Reciprocal trigonometrical levelling formula

Reciprocal trigonometric levelling

When two stations A and B are far apart, or are on opposite sides of a wide obstacle (river, valley), the effects of curvature and refraction cannot be neglected. The vertical angles are then measured from both stations, at nearly the same time, so that these errors cancel. From A the angle of elevation $\alpha$ to B is measured, and from B the angle of depression $\beta$ to A is measured, with the signal at each station at the same height as the instrument axis (otherwise the angles are corrected for the difference).

                                     B ---- horizontal at B
                                _.-' |      (beta = depression to A)
                           _.-'      |
                      _.-'           |  h
                 _.-'                |
     A  _.-'  alpha                  |
     --------------------------------C   (level line through A)
     |<-------------- D ------------>|

Expression for elevation difference

Let $D$ = horizontal distance AB, $\phi = D/R$ the angle between the horizontals at A and B (R = radius of the earth), and $\varepsilon$ the angular effect of refraction, which is assumed equal at both ends because the sights are taken at the same time. If $\theta$ is the true angle of elevation of the chord AB at A, the true angle of depression of the chord at B is $\theta + \phi$, because the horizontal at B is turned through $\phi$ relative to that at A. Refraction raises the apparent position of the target at both ends, so the observed angles are

$$\alpha = \theta + \varepsilon, \qquad \beta = (\theta + \phi) - \varepsilon$$

Adding, refraction cancels:

$$\alpha + \beta = 2\theta + \phi \quad\Rightarrow\quad \theta + \frac{\phi}{2} = \frac{\alpha+\beta}{2}$$

The difference in elevation $h$ is measured along the vertical at B, from the level line through A (point C). The chord AC lies $\phi/2$ below the horizontal at A, so $\angle BAC = \theta + \phi/2$ and

$$\boxed{h \approx D\tan\left(\theta + \frac{\phi}{2}\right) = D\tan\frac{\alpha + \beta}{2}}$$

Both refraction (through $\varepsilon$) and curvature (through $\phi$) drop out, so no separate correction for curvature and refraction is needed.

If the signal heights are not equal to the instrument heights, the observed angles are first corrected for the difference (or the difference in heights is allowed for in $h$). Then $RL_B = RL_A + h$.

The principal advantage of the method is that the uncertain effects of curvature and refraction are eliminated by taking the mean of the two reciprocal observations.