Navigation Notes

NAVIGATION

5

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

BBA Earth is considered as a Sphere.

D

Great Circle. Its a circle which when drawn across the earth cuts the earth into two equal hemispheres.

Properties.

(a) Its 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 centre1 Nm =6045 feet at Equator geocentric1 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.

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.

True NorthCompass NorthMagnetic NorthVariation WDeviation EFig: Deviation and Variation

POBLEMS ON DIRECTION (Black-Given, Red-Determined) (East is +, W is -ive)

CDMVT

12695E 2744W270

20273W0244E020

3 051+30544W050

4300 -729310E303

5045-4041+6047

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. Dont 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 = 10800Hence 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 CosLat545= 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. 17350W 180 168E 610 12A. 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 E8. 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.

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 (Applicable for Mercator Chart Only)

Conversion Table

1 Nm = 6080 ft1 m = 3.28 ft; 1 ft = 12 in; 1 in = 2.54 cms1 Km = 3280 ft1 SM = 5280 ft

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 EDA. 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 62N = 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 = 1/1631716Note : For Mercators Chart SAL= SAE x Sec Lat or SAE x 1/Cos Lat.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 60N 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 7531

12. If the scale at 5720N is 1:1091000 what is the meridian spacing in cm between one deg longitude.

A. Dep =1 x 60 x Cos5720 = 32.385 Nm = 5997709 cm (1Nm = 1.852Km)

13. You are flying east along a parallel 60N and cover 10 inches distance on the chart every hour. The scale at 25S 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 40N is 10 cm on a Mercator chart and 90 km on earth. Find scale at equator.A.Scale at 40N = 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 56N. Find the Chart length from 2845N 11330W to 2845N 9815W.

A. Dep = 1515 x 60 =915 Nm, 915 x 6080 x 2 = 66758400/1000000 corresponding to 66.75 .

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.

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.

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.

Lamberts Conical Orthomorphic (Conformal) between two Standard Parallels

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

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.

Scale ExpansionQ. 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.

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. Its 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 = 050 045 RL Brg of B from A = 050 + 180 = 230 B A GC Brg of B from A = 230 + 5 = 235

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

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 8W. 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 = dlong Sin 30 or dlong = 2 CA Sin 30 = (2 x 3) 0.5 = 12. Hence B lies 12 apart ie 12-8 = 4E

Q. A & B are on parallel of 30 N. GCB of A from B is 266. Longitude of A is 10W. 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 = dlong Sin 30 or dlong = 2 CA Sin 30 = (2 x 4) 0.5 = 16. Longitude of B = 16-10 = 6 E

WIND TRIANGLE

DRIFT (p)HDG & TASW/VTRK & 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.

Trk/GSW/VHdg/TASHdg/TASTrk/GSW/VTMG Stbd TMG PortTr ReqdTMG. 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.

Winds 90 to Trk (GS < TAS) Winds 90 to Hdg (GS > TAS)

Problems on Wind Triangle.

TRKTETMGHDGDRIFT

(a)1018 P0930903 S

(b)1805 S1851832 S

(c)2766 P2702755 P

(d)04510 S0550594 P

(e)23415 P2192254 P

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.

LegTASDISTW/VGSTIMEWIND EFFECT (TIME x W/V)

A-B300700+303302:07+63.5 Nm (30 x 2:07)

B-C200280-101901:28-14.7 Nm (-10 x 1:28)

C-D350300+403900:46+30.7 Nm (+40 x 0:46)

5:26 + 79.5 NmTotal Time = 2:07 + 1:28 + 0:46 = 4:21, Total Wind Effect = +63.5 -14.7 + 30.7 = 79.5Hence 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 = 268.33Total = 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:-

LegTASDISTW/VGSTIMEWIND EFFECT (TIME x W/V)

A-B160600+301903:09+94.5 Nm (30 x 3:09)

B-C290500-152751:47-26.75 Nm (-15 x 1:47)

C-D380200+454250:28+21 Nm (+45 x 0:28)

