October 31, 2017 | Author: Vivek Baskar | Category: Latitude, Orbit, Longitude, Twilight, Equator

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Cpl practice questions...

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1 NAVIGATION EARTH. It is an oblate spheroid whose major axis AB is 12748 km and minor axis CD is 12705km ie a difference of 43 km. Compression Ratio = diff of two axis/larger axis C A

B

CR = 43/12748 ~ 1/299 or 1/300 approx, Polar axis is Shorter by 1/300 times the equator. For all practical purposes Earth is considered as a Sphere.

D Great Circle. It’s a circle which when drawn across the earth cuts the earth into two equal hemispheres. Properties. (a) It’s the shortest path between two poles. (b) On earth sphere it appears as a straight line. (c) Only one GC can be drawn between two points unless they are diagrammatically opposite each other, in that case infinite number can be drawn. (d) Radio signal follow Great Circle Path. Latitude. The angle subtended by the shorter arc of the meridian at the centre of the earth from the Equator to the point to be identified is termed as latitude of that point. It is denoted as N or S depending on whether the point lies North or South of the Equator also known as Geocentric Latitude, whereas, Geodetic Latitude is the angle between the normal to the observers horizontal plane and the equatorial plane. Longitude. The angle subtended by shorter arc of equator at the centre of the earth from prime meridian to the point to be identified is termed as longitude and is denoted East or West depending on whether the point lies East or West of the prime meridian. Nautical Mile. It is the arc length subtended by 1 minute angle at the surface of the earth. The angle is measured from the Geocentric Latitude. Length of Nm is more at poles when measured from geographic centre. 1 Nm = 6017 feet at Poles- geographical centre 1 Nm =6045 feet at Equator – geocentric 1 Nm = 6080 feet subtended at 45 deg Latitude. 1 Deg = 60 Nm hence distance covered around the globe =360 x 60 = 21600 Nm (Great Circle path) Kilometre : It is 1/10,000 th part of the distance from pole to the equator. 1 km =3280 ft. 1 Nm =6080/3280 = 1.854 km and 1 km = 3280/6080 = 0.54 Nm Statute Mile. By statute it is 5280 ft.

2 Rhumb Line : A line which cuts all meridians at equal angles. It spirals to the poles at angles less than 90°. ATS routes are rhumb line tracks. Equator is a RL as well as GC. Parallels of Lat are RL tracks. Meridians are GC track (Flying a constant direction). Direction. It is measured with respect to the North, clockwise. All meridians point to North. True Direction. A direction which is measured with respect to True North the symbol is and is annotated as 005 T or 005 (T).The geographical north is not aligned with the magnetic earth since the earth behaves as a bar magnet with its axis slightly aligned from the geographical axis. The exact position of North an South pole with respect to the bar magnet is defined and known, however the position of magnetic north depends on where the observation is taken from vis a vis the position of True North. Magnetic North. It is the direction which points to the magnetic north. The angular difference between the True North and Magnetic North is known as Variation and is denoted as E or W depending on whether the Magnetic North lies east or west of true North. Simply, if Magnetic North lies East of True North the Variation is East and if Mag N lies W of True North Var is West. Thumb Rule - Var E Mag Least, Var W Mag Best. It is applicable in any hemisphere. The symbol for Magnetic north is

and is denoted by 045 M or 045 (M).

Isogonal : Lines joining places of equal Variation. Agonal : Lines joining places of zero Variation. Compass North. The direction measured wrt to Compass North is called Compass direction. The symbol for Compass North is

and is denoted as 005 (C) or 005 C.

Deviation. It is the angular difference between Magnetic North and Compass North and is termed as Easterly if Compass North lies East of Magnetic North and Westerly if Compass North lies West of Magnetic North. Deviation is obtained from Compass Card and it varies from aircraft to aircraft due to inherent magnetic fields present in the ac and is different for different headings. Thumb Rule - Dev E Com Least, Dev W Comp Best. It is applicable in any hemisphere. Any line drawn on a map represents a True Track.

3 Compass North

Magnetic North

True North

Variation W Deviation E

Fig: Deviation and Variation

POBLEMS ON DIRECTION (Black-Given, Red-Determined) (East is +, W is -ive) 1 2 3 4 5

C 269 027 051 300 045

D 5E 3W +3 -7 -4

M 274 024 054 293 041

V 4W 4E 4W 10E +6

T 270 020 050 303 047

4 SCALE FACTOR AND DEPARTURE Departure (Nm) = dlong x 60 x Cos Lat based on this formula distance in Nm along any latitude can be determined. As latitude increase for same dlong distance reduces. Therefore distance is maximum at the equator and zero at poles, also this is a Cosine function. If the Longitudinal change is known distance traveled along a particular latitude can be determined. It is applicable to both the hemispheres. While flying on Easterly hdg and crossing the ante meridian the value of longitude will change to westerly and vice versa. Hence to determine the longitude the following can be resorted to (360{longitude at origin + change in longitude}) in case we cross the ante meridian. Don’t forget to change the Easterly or Westerly Longitude depending whether the aircraft is traveling from East to West or vice versa. For eg an ac flying from 172 E on an easterly heading changes longitude by 12 degrees (dlong) then the position of the aircraft is (360-{172+12}) = 360 – 184 =176 degrees W. Problems on Departure 1. At what latitude a distance of 100 nm will involve a dlong of 30 degrees? A. Dep =100 nm, dlong =100nm, Lat=?, Dep=dlong x 60 x Cos Lat or 900=30 x 60 x Cos Lat or Cos Lat = 0.5 or Lat = Cos inv 0.5 = 60 Deg 2. How long will it take to go around the earth at 60 deg lat at G/S of 600 Kts? A. Dep =? Lat = 60 deg, dlong =360, G/S = 600 K, Dep=360 x 60 x Cos 60 = 10800 Hence time taken =10800/600 =18 hrs, whereas, the time taken to go round the equator is 36 h. 3. A is 60N 168E, B is 545 nm due East of A, what is its longitude. (See figure below) A. Dep = 545 nm, Lat = 60 deg, dlong = ? Substituting in Dep=dlong x 60 x CosLat 545= dlong x 60 x 0.5 or dlong = 545/30 = 18 deg 10 min =17350W 4. Find dep between 32 30 N, 20 42 E and 32 30 N, 89 26 E.

173°50W

180° 6°10’ 12°

168°E

A. dlong = 68° 44´ x 60 x Cos 52° 50´ =3478 Nm 5. Find RL distance between 42° 42’ N, 32° 42’ E & 42° 42’ N 69° 42’ W. A. 102° 24’ x 60 x Cos 42° 42’ = 4515.3 Nm 6. Aircraft takes off from A 60° 29’ N 177° 23’ E and flies a RL track of 090° for 600 nm. Find lat & long of destination. A. 600 ÷ 60 =10 ÷ Cos 60° 29’ = 20° 18’, 177° 23’ + 20° 18’ = 197 41’, 360° - 197 41’ =162° 19’ W 7. Aircraft takes off from A 40° 40’ N 176° 30’ W and flies a RL track of 270° for 600 nm. Find lat & long of destination. A. 600 (60 x Cos 40° 40’ = 13° 11’ + 176° 30’ = 189° 41’, 360° - 189° 41’ =179° 19’ E 8. An aircraft flying for 360 nm undergoes long change of 7° 23’. Find lat. A. 360 ÷ (60 x 7° 23’) = 0.8126, Cos inv 0.8126 = 35° 30’ N/S.

5

6 Scale. It is defined as the ratio of Map Distance (MD) to Earth Distance (ED). Scale = MD/ED. Scale is large (1/!00) or small (1/1000). Large scale map on a unit area smaller distance is shown, whereas in small scale map on a unit area large earth distance is shown. On a map Scale is represented by three methods:(a) Representative Fraction, a fraction whose numerator is always 1, e.g. 1:10000. (b) Statement in words, e.g. 1cm = 100 nm. (c) Graduated Scale

Km or Nm

Scale at Lat (SAL) = Scale at Equator (SAE) x Secant Lat or SAL = SAE/Cos Lat (A pplicable for Mercator Chart Only) Conversion Table 1 Nm = 6080 ft 1 Km = 3280 ft 1 SM = 5280 ft

1 m = 3.28 ft; 1 ft = 12 in; 1 in = 2.54 cms

Problems on Scale 1. Given MD = 20 cm; ED = 100 nm; Find Scale. A. Scale = MD/ED = 20/(100 x 6080 x 12 x 2.54) (1 Nm =6080 ft, 1 in =2.54 cms) or Scale = 1: 926592 2. Given MD = 20.5 in; ED = 600 km; Find Scale. A. Scale = MD/ED = 20.5 x 2.54 cm/600 x 1000 x 100 cm = 1:1152295 3. Given MD = 5 in; Scale = 1:2,000,000; Find ED A. 1/2,000,00 = 5/ED or ED = 5 x 2,000,000 in = 10,000,000/12 x 6080 Nm = 137 Nm. 4. An ac at a G/S of 300K covers map distance of 15 cms in 24 mins. Find Scale. A. Distance covered = 300/(24/60) = 80 Nm. Hence Scale = 15/(80 x 6080 x 12 x 2.54) = 1:988365 5. An ac covers a distance of 5 in on a chart (scale of 1: 1,000,000) in 20 min. Find G/S. 1/1,000,000 = 5/ED or ED = 5,000,000 in = 5000000/(6080 x 12) = 68.5 Nm in 20 mins. Hence Speed = (68.5 x 60)/20 =205.5 K 6. On a chart SAL 62°N = 1/1000000, Find (a) SAE (b) Scale at 40° N. A. (a) SAE = SAL x 1/Cos Lat or 1/1000000/Cos 62 = 1/ 2130054 (b) SAL = SAE x 1/Cos 40° = 1/2130054 x 1/Cos 40 = 1/1631716 Note : For Mercators Chart SAL= SAE x Sec Lat or SAE x 1/Cos Lat.

7 Thumb Rule. For conversion of SAE to SAL, multiply denominator by Cos Lat and for SAL to SAE divide denominator by Cos Lat. 7. On a mercator chart SAE = 1: 1450000, at what Lat scale will be 1: 1000000. A. SAL = SAE x 1/Cos Lat or 1/1000000 = 1/1450000 x 1/Cos Lat or 1450000 x Cos Lat = 1000000 or Cos Lat = 1000000/1450000 = 0.689, Hence Cos inv 0.689 = 46° 24’ N/S. 8. On a chart SAE = 1:1000000. On this chart two points A & B are 10° apart at 54° N. Find difference in longitude. A. SAL 54° N = 1000000 x Cos 54 = 1: 587785. MD/ED = 1/587785 or 10/ED = 1/587785 or ED = 5877850 in = 80.56 nm. Dep = 80.56 = dlong x 60 x Cos 54° = 80.56/ (60 x Cos 54) = 2° 17’ 9. Distance between two points at 45° N is 10 cm and ED is 100km. Find SAE. A. SAL at 45° N is MD/ED =10/ 100 x 100 x 100= 1: 1000000. SAE =1:1000000/Cos 45= 1:1414213 10. A & B are located at 50° N and are 1° 42’ long apart. Distance between them is 8 cm. Find SAE and SAL at 50° S. A. Dep = dlong x 60 x Cos 50 = 65.56 nm. Scale at 50° = MD/ED = 8 CM /65.56 X 6080 X 12 X 2.54 = 1: 1518684. SAE = SAL/ Cos Lat = 1: 1518684/Cos 50 = 1/2362653 11. The scale at 60°N is 1/2000000 on a Mercator Chart. At what latitude will you find the scale 1/1000000. A. SAE = 20000000/Cos60 =4000000, SAL = 1000000, SAE = 4000000 Hence Cos Lat =.25 or 75°31’ 12. If the scale at 57°20’N is 1:1091000 what is the meridian spacing in cm between one deg longitude. A. Dep =1 x 60 x Cos57°20’ = 32.385 Nm = 5997709 cm (1Nm = 1.852Km) 13. You are flying east along a parallel 60°N and cover 10 inches distance on the chart every hour. The scale at 25°S is 1:1000000. Find GS. A. SAE = 1000000/Cos 25 = 1:1103378, SAL = 1103378 x Cos 60 =551688.95. So 10” = 5516889” on chart which corresponds to 5516889/(6080 x 12) = 75Nm Hence speed is 75 K. 14. The distance between A & B both at 40°N is 10 cm on a Mercator chart and 90 km on earth. Find scale at equator. A.Scale at 40°N = MD/ED = 10cm/90,000,00 cm (90km)= 1:9000000. SAE = 9000000/Cos40=1174866 15. On a Mercator Chart if scale is 1:1M at 56°N. Find the Chart length from 2845N 11330W to 2845N 9815W. A. Dep = 15°15’ x 60 =915 Nm, 915 x 6080 x 2 = 66758400”/1000000 corresponding to 66.75” .

