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Subtense bar principle, increasing precision

Short answer

A horizontal subtense bar of length l subtends a small angle β, giving D=(l/2) cot(β/2). Precision improves by using a 1″ theodolite with repeated measurements, a longer invar bar, serial measurement of short sections, an auxiliary base, and keeping lines under 150 m with exact perpendicularity.

Write working principle of a subtense bar. How precision can be increased by using subtense bar for computed distance.

Working principle

A horizontal subtense bar of fixed length ll (usually 2 m) is placed at the far end of the line, perpendicular to it, and the small horizontal angle β\beta it subtends at the theodolite is measured. The horizontal distance is D=l2cot⁡β2≈l/βD=\frac{l}{2}\cot\frac{\beta}{2}\approx l/\beta.

Increasing the precision

The error in distance for a small error dβd\beta in the angle is

dD=−D2l dβdD=-\frac{D^2}{l}\,d\beta

so the error increases with the square of the distance, and the relative error dD/D=D dβ/ldD/D=D\,d\beta/l grows as DD increases. The precision is improved by:

  1. Measuring β\beta more accurately: use a 1" theodolite, repeat the angle several times (the error reduces by n\sqrt{n} for nn repetitions) and observe on both faces.
  2. Using a longer bar (for example 4 m, or two bars end to end), which gives a larger angle.
  3. Serial method: divide a long line into short sections and measure each section separately with the bar, so each angle is large.
  4. Auxiliary base method: at one end of a long line, set out a short auxiliary base perpendicular to it and measure that base with the subtense bar; then measure the angle the auxiliary base subtends at the other end and compute the long line from it.
  5. Keeping lines short (below about 150 m), with the bar exactly horizontal and perpendicular to the line and the targets steady.
  6. Using an invar bar of accurately known length, so that its length hardly changes with temperature.

Example: for D=100D=100 m, l=2l=2 m, dβ=1′′=4.85×10−6d\beta=1''=4.85\times10^{-6} rad, dD=1002×4.85×10−6/2=0.024dD=100^2\times4.85\times10^{-6}/2=0.024 m (about 1 in 4000).

✓ Answer checked by Claude Opus 5.5 (high effort)

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