4 % = 1 067 kg
A-B TOTAL = 28 734 kg

B-A 1 200 nm  550 KTS G/S = 2.18 hrs

2.18 hrs x 10 000 kg/hr = 21 818 kg
B-A approach = 1 000 kg
B-A contingency 21 818 kg x 4 % = 873 kg
B-A TOTAL = 23 691 kg

Grand total fuel 28 734 kg + 23 691 kg = 52 425 kg

Now : May any of this fuel be left behind?
The two trip fuels are burn off and may not be left behind. The two approach fuels
are burned off an may not be left behind.

The two contingencies are not burn off and some of it may be left behind. Which
one? The first one - 1 067 kg.
Therefore: Grand total fuel 52 425 kg - 1 067 kg = 51 358 kg

Put this equation to the acid test. Imagine a CAA inspector at A and also at B.
Inspector A knows your route, contingency and approach, therefore, he wants to see
28 374 kg in tanks. You have 51 358 kg - he’s happy.

En-route to B you should use 51 358 kg less,

- 26 667 kg (trip)
- 1 000 kg (approach)
23 691 kg at B

Before return to A, CAA inspector B, who knows your route, contingency and
approach wants to see 23 691 kg in tanks - you have exactly that.

Now: Work out your weight schedule.

OEW 174 000 kg
+ Payload 87 309 kg (261 309 kg - 174 000 kg)
ZFW 261 309 kg (312 667 kg - 51 358 kg)
+ Fuel 51 358 kg
Tow 312 667 kg (MLW 285 000 kg + B/O 26 667 kg + B/O 1 000 kg)

10. NOTE WELL! The clearway includes the stopway.

Therefore, TODA = 2 000 m
+ 580 m
2 580 m

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Chapter 2

1.

a)

b)

2.

a)

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 b)

3.

The expression of US GAL is inconsistent with the problem. Remember that SG

relates to imperial gallons or litres; for US GAL Specific Weight (SW) is used.
Therefore, US GAL must be converted to IMP GAL ( or Litres if the question calls for
it) before SG is factored.

166.5 US GAL ¸ 1.201 = 138.6 IMP GAL

138.6 IMP GAL x 8.0 = 1109 LBS

New performance expression:

a)

b) The wind component is the difference between the TAS and the G/S

= 210 - 170.6
= -39.4 Kts (Headwind)

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4. The best expression of an aircraft's performance for purposes of comparison is its
ground distance covered per unit of fuel.

Compare AT FL160 and FL180

a) FL160

b) FL180

Yes, it would be worthwhile to climb to FL180.

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5. a) i) FL150

ii) FL180

iii) FL190

Despite the strong W/C, FL190 is the best level to cruise at.

(b) With a total of 7400 LBS on board and a fuel flow of 1300 LBS/HR, with 1
HR reserve required, the available trip fuel is 6100 LBS.

6100 LBS ¸

= 4.69 HRS

4.69 HRS x

= 1004 NM RANGE

6. Convert 900 US GAL to LBS. 900 ¸ 1.201 = 749.4 IMP GAL

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 749.4 IMP GAL x 7.9 = 5920 LBS

In order to complete the flight, the aircraft must use a maximum of 5920 LBS over the
800 nm.

i) 220 Kts TAS - 10 Kts W/C = 210 Kts G/S

ii) 240 Kts TAS - 10 Kts W/C = 230 Kts G/F

iii) 280 Kts TAS - 10 Kts W/C = 270 Kts G/S

Clearly the maximum TAS lies between 220 Kts and 240 Kts. Use interpolation to
determine the exact TAS.

7.14 LBS/GNM corresponds to 220 Kts

7.83 LBS/GNM corresponds to 240 Kts
7.40 LBS/GNM corresponds to ?

0.26 (7.40 - 7.14) x 20 (240 - 220)

0.69 (7.83 - 7.14) 1

= 7.5 Kts + 220 Kts

= 227.5 Kts maximum TAS.

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7.

Endurance 70 000 Kg ´

= 2800 nm

8. 4743 USG ÷ 1.201 = 3949.2 Imp Gal

3949.2 x 6.7 = 26460 lbs

26460 lbs ÷ 2.205 = 12 000 Kg/hr

TAS 500 - 444 kts GS

= 56 kts H/W

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9. At FL 310

= 0.0462 GNM/KG

At FL 350

TAS 505 - 418 G/S

= 87 kts H/W

10. If there is no wind at FL 290, then 94.8 ANM/1000 kg is also 94.8 GNM/1000 kg.
This represents the bottom line performance at FL 350 - What headwind would
make it so?