8 MAPS & CHARTS (PROJECTIONS) Map. It contains all geographical features like roads, rivers, mountains etc depending on the scale of the map. It depends on the scale of the map. Chart. It contains limited information for which the chart has been made e.g. enroute chart, approach chart etc. Large Scale Map. On a unit area of the map smaller earth distance is shown e.g. 1: 100 scale map is a bigger fraction than 1: 1000 wherein the details are more but the area depicted is smaller. Small Scale Map. On a unit area of the map large earth distance are shown e.g. 1:1000 meaning greater details are available for the same size of the map sheet as compared to a large scale map. Types of Map. There are two methods to construct a map, they are (a) Perspective and (b) Non Perspective (These are drawn mathematically). Construction. The perspective method of constructing a map involves projection of the graticule of the earth on a sheet of paper with the help of a light source placed at the appropriate place. The non perspective method involves mathematical reduction of spherical globe on a plain sheet of paper. Types of Projection. These are three types (a) Cylindrical (b) Conical and (d) Azimuthal or Zenithal. Cylindrical Projection

____ Rhumb Line

____ Great Circle

Construction. Light source is at the centre and point of tangency of the cylinder superimposed on the global sphere is at the centre (equator). Properties. The following properties emerge as a result of placing a cylinder on the global sphere with point of tangency at the Equator:1. Meridians are straight lines equidistant from each other. 2. Parallels of latitude are also straight lines but not equidistant from each other. Distance between them progressively increases from equator to poles. 3. Convergence (n=0) is zero. A straight line on this map is a Rhumb Line. A Great Circle is a curved line concave to the equator and convex to the poles. Convergency on earth is angle of inclination between 2 meridians at a given latitude and is = dlong x Sin Lat.

9 4. It is not an orthomorphic projection (Orthomorphism is the property of a projection in which bearings are correct in all directions within vicinity of the point). Note: For a projection to be orthomorphic the following conditions are to be satisfied:(a)

Meridians and Parallels of Latitude should cut each other at right angles (90°).

(b)

Scale should be constant within the vicinity of a point.

. MERCATOR PROJECTION 1. Mathematical modifications/corrections are carried out to make a Cylindrical Projection orthomorphic in that scale in E-W direction is varied at the same rate of scale expansion in the N-S direction, with increase in Latitude. 2. The scale varies as the Secant of the Latitude and is represented by the formulae Scale is correct only along the equator. Scale at any Lat (SAL) = Scale at Equator(SAE) x Secant of Lat or SAE/Cos Lat 3. Appearance & Properties are similar to Cylindrical Projection which are:(a) Rhumb Line(RL) is a straight line. (b) Great Circle (GC) curved concave to the RL. (c) Meridians cut parallels of Lat at 90°. Chart convergence is equal to earth convergence only at Equator, otherwise it is zero. Limitations. The limitations are:(a) This projection cannot be used for Polar Regions. (b) It can only be used effectively upto 70-75° N/S beyond which scale expansion is very large. (c) Adjacent sheets fit together in E-W direction not N-S. Usage. These charts are used for flying on Rhumb line tracks and are also used for Met Charts.

10 CONICAL PROJECTION

Lat of Origin is midway between two Std Parallels

Construction. The point of tangency is a particular latitude which can be selected by changing the Cone angle or Apex angle. Properties/Appearance. The following properties emerge for this type of a perspective projection:1. Meridians are straight lines converging to the nearest pole. 2. Parallels of Latitude arc of concentric circle not equidistant from each other. Distance between them increases away from Lat of origin on either side. 3. Rhumb line is a curved line concave to the nearest pole or great circle and convex to the equator. 4. A Great Circle is a straight line. 5. Convergence, which is angle of inclination between two meridians on a projection and is denoted by the symbol n = c/dlong, where c =convergency and dlong = difference in longitude, is less than 1. (Convergence is ratio of convergency to dlong on that map) 6. It is not an orthomorphic projection.

11 Lamberts Conical Orthomorphic (Conformal) between two Standard Parallels

Construction. It is a conical projection between two std parallels. mathematically modified to make it orthomorphic.

Base is perspective but

Properties. 1. Its an orthomorphic only between two std parallels. 2. Rhumb line is a curved line concave to the lat of origin/great circle/nearest pole and convex to the equator. 3. Great Circle is a curved line concave to the lat of origin, but for practical purposes it is nearly a straight line, as indicated below.

Scale. It is almost constant within the two std parallels. Away from lat of origin scale expansion takes place, but this scale expansion is negligible within the std parallels and is approx 1%. Outside the std parallels scale expansion is very large.

12

Scale Expansion

Q. On Lamberts Conical Projection scale is almost correct between Std Parallel and convergence is correct at Lat of Origin. Q. While measuring track on LC Projection protractor is placed at (a) Lat of Dep (b) Lat of arr/dest (c) Mid way Lat (d) any of the Latitudes. A. (c) Usage. All Jeppesen charts are Lambert Conical Projection. ZENITHAL OR AZIMUTHAL PROJECTION

Properties 1. Meridians are straight lines converging at poles. 2. Parallels of Lat are concentric circles not equidistant from each other, the distance between them increases from poles to the equator. 3. Rhumb Line is a curved line concave to the Great circle/nearest pole. 4. Great Circle is a straight line. 5. Its not an orthomorphic projection.

13 POLAR STEREOGRAPHIC PROJECTION

Properties. The point of tangency is at poles and light source is placed at the opposite pole, appearance is similar to Zenithal projection. Scale expands at the rate of Sec2 Lat. CONVERGENCY Convergency. It’s the inclination between two meridians or angular difference between two meridians. At equator the angle is Zero and at poles it is 1° (dlong). Therefore it varies as a function of Sin. Convergence at equator = 0, Conversion at poles = dlong and Conv at any Lat =dlong Sin Lat or Conv at Lat = dlong x Sin mean Lat. Q. GC Brg of A from B = 045°, CA = 5°. What is RL Brg of B from A in Northern hemisphere. A. RL Brg of A from B = GC Brg of A from B + CA = 045° + 5° = 050° RL Brg of B from A = 050° + 180° = 230° GC Brg of B from A = 230° + 5° = 235°

045° B

A

Q. RL Brg of X fom Y is 060°, CA = 6°, NH. Find GC Brg of X from Y, Y from X and RL Brg of Y from X. A. RL Brg of X from Y = 060° Therefore GC Brg of X from Y = 060-CA =060-6 =054° RL Brg of X from Y = 060 + 180 = 240° & GC Brg of Y from X = 240 + 6 = 246° Q. GC Brg of X from Y is 300°. CA = 7°, NH. Find GC Brg of X from Y, of Y from X & RL Brg of Y from X. A. GC Brg of X from Y = 300°, Hence RL Brg of X from Y = 300° - 7° = 293°, RL Brg of Y from X = 293 -180 = 113°, GC Brg of Y from X = 113-7 = 106° Q. RL Brg of A from B is 220°, GC Brg of A from B is 216° Find CA, Hemisphere, GCB of B from A. A. Difference of RLB & GCB = CA = 220-216 = 4°, RLB of B from A = 220-180 = 040°, Hence GCB of B from A = 040 + 4 = 044° in SH 216°

14 Q. GCB of A from B is 340°. RLB of A from B is 345°. Find, CA, Hemisphere & GCB of B from A. A. CA = 345-340 =5°, RLB of B from A = 345-180 = 165°, GCB of B from A = 165 + 5 = 170°, SH. Q. GCB of B from A is 070°.RL Brg of A from B is 256°. Find CA, Hemisphere & GCB of B from A. A. RLB of B from A = 256 -180 = 076°. CA = 076 – 070 = 6°, NH, GCB of B from A= 256 + 6 = 252° Q. GCB of A from B is 234°. GCB of B from A is 066°. Find, CA, Hemisphere & RLB of B from A. A. In these problems add both the GCBs and if difference is more than 180° then location is in NH, if difference less than 180° then in SH. Next, the value obtained after subtracting the GCB is to be subtracted from 180, Divide the absolute value by 2 to get CA. GCB of A from B = 234°. GCB of B from A = 066°, 234 -066 = 168° < 180 so in SH, 180 -168 =12, 12/2 = 6° = CA. RLB of B from A = 066-6 = 060°. Q. GCB of P from Q is 130°. GCB of Q from P is 318°. Find, CA, Hemisphere & RLB of Q from P. A. 318-130 = 188, Hence NH, 188 – 180 = 8°, CA = 8/2 = 4°, RLB of Q from P = 318-4= 314°. Q. A & B are on parallel of 30° N. GCB of B from A is 087°. Longitude of A is 8°W. Find long of B. A. RLB of B from A = 090° (since on same parallel of Lat), hence CA = 090-087 =3°. Also Convergence = 2 CA = dlong Sin Lat or CA = ½ dlong Sin 30° or dlong = 2 CA ÷ Sin 30 = (2 x 3) ÷ 0.5 = 12. Hence B lies 12° apart ie 12-8 = 4°E Q. A & B are on parallel of 30° N. GCB of A from B is 266°. Longitude of A is 10°W. Find long of B. A RLB of A from B = 270° (since on same parallel of Lat), hence CA = 270-266 =4°. Also Convergence = 2 CA = dlong Sin Lat or CA = ½ dlong Sin 30° or dlong = 2 CA ÷ Sin 30 = (2 x 4) ÷ 0.5 = 16. Longitude of B = 16-10 = 6° E

15 WIND TRIANGLE

HDG & TAS W/V DRIFT (p) TRK & G/S

When Trk is left of Hdg it is Port Drift and vice versa

Drift. It is the angular difference between the Heading and the Track. When Track is right of heading it is called Starboard Drift and when Track is left of heading it is called Port Drift. TMG. Track Made Good is the physical path followed on the ground and may differ from Track required due to inadequate drift correction. The angle between the Track Required and TMG is called Track error. When TMG is right of Tr reqd then it is called Stbd TE and when MG is Port of Track required it is called Port TE. Tr Reqd

TMG Port Hdg/TAS

TMG Stbd Hdg/TAS

W/V

W/V Trk/GS

Trk/GS Winds 90° to Trk (GS < TAS)

Winds 90° to Hdg (GS > TAS)

Problems on Wind Triangle. (a) (b) (c) (d) (e)