* Data CalculatedMean 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 respectivelyQ. 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) Original ETA (b) Revised ETA (c) New distance covered with revised speed (d) Time to drop speed (e) 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

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 = 920Heading correction at B, (a) to destination.< EBD=< BAD+ < BDA=60 x 3+ 60 x 3 20 60=9 + 3=12(b)To return to track.=2 x BAD=2 x 9=18 (Heading altered back a C)

Examples on 1:60 rule

= S/R x 60 (to be used when is < 20)

1. After flying for 240 nm an aircraft is 12 nm right of track. What is the drift.

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

Dist =240 nm

12 nm off track 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 300 Nm 3 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

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 Suns 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 Earths motion, as we shall see.

Fig 14.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 Earths axis at about 66 , 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 Earths axis is at right angle to the Sun, day and nights are equal the Spring and 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 Keplers 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.

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 23 and 66 from the plane of rotation.8. Position of sun varies from 23 N to 23 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. Its 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 23:56A Apparent not fixed due to orbit and revolution of earth around the sun 23:44 to 24:14M 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 Suns 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.

TIME

There are of four types namely, LMT, UTC, Zone Time and Standard Time.

Local Mean Time (LMT) Its 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 earths rotation.

Q. LMT at 35E is 1300h. Find LMT at (a) 102E (b) 40W

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 7 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.

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 ante-meridian 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 whos Standard Time you want is listed in the Air Almanac: apply the correction to GMT, and the date will take care of itself. Going on dateline from,East to West gain a day subtract a dateWest to East lose a day add a date

Prime Meridian0600 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 behindA. (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).

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 180is 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

Q2. Aircraft is approaching Station on a radial 150is asked to approach on radial 360. Find out (a) Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading.

A. (a) 60 (b) 300 (c) 30 (d) 060

Q3. Aircraft is homing on to a Station on a radial 010is asked to approach on radial 330. Find out (a) Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading

A. (a) 90 (b) 240 (c) 50 (d) 270

Q4. Aircraft on outbound radial 090 Aircraft is asked to track out on a radial 120. Find out (a) Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading (e) Which side to turn.

A. (a) 60 (b) 180 (c) 90 (d) 120 (e) Right

Q5. Aircraft is approaching Station on a radial 090 with 10 S drift is asked to approach on radial 110. Find out (a) Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading.

A. (a) 60 (b) 230 (c) 30 (d) 060

Q6. Aircraft is approaching Station on Hdg 270 with 10 P drift is asked to approach on radial 075. Find out (a) Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading.

A. (a) 30 (b) 285 (c) 15 (d) 330

PAYLOAD

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

ZERO FUEL WT (ZFW)MANUFACTURERS WT ENGINE + AVIONICSBASIC WT OR APS WTDRY OPERATING WTPAYLOAD (PAX + CARGONEWS PAPER + CATERINGCREW + BAGGAGEAIRFRAME + ENGINE + AVIONICSFOB (FLT FUEL +RESERVE)TAKE OFF WT (TOW)AICRAFT AUW BREAKDOWN

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.

FOB TOW

PayloadTOW = BASIC WT + PAYLOAD + FUEL ON BOARD (FOB) 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:-

MTOWRTOWMLWMZFW+FF (Flight Fuel) +FOB= TOW =TOWChoose 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.MTOWRTOWMLWMZFW83000*66000 +17200 + FOB83200 =TOW* LOWEST VALUEPAYLOAD = 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.MTOWRTOWMLWMZFW8200064500 +16000 + FOB=80500* =TOW* LOWEST VALUEPAYLOAD = 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.MTOWRTOWMLWMZFW12000090000 +15000 85000 + 17000=104000 =102000** LOWEST VALUEPAYLOAD = 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.MTOWRTOWMLWMZFW20000*18000 + 3000 17000 + 4600=21000 =21600 * LOWEST VALUEPAYLOAD = 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

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. MTOWRTOWMLWMZFW71506900 + 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 + ReserveFuel 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:

MTOWRTOWMLWMZFW6180*5740 + 468 5100 + 1016=6208 =6196* LOWEST VALUE PAYLOAD = MTOW (BASIC WT + FOB) = 6180 - (4400 + 1016) = 764 kgs

Payload from N to M:

MTOWRTOWMLWMZFW5800*5460 + 298 5180 + 548=5758 =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.

