Chain Survey

Chain and tape surveying measures horizontal distances and simple areas. Learn equipment, ranging, offsets, tape corrections (temperature, tension, sag, slope, standardisation), errors, and area computation by the trapezoidal and Simpson's rules.

The simplest survey uses only linear measurements: a network of triangles measured with a chain or tape, plus perpendicular offsets to features. It's still used for small, open sites and teaches the fundamentals of error and correction.

Precise taping requires corrections because a steel tape changes length with temperature, pull and sag, and slope distances must be reduced to horizontal.

An analogy

Like measuring a room with a tape measure — but outdoors, over hundreds of metres, where heat, slope and a sagging tape all creep into the result.

Equipment and procedure

Chains (20/30 m) or steel/fibreglass tapes, ranging rods, arrows, pegs, plumb bobs, cross-staffs or optical squares for offsets. Establish main survey lines forming well-conditioned triangles (angles 30°–120°), range them straight, measure lengths and take offsets to details.

Errors and corrections

Systematic errors (wrong tape length, temperature, slope, sag) accumulate and are corrected; random errors partly cancel. Standardise the tape against a known length; apply corrections with the right signs — sag and slope always shorten the true horizontal distance.

Areas

Offsets at regular intervals from a baseline give irregular boundary areas via the trapezoidal rule or Simpson's rule (more accurate for curved boundaries). Areas of triangles from three sides use Heron's formula.

Precision and limits

Ordinary chaining gives about 1:500–1:1000 accuracy; precise taping with corrections reaches 1:10 000+. For larger or accurate work, EDM/total stations and GNSS have replaced chaining, but correction principles carry over (atmospheric corrections, slope reduction).

Formulas

Worked example: Correcting a tape length

Given: 30 m steel tape, measured 30.000 m; T = 35 °C (std 20 °C), W = 7.5 N, P = 100 N; Height difference 2 m

Find: Corrections and true horizontal length.

Answer: ≈ 29.932 m

Worked example: Area by offsets

Given: Offsets every 10 m: 0, 4, 6, 5, 3 m

Find: Area by trapezoidal and Simpson's rules.

Answer: 165 m² / 170 m²

Common mistakes

Applications

Key takeaways

Going further