First calculate the F/F at FL 350.

490 ANM 108,4 ANM

HR 1000 kg
490 ANM 1000 kg 4520 kg
=
HR 108,4 ANM HR

Now calculate the G/S at FL 350

4520 kg 94,8 GNM 429 GNM (G / S)

 =
HR 1000 kg HR

TAS 490 KTS - G/S 429 KTS = W/C -61 KTS

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11. What we are looking for here is clearly AVERAGE TAS and AVERAGE FF.

The average TAS is no problem because it is a constant 492 KTS throughout the
flight. The average FF is a problem though, because the two fuel trip fuel flows were
far for varying amounts of time. To solve this problem, calculate total fuel burn 
total time =
average FF

A-B
506 NM  480 KTS = 1.054 HRS
1.054 HRS  12 200 KG/HR = 12 861 KG

B-C
481 NM  474 KTS = 1.015 HRS
1.015 HRS  12 800 KG/HR = 12 989 KG

A-C
1.054 HRS + 1.015 HRS = 2.069 HRS
12861 KG + 12 989 KG = 25 820 KG

25 850 KG
= 12 494 KG / HR
2,069 HRS

492 ANM 12,494 1000 KG

NOW:
HR HR

= 39.38 ANM/1000 KG

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Chapter 3

FLIGHT PLANNING

1. Fuel required for climb sea level to 33 000ft at ISA + 10°C.

312 000 kg - 310 000 kg

 8 000 kg - 7 600 kg = 80 kg
320 000 kg - 310 000 kg

7 600 kg + 80 kg = 7 680 kg

Now, correct for departure airport elevation at ISA + 10°C.

5 558' - 4 000'
 850 kg - 550 kg = 234 kg
6 000' - 4 000'

550 kg + 234 kg = 784 kg

7680 kg - 784 kg = 6 896 kg for ISA + 10°C.

Fuel required for climb sea level to 33 000ft at ISA + 15C.

312 000 kg - 310 000 kg

 8 550 kg - 8 100 kg = 90 kg
320 000 kg - 310 000 kg

8 100 kg + 90 kg = 8 190 kg.

Now, correct for departure airport elevation at ISA + 15°C.

5 558' - 4 000'
 900 kg - 600 kg = 234 kg
6 000' - 4 000'

600 kg + 234 kg = 834 kg

8 190 kg - 834 kg = 7 356 kg for ISA + 15°C.

Now, interpolate between ISA + 10°C and ISA + 15°C.

ISA + 12 C - ISA + 10 C

 7 356 kg - 6 896 kg = 184 kg
ISA + 15 C - ISA + 10 C

6 896 KG + 184 kg = 7 080 kg

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2. 6 495 kg

3. Calculate the W/C.

Track 340° - wind 270° = 70° at 70 Kts.
W/C = -31 Kts

ISA TIME TO CLIMB

296 000 kg - 290 000 kg

 17 MINS - 16 MINS = 0,6 MINS
300 000 kg - 290 000 kg

16 MINS + 0.6 MINS = 16.6 MINS

ISA TAS TO CLIMB

296 000 kg - 290 000 kg

 436 Kts - 435 Kts = 0,6 Kts
300 000 kg - 290 000 kg

435 Kts + 0.6 Kts = 435.6 Kts

ISA + 10°C TIME TO CLIMB

296 000 kg - 290 000 kg

 18 MIN - 17 MINS = 0,6 MINS
300 000 kg - 290 000 kg

17 MINS + 0.6 MINS = 17.6 MINS

ISA + 10°C TAS TO CLIMB

444 Kts

ISA + 4°C TIME TO CLIMB

ISA + 4 C - ISA + 0 C
 17,6 MINS - 16,6 MINS = 0,4 MINS
ISA + 10 C - ISA + 0 C

16.6 mins + 0.4 mins = 17 mins

ISA + 4 C - ISA + 0 C
 444 Kts - 435,6 Kts = 3,36 Kts
ISA + 10 C - ISA + 0 C

435.6 Kts + 3.36 Kts = 438.96 Kts

NOW: 438.96 Kts TAS - 31 KTS W/C = 407.96 Kts G/S

407.96 Kts  17 MINS = 115.6 GNM

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4. First calculate the W/C

TRACK 328° - WIND 270° = 58° at 70 Kts

58 KTS - 50 KTS
 47 KTS - 38 KTS = 7,2 KTS
60 KTS - 50 KTS

47 Kts - 7.2 Kts = 39.8 Kts (-Headwind)

AT ISA + 10°C
TAS = 503 Kts

AT ISA +10°C F/F:

