Air navigation for the partial

Welcome to the first of three lectures for the navigation-systems partial exam of Transport Aeri i Sistemes de Navegació. This part covers air navigation: the vocabulary, the charts, the wind triangle, and every navigation problem type that has appeared in past partials. Parts two and three cover the telecommunication fundamentals: signals and systems, noise, data transmission, analogue to digital conversion, propagation and antennas. Every worked answer in these videos has been recomputed and checked against Professor Barlabé's answer key where one exists; the 2022 and 2023 papers have no key, so those answers were recomputed independently.

Let us start with the rules of the game, as announced by Professor Barlabé on the twenty-second of October. The exam covers everything from the first topic up to analogue to digital and digital to analogue conversion, so passband signals are out. It lasts about an hour, and you may bring only a normal scientific calculator: no tables and no formula sheet, so every formula in these videos must live in your head. The past papers vary: 2016 had one navigation problem and nine short questions, 2022 one navigation problem, a radio-link problem and seven short questions, and 2023 two navigation problems, a Friis link budget and nine short questions, so be ready for any mix. This first partial is worth 20 % of the course, and anyone may retake it on the day of the final exam to pass or to improve the mark.

Navigation, in Barlabé's words, is the art of getting from one place to another without getting lost. Historically it began with observation of landmarks and stars, and with dead reckoning, which is estimating position from heading, speed and time. Then came radio: direction finding, and radio beacons such as the non-directional beacon with its automatic direction finder, the VOR, the DME and the military TACAN. Air-traffic control added radar: primary radar sees reflections, secondary radar interrogates the aircraft transponder. Satellite systems such as GPS, GLONASS and Galileo are broadcast systems, while Argos or Inmarsat C work point to point; and there is a long list of aids such as ILS, collision avoidance, radio altimeters, ADS-B and ACARS. Finally, augmentation systems such as EGNOS or differential GPS improve accuracy, and inertial platforms navigate without any telecommunication at all.

Positions on the Earth are given by latitude, the angle north or south of the equator, and longitude, the angle east or west of the Greenwich meridian. One minute of arc of a great circle, for example of a meridian, measures about one nautical mile, and the nautical mile itself is defined as exactly 1 852 m. So one degree of latitude is 60 nmi, and a speed of one knot means one nautical mile per hour. This gives a very fast trick for problems where an aircraft flies along a meridian: the change of latitude in minutes is directly the distance in nautical miles. Be careful along a parallel, though: meridians converge towards the poles, so one minute of longitude measures only the cosine of the latitude in nautical miles.

A chart represents a sphere on a flat sheet of paper, so it needs a projection, and no projection preserves everything. The loxodrome, or rhumb line, is the curve that cuts every meridian at the same angle: flying it means keeping a constant heading. On a Mercator chart, a conformal cylindrical projection, the rhumb line is a straight line, which is why sailors loved it. The orthodrome, or great circle, is the arc of the largest circle through the two points, shorter than 180 degrees, and it is the shortest route between them. Look at Barcelona to New York on the Mercator chart: the great circle bows north, towards the pole, while the rhumb line stays straight but is longer. On the Lambert conformal conic projection, the great circle is approximately a straight line, which is why aeronautical charts use it, and polar regions use a polar stereographic projection. A theory question that students reported from the partial asks exactly this difference between a loxodromic and an orthodromic line, so learn the two definitions word for word.

Now the vocabulary, which the exam uses in Catalan, so I will give both names. The heading, rumb in Catalan, is the direction in which the nose of the aircraft points, and it is what the compass shows the pilot. The track, or derrota, is the projection of the real trajectory on the ground: where the aircraft actually goes. The drift, or deriva, is the angle between heading and track, and it is caused by the wind. The bearing, demora, is the direction in which you see an object, measured clockwise from north, while the marcació, the relative bearing, is measured from the longitudinal axis of the aircraft. All these angles go clockwise from north, as on a compass rose, with three digits: east is zero nine zero, south one eight zero and west two seven zero, and in aviation they are magnetic, although Barlabé notes that in Catalonia the magnetic declination is now negligible. The magnetic declination is the angle between true and magnetic north, counted positive when magnetic north lies east of true north and negative when it lies west; runway numbers and VOR radials are magnetic, which is why aeronautical charts print the declination.

