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Reciprocal leveling, RL of hill top relation

Short answer

Reciprocal trig levelling is used when stations are far apart or separated by a river or obstacle and curvature-refraction errors must be eliminated. If P is not in the vertical plane of X-Y, measure base XY and horizontal angles at X and Y; compute XP and YP by sine rule, then RL P = axis RL + horizontal distance times tangent of vertical angle.

In what situation reciprocal trigonometrically levelling is conducted? Derive a relation for determining the reduced level of a hill top when two instrument stations and target point are at different vertical plane.

Situations for reciprocal trigonometric levelling

Reciprocal trigonometric levelling is conducted when:

  • two stations are far apart or are separated by a wide obstacle such as a river, valley or lake, so that ordinary (spirit) levelling between them is not practical, and
  • the effects of curvature and refraction are large and must be eliminated.

Vertical angles are measured from both stations at the same time, using the same type of instrument and the same height of signal, so that the errors of curvature and refraction are equal and opposite and cancel in the mean:

h=Dtan⁡α+β2h = D\tan\frac{\alpha+\beta}{2}

where α\alpha is the angle of elevation from the lower station and β\beta is the angle of depression from the upper station.

Hill top when the stations and the target are in different vertical planes

If the instrument stations X and Y are not in the same vertical plane as the object P, a base line XY is measured and horizontal angles are observed at X and Y. The horizontal distances XP and YP are found by the sine rule in the plan triangle XYP, and the height of P is found from the vertical angles.

                 P (top)
                /  \
          XP   /    \  YP
              /      \
             /        \
            X -------- Y
         alpha   b    beta

Derivation

Let the base line be b=XYb = XY, the horizontal angles α=∠YXP\alpha = \angle YXP and β=∠XYP\beta = \angle XYP, so ∠XPY=180∘−(α+β)\angle XPY = 180^\circ - (\alpha + \beta).

XP=bsin⁡βsin⁡(α+β)YP=bsin⁡αsin⁡(α+β)\begin{aligned} XP &= \frac{b\sin\beta}{\sin(\alpha+\beta)} \\ YP &= \frac{b\sin\alpha}{\sin(\alpha+\beta)} \end{aligned}

Let θX\theta_X and θY\theta_Y be the vertical angles (elevation) of P from X and Y, and hXh_X, hYh_Y the heights of the instrument axes above their stations. If a BM of known RL is sighted from X and Y with back-sights BSXBS_X, BSYBS_Y:

RLaxis X=RLBM+BSX,RLaxis Y=RLBM+BSYRLP=RLaxis X+XPtan⁡θX(from X)RLP=RLaxis Y+YPtan⁡θY(from Y)\begin{aligned} RL_{\text{axis X}} &= RL_{BM} + BS_X, \qquad RL_{\text{axis Y}} = RL_{BM} + BS_Y \\ RL_P &= RL_{\text{axis X}} + XP\tan\theta_X \quad (\text{from X}) \\ RL_P &= RL_{\text{axis Y}} + YP\tan\theta_Y \quad (\text{from Y}) \end{aligned}

For zenith angle zz, θ=90∘−z\theta = 90^\circ - z. The mean of the two values is the RL of P. If the stations are at different levels, the difference of the axis RLs is already included through the BM sights. When the sights are long, the correction for curvature and refraction 0.0673 d20.0673\,d^2 (d in km) is added to each height.

✓ Answer checked by Claude Opus 5.5 (high effort)

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