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) 1 USG = 100/26.4 L = 3.785, 100 USG = 378.7 l Litres (b) Kgs = l x SG also 100 USG = 378.71 litre Therefore 100 USG = 378.5 x 0.78 = 295 Kgs (c) 100 USG = 295 kgs or 295 x 2.2046 = 650 lbs (d) 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

TASFuel Cons (Gals/Hr) GS/FC(a)200250180/250 = 0.72(b)220265200/265 = 0.75(c)240280220/280 = 0.78(d)260295240/295 = 0.81(e)285315265/315 = 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

TASFuel Cons FC (IG/hr)(a)1801.25 USG/Nm1.25 x 180 =225 USG/Hr = 225/1.2 = 187.5 IG/hr(b)2404.1 IG/min4.1 IG/min x 60 = 246 IG/hr (c)220815 Kg/hr815/0.8 = 1018.75 l/hr = 1018.72 x 0.22 = 224.125 IG/hr(d)16012.7 L/min12.7 x 60 = 762 l/hr x 0.22 =167.64 IG/hr(Most Efficient)(e)1901.13 IG/min1.13 x 190 = 214.7 IG/hr (f)20013.41 Kg/m13.41 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.

Q. In the following table find the most fuel efficient figure when SG = 0.8

TASFuel Cons (IG/hr) FC (IG/Nm)(a)180187.5187.5/180 = 1.0416(b)240246 246/240 = 1.025 (c)220224.125224.125/220 = 1.018 (Most Efficient)(d)160167.64167/160 = 1.0477(e)190214.7214.7/190 = 1.1263 (f)200221.65221.65/200 = 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

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 PointA CP B W/V 090/20

Increased GS Home (TAS + WV) 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 TR TAS W/V GS

H CP to A 270 150* 090/20 170

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

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

TR TAS W/V GS

H CP to A 090 270* 090/25 245

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


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. TR TAS W/V GS

O A to PNR 270 330 090/25 355

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.

(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)

TRKTASW/VGSHDGDRIFT

270190330/101852733 P*

350200270/251863446 S

135240050/372341269 S

293137330/171232974 P

* TRK less than HDG Drift is Port and vice versa

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. TR TAS W/V GS

H CP to A 270 180* 330/18 188

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


(b) TR TAS W/V GS

O A to PNR 090 210 330/18 218

H PNR to A 090 210 330/18 200

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

(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

TR TAS W/V GS Dist FTime FC F Used A to PNR 090 210 330/18 218 1400 6:25 100 642

PNR to A 270 180 330/18 170 1400 8:14 80 659

Total Fuel Used = 642 + 659 = 1301 (Adopt Unitary method)

If flight fuel is 1301 then distance to PNR is 1400If flight fuel is 1 then distance to PNR is 1400/1301If 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. TR TAS W/V GS

H CP to A 150 250* 160/37 213

O CP to B 330 250* 160/37 286

G A to CP 330 300 160/37 336


Time to CP = Distance to CP/ GS out = 940/336 = 02h:48m 0n 4 Engine A 940 Nm CP 1060 Nm B

(b) TR TAS W/V GS DIST TIME FCF USED

O A to PNR 330 300 160/37 336 1400 4:10 2001190

H PNR to A 150 250* 160/37 2131690 If fuel is 2880 PNR is 2000, for fuel 1900 PNR = 2000/2880 x 1900 = 1320 Nm/3:55m 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. TR TAS W/V GS

