Section 4: Soil and Water Conservation Engineering
Surveying and Levelling
Subtopic: Measurement of distance and area
Question Statement:
When a surveying tape of total weight $W$ and length $L$ is supported only at its two ends under a tension pull $P$, the sag correction $C_{\text{sag}}$ is given by:
Options:
(A) $C_{\text{sag}} = -\frac{W^2 L}{24 P^2}$
(B) $C_{\text{sag}} = +\frac{W^2 L}{24 P^2}$
(C) $C_{\text{sag}} = -\frac{W L^2}{12 P}$
(D) $C_{\text{sag}} = +\frac{W L}{12 P^2}$
Answer Key:
A
Step-by-Step Solution & Explanation
Due to tape weight, the tape sags into a catenary curve, making the horizontal straight chord distance shorter than the curved arc length read on the tape. Hence, the correction is always negative (subtractive):
$C_{\text{sag}} = -\frac{W^2 L}{24 P^2} = -\frac{w^2 L^3}{24 P^2}$
where $w = W/L$ is tape weight per unit length. Option A is correct.