Examples on Tape Corrections - Errors in Chain Surveying, Surveying and Levelling | Surveying and Levelling Notes- Agricultural Engg - Agricultural Engineering PDF Download

Examples on Tape Corrections

Examples 1: - A line was measured with a steel tape which was exactly 
30m long at 18^oC and found to be 452.343 m.  The temperature during measurement 
was 32^oC.  Find the true length of the line.  Take coefficient of expansion of the tape  per  ^oC=0.0000117.

            Temperature correction per tape length = C_t

                                  = α  (T_m - T_o) l

Here     l = 30 m:  T_o =18^oC; T_m = 32^oC;

            α = 0.0000117

            ~          C_t = 0.0000117 (32-18) 30

                            = 0.004914 m (+ ve)

Hence the length of the tape at 32^oC = 30 + C_t

                                   = 30 + 0.004914 = 30.004914 m.

            Now true length of a line = L’ / L x its measured length.

L = 30 m: L’ = 30.004914 m; measured length = 452.343 m.

~ True length = 30.004914 / 30 x 452.343 = 452.417 m.

Example 2: - A line was measured with a steel rape which was exactly 30 m 
at 18^oC and a pull of 50 N and the measured length was 459.242 m.  Temperature during measurement was 28^oC and the pull applied was 100 N.  The tape was uniformly supported during the measurement.  Find the true length of the line if the cross-sectional area of the tape was 0.02 cm², the coefficient of expansion 
per ^oC = 0.0000117 and the modulus of elasticity = 21 x 10⁶ N per cm².

Temperature

            Correction per tape length                                = α ( (T_m – T_o)L

                                                                                    = 0.0000117 x (28 -18) 30

                                                                                    = 0.00351 m (+ ve)

            Sag correction per tape length                          = 0

            Pull correction per tape length                         = (P_m - P_o)L / AE

                                                                                    = (100 – 50)30 / 0.02 x 21 x 10⁶    

                                                                                    = 0.00357 m (+ve)

            ~ Combined correction                                    = 0.00351 + 0.00357 m.

                                                                                    = 0.00708 m

            True length of tape                                           = 30.00708 m

            True length of the line                                      = 30.00708 / 30 x 459.242

                                                                                    = 459.350 m.

Example 3: - A 50 m tape is suspended between the ends under a pull of 
150 N.  The mass of the tape is 1.52 kilograms.  Find the corrected length of the tape.

            Correction for sag                                            = C_s = l₁ (Mg)² / 24 P²

                 l₁ = 50 m; M = 1.52 kilograms; P = 150 N.

                        ~ C_s = 50 x (1.52 x 9.81)² / 24 x 150² = 0.0206 m.

            ~ Corrected length of the tape                          = l – C_s

                                                                                    = 50 – 0.0206

                                                                                    = 49.9794 m.

Example 4: - The downhill end of the 30 m tape is held 80 cm too low.  What is the horizontal length?

            Correction for slope = h² / 2l

            Here h = 0.8 m; l = 30 m

            ~ The required correction                                = 0.8² / 2 x 30 = 0.0167 m.

Hence the horizontal length                                         = 30 – 0.0167

                                                                                    = 29.9833 m

Example 5: - A 100 m tape is held 1.5 m out of line.  What is the true length?

            Correction for incorrect alignment                   = d² / 2l ( - ve)

            Here d = 1.5 m; l = 100 m.

            ~ Correction = 1.5² / 2 x 100 = 0.011 m.

            ~ True length = 100 – 0.011 = 99.989 m.