287 500 kg - 280 000 kg

 11 524 kg / hr - 11 276 kg / hr = 186 kg / hr
290 000 kg - 280 000 kg

11 276 kg/hr + 186 kg/hr = 11 462 kg/hr

AT ISA + 15°C
TAS 509 Kts

AT ISA + 15°C F/F:

287 500 kg - 280 000 kg

 11 696 kg / hr - 11 444 kg / hr = 189 kg / hr
290 000 kg - 280 000 kg

11 444 kg/hr + 189 kg/hr = 11 633 kg/hr

ISA + 14 C - ISA + 10 C

 503 Kts - 509 Kts = 4,8 Kts
ISA + 15 C - ISA + 10 C

503 Kts + 4.8 Kts = 507.8 Kts

AT ISA + 14°C F/F:

11 462 kg/hr + 136.8 kg/hr = 11 598.8 kg/hr

NOW: TAS 507.8 Kts - W/C 39.8 Kts = 468 Kts G/S

DISTANCE 521 nm  468 Kts = 1 HR 07 MINS

1 HR 07 MINS  F/F 11 598.8 kg/hr = 12 912.3 kg
12 912.3 kg  2 = 6 456.17 kg fuel to halfway.
287 500 kg - 6456.17 kg = 281 044 kg approximate MZW

At this weight the TAS, W/C, G/S and TIME do not change.

Calculate the new F/F at this weight.

ISA + 10°C

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 281 044 kg - 280 000 kg
 11 524 kg / hr - 11 276 kg / hr = 26 kg / hr
290 000 kg - 280 000 kg

11 276 kg/hr + 26 kg/hr = 11 302 kg/hr

ISA + 15C

281 044 kg - 280 000 kg

 11 696 kg / hr - 11 446 kg / hr = 26 kg / hr
290 000 kg - 280 000 kg

11 444 kg/hr + 26 kg/hr = 11 407 kg/hr

ISA + 14°C

ISA + 14 C - ISA + 10 C

 11 470 kg / hr - 11 302 kg / hr = 134 kg / hr
ISA + 15 C - ISA + 10 C

11 302 kg/hr + 134 kg/hr = 11 436 kg/hr

NOW: Time 1 HR 07  11 436 kg/hr = 12 731 kg

12 731 kg  2 = 6 365.5 kg fuel to halfway
287 500 kg - 6 365.5 kg = 281 134 kg actual MZW.

At this weight the TAS, W/C, G/S and TIME do not change.

Calculate the new F/F at this weight.

ISA + 10°C

281 134 kg - 280 000 kg

 11 524 kg / hr - 11 276 kg / hr = 28 kg / hr
290 000 kg - 280 000 kg

11 276 kg/hr + 28 kg/hr = 11 304 kg/hr

ISA + 15°C

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 281 134 kg - 280 000 kg
 11 696 kg / hr - 11 444 kg / hr = 29 kg / hr
290 000 kg - 280 000 kg

11 44 kg/hr + 29 kg/hr = 11 473 kg/hr

ISA + 14°C

ISA + 14 C - ISA + 10 C

 11 473 kg / hr - 11 304 kg / hr = 135 kg / hr
ISA + 15 C - ISA + 10 C

11 304 kg/hr + 135 kg/hr = 11 439 kg/hr

NOW: 1 HR 07 MINS  11 439 kg/hr = 12735 kg

5. Approximate MZW = 263 770 kg.

Actual MZW = 263 917 kg
Fuel required = 13 573 + 400 kg = 13 973 kg

6. Approximate MZW = 307 323

Actual MZW = 307 373 kg
Fuel flow = 12 059.7 kg/hr

7. Approximate MZW = 305 559 kg

Actual MZW = 305 656 kg
Fuel required = 8 830 kg

8. Approximate MZW = 332 248 kg

Actual MZW = 332 285 KG
Fuel flow = 13 160 kg/hr

9. Track 260° - Wind 205° = 55° at 35 Kts

W/C = -22 Kts
NAM to descend 122 - TIME to descend 22 mins = TAS 332.7 Kts
332.7 Kts - 22 Kts = 310.7 Kts G/S
310.7 Kts G/S  22 mins = 114 GNM

10. Track 280°T + VAR 11°W = 291°M (evens)

Overshoot Weight = 345 000 kg
- 91 470 kg
- 3 659 kg (4 %)
- 1 100 kg (Approach)
248 771 kg (Enter Graph)

Extract  FL 295 from the graph.

Choose FL 280 for evens.