TRK 101 180 276 045 234

TE 8P 5S 6P 10 S 15 P

TMG 093 185 270 055 219

HDG 090 183 275 059 225

DRIFT 3S 2S 5P 4P 4P

16 Mean Winds. To calculate mean winds in a multiple leg, it is imperative to determine the GS and time for each leg. Alongside calculate the Wind Effect by multiplying Time on respective leg with corresponding wind velocity on that leg. Calculate total time by adding time taken on each leg, sum up the wind effect and divide by the sum of time taken on each leg. This is illustrated in the example below. Similarly to calculate TAS, multiply TAS of respective legs with corresponding wind velocity, total for all the legs and divide by the total time flown. This would give Mean TAS. Thereafter Mean GS = Mean TAS ± Mean Wind. Leg A-B B-C C-D

TAS 300 200 350

DIST 700 280 300

W/V +30 -10 +40

GS 330 190 390

TIME WIND EFFECT (TIME x W/V) 2:07 +63.5 Nm (30 x 2:07) 1:28 -14.7 Nm (-10 x 1:28) 0:46 +30.7 Nm (+40 x 0:46) 5:26 + 79.5 Nm Total Time = 2:07 + 1:28 + 0:46 = 4:21, Total Wind Effect = +63.5 -14.7 + 30.7 = 79.5 Hence Mean Winds experienced from A to D = 79.5 ÷ 4:21 = + 18.27. Similarly Mean TAS is calculated :Mean TAS (A-B) = 300 x 2:07 =635, (B-C) = 200 x 1:28 = 293.33 & (C-D) = 350 x 0:46 = 268.33 Total = 635 + 293.3 + 268.3 = 1196.6, this divided by total time 4:21 (1196.6 ÷ 4:21) =275 Kts is Mean TAS. Mean TAS + Mean Wind ( 275 + 18.27) = 293.3 Kts = Mean GS Q. Find the Mean GS from the data given below in black:Leg TAS DIST W/V GS TIME WIND EFFECT (TIME x W/V) A-B 160 600 +30 190 3:09 +94.5 Nm (30 x 3:09) B-C 290 500 -15 275 1:47 -26.75 Nm (-15 x 1:47) C-D 380 200 +45 425 0:28 +21 Nm (+45 x 0:28) * Data Calculated Mean Winds = 88.25/5:26 = +16.24; Mean TAS = (160 x3:09) + (290 x 1:49) + (380 x 0:47) = 1212. This divided by Time (5:26) = 222.38, Hence Mean GS = 222.38 + 16.24 = 239 Kts RELATIVE MOTION Thumb Rule. While solving problems on relative motion, the following must be kept in mind:1. When aircraft are flying in same direction first calculate the relative speed (Difference in the two also called overtake speed) then divide by distance to obtain time taken to overtake.. For example Overtake = 40 Kts, Distance = 80 Nm, then time taken to overtake = 80/40 = 2h. Similarly when 4 nm behind, time taken =76/40 = 1:54 h and time taken when 4 nm ahead after overtaking = 84/40 =2:06 h 2. Time of Crossing = Relative Distance ÷ Relative speed (Add speeds when aircraft approaching each other and subtract when flying one behind the other. Q. At 0900 h Aircraft X is behind Y by 80 Nm while flying on same track. GS of X= 240 K and that of Y= 200K. Find when will X overtake Y and when will X be 4nm short & ahead of Y. A. Example worked out above, time will be 1100 h, 1054 h and 1106h respectively

17 Q. At 0700 h, while flying on same track aircraft A is behind B by 120 Nm. GS of A = 300K; B=250K. Both are flying to point P which is 1200 Nm from present position of aircraft. Find when will (a) A overtakes B, (b)A is 5nm short of B, (c) 5nm ahead of B. (d) At what distance from P, A will overtake B. A. (d) 300 x 2:24 = 720 Nm, hence distance from P =1200-720 = 480 Nm. Q. At 0900, Flying on same track aircraft P is behind Q by 50 nm. GS of P = 200K, Q=160K, Find (a) P will overtake Q, (b)P will be 6nm short of Q, (c) 6nm ahead of B. A. Overtake = 40 K, Distance = 50 Nm, (a) Time taken to overtake = 50/40 = 1:15, i.e. 1015 h. (b) 44/40 = 1:06 h = 1006 (c) 54/40 = 1:21 = 1021 h. Q. At 500nm from destination aircraft is asked to delay ETA by 8 min. At what time and distance should aircraft reduce speed to 150 K if it was flying at 180 K. Present time is 1200h. A. In this problem we need to determine:(a) (b) (c) (d) (e)

Original ETA Revised ETA New distance covered with revised speed Time to drop speed Distance to drop speed.

Original Speed = 180 K, Revised Speed = 150 K, Original ETA = 500/180 =02:47 = 1447h, Revised ETA = 02:46 + 0:08 =02:54 = 1455h. New distance with revised speed = 150 x 2:54 = 435 nm. If aircraft was at 435 nm from destination, it would have reached destination at correct ETA, the balance 65 Nm (500-435 = 65Nm) can be construed as if one aircraft behind the other at higher speed at 500 nm and overtakes at 435 Nm with an overtake of 180 -150 = 30 Kts. Now distance = 65, Overtake = 30, time taken = 65/30 = 2:10 h, (a) 1410h (b) Distance to drop speed = 2:10 x 180 = 390 Nm Q. At 800nm from destination aircraft is asked to delay ETA by 15 min. At what time and distance should aircraft reduce speed to 360 K if it was flying at 420 K. Present time is 1200h. A. Original ETA = 800/420 =1:54, Revised ETA = 1:54 + 0:15 = 2:09, New Distance with revised Speed = 360 x 2:09 =774 Nm, Distance = 800 -774 = 26 Nm, Overtake = 60, Time = 26/60 = 0:26 (time to drop speed) i.e 1226h. Distance to drop speed = 0:26 x 420 = 182 Nm. Hence 800- 182 = 618 Nm Q. At 600 nm from destination an aircraft is asked to reach early by 10 mins. At what time and distance it should increase its speed to 240 K from 160 K. Present time is 1200h. A. Original ETA =600/160 = 3:45, Revised ETA = 3:45 -0:10 = 3:35, New Distance with revised speed = 3:35 x 240 = 860 nm. This can be compared to an aircraft with overtake of 80 kts behind by 260 Nm 260 Nm 600 Nm Time taken to cover 260 Nm at 80 K = 260/80 = 3:15 At time =1200 + 03:15 = 1515 (Time to increase speed); Distance to increase speed = 3:15 x 160 = 520

The One-in-Sixty Rule

18 1. The one-in-sixty rule is based upon the fact that one nautical mile subtends an angle of one degree at distance of 60 nautical miles, i.e. 5 miles subtend 5 degrees etc.

One-in-Sixty Rule. 2. In applying the rule, the triangle relevant to the navigational problems is identified, and the ratio of the length of the long side to 60 is established. This ratio may then be applied to the angle to reveal the length of the side opposite to its or conversely, to the opposite side to reveal the angle it subtends.

Track error =

60 x 3 = 9° 20

Heading correction at B, (a) to destination. = = = = (b)

< EBD < BAD 60 x 3 20 9° + 3° 12°

+ < BDA + 60 x 3 60

2x

= =

2 x 9° 18° (Heading altered ‘back a’ C)

19 Examples on 1:60 rule θ = S/R x 60 (to be used when θ is < 20°) 1. After flying for 240 nm an aircraft is 12 nm right of track. What is the drift. Dist =240 nm 12 nm off track

θ = S/R x 60 =(12/240) x 60 = 3° Drift = 3° S

2. After flying for 480 nm aircraft is 20 nm port of track. If remaining distance to destination is 300 nm, what is approx heading to reach destination if ac was flying a heading of 045° A. 480 Nm 3°

300 Nm 4° 20 Nm off track 3° 045° Heading to Alter = 045° + 3° + 4° = 052°

3. After 2 hours at GS 180 K, aircraft is 12 nm left of track. If remaining distance is one hr at same GS, Find drift if ac was flying a course of 200° A. Distance covered = 360 nm, 12 nm port of track hence drift = (12/360) x 60 = 2°

20

SOLAR SYSTEM : TIME 1. The Solar System consists of the Sun, nine major planets of which the Earth is one, and about 2,000 minor planets or asteroids. All members of the solar system are controlled by the Sun which is distinguished by its immense size and its radiation of light and heat ; for all practical purposes, it may be considered as the stationary centre round which all the planets revolve. 2. Unlike the Sun, the planets and their satellites are not self-luminous, but reveal their presence by reflecting the Sun’s light. The planets revolve about the Sun in elliptical orbits, each one takes a period of time about the job: Mercury takes 88 days, for example, while Pluto which is rather a long way from the parent body, is thought to take about 248 years. The planetary satellites in the meantime are revolving about their own parents. 3. Certain laws relating to the motion of planets in their orbits were evolved by the astronomer Kepler, who died in abject poverty as a reward. (a) Each planet moves in an ellipse, with the Sun at one end of its foci. (b) The radius vector of any planet sweeps out equal areas in equal intervals of time. These are the important laws for our purpose in studying the Earth’s motion, as we shall see. 4. The Earth rotates on its axis in a West to East direction, resulting in day and night. It revolves round the Sun along a path or orbit which is inclined to the Earth’s axis at about 66 ½°, resulting in the seasons of the year. When the Earth is inclined towards the sun, we get the Summer Solistice (about June 21); when the axis is away from the Sun, we get the Winter Solistice (about Dec 22). When the Earth’s axis is at right angle to the Sun, day and nights are equal the Spring and

Fig 1 Autumn Equinox (March 21 and Sept 23). The point where the planet is nearest to the Sun is called perihelion, and where farthest aphelion; it is worth noting that in obeying Kepler’s second law, the speed of the Earth at perihelion is faster along its orbit than at aphelion. 5. Orbital velocity of the earth is not constant during its orbit, velocity is more when earth is closer to the sun and minimum when it is the farthest.

21 6. Earth rotates around its axis and revolves around the sun. Rotation gives us day and night, revolution gives us the year and inclination of the earths axis in its plane of rotation gives the seasons. 7. Inclination of the earth from its axis 23½° and 66½° from the plane of rotation. 8. Position of sun varies from 23½° N to 23½° S. This is called Declination (latitude of any heavenly body with respect to an observer). The northern most point corresponds to Tropic of Cancer and southernmost Tropic of Capricorn. Q. Sun will appear at the same latitude (a) once a year (b) twice a tear (c) every day (d) none of the above. A. (b) 9. The position when the earth is nearest to sun is called Perihelion and furthermost is called Aphelion. 10. The position when the earth is equidistant from the sun is called Equinox. Q. At what position of the sun you will have equal Day and Night? A. At Equinox, 21 Mar & 23 Sep. 11. Year. There are two types of Year, Sidereal and Tropical. Sidereal Year is the time interval elapsed between two successive conjunctions of earth, sun and a fixed point in space. Tropical Year is time interval elapsed between two successive conjunction of earth, sun and a fixed point in Aries. This is also known as Calendar Year. 12. Calendar Year. It takes 365 days 5hours 48 min 42 sec for the earth to go around the sun. Thus every 4 years adds to one day extra which is compensated by the leap year. Every 100 year is not a leap year. To compensate for 11 min 18 sec every 400 year is a leap year. 13. Sideral Day. It’s the time interval elapsed between two successive transits of a fixed point in space over an observer meridian or its time interval elapsed between two successive transits of a fixed point in Aries over an observer meridian. It is 23h 56 min since taken with reference to a star wherein the revolution and rotation of the earth does not matter. 14. Apparent Solar Day. It is the time interval elapsed between two successive conjunction of true Sun in space over an observer meridian. 15. Mean Solar Day. It is with respect to an imaginary Sun which goes around the earth nover equator at a constant velocity of 15°/hr. S Sidereal it is not wrt Sun 23:56 A Apparent not fixed due to orbit and revolution of earth around the sun 23:44 to 24:14 M Mean At a constant velocity of 15°/hr. Twilight 16. When the Sun is below the horizon, an observer will still receive light which has been reflected and scattered by the atmosphere. It is divided into three stages; Astronomical (Sun 12° to 18° below Horizon. It is completely dark with no natural light at 18°) ; Nautical (6° - 12° below the Horizon, and has to do with the sea horizon being indistinct, and artificial light is still required) and Civil Twilight when the Sun’s centre is actually between 1 and 6 below the horizon, when work is possible without artificial light, and the stars are nor clearly visible. This last is the one we are concerned with.