Every navigation problem in this course reduces to one triangle of vectors. The air vector points along the heading, with a length equal to the true air speed. The wind vector has the wind speed as its length, but here is the classic trap: a wind is named by where it comes FROM, so a wind from 360 blows towards 180, and the vector must point towards 180. Adding the two gives the ground vector, whose direction is the track and whose length is the ground speed, which is exactly what the GPS shows. The safest way to add vectors with a calculator is by components: with x pointing east and y pointing north, a vector of size V at angle theta has an east component V sine theta and a north component V cosine theta. To get an angle back, take the arc tangent of east over north and correct the quadrant; alternatively, solve the triangle with the sine and cosine laws, which the problem sheet prints at the bottom of the page and which you must now know by heart.

Watch the triangle being built for the first problem of the collection: heading 045 at 100 kts, with a north wind, from 360, at 20 kts. First the air vector: 100 kts along zero four five, to the north-east. At its tip we add the wind: 20 kts pointing south, towards 180, because the wind comes from the north. The vector that closes the triangle, from the origin to the final tip, is the ground vector. The north wind pushes the aircraft to the right of its heading, so the track is larger than 045, and the headwind component makes the ground speed smaller than 100 kts. The numbers come out as a track of 054, a ground speed of 87 kts, and a drift of 9 degrees to the right.

Let us solve it by components, the method that never fails. The air vector at 045 and 100 kts has east and north components of 70.7 each. The wind comes from 360, so it blows towards 180: its components are zero east and minus 20 north. Adding them, the ground vector is 70.7 east and 50.7 north. The track is the arc tangent of 70.7 over 50.7, which is 54.4 degrees, so track 054, exactly the official answer. And as a bonus, the ground speed is the square root of the sum of squares: 87 kts.

This is the inverse problem: we know the air vector and the ground vector, and we want the wind. The air vector is heading 000 at 100 kts: zero east, 100 north. The GPS gives the ground vector: 110 kts on track 010, which is 19.1 east and 108.3 north. Since ground equals air plus wind, the wind is ground minus air: 19.1 east and 8.3 north. Its size is 20.8 kts and it points towards 066 degrees. But meteorologists and pilots name the wind by where it comes FROM, so we add 180: the wind is from 246 at about 21 kts, a south-westerly that pushes us to the right and slightly from behind.

Here the track is imposed, zero six zero, and we must find the heading that makes the wind push us exactly onto it. The wind comes from 315, the north-west, so it blows towards 135 and pushes the aircraft to the right of the desired track. Only the crosswind component matters for the heading: 20 kts times the sine of the 75 degree angle between the wind and the track, which is 19.3 kts. The air vector must cancel this crosswind, so the wind correction angle is the arc sine of 19.3 over 100, which is 11.1 degrees. Because the wind pushes us right, we turn the nose left, into the wind: heading 060 minus 11.1 gives 048.9, so heading 049, which is the official answer. As a check, the along-track components give a ground speed of about 103 kts, since this wind has a small tailwind component.

In the 2023 paper both the track and the ground speed are imposed, so we solve for the full air vector. The desired ground vector, 200 kts on 130, has components 153.2 east and minus 128.6 north. The wind from 020 blows towards 200, giving minus 8.6 east and minus 23.5 north. The air vector is ground minus wind: 161.8 east and minus 105.1 north. Its direction is 123 degrees, south-east, so the pilot flies heading 123; notice the arc tangent alone would give an angle in the wrong quadrant, which is why you must look at the signs. Its size is the true air speed to set: about 193 kts, a little less than 200 because this wind has a small tailwind component along 130.

This one rewards knowing the definition of the nautical mile. The longitude does not change and the latitude increases, so the aircraft flies due north along the meridian: heading 000, also written 360. The change of latitude is 1 degree 25 minutes, which is 85 minutes of arc. Since one minute of latitude is one nautical mile, the distance is exactly 85 nmi, without any trigonometry. Fifteen minutes is a quarter of an hour, so the speed is 85 divided by 0.25, which is 340 kts.