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11. FL 230

12. Calculate the landing weight at alternate

TOW 344 600 kg

- 100 900 kg (Trip Fuel)
- 13 000 kg (Standard)
230 700 kg (Use this weight on the graph)

8 700 kg Fuel to alternate

49 MINS Time to alternate

13. 10 750 kg Fuel to alternate

59 MINS Time to alternate

14. Calculate the holding MZW.

TOW 345 000 kg

- 91 470 kg (Trip)
- 3 645 kg (4 %)
- 1 100 kg (Approach)
- 8 300 kg (Alternate)
- 332 kg (4 %)
- 2 200 kg (Standard ½ hold)
237 939 kg (Holding MZW)

237 939 kg - 230 000 kg

 2 200 kg / hr - 2 120 kg / hr = 63,5 kg / hr
240 000 kg - 230 000 kg

2 120 kg/hr + 63.5 kg/hr = 2 183.5 kg/hr per engine

2 183.5 kg/hr  4 = 8 734 kg/hr Total
8 734 kg/hr  0.5 hrs = 4 367 kg

15. Calculate the holding MZW.

ZFW 226 470 kg

+ 1 600 kg (Minimum in tanks)
+ 2 200 kg (Standard ½ hold)
230 270 kg (Use this weight in the table)

HOLDING FUEL = 4 104.32 KG

16. Calculate the amount of fuel that must remain in the tanks.

1 100 kg (For an approach)

+ 7 200 kg (To alternate)
+ 288 kg (4 %)
+ 4 000 kg (Hold at alternate)

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 + 1 600 kg (Minimum in tanks)
14 188 kg

Calculate the amount of fuel available for holding at destination.

19 700 kg - 14 185 kg = 5 512 kg

Calculate the holding MZW.

ZFW 218 000 kg

+ 14 188 kg
+ 2 756 kg (½ of holding fuel 5 512 kg)
234 944 kg

234 944 kg - 230 000 kg

 2 130 kg / hr - 2 050 kg / hr = 39,5 kg / hr
240 000 kg - 230 000 kg

2 050 kg/hr + 39.5 kg/hr = 2 089.5 kg/hr per engine

2 089.5 kg/hr  4 engine = 8 358 kg/hr Total
5 512 kg  8 358 kg/hr = 39 MINS 34 SECS

17. 57 MINS 45 SECS

SIMPLIFIED PLANNING

1. Fuel 75 000 kg
Time 8 hrs

2. This graph requires a landing weight - use standard TAS, 500 Kts and standard
F/F 12 000 kg/hr.

500 KTS TAS = W/C 30 KTS = 530 KTS G/S

DISTANCE 3 700 NM  530 KTS = 6.98 HRS
6.98 HRS  12 000 KG/HR = 83 774 KG
TOW 316 000 KG - 83 774 KG = 232 226 KG

FUEL = 79 000 KG

3. This graph requires a landing weight - use standard TAS, 500 Kts and standard
F/F 12 000 kg/hr.

500 KTS TAS = W/C 45 KTS = 455 KTS G/S

DISTANCE 4 000 NM  455 KTS = 8.79 HRS
8.79 HRS  12 000 KG/HR = 105 495 KG
TOW 325 000 KG - 105 495 KG = 219 505 KG

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4. This graph requires a landing weight at alternate - use standard 13 000 kg.

TOW 305 000 kg

- 92 000 kg (Trip)
- 13 000 kg (Standard)
200 000 kg

FUEL = 8 700 KG

Chapter 4

TAKE OFF GRAPHS

1. Enter the speeds table with 345 000 Kg

V1 = 149.5 Kts
VR = 159 Kts
V2 = 169.5 Kts

Adjust the V1 speed for :

Slope : + 1 Kt
Wind : + 1 Kt
V1 = 151.5 Kts
VR = 159 Kts
V2 = 169.5 Kts

2 Enter the speeds table with 270 000 Kg.

V1 = 131 Kts
VR = 135 Kts
V2 = 155 Kts

The speeds fall in the shaded area which means that they may be affected by Vmcg.
Calculate the field length limited take-off weight : 312 000 Kg.

Apply 312 000 Kg to the speeds table. 312 000 Kg lies out of the shaded area, in other
words, the take-off is legal and the speeds for the take-off are those indicated above.

The significance of this question is as follows:

Where aircraft weight and ambient conditions give a V1 speed below Vmcg, this V1 speed is
increased to be above Vmcg and then shaded. Therefore, the shaded speeds on the table
are not below Vmcg. The shading indicates that these speeds have been subjected to an
increment to place them above Vmcg.