22 TIME There are of four types namely, LMT, UTC, Zone Time and Standard Time. Local Mean Time (LMT) It’s the time kept with respect to position of the sun at anti-meridian of an observer. At places east of any observer the LMT will be ahead and west of observer LMT will be behind due to earth’s rotation. Q. LMT at 35°E is 1300h. Find LMT at (a) 102°E (b) 40°W A. (a) dlong/15 = (102-35)/15 = 4h 28 m, hence LMT at 102E = 1300 + 4:28 = 1728 (b) dlong = (75/15) =5h, hence LMT = 1300- 5 = 0800h Coordinated Universal Time (UTC) It is the LMT prevailing at prime meridian or time kept with respect to antemeridian of Prime Meridian (observer is sitting at Prime Meridian). Q. LMT at 000E is 1200h on 28 Feb 04. What will be the UTC at 180W? A. UTC WILL NOT CHANGE AT ANY LONGITUDE IT REMAINS THE SAME. Q. LMT at 40N 60E is 1100 h. Find (a) UTC (b) LMT at 60S 120E (c) 60S 30W. A. (a) UTC at 60E = LMT - dlong/15 = 1100 -60/15 = 0700h, (b) UTC = LMT 120E – dlong/15 or LMT 120E = UTC +120/15 =0700 + 8h = 1500 h (c) UTC = LMT 30W + dlong/15 or LMT 30W = UTC – 30/15 = 0700 – 2 = 0500 h Q. An ac takes from place X (30N 170W) for Y (50S 160E). Total flight time is 08 Hrs. Time of departure is 2200h on 06 Jun (LMT). Find ETA at destination in LMT. A. UTC = LMT + 170/15 = 2200 + 11h20m = 0920 (07 Jun), After 8 hrs of flying UTC is 0920+8 =1720 (07 Jun), Now LMT (160W) =UTC + C = 1720 (07) + 160/15 =1720 +10:40 = 0400 (08 Jun) Q. LMT at 45N 100E on 17 May is 0512. Find UTC & LMT at 60N 120W. A. UTC = LMT (100E) – dlong/15 = 0512 – 6:40 = 22:32 (16 May), UTC = LMT (120W) + dlong/15 = LMT+8 or LMT = 22:32 - 8h = 14:32 (16 May) ZONE TIME The earth is divided into 24 hr zones, alphabetically assigned, beginning from “A” to “Z” except “I” & “O” The longitudes on earth measuring 360° are divided into 24 zones, each of 15° corresponding to 1 hour of time. The zones east of prime meridian are assigned negative signs while zones lying west of Prime meridians are assigned positive sign. Each zone of 15° is further divided into 7½° either side of the prime meridian which corresponds to 30 mins of time. For example India lies at 82° 30´ which when divided by 15 gives us 5h30m and that is the time we are ahead of UTC.

23 Zone Number. It is a number which is to be added algebraically in zone to get UTC. For example if at 82E the zone time is 1200, then UTC = ZoneTime ± Zone Number = 1200 -82/5 (5)=0700. Thumb Rule is divide the longitude by 15 if remainder is ≤ 7.5 use lower zone else use higher zone. Q. Find the zone number of (a) 120W (b) 127.5W (c) 130E A. (a) 120/15 = +8 (since West) (b) 127.5/15 = +8 Remainder 7.5 (hence same zone) (c) 130/15 = -8 Remainder 10, Hence higher Zone i.e -9 Q. At 160E difference between LMT and Zone Time is (a) LMT will be ahead by 40 min than zone time (b) LMT will be behind by 40 min (c) LMT will be ahead by 20 min (d) LMT will be behind by 20 min. A. Zone Number = 160/15 = -10 + Rem 10 hence = -11. LMT at 160E =160/15 = 10h 40m. Zone Time =1100 hrs LMT = 10h40m hence 20 minutes behind time (d) is the correct choice. Standard Time Time maintained with respect to a specific meridian or longitude is called Standard Time. IST is maintained with respect to 82° 30´ longitude. Q. Find LMT, UTC, Zone Time, IST at 8250N 8345E. A. UTC = LMT –C or LMT =UTC + 83° 45´/15 =UTC + 5h 35m or 0535 h (since UTC =0000) Zone Time =0600, IST =0530h UTC = 0000 International Date Line (IDL). When traveling Westward from Greenwich, an observer would eventually arrive at longitude 17959W, where the LMT is about to become 12 hours less than UTC. An observer traveling Eastward from Greenwich would eventually arrive at 17959E where the LMT is about to become 12 hours more than UTC. Thus there is a full day of 24 hours difference between the two travelers, although they are both about to cross the same meridian. When the antemeridian of Greenwich is crossed, one day is gained or lost, depending on the direction of travel: the Dateline is the actual line where the change is made, and is mainly the 180 meridian, with some slight divergences to accommodate certain groups of South Sea Islands and regions of Eastern Siberia. The problem readily resolves itself in flying - your watch is always on UTC: the place who’s Standard Time you want is listed in the Air Almanac: apply the correction to , and the date will take care of itself. GMT

Going on dateline from, East to West – gain a day subtract a date West to East – lose a day add a date

24

Prime Meridian 0600 UTC Dec 10

Q. While crossing IDL from East to West (a) LMT will be ahead, date will be ahead. (b) LMT will be behind, date will be ahead. (c) LMT will be behind, date will be behind (d) LMT will be ahead, date will be behind A. (d) Twilight Period. It is the period before sunrise and after sunset when diffused light of Sun is available.. Sensible Horizon. Horizon which is visible to the naked eye. Visible Horizon. A horizon which is not visible is called visible horizon. It is belo the sensible horizon. Note. When a body rises above the visible horizon it is said to be visible and it is said to be set when it is below the visible horizon. Twilight period in Air almanac is with respect to Civil Twilight. Q. Sunrise and moonrise table on Air Almanac are given in (a) UTC (b) Zone time (c) LMT. A. (c).

25 RADIAL INTERCEPT Procedure to adopt for intercepting a final track radial is:(a) Angle should be either 30/60/90° with respect to the final track/radial. (b) NO DRIFT IS TO BE APPLIED DURING INTERCEPT. (c) To determine the angle of intercept, find how many degrees the ac has to turn to intercept the radial. Double the number of degrees to turn and the figure closest to 30/60/90 will be the intercept angle e.g. if the number of degrees to turn is 25° then 25 x 2 = 50 which is closer to 60° intercept. Q1. Aircraft is approaching Station on a radial 180°is asked to approach on radial 155°. Find out (a) Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading. (a) Difference = 25° x 2 = 50° Hence 60° (b) On radial 155° Hdg to Station is 335° + 60° = 035° (c) Aircraft on Hdg 000°, hence degree to turn = 35° 360°

355°

(d) RBI will read 60° to left i.e. 300°

26 PAYLOAD Payload. It is the load which can be carried in the form of passengers and cargo.

AICRAFT AUW BREAKDOWN

FOB (FLT FUEL +RESERVE)

TAKE OFF WT (TOW)

PAYLOAD (PAX + CARGO NEWS PAPER + CATERING

ZERO FUEL WT (ZFW) DRY OPERATING WT

CREW + BAGGAGE

BASIC WT OR APS WT

AIRFRAME + ENGINE + AVIONICS

MANUFACTURERS WT

MTOW. It is the Take Off weight given by the manufacturer which cannot be exceeded in any circumstances. This caters for the best operating conditions i.e. runway length, elevation, density altitude, runway gradient, runway condition and winds etc. It is also known as Max Gross take off wt. RTOW. (Restricted/regulated/rated) This is the take off weight restricted due to prevailing conditions at the places of departure. TOW FOB TOW = BASIC WT + PAYLOAD + FUEL ON BOARD (FOB) Payload PAY LOAD = TOW – (BASIC WT + FOB)

Basic Wt

MZFW = PAYLOAD + BASIC WT MLW (Max Landing Wt) It is the maximum weight at which a landing can be made at a destination without imposing any structural damage to the aircraft. MZFW (Max Zero Fuel Wt). When wing tanks are empty there is a maximum permissible weight of an aircraft including all its contents. Exceeding this weight causes unacceptable load to the structure of the aircraft. Above this weight, if any load is taken onboard it can be fuel only. Numericals on Payload. To solving any problem on Payload the following procedure is adopted:Step 1. Make a table as given below and enter relevant information as given in the problem:MTOW

RTOW

MLW +FF (Flight Fuel) = TOW

MZFW +FOB =TOW

27 Choose the lowest value obtained out of MTOW/RTOW/MLW OR MZFW. Then calculate Payload by substituting this value of TOW in Payload = MTOW – (BASIC WT + FOB) Q. MTOW = 83000 lbs, MLW = 66000 lbs, Basic Wt = 52000, FF = 20000 lbs, Reserve = 2800 lbs. Find Payload. A.

MTOW 83000*

RTOW

MLW 66000 +17200 83200

MZFW + FOB =TOW

* LOWEST VALUE PAYLOAD = TOW – (BASIC WT + FOB) = 83000 - (52000 + 20000) = 11000 lbs Q. MTOW = 82000 lbs, MLW = 64500 lbs, Basic Wt = 50000, FOB = 20000 lbs, Reserve = 3000 lbs. Find Payload. A.

MTOW 82000

RTOW

MLW 64500 +16000 =80500*

MZFW + FOB =TOW

* LOWEST VALUE PAYLOAD = TOW – (BASIC WT + FOB) = 80500-(50000 + 19000) = 11500 lbs Q. In the above question can you carry additional fuel without affecting payload? A. Yes, (82500-80500=1500) but the fuel carried has to be consumed (burn off/dump) prior to landing, Q. MTOW = 120000 lbs, MLW = 90000 lbs, MZFW = 85000, Basic Wt = 76400, Trip Fuel = 15000, Reserve = 2000 lbs. Find Payload. A.

MTOW 120000

RTOW

MLW 90000 +15000 =104000

MZFW 85000 + 17000 =102000*

* LOWEST VALUE PAYLOAD = TOW – (BASIC WT + FOB) = 10200 - (76400 + 17000) = 8600 lbs Q. In the above question find payload if Flight Fuel is reduced by 1000 lbs and increased by same amount? A. Payload in both cases will remain the same since value of MZFW has been applied. Q. MTOW = 20000 lbs, MLW = 18000 lbs, MZFW = 17000, Basic Wt = 14000, Trip Fuel = 3000, Reserve = 1600 lbs Find (a) Payload (b) Payload if aircraft consumed 700 lbs reserve before landing (c) Find Payload if FOB is reduced by 700 lbs. A.

