BASICS NAVIGATION
Introduction
The rule JAR-FCL 1.130, the candidate to PPL must demonstrate the
appropiate theoric knowlege to licence private pilot.
The requirements theoric knowledge to get PPL are: air laws, general
knowledge aircraft systems, planning and performance, humans limits,
meteorologic, operatives procedurs, comunications rules, aerodinamic and
general navigation.
The air navigation consit to fly with a direction, altitude, fuel consumption in a
derminated time over a determinate leg.
In the navigation there is a factors as the wind and its effect over direction and
speed and therefore effect over the time and consumption in leg.
To know what direction put to fly on leg, how long it take, the fuel burnt and
effect wind over the track of plane in leg it is neccesary have knowledge
basics navigation.
BASICS
OF NAVIGATION
.Concepts wind, heading, and aircraft speed
.Kinds speed
. Direction
. Triangle of speed
. Drift correction
.ETAS, TAS, and GS
.Calculate: correction drift, heading, ground speed, estimated
time arrival and fuel consumption of the one leg
.Correction route
Concepts wind, heading and
aircraft speed
Wind


The movemt airmass over Earth’s surface. Can procced from any direction an
it have any speed. That movement is the wind.
The wind can to measure acording to come from direction and speed. For
example 050º / 30 kt
.The heading:
Is the horizontal direction to point out the aircraft.
The heading can measured with actual azimut of longitudinal axis with
reference a point determinate, as are geographic north, magnetic north or
compass north.
.Aicraft speed:
The movement aircraft through air mass is due to traction of propeller giving it
an speed to the movement. The headwind or tailwind to affect over ground
speed.
Kinds of airspeed
IAS: Indicated airspeed  is the speed that indicates the anemometer without making any
correction
εi : Error of instrument  is the error produced by plant, scale and temperature over metals and
capsule. The value is almost negligible and would be on the plane's manual.
BAS: Básic Airspeed  IAS corrected by εi BAS = IAS – (± εi)
εp : Error of position  is the error produced by the static shots of the position and attitude of the
aircraft in flight. This error is almost the same for all aircraft of the same model and found in the
manual of the plane.
CAS: Calibrate airspeed  is the BAS corrected by εp, or the IAS corrected by εi/p
CAS = BAS – (± εp) or CAS = IAS – (± εi/p)
εc : Error of compresibility  produced by the dynamic pressure, it is always positive and increases
with Mach No, it's the same for all aircraft and airspeed
DAS: density airspeed  is the CAS corrected CAS is the corrected by a factor of density of the
atmosphere DAS = CAS x f f = 1/ √ρ/ρo ρ (air density in actual level) ρo (density of level
sea)
EAS: equivalent airspeed  CAS corrected by εc
ερ: is because the anemometers are calibrated to measure speeds, when the relative density of air is 1,
id est at sea level on a day ISA
TAS: True air speed  is the EAS corrected by factor density or CAS correcting by εC/ρ
TAS = EAS x f
o TAS = EAS – (± ερ)
or
TAS= CAS ± εC/ρ
Cs sor LSS : Local sound speed  incrase with temperature. One way to calculate it is
(√273 + ( ± ºC en TAT ) ) x 38.96 = C
MACH: nº Mach M= TAS / Cs
ETAS: Efective true air speed  TAS is the corrected for the effect of the component of cross wind.
It may be corrected when the angle of drift correction (dc) is higher than 10º. If the angle (dc) is
lower, the value ETAS is almost equal to that of TAS.
GS: Ground speed  is the TAS or ETAS corrected by the wind. From this speed can be determined
by the time it would take to fly a route or the distance flown from a time. Speed = space / time
Altittude and
temperature
IAS
±εi/p
CAS
+εc
EAS
±ερ
TAS
WIND
ETAS
GS
The εc ≥ 0
If CAS ≤ 250kt
CAS= EAS
IAS
CAS = IAS – (±εi/ρ)
EAS = CAS – (+ εc)
TAS = EAS – (±ερ)
Direction
Direction is postion’s point whith reference another point in the space
determinate by the way and line that join both.
B
A
True North ( TN ): points intersection of the Earth’s surface with its axis of
rotation is called Geographic North. The bearing or the routes they use as a
benchmark geographic north is called the true route or course.
Magnetic North ( MN ): Point the Earth’s surface where to join the magnetic
field lines burned in the south magnetic pole . The compass use as reference
this lines of force with absence of the magnetic impact of the radio’s plane.



Compass North (CN): It is the north magnetic corrected by the magnetic
effects that make the avionics equipment.
Desviation (Δ): Deviations are produced by the equipment of avionics, are
added to or subtracted away from magnetic north to get compass course
Declination (δ): Is the angle of the meridians that form the magnetic north
and true.
Both, the Decline and Desviations as the change if they are East are positive
value and if this is West are negative value.

To calculate the course track of the plane, use the formulas.

