Correction Due to Sag | Taping Corrections | Surveying
The correction for sag is equal to the difference in length between the arc and its subtended chord and is always negative. As the sag correction is a function of the weight of the tape, it will be greater for heavy tapes than light ones.
A tape not supported along its length will sag and form a catenary between end supports. The correction due to sag must be calculated separately for each unsupported stretch separately and is given by:
𝐶_𝑠=(𝜔^2 𝐿^3)/(24𝑃^2 )
𝐶_𝑠=(𝜔^2 𝐿^3)/(24𝑃^2 ) →𝐶_𝑠=(𝑊^2 𝐿)/(24𝑃^2 )
where:
𝐶_𝑆=correction due to sag
𝜔=weight per unit length
=𝑊/𝐿
𝑊=weight of unsupported length of tape
𝑃=pull during measurement or laying out
𝐿=distance between supports
Ex1. A 30-m tape is supported only at its end and under a steady pull of 8 kg. If the tape weighs 0.91 kg, determine the sag correction and the correct distance between the ends of the tape.
Ex 2. A 50-m steel tape weighs 0.04 kg/m and is supported at its end points and at the 8-m and 25-m marks. If a pull of 6 kg is applied, determine the following:
Correction due to sag between the 0m and 8-m marks, 8-m and 25-m marks and 25-m and 50m marks.
Correction due to sag for one tape length.
Correction distance between the ends of the tape.