Chapter 14

Convergency and Conversion Angle

Earth convergency · Great circle vs rhumb line · Conversion angle · Practical examples

1. Introduction

Meridians are great semi-circles joining the poles. Because they converge towards each pole, two meridians that are parallel at the equator will meet at the pole at an angle exactly equal to their difference in longitude. At intermediate latitudes, the angle of inclination between them is intermediate — greater than zero (at equator) but less than their d.long (at the pole).

Figure 1 — Meridians diverge from the North Pole, are parallel at the Equator, then converge again to the South Pole

Figure 2 — At the Equator the two meridians are exactly parallel to each other

Figure 3 — At the Pole the two meridians meet at an angle equal to the change in longitude

Figure 4 — At intermediate latitudes there is an intermediate angle of inclination

2. Convergency

Convergency (also called Earth Convergence) is defined as the angle of inclination between two selected meridians measured at a given latitude.

At the equator (lat 0°) convergency is zero; at the poles (lat 90°) convergency equals the change in longitude. The ratio convergency/ch.long follows the sine of latitude:

Convergency Formula (same latitude)

Convergency = Change in Longitude × sin(Latitude)

Figure 5 — Convergency as the angle of inclination between two meridians — includes the summary table showing convergency at equator (0) and pole (1×ch.long)

Figure 6 — Convergency at latitude 30°N — tangents drawn to each meridian show the angle of inclination

Example

Change of longitude = 40°; Latitude = 30°N (sin 30° = 0.5)
Convergency = 40° × 0.5 = 20°

3. Convergency and the Great Circle Track

True North at any position is defined by the local meridian. If you fly between two meridians, your local True North direction has changed — by exactly the amount of convergency. This is why:

A rhumb line must constantly curve on the Earth to cut all (converging) meridians at the same angle.
A great circle (the shortest path) changes its measured track direction by the amount of convergency between the start and end meridians.

Convergency Formula (different latitudes)

Convergency = Change in Longitude × sin(Mean Latitude)

Mean latitude is approximated by mid-latitude. The difference is negligible for typical distances.

Figure 7 — Great circle cuts two meridians — the change in track direction equals the angle of inclination of the meridians (convergency)

D-I-I-D Rule — Track Angle Change Direction

Hemisphere	Eastward track	Westward track
Northern	Track angle Increases	Track angle Decreases
Southern	Track angle Decreases	Track angle Increases

Equivalently: going towards the nearer pole → track increases; going away → track decreases.

4. Worked Example: Q1

Question 1

The initial great circle track from A (4000N 00200W) to B (5000N 01000E) is 060°(T). What is the initial great circle track from B to A?

Figure 8 — Q1: Meridians converging northwards — initial great circle track of 060°(T) from A to B; checking latitude/longitude confirms the diagram orientation

Step 1 — Convergency:
Ch.Long = 002°W to 010°E = 12°. Mid-lat = (40°N + 50°N) ÷ 2 = 45°N. sin 45° = 0.7071.
Convergency = 12 × 0.7071 = 8.5°

Step 2 — Track at B (going A→B):
Going eastward in the Northern hemisphere → track increases.
Track at B = 060° + 8.5° = 068.5°(T)

Step 3 — Reciprocal at B (going B→A):
Track B→A = 068.5° + 180° = 248.5°(T)

Alternatively: parallel the meridian at A across to B — the convergency angle adds to the track at B, giving the same result of 248.5°.

5. Worked Examples: Q2 and Q3 (Summaries)

Q2 — Finding longitude and mid-track bearing

C (36°N 015°E) to D (42°N, unknown longitude). Initial GC track = 300°(T), final GC track at D = 295°(T).

Convergency = change in track = 300° − 295° = 5°. Mid-lat = 39°N, sin 39° = 0.629.
Ch.Long = 5° ÷ sin 39° = 8°. Track decreasing westward ∴ D is west of C → Long D = 007°E.

Track at 011°E (halfway between 015°E and 007°E) = halfway between 300° and 295° = 297.5°(T).

Q3 — Southern hemisphere

H (40°S 170°W) to G (45°S 174°E). Initial GC track from H = 250°(T). Find initial GC track from G to H.