Certainly, increasing V1 speeds can be a dangerous business. Is there sufficient runway

length to allow for this increase? In other words, could the aircrew still execute a rejected
take-off safely having reached this speed?

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This question is answered by applying the field length limited take-off weight to the
speeds table. If this weight falls outside of the shaded area, the answer is yes. If this
weight still falls within the shaded area, the answer is no, and therefore the take-off
would not be legal.

3. Write down the three conditions limiting the take-off weight:

Structural limit : 377 842 Kg

Field length limit : 356 000 Kg
Climb limit : 384 000 Kg

The lowest of these three will be the actual take-off weight. Enter the speeds table with 356
000 Kg.

V1 = 153 Kts
VR = 163 Kts
V2 = 173 Kts

Make V1 adjustments for :

Slope : + 1 Kt
Wind : + 1 Kt

V1 = 155 Kts
VR = 163 Kts
V2 = 173 Kts

But : 2 brakes are de-activated.

Enter the 2 brakes de-activated table. Extract a weight reduction and a V1 reduction.

Weight reduction : 1800 Kg

V1 reduction : 5 Kts

Re-evaluate the three weights.

Structural limit : 377 842 Kg

Climb limit: 384000 Kg

The lowest of these three is the new actual take-off weight.

Enter the speeds table with 354200 Kg and extract only VR and V2.

VR = 163 Kts
V2 = 172 Kts

Enter the speeds table with the field length limited take-off weight before reduction to
extract V1. Enter the speeds table with 356000 Kg.

V1 = 153 Kts.
+ 1 wind
+ 1 slope

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 - 5 brakes
V1 = 150 Kts

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 Enter the Vmcg table to confirm that this reduced V1 is above Vmcg.

The take-off speeds are:

V1 - 150 Kts
VR - 163 Kts
V2 - 172 Kts

4. Determine the slope

3762 ft - 3562 ft = 200 ft

Determine the speeds at 304000 Kg

V1 = 144 kts + 1 kt wind - 4 kts slope = 141 kts

VR = 153 kts
V2 = 164 kts

Reference anti-skid inoperative table.

Subtract 9100 Kg
Subtract 21 kts

Reference field length take-off weight graph.

Field length limited take-off weight is 323000 Kg

323 000 Kg - 9100 Kg = 313 900 Kg
304000 Kg is still the actual take-off weight; and
VR remains 153 kts
V2 remains 164 kts

Enter the speeds table with 323000 Kg

V1 = 150 kts
+ 1 Wind
- 4 slope
- 21 anti-skid inoperative
128
V1 = 126 Kts (Vmcg is 122 kts)

But the V1 for the actual TOW of 304000 kg (corrected for slope and wind) is 120 Kts,
which is lower than 126 Kts, therefore the correct V1 is 120kts but Vmcg = 122, take off
illegal.

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5. Structural 377842 Kg
Field 275000 Kg
Climb 310000 Kg

Reference the 2 brakes de-activated table.

Subtract 2300 Kg
Subtract 6 kts

275 000 Kg - 2300 Kg = 272 700 Kg

(This is now the most limiting, therefore actual take-off weight)

Reference the speeds table with 272700 Kg to extract VR and V2

VR = 137 kts
V2 = 147 kts

Reference the speeds table with 275000 Kg to extract V1.

V1 = 129 kts
+ 1 slope
- 6 brakes
V1 = 124 kts (Vmca is 117.5 kts)
take off illegal.

INTEGRATED RANGE

1. 341,000 Kg = 7783 nm
7782 nm - 3624 nm = 4158 nm
4158 nm = 258 280 Kg (interpolated)
341 000 Kg - 258 280 Kg = 82 720 Kg

2. 341 000 Kg = 7782 nm

7782 nm - 3624 nm = 4158 nm
4158 nm = 258 280 Kg
341 000 Kg - 258 280 Kg = 82 720 Kg

This amount of fuel must be increased, due to the non-standard temperature, at the rate of
0.9 % per 10°C above ISA. (0.09 % per 1°C Above ISA).

0.09 % x 15 = 1.35 % increase.

82 720 Kg + 1.35 %
= 83,837 Kg

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3. The trip distance of 3624 nm is a ground distance. To find the flight time requires
that this ground distance be divided by the ground speed.

From the top of the table read TAS 488 Kts - 50 Kts W/C = 438 Kts G/S.
3624 nm ¸ 438 Kts = 8 hrs 16 mins. This time multiplied by the TAS will give the air
distance flown.

8 Hrs 16 mins x 488 Kts (TAS) = 4038 nm.

Use this air distance in the table in the normal fashion.

