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BASICS NAVIGATION

BASICS NAVIGATION. Introduction. The rule JAR-FCL 1.130, the candidate to PPL must demonstrate the appropiate theoric knowlege to licence private pilot.

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BASICS NAVIGATION

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  1. 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.

  2. 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

  3. 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

  4. .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.

  5. 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

  6. Altittude and temperature ±εi/p +εc ±ερ IAS CAS EAS TAS WIND ETAS GS The εc ≥ 0 If CAS ≤ 250kt CAS= EAS IAS CAS = IAS – (±εi/ρ) EAS = CAS – (+ εc) TAS = EAS – (±ερ)

  7. Direction Direction is postion’s point whith reference another point in the space determinate by the way and line that join both. B A

  8. 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.

  9. 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.

  10. 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.

  11. 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

  12. TN MN MN TN MC MC δE TC δw TC Reference line Reference line Figure 2 Figure 1 TN TN MN CN MN MC CN CC ΔE MC CC δw Δw TC TC δw Reference line Reference line Figure 3 Figure 4

  13. 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 A 050º /30 kt 23 kt B R B T1 19 kt H 000º TAS 100 kt TC 347º GS 78 kt T2

  14. 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

  15. Ttrigonometric basics knowledge

  16. 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 = ETAS ± wind component GS = ETAS ± wind component

  17. 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 ?

  18. 1º drift correction Development of calculating the intensity of the cross wind component. TC 000º A Wind 050º /30kt α 40º B C 50º 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

  19. 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)

  20. 2º calculate Headwind The wind that comes from the front quadrant is negative The wind that comes from the rear Quadrant is positive 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 D= Cos 50 x 30 = 19,2 kt as it is headwind will be -19 kt Vector wind 050 º/ 30kt β β

  21. 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

  22. 4º Check drift correction (dc) with ETAS ETAS Cos ( dc) =  TAS 97 Cos (dc) =  100 Cos 0.2371  invers  14 dc = 14º TH = 014º

  23. 5º Calculate the ,megnetic heading MC= TC – ( -δw) MH = TH –(-δW) TH= 14 MH= 14 –(-7)  MH= 021º TN MN MH 021 º -7º TH 014º

  24. 6º Time used to fly 50 NM and fuel burned Space Space 50 NM Speed =  Time =  Time = = 0,6410 hours Time Speed 78 kt 0,6410 hours x 60 minutes = 38,46 minutes. Fuel burned 0,6410 hours x 5 gal imp = 3,2 gal imp

  25. 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 B f g B Â Chf i h c B Ĉ Â CHi: Compass heading inicial CHf: Compass heading final CHf = CHi ± Ĉ Ĉ = Â + B CHf ± (Â + B)

  26. i i •  =  Sin  =  c = • c Sin • i i • B =  Sin B=  h = • h Sin B

  27. 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?

  28. dc 10 •  =  Sin  =  Sin  = = 0.1428  revers SIn = 8,2 º • ac 70 • dc 10 • B =  Sin B=  Sin B = = 0,03846  revers Sin = 2,2º • cb 260 CHf = CHi ± Ĉ  Ĉ =  + 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

  29. Author: Javier Pérez Mate Comercial pilot CPL Center study: Aeronautic formation school Aerofan FTO TRTO javinet20@hotmail.com Madrid Spain

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