Question

FIELD WORK 4: TRAVERSE AND TACHEOMETRIC SURVEY a. Write the steps to be followed for traverse and tacheometric survey. b. The table below shows the field data of the tacheometric survey. Inst. Inst. Station height (hi) Staff Station Horizontal Angle Staff Readings Lower Mid Upper Vertical Angle WCB A 1.50 B 1.800 2.100 2.400 -06° 30'40" 160° B 1.40 с 2.200 3.200 +12° 45'35" | ABC 200° (clockwise) с 1.45 D 2.800 3.200 3.600 +05° 20'10"| BCD 140° (unticlockwise) If coordinate of A ( 100.00 E, 50.00 N) and RL of A = 50 m. Find ; i. Percent slope in between points A to D. ii. WCB and RB of line AD.

Answer 1

1. Steps to be followed in traverse and tacheometry survey are:

a) Set the compass on first point of known elevation and measure bearing of line joining first and second point.

b) Set tacheometer on first point of known elevation and sight the staff at second point and note down upper middle and lower stadia reading.

c) Set instrument on second point orient towards first point and make horizontal circle zero and rotate telescope to third point to measure angle between first and second line.

d) Measure the upper, middle and lower stadia readings on staff.

e) Repeat steps b,c and d for next all the points.

2. Complete solution is given below.

18 To find ekvations of all the points ve angle 6° 30' 40" for Inst at A and sighting B. ISA A From diagram Kstc 100 (2.4-18) +0 . DAB Gom Dxtamo Go x tan 6 30 461 = 6.85m •RL of B = RL of At 1.5-y-2.1 = 50 +1.5 - 6.85-2.1 RL of B 42.55 m

* consider B toc Id2.7 ү ) 12°45'35" 1.4 DBC B (3.272.2)- 2,7 = * Mid hair reading * Dec 10 ( 3.2 – 2.2) TOBC 100m) v= Dx tano = 100 x tan 12°45'254 <V = 22.64m) * From diagram ARL of c RL of B F104 + r -diy 42.55 +1.4+ 22.64 - 2.7 [RL of c = 63.89 m

* consider c to D The V 50 20' 10" 1.451 -Dco с * Depa 100 ( 3.6 -2.8) KDco = som> tano = 808 ton 5 20'10" TY 7.44m From Diagram RL of sa RL of c + 1.45 tv - 3.2 11 63.89 +1.45+ 7.47 - 3. 2 JRL of D = 69.61 m

27 Consider the plan of given traverse N Co-ordinates 160° EA 100.00 01 NA = 50.00 Gom 2000 100 m 14022 03 w og P

& From above diagram -49, = 160°-90" - 70" 7° ) . 92 70° + goo° -18° 2 9 + dom° — 180° Көх = 90° е, = 4 140° — 0, 14 o° — 0° Коз so°) w Х and y distance of each wine Xде = & Co, et = 20.52 Уло 60 sino, S6, 38 Xдс 10 Сед 02 (І YBC= 100 sünoz = loo Хар : 8ъ Сол өз - 5. 42 Yco 80 Sun 93 = 6, 28

* co-ordinates of all the points B; /EB = EAF XAB = 120.52 100 + 20.52 WB 56.38 = -6.387 NA YAB c! lec 120.52 to 120.52 Wc -6.38 -100 = -106.38 69.1 V Ep = 120.52 51.42 61. 28 -167.66 No = -106.38 ci) percent slope / A to D DAD = (EA-ED)2 + (NA-ND)? (100-69.1)? +(50 +167.66)2 219.84 m IDAD

he distance to A and D = Rh of D- RL of A VAD 69.61-50 = 19.61 Slope = #ls — 19.6/ 219.84 = 0.0892 % slope = 8.92% i KCB and RB of line AD tant OAD NAND AD EA-EA tant ( 50 + 167.66 100 - 69.1 OAD = 81.92°

Actual plan from obtained co-ordinates OAD B to i. From above diagram WCB of AD = 270° - OAD 270° 81.92 188.080 RB of AD = s 8:08 W