H CP to A 090 180* 270/25 205

O CP to B 270 180* 350/38 170

G A to CP 270 200 270/25 175



(b) TR TAS W/V GS DIST TIME FC F USED

O A to CP 270 200 270/25 175 740 04:14 110 465

H CP to A 090 180 270/25 205 740 03:37 90 325 Total Fuel Used : 790; Bal = 810 PNRTR TAS W/V GS DIST TIME FC F USED CP to B (4 Eng) 270 200 350/38 190 610 03:13 110 353 B to CP (3 Eng) 270 180 350/38 183 610 03:10 80 300 Total Fuel Used : 653If flight fuel is 653 then distance to PNR is 610If flight fuel is 1 then distance to PNR is 610/653If 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:59mDPNR = 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. TR TAS W/V GS

H CP to A 270 150* 280/20 130

O CP to B 090 150* 330/20 159

G A to CP 090 180 280/20 200



(b) TR TAS W/V GS DIST TIME FC F USED

O A to CP 090 180 280/20 200 562 02:49 110 309

H CP to A 270 150 280/20 130 562 04:19 90 389 Total Fuel Used : 698; Bal = 402 PNRTR TAS W/V GS DIST TIME FC F USED CP to B (4 Eng) 090 180 330/20 189 688 03:38 110 400 B to CP (3 Eng) 270 150 330/20 139 688 04:56 90 445 Total Fuel Used : 845If flight fuel is 845 then distance to PNR is 688If flight fuel is 1 then distance to PNR is 688/845If 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:44mDPNR = 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 TR TAS W/V GS TIME

O A to PNR 250 210 270/30 181 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 mPNR = EOH/(O+ H) or 785= (E x 181 x 238)/ 419 or E = (785 x 419)/ (181x 238) = 7h 38mE = 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 TR TAS W/V GS

O A to PNR 155 220 240/30 215

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 mPNR = EOH/(O+ H) or 1080= (E x 215 x 221)/ 436 or E = (785 x 419)/ (181x 238) = 9h 55mE = 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. TR TAS W/V GS

H CP to A 312 160* 260/40 132

O CP to B 132 160* 350/60 203

G A to CP 132 190 260/40 212


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. TR TAS W/V GS

H CP to A 090 150* 330/20 159

O CP to B 270 150* 330/20 139

G A to CP 270 180 330/40 169


Time to CP = Distance to CP/ GS out = 640/169 = 03h:47m.

(b) DPNR &TPNR

TR TAS W/V GS

O A to PNR 270 180 330/20 169

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 600 nm B

W/V 270/30 030/17

TR TAS W/V GS DISTTIME(min)

A-X (3)090 190 270/30 220 1000273

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 375 (375 + 182) = 557

000 375 557 A X BSubtract 472 1 98 000 -472 +177 +577 CP lies in this LegTime 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 NmTCP = 727/250 =2h:54m (GS out 4 Eng = 220 +30 K tail wind)

Similar calculations can now be made for PNR with Fuel considerations.

PNR

TR TAS W/V GS DISTTIME FC F USED BAL

A-X (4)090 220 270/30250 1000 4:00 100 400

X-A (3) 270 190 270/30 160 1000 6:15 80 500 300 (1200 (500+400)) 900X-B (4) 090 220 030/17 211 600 2:50 100 284

B-X (3) 270 190 030/17 198 600 3:01 80 241 675 (1200 (284+241)) 525 A 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 ATime 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 B

W/V 280/22 330/20

TR TAS W/V GS DISTTIME(min)

A-X (3)270 150 280/22 128 900422

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) 090 150 280/22 172 900 314 (415 + 314) = 729A-X (4) 270 180 280/22 & 330/20 = 158/169 0 314 729 A X BSubtract 897 475 000 -897 -161 +729 CP lies in this LegTime 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.DCP = Distance of leg in which CP lies x 161 /(729 +161) =(1100 x 161)/890 = 199 + 900 =1099 NmTCP = 199/169 + 900/ 158 =1:11 + 5:42 = 6h:53m (GS out 4 Eng = 158 from A-X & 169 from X-B)


