Trigonometric levelling

In this method the difference in elevation of the points is determined from the observed vertical angles and measured distances.

1. introduction Trigonometric leveling

2. Introduction This is an indirect method of levelling . In this method the difference in elevation of the points is determined from the observed vertical angles and measured distances. The vertical angles are measured with a transit theodolite and The distances are measured directly (plane surveying) or computed trigonometrically (geodetic survey). Trigonometric levelling is commonly used in topographical work to find out the elevation of the top of buildings, chimneys, church spires, and so on. Also, it can be used to its advantage in difficult terrains such as mountaineous areas. Depending upon the field conditions and the measurements that can be made with the instruments available, there can be innumerable cases.

3. Base accessible Determination of elevation of object when the base is accessible – the object is Vertical It is assumed that the horizontal distance between the instrument and the object can be measured accurately. In Fig. 1, let B = instrument station F = point to be observed C = centre of the instrument AF = vertical object D = CE = horizontal distance between the object and instrument h 1 = height of the instrument at B h = height FE S = reading on the levelling staff held vertical on the Bench Mark (B.M)  = angle of elevation of the top of the object

4. From triangle CFE, tan  = FE / CE FE = CE tan  h = D tan  If the reading on the staff kept at the B.M. is ‘S’ with the line of sight horizontal, R.L. of F = R.L. of B.M. + S + h = R.L. of B.M. + S + D tan  The method is usually employed when the distance between the instrument and the object is small. However, if the distance is large, the combined corrections for curvature and refraction should also be applied. The combined correction for curvature and refraction is given by C = 0.06728 D 2 metres , when D is in kilometers. R.L. of F = R.L. of B.M. + S + D tan  + C

5. Determination of elevation of object when the base is inaccessible – the Instrument Stations and the Elevated Object are in the Same Vertical Plane If the horizontal distance between the instrument and the object cannot be measured due to obstacles etc., two instrument stations are used so that they are in the same vertical plane as the elevated object. Fig. 2

6. Procedure Set up the theodolite at O 1 and level it accurately with respect to the altitude bubble. Direct the telescope towards O 2 and bisect it accurately. Clamp both the plates. Read the vertical angle  1 . Transit the telescope so that the line of sight is reversed. Mark the second instrument station O 2 on the ground. Measure the distance O 1 O 2 accurately. Repeat steps (2) and (3) for both face observations. The mean values should be adopted. With the vertical vernier set to zero reading, and the altitude bubble in the centre of its run, take the reading on the staff kept at nearby B.M. Shift the instrument to O 2 and set up the theodolite there. Measure the vertical angle  2 to F with both face observations. With the vertical vernier set to zero reading, and the altitude bubble in the centre of its run, take the reading on the staff kept at the nearby B.M.

7. Assuming the instrument stations and the object to be in the same vertical plane, the following two cases arise. * Instrument axes at same level * Instrument axes at different level

8. Instrument axes at same level In Fig. 2, Let h = FA’  1 = angle of elevation from O 1 to F  2 = angle of elevation from O 2 to F S = staff reading on B.M., taken from both O 1 ’ and O 2 ’, the reading being the same in both the cases. d = horizontal distance between the two instrument stations. D = horizontal distance between O 1 and F From triangle O 1 ’A’F, h = D tan  1 -------------------------- ( i ) From triangle O 2 ’A’F, h = (D + d) tan  2 ------------------- (ii) From Eqs . ( i ) and (ii) D tan  1 = (D + d) tan  2 or D (tan  1 - tan  2 ) = d tan  2

9. or D = Hence, h = D tan  1 = R.L. of F = R.L. of B.M. + S + h

10. Instrument axes at different level Depending upon the terrain, three cases arise: A. Instrument axis at O 2 higher that that at O 1 (Fig. 3)

11. h 1 - h 2 = A’A” = S 2 – S 1 = S From triangle O 1 ’A”F, h 1 = D tan  1 ------------ ( i ) From triangle O 2 ’A”F, h 2 = (D + d) tan  2 ---- (ii) Subtract Eq. (ii) from Eq. ( i ) to get h 1 - h 2 = D tan  1 - (D + d) tan  2 S = D tan  1 – D tan  2 – d tan  2 = D (tan  1 –tan  2 ) - d tan  2 or D (tan  1 - tan  2 ) = S + d tan  2

12. or D = But, h 1 = D tan  1 or h 1 = and R.L. of F = R.L. of B.M. + S 1 + h 1

13. B. Instrument axis at O 1 higher than that at O 2 (Fig. 4)

14. h 2 – h 1 = S 1 – S 2 = S From triangle O 1 ’A”F, h 1 = D tan  1 ---------------- ( i ) From triangle O 2 ’A”F, h 2 = (D + d) tan  2 ------- (ii) Subtract Eq. ( i ) from Eq. (ii) to get h 2 – h 1 = (D + d) tan  2 - D tan  1 S = D tan  2 – D tan  1 + d tan  2 = D (tan  2 –tan  1 ) + d tan  2 or D =

15. But h 1 = D tan  1 or h 1 = Hence, R.L. of F = R.L. of B.M. + S 1 + h 1

16. C. Instrument axes at very different levels (Fig. 5 and 6) If the difference in elevation (S 2 – S 1 ) between the two instrument stations is too large and cannot be measured on a staff at the B.M., then the following procedure is adopted:

17. Set up the instrument at O 1 and measure the vertical angle at the point F (Fig. 5) Transit the telescope and establish point O 2 , at a distance d from O 1 . Shift the instrument to O 2 and measure the vertical angle at the point F. Observe the staff reading r on the staff at O 1 (Fig. 6). Let S be the difference in level between the two axes at O 1 and O 2 .  S = h 2 – h 1 We know that

18. D = and h 1 = Height of station O 1 above the axis at O 2 = h – r = d tan  - r S = d tan  - r + h’ Hence, R.L. of F = R.L. of B.M. + S 1 + S + h 1 = R.L. of B.M. + S 1 + (d tan  - r + h’) + h 1

19. Determination of elevation of object when the base is inaccessible – the Instrument Stations and the Elevated Object are not in the Same Vertical Plane Let P and R be the two instrument stations not in the same vertical plane as that of Q. The procedure is as follows: Set the instrument at P and level it accurately with respect to the altitude bubble. Measure the angle of elevation  1 to Q. Sight to the point R with reading on horizontal circle as zero and measure the angle RPQ 1 , i.e , the horizontal angle  at P. Take a backsight s on the staff held at B.M. Shift the instrument to R and measure  2 and  there.

20. In Fig. 7, AQ’ = horizontal line through A Q’ = vertical projection of Q Thus, AQQ’ is a vertical plane Similarly, BQQ” is a vertical plane Q” = vertical projection of Q on a horizontal line through B PRQ 1 = horizontal plane Q 1 = vertical projection of Q R = vertical projection of B on a horizontal plane passing through P  and  = horizontal angles  1 and  2 = vertical angles measured at A and B respectively.

21. From triangle AQQ’ , QQ’ = h 1 = D tan  1 ---------- (1) From triangle PRQ 1 ,  PQ 1 R = 180  - (  +  ) =  - (  +  ) From the sine rule, = = =  PQ 1 = D 1 = ------------------ (2) and RQ 1 = D 2 = --------------------- (3)

22. Substituting the value of D in (1), we get h 1 = D 1 tan  1 =  R.L. of Q = R.L. of B.M. + s + h 1 As a check, h 2 = D 2 tan  2 = If a reading on B.M. is taken from B, the R.L. of Q can be known by adding h 2 to R.L. of B.