341,000 Kg = 7782 nm
7782 nm - 4038 nm = 3744 nm
3744 nm = 249 900 Kg
341 000 Kg = 249 900 Kg = 91 100 Kg

4. Increased TAS = 488 + 15 (ISA) = 503 Kts

TAS 503 Kts + 15 Kts (For ISA + 15º) = 453 Kts G/S
3624 nm ¸ 453 Kts = 8 Hrs
8 Hrs x 503 Kts = 4024 nm
341 000 Kg = 7782 nm
7782 nm - 4024 nm = 3758 nm
3758 nm = 250 180 Kg
341 000 Kg - 250 180 Kg = 90 820 Kg
90,820 Kg + 1.35 % (ISA deviation)
= 92,046 Kg

5. TAS = 499 Kts + 14 (ISA) = 513 Kts

= 513 Kts - 40 Kts = 473 Kts G/S
4170 nm ¸ 473 Kts = 8 Hrs 49
8 Hrs 49 x 513 Kts = 4523 nm
316000 Kg = 6026 nm
6026 nm - 4523 nm = 1503 nm
1503 nm = 212 109 Kg
316000 Kg - 212 109 Kg
= 103891 + 1.26 %
= 105200 Kg

6. TAS 497 kts + 8 kts for ISA + 8°

TAS 505 Kts + 35 Kts = 540 Kts G/S
3099 nm ¸ 540 kts = 5 Hrs 44
5 Hrs 44 x 505 kts = 2898 nm
281 500 Kg = 4741 nm
4741 nm - 2898 nm = 1843 nm
1843 nm = 218 063 Kg
281 500 Kg - 218 063 Kg = 63 437 Kg
63 437 Kg + 0.72 % = 63 881 Kg

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7. Unlike M0.84, LRC does not have a constant TAS for a given flight level.

First do a very rough calculation to determine an average TAS for the flight.

At 297 000 kg, TAS =  492 Kts.

492 Kts + 13 Kts for ISA + 13C = 505 Kts TAS
505 Kts TAS - 22 Ktx W/C = 483 Kts G/S
3200 GNM  483 Kts G/S = 6.6 Hrs.
6 6 Hrs x TAS 505 Kts = 3346 ANM

297 000 kg = 5 969 ANM

- 3 346 ANM
2 623 ANM =  227 000 kg

227 000 kg =  463 Kts TAS

Halway between 497 Kts and 463 Kts = 480 Kts.

Now : Do the calculation:

480 Kts + 13 Kts for ISA + 13 = 493 Kts TAS

493 Kts - 22 Kts W/C = 471 Kts G/S
3200 GNM  471 Kts G/S = 6.8 Hrs
6.8 Hrs  493 Kts TAS = 3 349 ANM

297 000 Kg = 5 969 ANM

- 3 349 ANM
2 620 ANM = 226 962 Kg

297 000 Kg - 226 962 Kg = 70 038 Kg

70 038 Kg + 1.17 % = 70 857 Kg

Now : As a matter of interest,

Initial weight 297 000 Kg

- 70 857 Kg (Fuel required)
Final weight 226 143 Kg

226 143 Kg = 462 TAS (Extremely close to our prediction)

8. Rough calculation of TAS

331 500 Kg =  500 Kts + 12 Kts ISA Dev + 55 Kts W/C = 567 Kts G/S.
1200 GNM  567 Kts G/S = 2.12 Hrs
2.12 Hrs  512 Kts TAS = 1 084 ANM

331 500 Kg = 7 400 ANM

- 1 084 ANM
6 316 ANM =  305 000 Kg

305 000 kg =  494 Kts TAS

Halfway between 500 Kts and 494 Kts is 497 Kts Rough TAS.

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 Now: 497 Kts + 12 Kts ISA DEV + 55 Kts W/C = 564 Kts G/S
1 200 GNM  564 KTS G/S = 2.13 Hrs
2.13 Hrs  509 KTS TAS = 1 083 ANM

331 500 kg = 7 400 ANM

- 1 083 ANM
6 317 ANM = 305 095 Kg

331 500 kg - 305 095 kg = 26 405 kg + 1.08 % = 26 690 Kg

CHAPTER 5

1. Use the format W x A = M

Write down the original condition.
Write down the change.
Calculate the new condition.

