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Subtense bar principle and formula

Short answer

A subtense bar is a horizontal 2 m bar with targets at ends. The horizontal angle β is measured; horizontal distance D = (l/2) cot(β/2). For an upward inclined sight, D is unchanged and vertical distance V = D tan α, where α is the vertical angle.

Describe the working principle of subtense bar and derive subtense bar formula for computing horizontal and vertical distances when line of sight is inclined upward.

A subtense bar is a horizontal bar of fixed length ll (usually 2 m) with targets at its ends. It is mounted on a tripod over the station, levelled, and set perpendicular to the line joining it to the instrument by sighting through a small alignment telescope on the bar. The small horizontal angle β\beta subtended by the two targets at the theodolite is measured (to seconds, by repetition), and the distance is computed from it. It is used to measure distances over difficult ground (hilly terrain, water, thick vegetation) without taping.

 Theodolite                       Subtense bar
     O  <------------ D ---------->  A |
       \  beta                       | l (2 m)
         \_____________________________ B

Working principle (plan view)

From the isosceles triangle OABOAB, with the bar normal to the line OM (M the middle of the bar):

l2=Dtan⁡β2 ⇒ D=l2cot⁡β2 ≈ lβ (β in radians)\frac{l}{2}=D\tan\frac{\beta}{2}\ \Rightarrow\ D=\frac{l}{2}\cot\frac{\beta}{2}\ \approx\ \frac{l}{\beta}\ (\beta\ \text{in radians})

Line of sight inclined upward by angle α\alpha (derivation)

The bar is always kept horizontal, and the theodolite measures the angle β\beta on its horizontal circle, i.e. the angle projected onto the horizontal plane. Hence the triangle OAB above is a horizontal triangle whatever the inclination of the sight, and the distance found from it is directly the horizontal distance:

D=l2cot⁡β2(independent of α)D=\frac{l}{2}\cot\frac{\beta}{2}\qquad(\text{independent of }\alpha)

The vertical angle α\alpha is then measured to the centre of the bar (the sighting target). From the vertical triangle formed by the instrument axis, the horizontal line and the inclined line of sight:

V=Dtan⁡α=l2cot⁡β2tan⁡αS=Dsec⁡α(inclined distance, if required)\begin{aligned} V&=D\tan\alpha=\frac{l}{2}\cot\frac{\beta}{2}\tan\alpha\\ S&=D\sec\alpha\quad(\text{inclined distance, if required}) \end{aligned}

If the bar is not exactly normal to the line of sight in plan but is out by an angle δ\delta, the effective length is lcos⁡δl\cos\delta and the distance becomes D=lcos⁡δ2cot⁡β2D=\frac{l\cos\delta}{2}\cot\frac{\beta}{2}.

The RL of the bar station is RL of instrument axis +V−h+V-h, where hh is the height of the bar's sighting target above the station.

✓ Answer checked by Claude Opus 5.5 (high effort)

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