TC – (- δW) = MC
MC – (-ΔW) = CC
TC – (δ E) = MC
MC – (ΔE) = CC
To pass from to
Course to Heading,
it must to correct the
crosswind
MN
TN
TN
MN
MC
MC
δw
δE
TC
TC
Reference line
Reference line
Figure 2
Figure 1
CN
MN
TN
TN
MN
CC
MC
CN
MC
δw
CC
TC
TC
δw
Reference line
Reference line
Figure 3
Figure 4
Triangle of speeds
The triangle of speed is used to determinate the efect wind while plane is flying.
It are three the parameters:
Speed aircraft vector: this vector has a amaunt of speed and direction. They are the
speed and heading’s aircraft.
Speed wind vector: is a module or vector whose magnitude is the wind speed and
direction of the place where the wind comes from
Ground speed aircraft : is a module or vector whose magnitude is the aircraft speed
and track.
A
B
23 kt
A
050º /30 kt
B
R
T1
19 kt
H 000º TAS 100 kt
TC 347º GS 78 kt
T2
Drift correction



When there is crosswind from the right, the plane is shifted to the left. To
fly line track the heading must be corrected to the right and vice versa.
The wind oblique has two components: a transverse and longitudinal
another. If the wind is perpendicular to our path, then drift will be
maximum, and if the wind parallel to the path drift will be 0, and intensity
in the tail or head will are maximum.
That is apparent in the values of: Cos 0 = 1 for the wind component
longitudinal and Seno 90 = 1 for the cross-wind component.
TRACK
TRACK
Ttrigonometric basics knowledge
ETAS, TAS and GS
ETAS: is the real effective rate, id est is the TAS corrected for the effect of
cross wind. And from which we get the GS.
When the (dc) is less than 10 degrees, TAS and ETAS are semejants.
If the value of (dc) is more than 10º then it should
be considered for calculating the ETAS to get GS.
wind
GS
wind component
GS =
= ETAS
ETAS ±± wind
component
Example
A plane at 5000 feet with TAS 100 kt, and has to fly a route of 000 º. It
affects a wind of 050 to 30 kt and fuel flow is 5 gal imp/ h. What true
heading and magnetic heading will be put to follow the route? How long will
it take to travel 50 nm? How many Gal Imp fuel burned in in leg?
TAS: 100 kt
TC: 000º
Decline δ: 7º W
Wind: 050º/30 kt
dc?
TH?
GS?
Time?
MH ?
1º drift correction
Development of calculating the intensity of the cross wind
component.
TC 000º
A
40º
α
Wind
050º /30kt
B
50º
C
β
A= sinβ x C  A= sin50 x 30= 23 kt
Is put the angular value that separates the wind
direction of the longitudinal axis to be
It checks the intensity of the crosswind component with 10% of TAS.
The intensity crosswind is 23 kt >10 kt. The 10 kt is 10% of 100 kt of TAS.
So we know that dc is ≥ 6
Making a triangle of speeds:
Crosswind
Sin (dc)=
TAS
23
Sin (dc)
= Sin 0.23 Revers = 13,29º
100
dc= 13º
By having a wind from the right, the plane will be
shifted to the left, therefore must be corrected to
the right what they have to add the 13º (dc) to TC
to get a TH.
With from right wind TH > TC  TH = 000º +
13º = 013º
(13º of moment , until check ETAS)
2º calculate Headwind
The wind that comes
from the front
quadrant is
negative
The wind that comes
from the rear
Quadrant is positive
Vector wind 050 º/ 30kt
Knowing β and knowing the intesity of C can we
Know the intensity of the wind on side D.
D
Cos β =
=  D = Cosβ x C 
C
β
β
D= Cos 50 x 30 = 19,2 kt as it is headwind will
be -19 kt
3º Calculate GS ( ground speed)
the GS can be calculated directly with the TAS when (dc) <10 º
In the example we have a (dc) = 13> 10 % of TAS so
we have to calculate the GS from the ETAS
Knowing and TAS (dc) we can get ETAS
ETAS

Cos (dc) =
TAS
ETAS = TAS x Cos (dc) 
ETAS = 100 x Cos 13
ETAS = 97 kt
GS = ETAS ± wind component
GS = 97 + (-19) = 78 Kt
4º Check drift correction (dc) with ETAS
ETAS

Cos ( dc) =
TAS
97

Cos (dc) =
100
Cos 0.2371  invers  14
dc = 14º
TH = 014º
5º Calculate the ,megnetic heading
MC= TC – ( -δw)
MH = TH –(-δW)
MH= 14 –(-7)
TH= 14

TN
MN
MH 021 º
-7º
TH 014º
MH= 021º
6º Time used to fly 50 NM and fuel burned
Space
Space
 Time =
Speed =
Time
50 NM
 Time =
Speed
0,6410 hours x 60 minutes = 38,46 minutes.
Fuel burned
0,6410 hours x 5 gal imp = 3,2 gal imp
= 0,6410 hours
78 kt
Correction route
Assuming an airplane flying a route and after some time the pilot realizes he
has traveled a distance and that is X miles away from the route.
Chi
A
f
Â
B
g
B
i
h
c
B
Â
CHf = CHi ± Ĉ
Ĉ=Â+B
Ĉ
CHi: Compass heading inicial
CHf: Compass heading final
CHf ± (Â + B)
i
- =  Sin  =
i
 c=
c
SinÂ
i
-B =  Sin B=
i
 h=
h
Sin B
Example
A plane flying from A to B, with a course of 090 º.
He has flown 70 NM and the pilot realizes that is 10 NM away
and right of the map.
With the new position, he have to travel 260 NM.
What course should make to reach B?
dc
- =  Sin  =
10
 Sin  =
ac
70
dc
-B =  Sin B=
= 0.1428  revers SIn = 8,2 º
10
 Sin B =
cb
= 0,03846  revers Sin = 2,2º
260
CHf = CHi ± Ĉ  Ĉ = Â + B  Ĉ = 8,2 + 2,2 = 10,4º
as the plane is shifted to the right of the route,
then he have to be corrected to the left which implies
that they must subtract the angle C.
CHf = 090º - 10º = 080 º
Author: Javier Pérez Mate
Comercial pilot CPL
Center study: Aeronautic formation school Aerofan FTO TRTO
javinet20@hotmail.com
Madrid Spain