MTOW 20000*

RTOW

MLW 18000 + 3000 =21000

MZFW 17000 + 4600 =21600

* LOWEST VALUE PAYLOAD = TOW – (BASIC WT + FOB) = 20000 - (14000 + 4600) = 1400 lbs (a) 1400 lbs (b) Will remain same (c) Payload can be increased by 700 lbs (1400+700) =2100

28 Q. Fuel Consumption = 120 lbs/hr; MTOW = 7150 lbs, MLW = 6900 lbs, MZFW = 6150 lbs, Basic Wt = 5000 lbs, Reserve = 160 lbs. Dist =960 Nm, TAS=180k, Head Winds of 20 Kts. Find payload in NIL wind conditions. A. Dist = 960 nm; TAS = 180 Kts, Time = 5.33 x (FF = 120 lbs/hr) = 640 lbs. MTOW RTOW MLW MZFW 7150 6900 + 640 6150 + 800 =7540 =6950* * LOWEST VALUE PAYLOAD = MTOW – (BASIC WT + FOB) = 6950 - (5000 + 800) = 1150 lbs Q. A flight is to be made from M to N and return to M carrying max payload in each direction. Fuel is not available at N. Distance M to N =80 Nm, Mean GS M to N = 70 kts, Mean GS N to M = 110 Kts, Mean Fuel consumption = 410 Kg/hr, MTOW at M = 6180 kg, MLW at M = 5740 kg MTOW at N = 5800 kg, MLW at M = 5460 kg, MZFW = 5180, Basic Wt = 4400 kgs, Res Fuel = 250 kgs. Calculate (a) Max payload which can be carried from M to N and from N to M. A.

Total Fuel = Fuel reqd from M to N + Fuel reqd from N to M + Reserve Fuel reqd: M to N = 80/70 = 1.42 x 410 = 468, Fuel reqd: N to M = 80/110 = 0 .727 x 410 = 298, FF = 468 + 298 = 766 kgs, Reserve = 250 kgs, Total Fuel = 1016 Kgs. Payload from M to N: MTOW 6180*

RTOW

MLW 5740 + 468 =6208

MZFW 5100 + 1016 =6196

* LOWEST VALUE PAYLOAD = MTOW – (BASIC WT + FOB) = 6180 - (4400 + 1016) = 764 kgs Payload from N to M: MTOW 5800*

RTOW

MLW 5460 + 298 =5758

MZFW 5180 + 548 =5728

* LOWEST VALUE PAYLOAD = MTOW – (BASIC WT + FOB) = 5728 - (4400 + 548) = 780 kgs Q. Given MTOW = 34,500 kgs, MZFW = 28,000, MLW = 31,000, Empty Wt = 17,500 kgs, TAS = 350 Kts, Fuel Consumption = 1450 Kg/hr. Reserve Fuel 1200 kgs for all flghts (assume not used) Fuel Tank Capacity = 10, 500 kgs. Find (a) Max Payload (b) In NIL wind condition distance upto which above payload can be carried. (c) Max distance you can fly in NIL winds. (d) What payload you can carry in part (c). A. Max payload = MZFW – Basic Wt = 28000-17500 = 10500 kgs. TOW =10500 + 17500 +FOB or 34500=28500 + FOB or FOB = 6500. Hence FF = 6500-1200 = 5300 kgs. Fuel Flow = 1450 kg/hr, TAS = 350 hence max distance = 5300/1450 x 350 = 1279 Nm. Tank capacity = 10500 kgs. Hence 10500 – 6500 = 4000 kgs of fuel can be carried in lieu of payload. So total fuel available = 5300 + 4000 = 9300 kgs. Hence max distance = 9300/1450 x 350 = 2245 Nm and pay load would reduce by 4000 kgs in lieu of fuel. Payload = 10500-4000 = 6500 kgs.

29 CONVERSION OF UNITS 1.

Litre (l) x Specific Gravity (SG) = Kg

2.

Imperial Gallon (UK Gal or IG) x SG x 10 = Pounds (lbs)

3.

Kg x 2.2046 = lbs

4.

IG x 1.2 = US Gal (USG)

5.

100 l = 22 IG = 26.4 USG.

Problems Q. Convert 100 USG to (a) litres (b) Kgs (c) lbs (d) IG. Given SG = 0.78 A. (a) (b) (c) (d)

1 USG = 100/26.4 L = 3.785, 100 USG = 378.7 l Litres Kgs = l x SG also 100 USG = 378.71 litre Therefore 100 USG = 378.5 x 0.78 = 295 Kgs 100 USG = 295 kgs or 295 x 2.2046 = 650 lbs 100 USG = 100/1.2 = 83.3 IG

Q. In the following table find the most fuel efficient figure when winds = -20 and SG = 0.8 (a) (b) (c) (d) (e)

TAS 200 220 240 260 285

Fuel Cons (Gals/Hr) 250 265 280 295 315

GS/FC 180/250 = 0.72 200/265 = 0.75 220/280 = 0.78 240/295 = 0.81 265/315 = 0.84 (Most Efficient)

A. GS/FC = Nm/Hr ÷ Gals/Hr = Nm/ Gals ie Nm per gallons, the highest figure will be most efficient. Winds are 20 kts Head Wind, GS will be TAS – 20 in Kts. Working is on the above tale in red. Q. In the following table find the most fuel efficient figure when SG = 0.8 (a) (b) (c) (d) (e) (f)

TAS 180 240 220 160 190 200

Fuel Cons 1.25 USG/Nm 4.1 IG/min 815 Kg/hr 12.7 L/min 1.13 IG/min 13.41 Kg/m

FC (IG/hr) 1.25 x 180 =225 USG/Hr = 225/1.2 = 187.5 IG/hr 4.1 IG/min x 60 = 246 IG/hr 815/0.8 = 1018.75 l/hr = 1018.72 x 0.22 = 224.125 IG/hr 12.7 x 60 = 762 l/hr x 0.22 =167.64 IG/hr(Most Efficient) 1.13 x 190 = 214.7 IG/hr 13.41 x 60 =804.6 Kg/hr /0.8 =1005.75 l/h = 221.6 IG/hr

A. In this problem convert all Fuel Consumption figures to IG/hr then compare which is the lowest as worked out in red in the table above.

30 Q. In the following table find the most fuel efficient figure when SG = 0.8 (a) (b) (c) (d) (e) (f)

TAS 180 240 220 160 190 200

Fuel Cons (IG/hr) 187.5 246 224.125 167.64 214.7 221.65

FC (IG/Nm) 187.5/180 = 1.0416 246/240 = 1.025 224.125/220 = 1.018 (Most Efficient) 167/160 = 1.0477 214.7/190 = 1.1263 221.65/200 = 1.108

A. Convert all figures into IG/Nm, the least figure will give the most fuel efficient figure as worked in in red in the table above. Q. Fuel efficiency is 10.13 kgs/nm, TAS = 310 kts, Winds = +45, SG = 0.81. Find Flow in IG/hr. A. 10.13 Kg/Nm = 10.13/.81 = 12.506 l/Nm = 12.506 x 355 (GS) = 4439.63 l/hr = 4439.63 x 0.22 = 976.5 IG/hr

31 CRITICAL POINT (CP) Critical Point. It is a point in between two places from where it takes same time to reach either of the point or it is a point enroute from which it takes equal time to either come back or go the destination. It is calculated with one engine failed or switched off. It is also called Equitime Point. 1200 Nm Mid Point CP

A

Increased GS Home (TAS + WV)

B W/V 090/20 Reduced GS Out bound (TAS – WV)

If distance D between two points A & B, TAS (with one engine failed) and W/V are known, GS out (O) and GS Home (H) can be calculated and the figure of PNR can be arrived at by substituting these values in the equation:Distance to CP = DH/(O +H) where D = Total Distance, O = GS outbound with one engine failed and H = GS Home with one engine failed. Thumb Rule:When you find Dist to CP, always calculate GS Out and GS Home with reduced TAS Time to CP = Distance to CP/ GS out with all engines running unless specified. Example. Distance A to B = 1200 nm; Tr = 090, W/V = 090/20, TAS (4/3 engines) = 180/150* Kts, Find Distance & Time to CP. Always draw a rough diagram and a table as indicated below before attempting any problem:Refer figure above

H

CP to A

TR 270

TAS 150*

W/V 090/20

GS 170

O

CP to B

090

150*

090/20

130

G

A to CP

090

180

090/20

160

To solve the problem (a) Calculate the GS from Nav Computer and enter the figures obtained. (b) Substitute these values in the Distance to CP and Time to CP formula (use bracket function in the calculator it is that much faster and easier to obtain the correct final figure) Distance to CP = DH/(O +H) = (1200 x 170)/(130 + 170) = 680 Nm Time to CP = Distance to CP/ GS out = 680/160 = 04h:15m

32 Q1. D = 2000 nm, Tr = 270, W/V = 090/25, TAS (4/3) = 330/270 Kts. Find Distance & Time to CP. A.

090/25 B

CP

A

H

CP to A

TR 090

TAS 270*

W/V 090/25

GS 245

O

CP to B

270

270*

090/25

295

G

A to CP

270

330

090/25

355

Distance to CP = DH/(O +H) = (2000 x 245)/(295 + 245) = 907 Nm Time to CP = Distance to CP/ GS out = 907/355 = 02h:33m POINT OF NO RETURN (PNR) Point of No Return (PNR). It is a point at maximum distance removed from base upto which an aircraft can fly and still be able to return within safe endurance of the aircraft. It is calculated primarily to cater for non-availability of destination. This is purely a function of endurance which is given by the equation Endurance = (FOB – Reserve)/ Fuel consumption. The distance to PNR is calculated by the formula :Distance to PNR = EOH/(O + H), where E = Endurance, O = GS outbound (with all engines operating) and H = GS Home (with all engines operating) unless specified.

PNR GS Out A GS Home

B

W/V

Example Q2. D = 2000 nm, Tr = 270, W/V = 090/25, TAS (4/3) = 330/270 Kts. Fuel on board = 1000 kgs, Reserve = 200 kgs, Fuel Consumption (4/3) = 180/150 kg/hr Find Distance & Time to PNR. A. O

A to PNR

TR 270

TAS 330

W/V 090/25

GS 355

H

PNR to A

090

330

090/25

305

To calculate Distance to PNR insert values of TAS with all 4 engines running in the table shown above. Next calculate the Endurance by determining Flight Fuel = (FOB – Reserve), divided by Fuel Consumption (all engines operating), then calculate the GS and enter values in Dist to PNR formula.