Ch.Long = 16°. Mid-lat = 42.5°S, sin 42.5° = 0.676.
Convergency = 16 × 0.676 = 10.8°.

Going westward in Southern hemisphere → track increases.
Track at G (going H→G direction) = 250° + 10.8° = 260.8°. Reciprocal at G = 260.8° − 180° = 080.8°(T).

6. Conversion Angle

For two points at the same latitude, the rhumb line track is constant and equals the direction of the parallel. The great circle track is symmetric about the mid-meridian — it leaves the departure point on a track less than the RL (closer to the near pole) and arrives at the destination on a track greater than the RL (same magnitude, other side).

Conversion Angle Definition and Formulae

CA = difference between GC and RL directions = ½ × Convergency

CA = ½ × Ch.Long × sin(Mean Latitude)

GC track is always closer to the nearer pole than the RL track. CA is the same magnitude at both ends.

Figure 10 — Conversion angle — the great circle (straight line on this projection) runs closer to the North Pole than the rhumb line (which appears curved). The CA at each end equals ½ × convergency.

Applying Conversion Angle

GC track at departure point = RL track − CA (in N hemisphere going east; GC curves pole-ward)
GC track at arrival point = RL track + CA
Radio bearings travel along great circle paths → apply CA to convert GC bearing to RL bearing before plotting
Formula: RL bearing = GC bearing ± CA (apply at the point where GC direction is measured)

7. Worked Examples: Q4–Q6 (Summaries)

Q4 — J to K on same latitude

J (5812N 004°W) ↔ K (5812N 006°E). Rhumb line track J→K? Initial GC track K→J?

a) Convergency: Ch.Long = 10°. Lat = 58.2°N. sin 58.2° = 0.851. Convergency = 10 × 0.851 = 8.5°.

b) RL track J→K: Same latitude → RL is the parallel → track = 090°(T).

c) GC track K→J: CA = ½ × 8.5° = 4.25°. GC at J going J→K = 090° − 4.25° = 085.75°. GC at K going J→K = 090° + 4.25° = 094.25°. Track K→J = reciprocal of GC at K = 094.25° + 180° = 274.25°(T).

Figure 11 — Q5 — INS waypoint changeover at WP2: great circle track increases eastward on each leg; at the changeover the track resets to the initial bearing of the new leg, producing an 8° decrease in track angle

Q5 — INS waypoints on same latitude

WP1 53°N 030°W → WP2 53°N 020°W → WP3 53°N 010°W. Track change at WP2?

Each leg: Ch.Long = 10°. Mid-lat = 53°N. sin 53° = 0.8.
CA = ½ × 10 × 0.8 = 4°. GC leaves WP1 on 086°T; arrives WP2 on 094°T.
At WP2 the INS resets to 086°T for the new leg → track change = 086° − 094° = 8° decrease. Answer: (d).

Figure 12 — Q6 — diagram showing A (55°N 000°) and B (54°N 010°E). GC from A to B at A = 092°(T); CA = 4°; RL track = GC + CA = 104°(T)

Q6 — Applying CA to get RL from GC

A = N55° 000°. B = N54° E010°. GC from A to B = 092°(T). RL track at A?

CA = ½ × 10 × sin 54.5° (mean lat) = ½ × 10 × 0.814 = 4°.
GC at A = 092°T (slightly south of east). RL runs further from pole than GC → RL > GC.
RL = GC + CA = 092° + 4° = 104°(T). Apply CA at the position where GC is measured. Answer: (c).

Practice Questions

All questions are computational. Show all working. Tap each question to reveal the solution.

Q1 The convergency of meridians through M and N (Southern hemisphere) is 12°. Rhumb line track M→N = 249°(T). Find: (a) GC track from M to N; (b) GC track from N to M.

CA = ½ × 12° = 6°.

(a) Going WSW-ish, in S hemisphere — track decreases going away from south pole.
GC at M = RL − CA = 249° − 6° = 243°(T).

(b) GC changes by full convergency between M and N.
GC at N (going M→N) = 243° + 12° = 255°(T). Track from N to M = reciprocal = 075°(T).

Q2 GC bearing of B (30°S) from A (30°S 165°E) = 100°(T). Find: (a) GC track B to A; (b) longitude of B.

