Chapter 31 — General Navigation Problems

Introduction

The General Navigation examination includes many questions that test geometric reasoning and problem-solving in a navigation context. These fall into four broad categories: triangular geometry problems, implications of the triangle of velocities, rhumb line track angle calculations, and timing-to-beacon problems.

📌 Key skill: Always draw a diagram — even a quick sketch. Most geometry problems become straightforward once you can see the triangle involved.

📌 Look for isosceles triangles — when two sides are equal (equal DME ranges, equal time legs) the angles are known.

📌 Remember: a right-angle triangle's angles sum to 180°. If you know two angles, you know all three.

Triangular Problems

Worked Example 1 — Relative Bearing to Beacon

The relative bearing to a beacon is 270°(R). Three minutes later, at a ground speed of 180 knots, the relative bearing has changed to 225°(R). What was the distance of the closest point of approach to the beacon?

a) 45 NM
b) 18 NM
c) 9 NM
d) 3 NM

Answer: (c) — 9 NM

Draw the situation with the aircraft flying north (so relative = true bearing). At position A, beacon is at 270°(R) = 90° to the left. The line from A to the beacon is perpendicular to the track → angle A inside the triangle = 90°.

3 minutes later at position B, bearing is 225°(R) → angle B inside the triangle = 180° − 225° = 45° (the supplement at that corner... actually the turn through the triangle).

Internal angles must sum to 180°: 90° + 45° + angle C = 180° → angle C = 45°. Since angles B and C are both 45°, the triangle is isosceles: AB = AC (the closest approach distance).

Distance AB = 3 min × 180 kts = 3/60 × 180 = 9 NM

Therefore AC (closest point of approach) = 9 NM ✓

Figure 1 — Triangular geometry for the relative bearing problem — isosceles triangle shows AB = AC = 9 NM at angles of 90°/45°/45°

Worked Example 2 — VOR/DME Position Fix

At 1000 hours an aircraft is on the 310° radial from a VOR/DME at 10 NM range. At 1010 the radial and range are 040°/10 NM. What is the aircraft's track and ground speed?

a) 080°(M)/85 knots
b) 085°(M)/85 knots
c) 080°(M)/80 knots
d) 085°(M)/90 knots

Answer: (b) — 085°(M)/85 knots

Radial 310°(M): the aircraft is ON the 310° radial, so the VOR-to-aircraft direction is 310°. The angle at the VOR (vertex A) between the two position lines: 310° → to the aircraft, and 040° → to the aircraft 10 min later. Angle between 310° and 040° = 360° − 310° + 40° = 90°.

Both ranges are 10 NM → both AB and AC = 10 NM → isosceles triangle. With 90° at the vertex, the two base angles B and C = (180° − 90°)/2 = 45° each.

Track: Direction from A to B is 310°(M), so direction from B to A is 130°(M). Angle at B = 45°. Track = 130° − 45° = 085°(M) ✓

Ground speed: By Pythagoras: BC² = 10² + 10² = 200 → BC = √200 = 14.14 NM in 10 minutes → GS = 14.14 × 6 = 85 knots ✓

Figure 2 — VOR/DME isosceles triangle — both ranges are 10 NM with a 90° angle at the VOR, giving two 45° base angles and a 085°(M) track

Triangle of Velocities — Wind Component Geometry

A common misconception is that a wind at right angles to heading or track produces zero headwind/tailwind component. This is not strictly true.

Worked Example 3 — Crosswind Headwind Component

An aircraft is tracking 090°(T) at TAS 180 knots. The wind from due North causes a 5 knot head wind component. What will the wind component be if the aircraft flies a track of 270°(T)?

a) 5 knots tail
b) zero
c) 5 knots head
d) not possible to tell

Answer: (c) — 5 knots head

This is a geometric result. Draw a triangle of velocities for Track 090°(T), wind from 360°(T):

The TAS vector (hypotenuse) must point in a slightly different direction from the track (090°) to account for drift. The GS is less than TAS because the wind's head component = GS = √(TAS² − crosswind²). The head component is not zero because the TAS vector has a slight northward heading to counter drift.