33 (a) Endurance = (1000 -200)/180 = 4h:26m (b) Dist to PNR = (4:26 x 355 x 305)/(355+305)=729 Nm (c) Time to PNR = 729/355 = 2h: 03m. STILL AIR RANGE (SAR) SAR. It is the maximum distance upto which an ac can fly out in NIL wind conditions consuming total fuel onboard. This is a theoretical figure to cross check whether Flight Plan is executable. SAR = (FOB x TAS)/ FUEL CONSUMPTION Q3. Find the SAR in the above question. FOB = 800, TAS = 330, FC = 180 kg/hr Hence SAR = (800 x 330)/ 180 = 1833 Nm. Practice on Nav Computer (Given TRK, TAS and W/V, Find GS, Hdg and Drift) TRK TAS W/V 270 190 330/10 350 200 270/25 135 240 050/37 293 137 330/17 * TRK less than HDG Drift is Port and vice versa

GS 185 186 234 123

HDG 273 344 126 297

DRIFT 3 P* 6S 9S 4P

Q4. Dist = 1400 Nm, Tr =090, W/V = 330/18 Kts, TAS (4/3) = (210/180), FC (4/3) = 100/80 Kg/hr FOB = 1600 kg, Reserve = 150 Kg. Find (a) Dist & Time to CP (b) Dist & Time to PNR (c) SAR. Assume one engine failed at PNR and ac returns on 3 engines. A. H

CP to A

TR 270

TAS 180*

W/V 330/18

GS 188

O

CP to B

090

180*

330/18

170

G

A to CP

090

210

330/18

218

(a) Distance to CP = DH/(O +H) = (1400 x 170)/(170 + 188) = 665 Nm Time to CP = Distance to CP/ GS out = 665/218 = 03h:02m (b) O

A to PNR

TR 090

H

PNR to A

090

TAS 210

W/V 330/18

GS 218

210

330/18

200

Endurance = (1600 -150)/100 = 14h:30m (b) Dist to PNR = (14:30 x 218 x 200)/(418)=1512 Nm (c) Time to PNR = 1512/218 = 6h: 56m. (c) SAR = (FOB xTAS)/FC = (1600 x 210)/100 = 3360 Nm

34

(d) Since one engine has failed we need to know how far we can go with 4 engines and come back with three engines operating. If X is the fuel consumed till PNR with 4 engines and Y with three engines, then total fuel consumed = X + Y Cruise 4 Engines PNR A

B Cruise 3 Engines

A to PNR

TR 090

TAS 210

PNR to A

270

180

W/V GS Dist FTime 330/18 218 1400 6:25 330/18

170

1400

8:14

FC 100

F Used 642

80

659

Total Fuel Used = 642 + 659 = 1301 (Adopt Unitary method) If flight fuel is 1301 then distance to PNR is 1400 If flight fuel is 1 then distance to PNR is 1400/1301 If flight fuel is 1450 then distance to PNR is (1400 x 1450)/1301 = 1560 Nm Hence Dist to PNR = 1560 Nm and time taken = 1560/218 = 7h:09m Q5. Dist = 2000 Nm, Tr =330, W/V = 160/37 Kts, TAS (4/3) = (300/250), FC (4/3) = 200/180 Kg/hr FOB = 2400 kg, Reserve = 500 Kg. Find (a) Dist & Time to CP (b) Dist & Time to PNR (c) SAR. Assume one engine failed at PNR and ac returns on 3 engines. A. H

CP to A

TR 150

TAS 250*

W/V 160/37

GS 213

O

CP to B

330

250*

160/37

286

G

A to CP

330

300

160/37

336

(a) Distance to CP = DH/(O +H) = (2000 x 286)/(286 + 213) = 940 Nm Time to CP = Distance to CP/ GS out = 940/336 = 02h:48m 0n 4 Engine A

940 Nm

(b)

CP 1060 Nm TR 330

TAS 300

W/V 160/37

B GS DIST 336 1400

TIME 4:10

FC 200

F USED 1190

O

A to PNR

H

PNR to A 150 250* 160/37 213 1690 If fuel is 2880 PNR is 2000, for fuel 1900 PNR = 2000/2880 x 1900 = 1320 Nm/3:55m

35 Endurance = (1600 -150)/100 = 14h:30m (b) Dist to PNR = (14:30 x 218 x 200)/(418)=1512 Nm (c) Time to PNR = 1512/218 = 6h: 56m. (c) SAR = (FOB x TAS)/FC = (1600 x 210)/100 = 3360 Nm Q6. Dist = 1350 Nm, TR =270 W/V 270/25 upto CP, thereafter 350/38. TAS (4/3) = 200/180, FC (4/3) = 110/90, FOB =1800, Res = 200. Find DCP, TCP, DPNR, TPNR & SAR (assume one engine failed at PNR) A. H

CP to A

TR 090

TAS 180*

W/V 270/25

GS 205

O

CP to B

270

180*

350/38

170

G

A to CP

270

200

270/25

175

(a) Distance to CP = DH/(O +H) = (1350 x 205)/(205 + 170) = 740 Nm Time to CP = Distance to CP/ GS out = 740/175 = 04h:14m (b) O

A to CP

TR 270

TAS 200

W/V 270/25

GS DIST TIME 175 740 04:14

H

CP to A

090

180

270/25

205

740

FC F USED 110 465

03:37 90 325 Total Fuel Used : 790; Bal = 810

PNR TR CP to B (4 Eng) 270 B to CP (3 Eng) 270

TAS 200 180

W/V 350/38 350/38

GS 190 183

DIST TIME FC 610 03:13 110 610 03:10 80 Total Fuel Used :

F USED 353 300 653

If flight fuel is 653 then distance to PNR is 610 If flight fuel is 1 then distance to PNR is 610/653 If flight fuel is 810 then distance to PNR is (610 x 653)/810 = 757 Nm Hence Dist to PNR = 757 Nm and time taken = 757/190 = 3h:59m DPNR = 740 + 757 = 1497 Nm, TPNR = 4:14 + 3:59 = 8h:13m (c) SAR = (FOB x TAS)/FC = (1800 x 200)/200 = 3273 Nm Q7. Dist = 1250 Nm, TR =090 W/V 280/20 upto CP, thereafter 330/20. TAS (4/3) = 180/150, FC (4/3) = 110/90, FOB =1800, Res = 700. Find DCP, TCP, DPNR &, TPNR (assume one engine failed at PNR) A. H

CP to A

TR 270

TAS 150*

W/V 280/20

GS 130

36 O

CP to B

090

150*

330/20

159

G

A to CP

090

180

280/20

200

(a) Distance to CP = DH/(O +H) = (1250 x 130)/(159 + 130) = 562 Nm Time to CP = Distance to CP/ GS out = 562/200 = 02h:49m (b) O

A to CP

TR 090

TAS 180

W/V 280/20

GS 200

DIST TIME 562 02:49

FC 110

F USED 309

H

CP to A

270

150

280/20

130

562 04:19 90 389 Total Fuel Used : 698; Bal = 402

TR CP to B (4 Eng) 090 B to CP (3 Eng) 270

TAS 180 150

W/V 330/20 330/20

GS 189 139

DIST TIME FC 688 03:38 110 688 04:56 90 Total Fuel Used :

PNR F USED 400 445 845

If flight fuel is 845 then distance to PNR is 688 If flight fuel is 1 then distance to PNR is 688/845 If flight fuel is 402 then distance to PNR is (688 x 402)/845 = 327 Nm Hence Dist to PNR = 327 Nm and time taken = 327/189 = 1h:44m DPNR = 562 + 327 = 889 Nm, TPNR = 2:49 + 1:44 = 4h:33m Q8. TR=250, W/V 270/30, TAS = 210, FOB = 1200, PNR =785 Nm. Find (a) FC (b) If CP is reached 45 minutes before PNR, find excess fuel carried. PNR O

A to PNR

TR 250

TAS 210

W/V 270/30

GS 181

TIME 4:19

H

PNR to A

070

210

270/30

238 3:18 TOTAL 7:37

Time to CP = Time to PNR – 0:45 = 7:37 – 0:45 = 6:52, Distance traveled in 0:45 = 181 x 0:45 =136 Nm, Hence distance to CP = 785-136 = 649; CP = DH/(O+H) or 649 = D x 238 /419 or D =1143. Time taken to cover 1143 Nm at 181 K = 1143/181 = 6h 19 m PNR = EOH/(O+ H) or 785= (E x 181 x 238)/ 419 or E = (785 x 419)/ (181x 238) = 7h 38m E = FOB/FC or FC = 1200/7:38 =157gph. Fuel Consumed for Flight of 6h 19 m = 6:19 x 157 = 991 gals. Fuel Carried = 1200, Excess Fuel = 1200-991 = 209 Q9. TR=155, W/V 240/30, TAS = 220, FC = 150 GPH, PNR =1080 Nm. Find (a) FOB (b) If CP is reached 1:15 minutes before PNR, find excess fuel carried. PNR O

A to PNR

TR 155

TAS 220

W/V 240/30

GS 215

37 H

PNR to A

335

220

240/30

221

Time to CP = Time to PNR – 1:15, Distance traveled in 1:15 = 215 x 1:15 =269 Nm, Hence distance to CP = 1080-268 = 811; CP = DH/(O+H) or 811 = D x 221 /436 or D =1599. Time taken to cover 1599 Nm at 215 K = 1599/215 = 7h 26 m PNR = EOH/(O+ H) or 1080= (E x 215 x 221)/ 436 or E = (785 x 419)/ (181x 238) = 9h 55m E = FOB/FC or FOB = 9:55 x 150 =1486 lbs. Fuel Consumed for Flight of 7h 26 m = 7:26 x 150 = 1115 gals. Fuel Carried = 1486, Excess Fuel = 1486-1115 = 371 lbs Q10. Dist = 1450 Nm, TR =132 W/V 260/40 upto CP, thereafter 350/60. TAS (4/3) = 190/160, Find (a) DCP, TCP, (b) DPNR if Fuel Consumption is increased by 8% (assume one engine failed at PNR) A. H

CP to A

TR 312

TAS 160*

W/V 260/40

GS 132

O

CP to B

132

160*

350/60

203

G

A to CP

132

190

260/40

212

(a) Distance to CP = DH/(O +H) = (1450 x 132)/(203 + 132) = 571 Nm Time to CP = Distance to CP/ GS out = 571/212 = 02h:42m. Since fuel is not given, CP & PNR are collocated. Now with 8% increase in fuel consumption there will be a 8% reduction in distance to PNR. Hence 8% of 571 = 46 Nm, so DPNR = 571-46 =525Nm and Time to PNR = 525/212 = 2h:26m. Q11. Dist = 1200 Nm, TR =270 W/V 330/20, TAS (4/3) = 180/150, FOB = 900, RES = 300; FC =110 Gal/hr. FC = 110 gals/hr. Find (a) DCP, TCP, (b) DPNR & TPNR (c) Is fuel sufficient for the flight, if not, how much less (d) If flight fuel is 981 gals calculate DPNR. A. H

CP to A

TR 090

TAS 150*

W/V 330/20

GS 159

O

CP to B

270

150*

330/20

139

G

A to CP

270

180

330/40

169

(a) Distance to CP = DH/(O +H) = (1200 x 159)/(139 + 159) = 640 Nm Time to CP = Distance to CP/ GS out = 640/169 = 03h:47m.

38 (b) DPNR &TPNR

O

A to PNR

TR 270

TAS 180

W/V 330/20

GS 169

H

PNR to A

090

180

330/20

189

Endurance = 600/110 = 5:27, DPNR = 5:27 x 189 x 169/358 =486, TPNR = 486/169 = 2:53 (c) Time to cover 1200 Nm = 1200/169 = 7h:06m, Fuel required = FC x 7:06 = 781 gals, FOB =600 gals, hence Fuel Less by 781-600 = 181 gals. (d) FOB = 781 + 300 (reserve) =1081, Endurance =781/110 = 7.1 hr, DPNR = 7.1x 169 x 189/358 = 634 Q12. Dist = 1600 Nm, TR =090 W/V 270/30 for first 1000 Nm for remaining distance 030/17, TAS (4/3) = 220/150, FC (4/3) = 100/80, FOB =1400, Res = 200. Find (a) DCP, TCP, (b) DPNR & TPNR (assume one engine failure at PNR) A. A

1000 nm

X

W/V 270/30

600 nm

B

030/17

TR

TAS

W/V

GS

DIST

TIME(min)

A-X (3)

090

190

270/30

220

1000

273

X-B (3)

090

190

030/17

181

600

199 (273 +199) = 472

B-X (3)

270

190

030/17

198

600

182

X-A (3)

270

190

270/30

160

1000

000 A Subtract 472 -472

375 X 1 98 +177

375 (375 + 182) = 557

557 B 000 +577

CP lies in this Leg Time taken for an aircraft to reach from A to B is 472 mins with existing winds. Time taken to return form B to A is 577 mins. At (A, X & B) we can calculate the residual time. The CP will lie in the leg where values are between negative and positive. DCP = Distance of leg in which CP lies x 472 /(472 +177) =1000 x 472/649 = 727 Nm TCP = 727/250 =2h:54m (GS out 4 Eng = 220 +30 K tail wind) Similar calculations can now be made for PNR with Fuel considerations.