A radial is the direction in which the wave travels away from the beacon, so the aircraft lies on that line, seen from the beacon. With the beacon at the origin, the first position is 10 nmi along 190: about minus 1.74 east and minus 9.85 north. The second position is 14 nmi along 210: minus 7.00 east and minus 12.12 north. The displacement in that minute is minus 5.26 east and minus 2.28 north. Its length is 5.73 nmi, flown in one minute, so the speed is 5.73 times 60, which is 344 kts. Its direction, in the south-west quadrant because both components are negative, is 246.6 degrees: track 247, matching the official answer.

Here there is no wind: it is pure vector addition, which is exactly what a drone flight plan is. The first leg, 100 m on 310, is minus 76.6 east and 64.3 north. The second leg, 300 m on 030, is 150.0 east and 259.8 north. So waypoint 2 lies 73.4 m east and 324.1 m north of the start. The distance home is the length of that vector, 332 m. The way home is the opposite vector, pointing south and slightly west, which gives heading 193, the official answer; a quick sanity check: waypoint 2 lies on bearing zero one three from the start, and the reciprocal of zero one three is one nine three.

The interception problem appeared as problem 1.2 of the collection and as question 1 of the November 2016 partial, worth two points. At time zero a target at 42 degrees north flies heading 225 at 500 kts. An interceptor at the same latitude, 300 nmi to the west, flies at 1 000 kts and must choose a heading that meets the target. Both fly straight lines at constant speed, so after the unknown time t the target has flown 500 t nautical miles and the interceptor 1 000 t. These two distances and the initial 300 nmi separation form a triangle, and the angle at the target between the line to the interceptor, due west at 270, and the target's heading, 225, is 45 degrees. Watch the two aircraft converge on the meeting point after about 14 minutes.

Barlabé's own hint in the FAQ is: it is just a triangle, use the cosine law for t and the sine law for the heading. The side opposite the 45 degree angle is the interceptor's path, so the cosine law gives 1 000 t squared equals 300 squared plus 500 t squared minus two times 300 times 500 t times cosine 45. Rearranging, we get a quadratic: 750 000 t squared plus 212 132 t minus 90 000 equals zero. Its positive root is t equal to 0.233 hours, about 14 minutes, the official answer. Now the sine law: the sine of the angle at the interceptor, between due east and its heading, equals 500 t times sine 45 divided by 1 000 t, which is 0.354, so the angle is 20.7 degrees. The target is south-east of the interceptor, so the interceptor turns 20.7 degrees to the right of 090, giving 110.7 degrees, which rounds to heading 111, as in the answer key.

Let me close part one with the traps that cost points. First, the wind: it is always named by where it comes from, so turn it around by 180 degrees before adding vectors. Second, always draw a sketch before calculating: it tells you the quadrant, and whether the drift is to the left or to the right. Third, work in components and decide the quadrant of the final angle from the signs, never from the arc tangent alone. Fourth, to correct a drift you point the nose into the wind, and a headwind lowers the ground speed while a tailwind raises it. Fifth, a radial is the direction in which the wave leaves the beacon, and a bearing is measured clockwise from north. Sixth, keep nautical miles and knots, remember that a degree of latitude is 60 nmi, and convert minutes to hours; and finally, give directions with three digits, as aviation does.

To summarise part one: you now know the navigation families, the two kinds of route on a chart, and the vocabulary of heading, track, drift and bearing. Every problem is the same triangle, ground equals air plus wind, solved by components or with the sine and cosine laws. We have solved every navigation problem from the collection and from the 2016, 2022 and 2023 partials, and every answer agrees with the official key where one exists, or was recomputed independently where there is none. Before the exam, redo problems 1.1 to 1.5 on paper without looking, and time yourself: each should take under ten minutes. In part two we move to telecommunications: signals, Fourier, linear systems, distortion, noise and the Shannon limit.