This document shows the numerical process used to solve the inverse geodesic problem with Vincenty: iterations of \(\lambda\), intermediate values (\(\sin\sigma,\cos\sigma,\sigma,\sin\alpha,\cos^2\alpha,\cos2\sigma_m\)), the corrections (\(u^2,A,B,\Delta\sigma\)), and the bearing calculations.
Origin: Les Sables-d'Olonne — φ₁ = 46.494953°, λ₁ = −1.792091°
Destination: Saint-François — φ₂ = 16.252360°, λ₂ = −61.273320°
We convert to radians (for trigonometric calculations):
$$\varphi_1 = 0.8114900154100151\ \text{rad}, \quad \varphi_2 = 0.2836571932194256\ \text{rad}$$ $$L = \Delta\lambda = \lambda_2-\lambda_1 = -1.0381432891827342\ \text{rad}$$Short explanation: U is the latitude “projected” onto an auxiliary sphere — it simplifies trigonometry on the ellipsoid.
Start: \(\lambda_0 = L\). At each iteration we compute a series of intermediate values and update \(\lambda\) using Vincenty’s formula. We repeat until the change in \(\lambda\) is smaller than \(10^{-12}\) rad.
Equations used at each iteration (for reference):
\[\sin\sigma = \sqrt{(\cos U_2 \sin \lambda)^2 + (\cos U_1 \sin U_2 - \sin U_1 \cos U_2 \cos \lambda)^2}\] \[\cos\sigma = \sin U_1 \sin U_2 + \cos U_1 \cos U_2 \cos \lambda\] \[\sigma = \operatorname{atan2}(\sin\sigma,\cos\sigma)\] \[\sin\alpha = \frac{\cos U_1 \cos U_2 \sin \lambda}{\sin\sigma},\quad \cos^2\alpha = 1-\sin^2\alpha\] \[\cos2\sigma_m = \cos\sigma - \frac{2\sin U_1 \sin U_2}{\cos^2\alpha}\] \[C = \frac{f}{16}\cos^2\alpha\left(4 + f(4-3\cos^2\alpha)\right)\] \[\lambda_{n+1} = L + (1-C)f\sin\alpha\left(\sigma + C\sin\sigma\left(\cos2\sigma_m + C\cos\sigma(-1+2\cos^2 2\sigma_m)\right)\right)\]In the table you can see, for each iteration: \(\lambda\), the change \(\Delta\lambda\), \(\sin\sigma\), \(\cos\sigma\), \(\sigma\), \(\sin\alpha\), \(\cos^2\alpha\), and \(\cos2\sigma_m\).
| Iter | λ (rad) | Δλ | sinσ | cosσ | σ (rad) | sinα | cos²α | cos2σₘ |
|---|---|---|---|---|---|---|---|---|
| 0 (start) | -1.0381432891827342 | - | — | — | — | — | — | — |
| 1 | -1.0404171135171536 | -0.00227382433441936 | 0.83951 | 0.54331 | 0.99322 | -0.67245 | 0.54739 | -0.21772 |
| 2 | -1.0404214142043005 | -0.00000430068714685 | 0.84310 | 0.53790 | 1.00175 | -0.67690 | 0.54204 | -0.21005 |
| 3 | -1.0404214223337993 | -8.1295e-09 | 0.84355 | 0.53705 | 1.00387 | -0.67722 | 0.54138 | -0.20935 |
| 4 | -1.0404214223491663 | -1.54e-11 | 0.84355 | 0.53705 | 1.00387 | -0.67722 | 0.54138 | -0.20935 |
| 5 (final) | -1.0404214223491954 | -2.91e-14 | 0.8435532581 | 0.5370455295 | 1.003865549518566 | -0.67721538895 | 0.54137931697 | -0.20935377160 |
Note how λ converges quickly (5 iterations here). The values of \(\sin\sigma\), \(\cos\sigma\), and \(\sigma\) also stabilize.
Using the final values:
$$u^2 = \cos^2\alpha \cdot \frac{a^2-b^2}{b^2} = 0.0036486241430452784$$ $$A = 1 + \frac{u^2}{16384}\left(4096 + u^2(-768 + u^2(320 - 175u^2))\right) = 1.000911532961068$$ $$B = \frac{u^2}{1024}(256 + u^2(-128 + u^2(74 - 47u^2))) = 0.0009104954804571988$$ $$\Delta\sigma = B\sin\sigma\Big[\cos2\sigma_m + \frac{B}{4}\big(\cos\sigma(-1+2\cos^2 2\sigma_m) - \frac{B}{6}\cos2\sigma_m(-3+4\sin^2\sigma)(-3+4\cos^2 2\sigma_m)\big)\Big]$$ $$\Delta\sigma = -0.00016088012080655317\ \text{rad}$$ $$s = b\,A\,(\sigma - \Delta\sigma) = 6\,388\,165.050133844\ \text{m}$$ $$s = 6\,388.165050133844\ \text{km}$$ $$s_{NM} = \frac{s}{1852} = 3\,449.3331804178424\ \text{NM}$$Vincenty also provides the initial bearing (\(\alpha_1\)) and the final bearing (\(\alpha_2\)). We compute them from the reduced latitudes and the final \(\lambda\).
Formulas (angles in radians):
$$\alpha_1 = \operatorname{atan2}\big(\cos U_2 \sin \lambda, \; \cos U_1 \sin U_2 - \sin U_1 \cos U_2 \cos \lambda\big)$$ $$\alpha_2 = \operatorname{atan2}\big(\cos U_1 \sin \lambda, \; -\sin U_1 \cos U_2 + \cos U_1 \sin U_2 \cos \lambda\big)$$Results (converted to degrees and normalized to 0°–360°):
$$\alpha_1 = 259.11026968403183^\circ \quad (\text{initial bearing})$$ $$\alpha_2 = 224.84728561996576^\circ \quad (\text{final bearing})$$Interpretation: 259.11° means that, from Les Sables-d'Olonne, one must sail approximately west-southwest to follow the geodesic toward Saint-François. The final bearing of 224.85° indicates the direction of the trajectory at the destination point (looking toward true north from the arrival point).
| Quantity | Value |
|---|---|
| Final λ | -1.0404214223491954 rad |
| σ | 1.003865549518566 rad |
| Δσ | -0.00016088012080655317 rad |
| A | 1.000911532961068 |
| B | 0.0009104954804571988 |
| u² | 0.0036486241430452784 |
| Distance | 6,388,165.05 m ≈ 6,388.17 km |
| Distance | 3,449.33 NM |
| Initial bearing α₁ | 259.11026968403183° |
| Final bearing α₂ | 224.84728561996576° |