39 PNR TR TAS W/V

GS

DIST TIME

FC F USED

A-X (4)

090

220 270/30

250

1000

4:00

100

400

X-A (3)

270

190 270/30

160

1000

6:15

80

300 (1200 –(500+400))

X-B (4)

090

220 030/17

211

600

2:50

100

500 900 284

B-X (3)

270

190 030/17

198

600

3:01

80

241 525

675 (1200 –(284+241))

A

BAL

X B PNR lies in this Leg

PNR lies between X & B since Balance of fuel is 300 gals after catering for return from X. Hence, if Flight Fuel is 525 then PNR is 600 Nm from X If Flight Fuel is 1 then PNR is 600/525 Nm from X If Flight Fuel is 300 then PNR is (600 x 300)/525 Nm from X = 343 Nm from X = 1343 Nm from A Time to cover 343 Nm @ GS =211 =343/211 = 1:37. Hence TPNR = 4:00 + 1:37 = 5:37 Q13. Dist = 2000 Nm, TR =270 W/V 280/22 for first 900 Nm for remaining distance 330/20, TAS (4/3) = 180/150, FC (4/3) = 100/80, FOB =1700, Res = 200. Find (a) DCP, TCP, (b) DPNR & TPNR (assume one engine failure at PNR) A. A

900 nm

X

1100 nm

W/V 280/22

B

330/20

TR

TAS

W/V

GS

DIST

TIME(min)

A-X (3)

270

150

280/22

128

900

422

X-B (3)

270

150

330/20

139

1100

475 (422 +475) = 897

B-X (3)

090

150

330/20

159

1100

415

X-A (3) A-X (4)

090 270

150 180

280/22 172 900 314 (415 + 314) = 729 280/22 & 330/20 = 158/169 0 314 729 A X B Subtract 897 475 000 -897 -161 +729 CP lies in this Leg Time taken for an aircraft to reach from A to B is 897 mins with existing winds. Time taken to return form B to A is 729 mins. At (A, X & B) we can calculate the residual time. The CP will lie in the leg where values are between negative and positive.

40 DCP = Distance of leg in which CP lies x 161 /(729 +161) =(1100 x 161)/890 = 199 + 900 =1099 Nm TCP = 199/169 + 900/ 158 =1:11 + 5:42 = 6h:53m (GS out 4 Eng = 158 from A-X & 169 from X-B) Similar calculations can now be made for PNR with Fuel considerations. PNR TR TAS W/V

GS

DIST TIME

FC F USED

A-X (4)

270

180 280/22

158

900

5:42

100

570

X-A (3)

090

150 280/22

172

900

5:14

80

511 (1700 –(570+419))

X-B (4)

270

180 330/20

169

1100

6:31

100

419 989 651

B-X (3)

090

150 330/20

159

1100

6:55

80

553 1204

496 (1700 –(651+553))

A

X

BAL

B PNR lies in this Leg

PNR lies between X & B since Balance of fuel is 511 gals after catering for return from X. Hence, if Flight Fuel is 1204 then PNR is 1100 Nm from X If Flight Fuel is 1 then PNR is 1100/1204 Nm from X If Flight Fuel is 511 then PNR is (1100 x 511)/1204 Nm from X = 467 Nm from X = 1367 Nm from A Time to cover 467 Nm @ GS =169 =467/169 = 2:46. Hence TPNR = 5:42 + 2:46 = 8:28 Q14. An aircraft flies from A-B on Tr = 090 for 600 Nm (W/V 030/20) and then proceeds to destination C on Tr = 120, D = 900 Nm (W/V 150/35) . TAS (4/3) = 240/210, FC (4/3) = 150/120, FOB =1600, Res = 200. Find (a) DCP, TCP, (b) DPNR & TPNR (assume one engine failure at PNR) A. A 600 nm

B

W/V 030/20

900 nm 150/35

C

A-B (3)

TR 090

TAS 210

W/V 030/20

GS 199

DIST 600

TIME(min) 181

B-C (3)

120

210

150/35

179

900

302 (181 +302) = 483

C-B (3)

300

210

150/35

240

900

225

B-A (3) A-B (4)

270 090

210 240

030/20 219 600 030/20 & 150/35 = 229/209

164 (225 + 164) = 389

41 0 A Subtract 483 -483

164 B 302 -138

389 C 000 +389 CP lies in this Leg Time taken for an aircraft to reach from A to B is 483 mins with existing winds. Time taken to return form B to A is 389 mins. At (A, X & B) we can calculate the residual time. The CP will lie in the leg where values are between negative and positive. DCP = Distance of leg in which CP lies x 138 /(389 +138) =(900 x 138)/536 = 236 + 600 =836 Nm TCP = 236/209 + 600/ 229 =1:07 + 2:37 = 3h:45m (GS out 4 Eng = 158 from A-X & 169 from X-B) Similar calculations can now be made for PNR with Fuel considerations. PNR TR TAS W/V

GS

DIST TIME

A-B(4)

090

240 030/20

229

600

2:37

150

393

B-A(3)

270

210 030/20

219

600

2:44

120

B-C (4)

120

240 150/35

209

900

4:18

150

329 722 646

C-B (3)

300

210 150/35

240

900

A

FC F USED

6:55

120

B

BAL

678

450 1096 C

PNR lies in this Leg PNR lies between B & C since Balance of fuel is 722 gals after catering for return from B Hence, if Flight Fuel is 1096 then PNR is 600 Nm from B If Flight Fuel is 1 then PNR is 600/1096 Nm from B If Flight Fuel is 678 then PNR is (600 x 678)/1096 Nm from B = 558 Nm from B = 1158 Nm from A Time to cover 558 Nm @ GS of 209 K =558/209 = 2:40. Hence TPNR = 2:37 + 2:40 = 5:17 Q15. On a flight from A to C via B. TAS on 4 engines is 360 K & in case of 3 engines it is 300 K. The route details are:Stage

Track

Wind Vel

Distance

A–B

180°

330°/ 35 K

900 Nm

B–C

210°

150°/28 K

1400 Nm

(a) Find distance and time to CP (the aircraft is required to return to B or C in case of engine failure). (b) If FOB is 2200 Kg, Reserve is 200 Kg, fuel consumption 250 Kg/hr (4Engine) and 220 Kg/hr (3 Engine), find Distance & Time to PNR (After take off A is not available and aircraft is to land at B, assume engine failure at PNR and return is on three engines)

42 A. A 330°/ 35 K B 180°/900 Nm

150°/28 K

C 210°/1400 NM

Between B – C

H

CP to B

TR 030

TAS 300*

W/V 150/28

GS 313

O

CP to C

210

300*

150/28

G

B to CP

210

360

150/28

345

A – B (4)

180

360

330/35

390

285

(a) Distance to CP (B-C) = DH/(O +H) = (1400 x 313)/(285 + 313) = 732 Nm Distance to CP = 732 + 900 = 1632 Time to CP (A-B) = Distance to CP/ GS out = 900/390 = 02h:18m (B-C) = Distance to CP/ GS out =732/345 = 2:07 Time to CP = 2:18 + 2:07 = 4h:25m PNR TR TAS W/V

GS

DIST TIME

A-B(4)

180

360 330/35

390

900

2:18

250

577

B-C(4)

210

360 150/28

345

1400

4:03

250

1014

C-B (3)

300

210 150/28

313

1400

4:28

220

984 1998

A

FC F USED

B

BAL 1423 (2000-577)

C PNR lies in this Leg

PNR lies between B & C since Balance of fuel is 1423 kgs after catering for return from B Hence, if Flight Fuel is 1998 then PNR is 1400 Nm from B If Flight Fuel is 1 then PNR is 1400/1998 Nm from B If Flight Fuel is 1423 then PNR is (1400 x 1423)/1998 from B = 997 Nm from B = 1897 Nm from A Time to cover 997 Nm @ GS of 390 K =997/390 = 2:53. Hence TPNR = 2:18 + 2:53 = 5h:11m

Q16. An aircraft is to fly from P to R via Q and return to Y via Q in case of engine failure, since P is not available. TAS on 4 engines is 500 K & in case of 3 engines it is 420 K. The route details are:-

43 Stage

TAS

Wind Vel

Distance

P–Q

420

-25

965 Nm

Q–R

420

-45

900 Nm

R–Q

420

+45

900 Nm

Q–Y

420

+30

240 Nm

(a) Find distance and time to CP(the aircraft is required to return to Y via Q in case of engine failure). (b) If FOB is 38000 Kg, Reserve is 6500 Kg, fuel consumption 6300 Kg/hr (4Engine) and 5600 Kg/hr (3 Engine), find Distance & Time to PNR (After take off P is not available and aircraft is to land at Y, assume engine failure at PNR and return is on three engines) 0 P 290 -290

32 (Time to Y) 148 965 Q 900 R 144 (Time to R) 0 240 -112 148 Y CP lies in this Leg

CP Stage

TAS

Wind Vel

GS

Distance

Flt Time (min)

P–Q

420

-25

395

965 Nm

146

Q–R

420

-45

375

900 Nm

R–Q

420

+45

465

900 Nm

144 290 116

Q–Y

420

+30

450

240 Nm

P – Q (4) 500 -25 Q – CP (4) 500 -45 TCP = 2:02 + 0:51 = 2h:53m

475 455

965 Nm 388 Nm

32 148 DCP (Q-R) = DH/(O +H) = 900 x 112/ (148 + 112) = 388 Nm from Q DCP = 965 + 388 = 1353 2:02 0:51

PNR Stage

TAS

Wind Vel

GS

Distance

Time FC

P–Q

500

-25

475

965 Nm

2:02 6300 12810

Q–Y

420

+30

450

240 Nm

0:32 5600

Q–R

500

-45

455

900 Nm

R–Q

420

+45

465

900 Nm

F Used

2987 15797 1:59 6300 12461 1:56 5600 10839 23300

If Flight fuel is 23300 then DPNR is 900 Nm If Flight fuel is 15703 then DPNR is (900 x 15703)/23300 = 607 Nm from Q DPNR = 965 + 607 = 1572 Nm Time to cover 607 Nm @ GS 455 K = 607/455 = 1:20 TPNR = 2:02 + 1:20 = 3h:22m

Bal

(31500-15797) 15703

44 CRITICAL POINT (CP) 1. CP is always half way when GS (Home = GS (Out) i.e. O = H, this happens during (a) Nil Winds and (b) Beam winds. 2. In case of HW, CP will always be more than half way i.e. into wind and in case of tail winds CP will be less than half way. 3. If HW component increases distance to CP will increase or it will move towards destination or it will move away from departure point. 4. In case tail wind component increases distance to CP will decrease or it will move closer to place of departure or away from departure point.. CP ALWAYS MOVES INTO WIND 5. For same HW component if TAS is reduced, distance to CP will increase and vice versa. 6. In case of HW, CP will be more than half way, if HW changes to tail wind, distance to CP will be less than half way by the corresponding distance if wind component remains same. Q. With 50K of HW, distance to CP is 1200 Nm. During actual flight wind component was found o be 50 K of tail winds. If total distance is 2000 Nm, new distance to CP will be (a) 1000 (b) 1200 (c) 800 (d) insufficient data cannot be calculated A (c) Q. If beam wind component doubles, distance to CP will be …..and time to CP will ….(a) same, same (b) same, decrease (c) decrease, same (d) same, increase. A. (d) Q. If fuel onboard or Flight Fuel increases distance to CP will remain the same. PNR Q. Distance to PNR will be maximum in (a) HW during outbound (b) Tail Wind during outbound (c) Tail wind during inbound (d) Nil Winds A. (d) Note. ANY KIND OF WINDS WILL CAUSE PNR TO REDUCE. Q. With a fuel of 10000 lbs, PNR calculated is 880 Nm, other factors remaining constant, if fuel is increased to 11000 lb, the distance to PNR will be (a) 928 (b) 968 (c) 950 (d) 920. A. (b) 10% increase 3. If fuel consumption is changed by certain percentage, distance to PNR will also change by corresponding percentage. Q. With 200 lbs/hr fuel consumption, PNR is 1000 Nm, if actual fuel consumption is found to be 220 lbs/hr , distance to PNR will be (a) 1100 Nm (b) 990 Nm (c) 900 Nm (d) 800 Nm A. (c) 10% Change

45 INERTIAL NAVIGATION SYSTEM 1. It is a DR Navigation System which gives Great Circle Tracks/ Distances and True Direction. It consists of two accelerometers which measure aircraft accelerations in N-S and E-W direction. 2. It has a Gyro Stabilised Platform, horizontally stabilised, to ensure accelerations are measured in the horizontal plane only. Three torque motors, two accelerometers and three rate gyros sensitive in each axis are mounted on the horizontally stabilized platform. The Zero position denotes the present position of the aircraft from which the georef coordinates are taken to initialize the system. A total of 9 way points can be fed into the system. 3. Control and Display Unit (CDU).