RL from A to B: same latitude → RL = 090°T.
CA = GC at A − RL = 100° − 090° = 10°. (GC > RL → track going away from south pole = S hemisphere going east.)

(a) GC at B going A→B = 090° + 10° = 100°T ✓. Track B→A = reciprocal of 100° track at B = ... Actually: GC at A going A→B = 100°T. GC at B going A→B = RL + CA = 090° + 10° = 100°T? No — CA is half convergency.
CA = 10°, so convergency = 20°. GC at B going A→B = 100° − (−10°)... Let me redo:
In S hemisphere going east (towards equator), track decreases. GC at A = 100°T, GC at B = 100° − 20° = 80°T. Reciprocal = 260°(T).

(b) CA = 10° = ½ × Ch.Long × sin 30°. Ch.Long = 10 ÷ (0.5 × 0.5) = 40°. Going east from A (165°E): Long B = 165°E + 40° = 205°E = 155°W. → 155°W.

Q3 RL from D (30°N 179°W) to C = 090°(T). GC track from C to D = 287°(T). Find: (a) GC track D to C; (b) approx. lat/long of C.

(a) RL = 090°T same latitude. GC from C to D at C = 287°T. Track D to C = GC from D to C at D.
GC at C going D→C direction = reciprocal of 287° = 107°T. GC track D→C differs by convergency from 107°T...
Simpler: GC from C to D (at C) = 287°T. D is due west (RL=090° from D means RL=270° going west to C... wait: RL from D to C = 090°T means going east, so C is east of D.
GC D→C at D = 090° − CA. GC C→D at C = 090° + CA, so 287° = reciprocal of 090°+CA = 090°+CA+180°−360° = ... Actually reciprocal of (090°+CA) = 270°−CA. 270°−CA=287° → CA=−17°... That's wrong.
Let me reconsider: GC from C to D = 287°T. So at C, the GC track going toward D = 287°T. Reciprocal = 287°−180° = 107°T = GC track from D looking back toward C. But we want GC from D to C = the initial track at D looking toward C = 107° − convergency (since going east in N hem, track increases; so going back westward from D to C it decreases).
Actually convergency change: track at D minus track at C going same direction... We need:
GC D→C at D = 287° − 180° = 107°T... and adjust:
The GC from D to C. At C the GC going C→D = 287°T. The GC at D going D→C is different by convergency.
Going west (270°-ish) in N hemisphere, track decreases. From D to C going east, GC at D < GC at C.
GC at C going D-direction = 287°T (i.e., at C heading toward D = 287°T = WNW).
GC at D going C-direction = 287° − convergency (going westward → track decreases).
Reciprocal of GC at D going C-direction = GC from D to C. But we need the value of convergency first — it's found from part (b).
The answer from the key is (a) 073°(T).

(b) RL D→C = 090°T (same lat, 30°N). GC C→D = 287°T. CA = difference between GC and RL at C = |287° − 270°| = 17°.
Wait: RL from C to D = 270°T (going west). GC at C = 287°T. Difference = 287°−270° = 17° = CA.
CA = 17° = ½ × Ch.Long × sin 30° = ½ × Ch.Long × 0.5. Ch.Long = 17 ÷ 0.25 = 68°.
C east of D: Long C = 179°W + 68° east = 179°W − 68°... going east from 179°W by 68° = 179°−68° = 111°W. → 30°N 111°W.
(a) GC D→C: CA = 17°. GC at D going D→C (east) = RL − CA = 090° − 17° = 073°(T). → 073°(T).

Q4 GC A→B at A = 227°(T); at B = 225°(T). What is the convergency and in which hemisphere?

Change in GC track = 227° − 225° = 2° → Convergency = 2°.

Track going ~SW, decreasing from A to B. Going westward (SW) in N hemisphere → track decreases. ∴ Northern hemisphere.

Answer: 2° convergency, Northern hemisphere

Q5 (a) At what latitude does convergency equal twice the convergency at 20°N? (b) Is there a latitude where it equals three times?

(a) Convergency ∝ sin(lat). We need sin(L) = 2 × sin(20°) = 2 × 0.342 = 0.684.
L = arcsin(0.684) ≈ 43°N.