Figure 3 — Track 090°(T) with northerly wind — the TAS vector (hypotenuse) is longer than the GS vector, giving a small but non-zero headwind component

Figure 4 — Track 270°(T) with northerly wind — a mirror image of the 090° case: the headwind component is identical (5 knots head), NOT a tailwind

Key Insight: Zero head/tail wind component does NOT occur when the wind is at right angles to heading or track. True zero wind component occurs when the wind is at right angles to the bisector between heading and track.

For practical purposes this is an academic distinction — in most real navigation the component is approximated as zero for crosswinds.

Rhumb Line Track Angle Calculation

To find the rhumb line track angle between two positions, treat the chart as a flat (Mercator-like) grid: north–south distances in nautical miles from change of latitude, east–west distances using the departure formula.

Worked Example 4 — Rhumb Line Track

What is the Rhumb Line track from A (45°00'N 010°00'W) to B (48°30'N 015°00'W)?

a) 315°(T)
b) 330°(T)
c) 215°(T)
d) 150°(T)

Answer: (a) — 315°(T)

N–S distance (change of latitude): 45°00'N to 48°30'N = 3°30' = 210 minutes = 210 NM (northward)

E–W distance (departure): Change of longitude = 5° = 300'. Using mid-latitude (≈46°45'N) or the simpler 45°N:

Departure = 300' × cos 45° = 300 × 0.7071 = 212 NM (westward)

Track angle: tan(x) = Departure / Δlat = 212/210 ≈ 1.01 → x ≈ 45.3°

The track is NW (north and west): starting from West (270°), add 45° northward = 270° − 45° = 225°... Actually: track from due west baseline: 270° − x does not work directly. Think of it geometrically: the aircraft goes 210 NM north and 212 NM west. From a northward reference, the track is 315° − 0.3° ≈ 315°(T) ✓

Method: the angle x from North = arctan(212/210) = 44.7°. Track = 360° − 44.7° = 315.3° ≈ 315°(T) ✓

Figure 5 — Rhumb line track calculation on a Mercator-style grid — N-S distance (210 NM) and E-W departure (212 NM) give a NW track of ~315°(T)

Rhumb Line Formula Summary

Step 1 — N-S distance: change of latitude in minutes = nm (1' lat = 1 NM)

Step 2 — E-W distance: Departure = Δlong (in min) × cos(mid-latitude)

Step 3 — Track angle: tan(x) = Departure / N-S distance → x = arctan

Step 4 — Convert x to a 360° bearing using the diagram quadrant (NE/NW/SE/SW)

Timing to a Beacon (Mach/TAS Adjustment)

When ATC requests a different arrival time, you must adjust TAS (hence Mach number) — but you must account for the wind component, which does NOT change just because you change speed.

Worked Example 5 — Mach Number Adjustment

Flying at TMN .77 in SAT −55°C. At 1000 hrs, 150 NM from DME, ETA is 1017. ATC requests arrival at 1021. What is the new TMN if you reduce speed at 150 NM to go?

a) M.72
b) M.62
c) M.67
d) M.59

Answer: (d) — M.59

Current ground speed: 150 NM in 17 min = 150/17 × 60 = 529 ≈ 530 kts GS

Current TAS: M.77 at SAT −55°C. LSS = 38.95 × √(273−55) = 38.95 × √218 = 38.95 × 14.76 = 575 kts. TAS = 0.77 × 575 ≈ 443 kts

Wind component: GS − TAS = 530 − 443 = +87 kts (tailwind component)

Required new GS: 150 NM in 21 min = 150/21 × 60 = 429 kts

⚠️ Common trap: Do NOT simply scale the Mach number (M.77 × 429/530 = M.62 is WRONG — ignores wind).

New required TAS: New GS − Wind component = 429 − 87 = 342 kts

New Mach number: New TMN = M.77 × (342/443) = M.77 × 0.772 = M.595 ≈ M.59 ✓

Critical rule for timing problems: The wind component does not change when you change Mach number. Always find the wind component from the current GS and TAS first, then use it to find the new required TAS, then convert to new Mach number. Never simply ratio the Mach numbers by the GS ratio.

Parameter	Current	New Required
Time to beacon	17 min (ETA 1017)	21 min (ETA 1021)
Ground Speed	530 kts	429 kts
Wind component	+87 kts (tail)	+87 kts (tail — unchanged)
TAS required	443 kts (M.77)	342 kts
Mach No.	M.77	M.59