78°23’36’’ W

24°15’07’’ N W P T

23 4

POS XTK/TKE HDG/GA

TK/GS

D I M

TK CHG

WPT

DIST/TIME

BATT

WARN

1

2

3

4

5

6

7

8

9

HOLD

0

WIND DSR TK/STS

TEST

INSERT

CLEAR

Fig 1 – Control and Display Unit 4. TK/GS(Track and Groundspeed). The INS computed track, usually referenced to magnetic north, is displayed to the nearest tenth of a degree in the left display and the groundspeed in knots in the right display. For example, a current track of 135 M and a groundspeed of 467 knots would appear as 135.0 and 0467. 5. HDG/DA(Heading and Drift Angle). The heading obtained from the angle between the platform frame and north reference is displayed to the nearest tenth of a degree in the left display. The angular difference between heading and track (drift angle) is displayed to the nearest tenth of a degree in the right display, preceded by the letter R or L to indicate whether drift is right or left. Thus, a heading of 137 M on a track of 135° M would be presented as 137.0 and L 02.0.

46 6. XTK/TKE (Cross Track Distance and Track Error Angle). Cross track distance is the distance by which the aircraft is displaced right or left of the desired great circle track and is displayed in the left display to the nearest tenth of a nautical mile. The track error angle is the angular difference, right or left, between the desired great circle track and the actual track being made, to the nearest tenth of a degree. If the aircraft were displaced 1½ nm to the left of the desired track of 135°M, the left display would read L 01.5. If the track being made good happened to be 130°M, the right display would read L 005.0. 7. POS (Present Position). The aircraft’s current latitude and longitude are shown in terms of Longitude and Latitude in the left and right displays, respectively. For example, 24°15’07’’N and 78°23’36’’W, 8. WPT (Waypoint Positions). The position of each inserted waypoint is shown as latitude in the left display and longitude in the right display by selecting WPT on the rotary selector switch and scrolling through the waypoint numbers with the waypoint selector wheel. 9. DIST/TIME (Distance and Time to the Next Waypoint). The distance in nautical miles from the present position to the next waypoint is shown in the left display and the time at present groundspeed to the nearest tenth of a minute in the right display. 10. WIND (Wind Speed). The INS is able to compute wind direction and speed and these are displayed in the left and right windows, respectively, to the nearest degree of arc and knot. 11. DSR TK/STS (Desired Track and Status). The Great Circle track from one waypoint to the next changes as the aircraft progresses between the two and INS computes the present desired magnetic track based upon distance from the waypoints, magnetic variation and the assumption that the aircraft is on track. This will appear in the left display to the nearest tenth of a degree and the right display will be blank. The status function is for use only whilst the INS is in ALIGN mode and it shows a numerical display in the right window that indicates the status of the alignment procedure. The display typically shows 99 at the start of alignment and counts down to 0, when alignment is completed and READY NAV is illuminated. Q. Way Point(WP) 4 is 60°N 90°W, WP3 is 60°N 70°W. Find distance shown at WP3 for WP4 (a) 605 Nm, (b) 600 Nm (c) 594 Nm (d) None. A. Dep = dlong x 60 x Cos 60 = 20 x 60 x 0.5 = 600 Nm (Rhumb Line Distance), remember GC distance is slightly lesser than Rhumb Line distance, Hence 594 Nm is the correct answer. Q. What is the track shown by INS on leaving WP3 in the above question (a) 270° (b) 279° (c) 261° (d) 099°. A. Conversion Angle = ½ dlong x Sin Mean Lat = ½ x 20 x Sin 60 = 9°. RL Track =270 + 9° = 279° which is the GC track. Q. What is the track shown before arrival at WP4 in the above question? A. It will be 9° less than 270 = 261° and curving left

270° 261°

Q At 80°W INS indicates 0 XTKE, what is the latitude shown by INS? A. Distance off from rhumb line can be calculated by taking half the CA i.e 4.5° and applying 1 in 60 rule. Distance at 80°W=300Nm =R, θ = 4.5° hence s = 22.5 Nm = 22.5’ (1’ = 1 Nm). Lat = 60°30’N

47 Q. WP7 is 52°27’S, 017°45’W, WP8 is 52°S27’S, 004°15’E. Find on leaving WP7,(a) DSR/TRK (b) Distance (i) 804 nm (ii) little less than 804 (iii) 494 Nm (iv) little less than 494 Nm (c) Initial true Hdg on leaving was 080 assuming constant drift find true hdg on reaching WP8. A. (a) CA =½ dlong x Sin Mean Lat = ½ x 22 x Sin 52° 27’ = 9°. RL Track =090 + 9° = 099° which is the GC track. (b) Dep = dlong x 60 x Cos 52° 27’ = 22 x 60 x 0.6 = 800 Nm (Rhumb Line Distance), remember GC distance is slightly lesser than Rhumb Line distance, (ii) is correct. (c) Initial Track was 099 but ac is flying 080 hence stb drift of 19°. At WP 8 Track would be 090 -9 -19 =062° INS Error. Distance from final ramp position to final INS position divided by time is termed as INS error Q. After 6h24m flight when at parking bay position 52°18.5’N, 4°45.9’E the ins shows a position of 52°20.7’N 4°40.3’e. Find INS Error 52° 20.7’N 04° 40.3E 2.2Nm

33° 3.42 Nm

Dep = dlong x Cos Lat = 5.6 x Cos 52°18.5’ =3.42 Nm, Similarly Dist covered in Northerly Hdg = 2.2 Nm, Tan(2.2/3.42) = Tan (.643) or = 33° Hypotenuse distance 52° 18.5’N = 4.066 Nm hence error =4/6:24 = 0.64 Nm/hr in 303° 04° 45.9’E

48 EXTENDED RANGE TWIN OPERATION (ETOPS) IT IS APPLICABLE TO TWIN ENGINE OPERATIONS ONLY Threshold Distance. The Maximum distance you can fly with one engine inoperative in 60 minutes (threshold time TT) for Class A aircraft and 120 min for Class B&C aircraft in nil wind condition is termed as threshold distance. It is calculated by TT x TAS, the TAS will vary between turbo-prop and jet ac, also the FL will vary. It is FL 80 for prop ac and FL 170 for turbo jet ac. The TAS would depend on the performance of the ac with single engine. Rule Time. The maximum diversion time that any point on the route of an ETOPS approved aircraft may be from a suitable aerodrome is referred to as the Rule time. The first clearance for ETOPS is 120 minutes, after safe operations for six months it may be extended to 138 minutes and after 12 months of safe operations further extended by 180 minutes. Rule Distance. The maximum distance an operator may plan any route from a suitable aerodrome is that which would be covered at the normal one-engine inoperative cruising speed, in still air, in the rule time Adequate Aerodrome. An aerodrome that is available at the anticipated time of use and is equipped with necessary ancillary services i.e. ATC, lighting, communications, weather reporting, nav aids and safety cover and having at least one let down aid available for an instrument runway. Suitable Aerodrome. A suitable aerodrome is an adequate aerodrome that is used as an en-route alternate or diversion aerodrome. It would normally be used only in the event of an engine failure or the loss of primary airframe systems. For an aerodrome to be considered suitable, weather probability should be more than 40% for the forecast if probability is less than 40% the aerodrome may be ignored for planning purposes.

49 GPS It utilises the range obtained from a constellation of satellites to fix position of an aircraft in the air or on ground. It is a satellite based navigational aid. Basic Data. Total number of Satellites : 21+ 3 (stand by) =24. Number of Orbital Paths : 06. Inclination of Orbital Path : 55° to the Equator Distance : 20, 200 kms above the surface of the earth. Time taken for each orbit : 12 hours. Number of satellites required for 2D fix : 3 Number of satellites required for 3D fix : 4 (also for initialisation of the system) Number of satellites required for RAIM (Remote Autonomous Integrity Monitoring) : 5 (This is similar to Built in Test Equipment BITE) Frequency : L1 -1575 MHz and L2 -1228 MHz in UHF Band. Principle. GPS transmits a PRN (Pseudo Random Noise) of 1 milli sec duration in UHF band. Each satellite has its own unique code. The information contained in PRN code is (a) Position of satellite (azimuth and angle) (b) Clock Time (c) Clock Error (d) Information on ionospheric condition (e) Supplementary information Two codes are transmitted, CA (Coarse Acquisition) and P (Precision) Code. Two services are provided by GPS. These are (a) Standard Position Service (SPS) using CA code which is available for civil use. (b) Precise Position Services (PPS) using CA & P Code. Pseudo Range. The receiver has an accurate crystal oscillator to provide time, however, the accuracy does not compare with that of the satellite clock. The receiver clock is deliberately kept in error by a small factor to ensure that correction process takes place in one way only. The initial calculated range is called Pseudo Range. For example if the receiver clock is 1 milli second fast the receiver will overestimate the range by 162 Nm. Therefore when the receiver sets about calculating the correct range it knows that it must reduce Pseudo Range. Errors of GPS Ephemeris Error. This error is due to the disturbed position of the satellite. Any deviation from its orbital path will induce this error. This is caused due to various reasons e.g. debris or gravitational effect of Sun, Moon and other planets. To obviate this error satellite position is checked every 12 hours and when necessary it is updated. The max permissible error is 2.5 m. Satellite Clock Error. The clock is also checked atleast every 12 hours. The max permissible error is 1.5 m.

50 Ionospheric Propagation Error. This is the most significant error in the system. The state of ionosphere is continuously checked at monitoring stations and the model is updated every 12 hours. The max permissible error is 5 m. Tropospheric Propagation Error. This is caused due to variation in pressure /temperature/density and humidity effects on EM waves. The max permissible error is 0.5 m. Receiver Noise Error. All radio receivers generate internal noise. In GPS receiver it can cause error in measurement of time. The max permissible error is 0.3 m. System Accuracy. The ICAO specifications require an accuracy of the SPS to be 30 m with a probability of 50% i.e. 30 m – 50% of the time. Multi-path Reception. Reflections from ground and parts of aircraft result in multi-path reception. GDOP. Geometric Dilution of Precision is caused due to poor cut between position lines. This is caused when the satellites are relatively closer to each other. Differential GPS. It is a means of improving the accuracy of the GPS by monitoring the integrity of the satellite data and warning the user of any error which may occur during flight. There are three kinds of DGPS in use. Accuracy is 3 m. (a) GBAS (Ground Based Augmentation System). This is Local Area Augmentation System (LAAS). (b) ABAS (Air Based Augmentation System). (c) SBAS (Space Based Augmentation System). This is Wide Area Augmentation System (WAAS).