PNR


A-X (4)270 180 280/22158 900 5:42 100 570

X-A (3) 090 150 280/22 172 900 5:14 80 419 511 (1700 (570+419)) 989X-B (4) 270 180 330/20 169 1100 6:31 100 651

B-X (3) 090 150 330/20 159 1100 6:55 80 553 496 (1700 (651+553)) 1204 A X 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 ATime 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 900 nm

W/V 030/20 150/35 C

TR TAS W/V GS DISTTIME(min) A-B (3)090 210 030/20 199 600181

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) 270 210 030/20 219 600 164 (225 + 164) = 389A-B (4) 090 240 030/20 & 150/35 = 229/209

0 164 389 A B CSubtract 483 302 000 -483 -138 +389 CP lies in this LegTime 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 NmTCP = 236/209 + 600/ 229 =1:07 + 2:37 = 3h:45m (GS out 4 Eng = 158 from A-X & 169 from X-B)


PNR


A-B(4)090 240 030/20229 600 2:37 150 393

B-A(3) 270 210 030/20 219 600 2:44 120 329 678 722B-C (4) 120 240 150/35 209 900 4:18 150 646

C-B (3) 300 210 150/35 240 900 6:55 120 450 1096 A B C PNR lies in this Leg

PNR lies between B & C since Balance of fuel is 722 gals after catering for return from BHence, 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 ATime 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:- StageTrackWind VelDistance

A B180330/ 35 K900 Nm

B C210150/28 K1400 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)

A. A 330/ 35 K B 150/28 K 180/900 Nm C 210/1400 NM

Between B C

TR TAS W/V GS

H CP to B 030 300* 150/28 313

O CP to C 210 300* 150/28 285

G B to CP 210 360 150/28 345

A B (4) 180 360 330/35 390

(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 DISTTIME FC F USED BAL

A-B(4)180 360 330/35390 900 2:18 250 577 1423 (2000-577)

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 B C PNR lies in this Leg

PNR lies between B & C since Balance of fuel is 1423 kgs after catering for return from BHence, 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 ATime 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:- StageTASWind VelDistance

P Q420 -25 965 Nm

Q R420 -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 32 (Time to Y) 148 P 965 Q 900 R 290 144 (Time to R) 0 -290 240 -112 148 Y CP lies in this Leg

CP

StageTASWind VelGS Distance Flt Time (min)

P Q420 -25395 965 Nm146

Q R420 -45375 900 Nm144290R Q 420 +45465 900 Nm116

Q Y 420 +30 450 240 Nm 32148DCP (Q-R) = DH/(O +H) = 900 x 112/ (148 + 112) = 388 Nm from QDCP = 965 + 388 = 1353

P Q (4) 500 -25 475 965 Nm 2:02Q CP (4) 500 -45 455 388 Nm 0:51TCP = 2:02 + 0:51 = 2h:53m

PNRStageTASWind VelGS Distance Time FC F Used Bal

P Q500 -25475 965 Nm2:02 6300 12810 (31500-15797)Q Y 420 +30 450 240 Nm0:32 5600 2987 15703 15797 Q R500 -45455 900 Nm1:59 6300 12461

R Q 420 +45465 900 Nm1:56 5600 10839 23300If Flight fuel is 23300 then DPNR is 900 NmIf Flight fuel is 15703 then DPNR is (900 x 15703)/23300 = 607 Nm from QDPNR = 965 + 607 = 1572 Nm Time to cover 607 Nm @ GS 455 K = 607/455 = 1:20TPNR = 2:02 + 1:20 = 3h:22mCRITICAL 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 calculatedA (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

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).

241507 N782336 WALERTBATTWARN123HOLD4567890CLEARINSERTWPTDIST/TIMEWINDDSR TK/STSTESTPOSXTK/TKEHDG/GATK/GS423TK CHGDIMWPT

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.

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 1 nm to the left of the desired track of 135M, the left display would read L 01.5. If the track being made good happened to be 130M, the right display would read L 005.0.7.POS (Present Position). The aircrafts current latitude and longitude are shown in terms of Longitude and Latitude in the left and right displays, respectively. For example, 241507N and 782336W, 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 l

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