(b) 3 × sin(20°) = 3 × 0.342 = 1.026 > 1.000. The sine function cannot exceed 1.000 → No, no such latitude exists.

Q6 (a) GC B→A = 268°T; GC A→B = 092°T. i) Hemisphere? ii) RL track A→B? (b) GC C→D = 063°T; RL D→C = 240°T. i) Hemisphere? ii) Approx. initial GC track D→C?

6a-i: GC at A = 092°T going A→B. GC at B going A→B = reciprocal of 268° = 088°T. Track decreased 092° → 088° = 4°. Going east (approximately), track decreasing → Southern hemisphere.

6a-ii: RL = (092° + 088°)/2 = 090°(T) (A and B on same latitude → RL is the parallel).

6b-i: GC C→D = 063°T (NE-ish). GC D→C at D must differ. RL D→C = 240°T → RL C→D = 060°T.
GC C→D = 063° > RL C→D = 060° → GC is 3° more than RL. CA = 3°. Going NE in N hemisphere, GC should be < RL (closer to pole = NW). GC > RL means going NE in S hemisphere.
∴ Southern hemisphere.

6b-ii: Convergency = 2 × CA = 6°. GC at C going C→D = 063°T. GC at D going C→D = 063° − 6° = 057°T (decreasing going east in S hem). GC D→C = reciprocal = 057° + 180° = 237°T. → 237°(T).

Q7 X = 64°S 011°50'W; Y = 64°S 005°10'W. Find: (a) convergency X–Y; (b) approx. initial GC track Y→X; (c) RL track X→Y.

(a) Ch.Long = 011°50'W to 005°10'W = 6°40' ≈ 6.67°. Lat = 64°S. sin 64° = 0.899.
Convergency = 6.67 × 0.899 ≈ 6°.

(b) Same latitude → RL X→Y = 090°T (going east). CA = 3°. Y is east of X.
GC track Y→X (going west in S hemisphere, track increases going west).
RL Y→X = 270°T. GC Y→X at Y = RL + CA = 270° − 3°... In S hemisphere going west: GC closer to S pole → GC track > 270°? No, closer to pole means track goes more southward...
GC at X going X→Y = RL − CA = 090° − 3° = 087°T. GC at Y going X→Y = RL + CA = 090° + 3° = 093°T.
GC Y→X = reciprocal of 093° at Y = 273°T. But wait: CA = ½ × 6° = 3°, so GC at Y going west = 270° − 3° = 267°T.
Hmm: going Y→X = westward. GC departs closer to south pole (south of RL) on first portion → GC at Y = 270° − 3° = 267°(T).

(c) Same latitude → RL X→Y = 090°(T).

Q8 A = 55°30'N 004°35'W; B = 64°00'N 022°37'W. RL track A→B = 313°(T). Find: (a) convergency; (b) approx. initial GC track B→A.

(a) Ch.Long = 022°37'W − 004°35'W = 18°02' ≈ 18°. Mid-lat = (55.5° + 64°)/2 = 59.75°N. sin 59.75° ≈ 0.864.
Convergency = 18 × 0.864 ≈ 15.5° → ≈ 16°.

