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Importance of trig levelling, inaccessible RL

Short answer

Trigonometric levelling is vital in rough terrain and for inaccessible points. When stations are in different vertical planes, horizontal distances are found by sine rule from base and horizontal angles; then RL of object = axis RL + distance × tan(vertical angle).

Discuss importance of trigonometrical levelling. Derive an expression to find the R.L. of an inaccessible object when the instrument stations are in the different vertical plane.

Importance of trigonometric levelling

Trigonometric levelling finds the difference in elevation between two points from the measured vertical angle and the horizontal (or slope) distance. It is important because:

  • it is the only practical method in rough, hilly or mountainous country (as in Nepal) where spirit levelling is slow and costly;
  • it gives the elevation of inaccessible points such as the top of a tower, chimney, hill, mountain peak, transmission tower, or the other bank of a river;
  • it is quick, needs few staff, and gives the elevations of triangulation stations and control points over a large area;
  • it is used to fix the heights of the stations of a traverse and to check the levels run by other methods;
  • the accuracy is sufficient for contouring, topographic survey and the height of structures.

Spirit levelling is more accurate and is preferred when high accuracy is required on gentle ground.

RL of an inaccessible object when the stations 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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