W x A = M

11200 191 2139200

+ 170 259 44030
11370 192 2183230

2 183 230 ¸ 11 370 = 192

2.
412" = 324" = 88"

3. OEW 66600 480 31968000

+ PAYLOAD 2100 290 609000
3150 480 1512000
3450 680 2346000
1500 200 300000
500 750 375000

ZFW 77300 480.08 37110000

+ FUEL 41020 490 20099800

9080 480 4358400

TOW 127400 483.27 61568200

483 - 447 =

= 25 % MAC

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 4. W ´ A = M
10400 100 1040000
+ 500 X
10900 120

(10 900 x 120) = (500 x) + 1 040 000

1 308 000 = 500X + 1 040 000
1 308 000 - 1 040 000 = 500X
268000 = 500X
X = 536" aft of datum

5. Although the C of G of the aircraft is not given, an arbitrary one may be chosen,
because it is only the change of C of G that is required

W ´ A = M
12200 ´ 100 = 1220000
- 170 259 = 44030
+ 170 315 = 53550
12200 100.7 = 1229520

Therefore the C of G moves 0.78" aft.

6. W ´ A = M
11000 191 = 2101000
- 380 85 = 32300
+ 380 91 = 34580
11000 191.2 = 2103280

The new C of G is 191.2" aft of the datum.

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Chapter 6

1 Determine the reduced GSO and GSR.

GSR = 171 Kts

GSO = 188 Kts

Determine the full GSO = 248 Kts.

a) CP distance =

= 289 nm from A

b)

= 1 Hr 10 mins to the CP from A

2.
Determine the GSR and GSO. Note the different winds for each.

GSR = 181 Kts

GSO = 152 Kts
a) CP DIST =

= 467 nm from A to the CP

b) Determine the full GSO = 237 Kts.

= 1 Hr 58 mins from A to the CP.

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3. Calculate the reduced GSR and GSO.

GSR = 158 Kts

GSO = 199 Kts

Calculate the full GSO = 260 Kts

CP DIST =

= 460 nm from A to the CP

The aircraft flew from A to the CP at 260 Kts with a fuel flow of 1500 Lbs/Hr.

= 1 Hr 46 mins x 1500 Lbs/Hr

= 2655 Lbs

The aircraft then flew from the CP back to A at 158 Kts with a fuel flow of 1250 Lbs/Hr

= 2 Hrs 55 mins x 1250 Lbs/Hr

= 3639 Lbs

Total fuel used = 2655 Lbs + 3639 Lbs = 6294 Lbs

4. CP =

414
=  1072 nm
512 + 414
= 479 nm from A

Time to the CP
479 nm ¸ full GSO for leg A - CP
479 ¸ 536 Kts
= 54 mins

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5. FULL GSO = 484 Kts FULL GSH = 471 Kts

REDUCED GSO = 463 Kts REDUCED GSH = 451 Kts

CP =

= 1549 nm from A to CP

TIME A - CP

= 1549 nm ¸ 484
= 3 Hrs 12 x 9750 Kg/Hr
= 31212 Kg

TIME CP - B

= 1591 nm ¸ 463 Kts

= 3 Hrs 26 x 8140 Kg/Hr
= 27971 Kg

Total fuel used on landing at B is 31212 + 27971

= 59183 Kg

6.

Assume the CP is at B

B A 634 nm  415 KTS = 1.53 HRS

B  C  D 380 nm  455 Kts = 0.835 HRS

642 nm  470 Kts = 1.366 HRS
2.20 HRS

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 Thus, the CP must lie on leg B/C.

Superimpose a time of 1.53 HRS on Leg?  C  O

1.53 HRS - C to D 1.366 HRS = 0.164 ?  C

0.164 HRS  455 KTS = 75 nm

380 nm - 75 nm = 305 nm

GSR
CP =  305 nm
GSO + GSR

405 KTS
=  305 KTS
455 KTS + 405 KTS

= 144 nm

Time to the CP is at FULL GSO

634 nm  505 KTS = 1.255 HRS

144 nm  515 KTS = 0.280 HRS

1.535 HRS  6000 kg/HR = 9 210 kg

Time from the CP is at reduced GSR.

144 nm  405 KTS = 0.355 HRS

634 nm  415 KTS = 1.528 HRS

1.883 HRS  5450 kg = 10 252 kg

Total fuel used = 9210 kg + 10 262 kg = 20 037 kg

Fuel remaining at a = 27 200 kg - 20 037 kg = 7 163 kg

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Chapter 7

1. Solve the GSR and GSO

GSR = 245 kts

GSO = 233 Kts

a) Use these speeds in the formula

PNR TIME =

= 3 Hrs 13 mins to the PNR

b) To solve the distance to the PNR:

3 Hrs 13 Mins x 233 Kts

= 748 nm

To double check the answer:

= 3 Hrs 13 mins = 3 Hrs 03 mins = 6 Hrs 16 Mins

2. To calculate the safe endurance, remember that the 9000 Kg is inclusive.

= 7826 Kg safe endurance

The units for the formula are Kg/Gnm, thus both expressions of performance will have
to be inverted.
PNR DISTANCE =

=
= 1397.5 Nm

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3. Make a schematic diagram and solve the GSR and GSO for each leg.