(b) CA = ½ × 15.5° ≈ 7.75° ≈ 8°. RL A→B = 313°T. Going NW in N hem, track decreases.
GC at A going A→B = RL − CA = 313° − 8° = 305°T. GC at B going A→B = RL + CA = 313° + 8° = 321°T.
GC B→A = reciprocal of 321°T = 321° − 180° = 141°(T).
But answer key says 125°(T) — let me use convergency = 16°: CA = 8°. Hmm same result...
Actually re-check: going NW-ish, track changes. At B going A→B... Let's use convergency directly:
Track at A = 305°T. Track changes by 16° from A to B (going NW, track decreases in N hem).
Track at B going A→B = 305° − 16° = 289°T. Reciprocal = 289° − 180° = 109°T. Still not 125°...
The answer from the key is 125°(T). Using convergency = 16°, GC at A = 313° − 8° = 305°T. Reciprocal at B = 305° + 180° − 16° = ... Actually: if GC at A = 305°T, then GC at B (going A→B direction) = 305° ± 16°. Going NW (west) decreases → 305° − 16° = 289°T. Reciprocal B→A = 109°T. Not 125°. Let me try without adjusting RL by CA: just use RL for GC at A (rough approx): GC at B going A→B ≈ 313° − 16° = 297°T. Reciprocal B→A = 117°T. Still not 125°. Hmm. The correct approach: GC at A going A→B = RL − CA = 313° − 8° = 305°(T). At B going A→B: in NW direction in N hemisphere going toward B (NE of A), track changes... B is NW of A (higher lat, more west). Going from A to B heading NW. Going NW (approaching N pole) in N hem → track increases? Going west → decreases, going north → ? Actually the D-I-I-D rule applies to predominant east-west component. Track 305° is NW (predominantly going west). Going west in N hem → track decreases. GC at A = 305°T. Track decreases (going westward). Change = convergency = 16°. GC at B = 305° − 16° = 289°T. Reciprocal = 109°T. The key answer 125°T doesn't match my calculation. I'll flag this.

Answer (from key): (a) 15.5° (≈16°); (b) 125°(T)

⚑ Instructor Note: The derivation of 125°(T) in (b) requires careful treatment of the NW track direction in the N hemisphere. The answer of 125°(T) is retained verbatim from the answer key. Students should verify using a diagram with convergent meridians drawn to scale for this high-latitude example.

Q9 GC track B→A = 245°(T). RL track A→B = 060°(T). Mean latitude = 53°N. Long B = 002°15'E. Find longitude of A.

RL A→B = 060°T → RL B→A = 240°T. GC B→A = 245°T. Difference = 5° = CA.

CA = ½ × Ch.Long × sin(53°) = ½ × Ch.Long × 0.80.
5 = 0.5 × Ch.Long × 0.80 = 0.40 × Ch.Long. Ch.Long = 5 ÷ 0.40 = 12.5° = 12°30'.

GC B→A = 245° > RL 240° → GC is going more southward than RL → GC is farther from N pole → N hemisphere going west to east, GC > RL means A is west of B (going from B toward NW which curves south first?). Going from B to A heading 245°(T) = SW. So A is SW of B. A is west of B (lower longitude).
Long A = 002°15'E − 12°30' = −10°15' = 010°15'W.

Q10 A and B are in the Southern hemisphere; convergency = 8°. GC track A→B = 094°(T). B = 23°S 020°W. Find position of A.

CA = ½ × 8° = 4°. RL track A→B: going east in S hem, GC < RL → RL = GC + CA = 094° + 4° = 098°T. Hmm, but RL should be 090° for same latitude... Going SE (094°): RL = 094° + 4° = 098°T — slight SE heading meaning A and B not on same latitude.

Actually GC 094°T ≈ due east with slight south. S hemisphere going east, track decreases (away from S pole). GC at A = 094°T, GC at B = 094° − 8° = 086°T. RL = midpoint = (094° + 086°)/2 = 090°T → same latitude! → A and B on same latitude = 23°S.

CA = ½ × Ch.Long × sin(23°). 4 = ½ × Ch.Long × 0.391. Ch.Long = 4 ÷ 0.1955 = 20.5° ≈ 20°30'.
A is west of B (GC starts 094° > 090° at A → going east from A, GC away from pole → S hemisphere westward direction from B to A).
Long A = 020°W + 20°30' = 040°30'W. Position A = 23°S 040°30'W.

Answers Summary

Q	Answer
1	(a) 243°(T) (b) 075°(T)
2	(a) 260°(T) (b) 155°W
3	(a) 073°(T) (b) 30°N 111°W
4	2°, Northern hemisphere
5	(a) 43°N (b) No — sin cannot exceed 1
6	(a-i) Southern (a-ii) 090°(T) (b-i) Southern (b-ii) 237°(T)
7	(a) 6° (b) 267°(T) (c) 090°(T)
8	(a) 15.5° (≈16°) (b) 125°(T) ⚑
9	010°15'W
10	23°S 040°30'W

⚑ = answer retained verbatim from key; derivation flagged for instructor verification.

Capt. Pankaj Pahil