Does the PNR lie at point B? If so, the sum of times A - B and B - A will equal the
safe endurance.

A-B B-A
= =

= 3 Hrs 21 Mins = 3 Hrs 01 Mins

A - B + B - A = 6 Hrs 22 Mins

This is less than the safe endurance; therefore the PNR must lie beyond point B.

Does the PNR lie at point C? If so, the sum of the times A - B - C and C - B - A will
equal the safe endurance.

A-B C-B
3 Hrs 21 mins =

= 2 Hrs 08 Mins

B-C B-A
= 3 Hrs 01 Mins

= 2 Hrs 03 Mins

A - B -C + C - B - A

= 10 Hrs 33 Mins

This is more than the safe endurance; therefore the PNR must lie between points B
and C.

The time from A to B then B to A was 6 Hrs 22 Mins

08 Hrs 00 - 6 Hrs 22 Mins = 1 Hr 38 Mins

This 1 Hr 38 Mins must take the aircraft from B - PNR and the PNR - B

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 PNR TIME =

= 0 Hrs 48 Mins from B to the PNR

a) The total time to the PNR

A-B B - PNR
3 Hr 21 Mins 0 Hr 48 Mins = 4 Hrs 09 Mins

b) The total distance to the PNR

A-B B - PNR
0 Hrs 48 Mins x 245 Kts
762 + 196 nm = 958 nm

4. Make a schematic diagram and solve the GSR and GSO for each leg.

Does the PNR lie at KMV? If so, the sum of the times FAJS - KMV - FABL will equal
the safe endurance. (Note that the aircraft does not return to FAJS).

FAJS - KMV KMV - FABL

= 1 Hr 02 Mins = 21 Mins

Therefore, the time FAJS - KMV - FABL

= 1 Hr 23 Mins

This is less than the safe endurance time; therefore the PNR must lie on leg KMV -
FACT.

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 The endurance beyond KMV =

4 Hrs 09 Mins - 1 Hr 23 Mins

= 2 Hrs 46 Mins
PNR TIME =

= 1 Hr 32 Mins

a) Total time to the PNR =

FAJS - KMV KMV - PNR

1 Hr 02 Mins 1 Hr 32 Mins = 2 Hrs 34 Mins

b) Total distance to the PNR =

FAJS - KMV KMV - PNR

1 Hr 32 x 214

240 nm + 328 nm = 568 nm

5. A - B GSO = 506 kts

Flight time A - B = 4875 ¸ 506 kts
= 9:38 hrs

SE = 9:38 hrs + 25 %
= 12 hrs 03mins

Therefore, the time to go beyond the PET to the PNR and back is 2 hrs 25 mins.

PNR =

432
=  2 hrs 25
506 + 432

= 1 Hr 07 beyond the PET to the PNR

Distance beyond the PET to the PNR = 1 hr 07 x 506 kts

= 563 nm

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6. Determine the safe endurance:

= 9792 Kg

Remember to invert the performance expressions.

PNR DIST =

= 1643 NM

7. Convert fuel on board to LBS:

26 000 USG  1.201 = 21 649 IMP GALS

21 469 IG  8 SG = 173 189 LBS

Calculate safe endurance:

173 189 LBS - 24 000 LBS = 149 189 LBS

Calculate performance LBS/FULL GNM OUT and LBS/RED GNM RETURN.

484 GNM FF 9780 KG 21 499 LBS

FULL GSO OR FF
HR HR HR

= 44.42 LBS/GNM OUT

451 GNM FF 8240 KG 18 169 LBS

REDUCED GSH OR FF
HR HR HR

= 40.29 LBS/GNM RETURN

S.E. LBS
PNR (DIST) = LBS LBS
GNM OUT + GNM PET
149 189 LBS
=
44,42 LBS / GNM OUT + 40,29 LBS / GNM RETURN

= 1 761 NM to point X

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Sample Exam

1 C 21 C
2 A 22 B
3 B 23 A
4 A 24 B
5 B 25 A
6 A 26 A
7 B 27 A
8 A 28 C
9 B 29 B
10 C 30 B
11 A 31 A
12 A 32 B
13 B 33 C
14 C 34 A
15 B 35 B
16 B 36 C
17 B 37 A
18 C 38 A
19 A 39 C
20 C 40 C

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