INTRODUCTION

Trigonometric leveling

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.

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 h1 = 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

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 D2 metres, when D is in kilometers. R.L. of F = R.L. of B.M. + S + D tan + C

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

Procedure Set up the theodolite at O1 and level it accurately with

respect to the altitude bubble. Direct the telescope towards O2 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 O2 on the ground. Measure the distance O1O2 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 O2 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.

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

Instrument axes at same level In Fig. 2, Let h = FA’ 1 = angle of elevation from O1 to F 2 = angle of elevation from O2 to F S = staff reading on B.M., taken from both O1’ and O2’, the

reading being the same in both the cases. d = horizontal distance between the two instrument stations. D = horizontal distance between O1 and F From triangle O1’A’F, h = D tan 1 -------------------------- (i) From triangle O2’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

or D =

Hence, h = D tan 1 =

R.L. of F = R.L. of B.M. + S + h

Instrument axes at different levelDepending upon the terrain, three cases

arise:A. Instrument axis at O2 higher that that

at O1 (Fig. 3)

h1 - h2 = A’A” = S2 – S1 = SFrom triangle O1’A”F, h1 = D tan1 ------------ (i)From triangle O2’A”F, h2 = (D + d) tan2 ---- (ii)Subtract Eq. (ii) from Eq. (i) to geth1 - h2 = D tan1 - (D + d) tan2 S = D tan1 – D tan2 – d tan2

= D (tan1 –tan2) - d tan2

or D (tan1 - tan2) = S + d tan2

or D =

But, h1 = D tan1

or h1 =

and R.L. of F = R.L. of B.M. + S1 + h1

B. Instrument axis at O1 higher than that at O2 (Fig. 4)

h2 – h1 = S1 – S2 = SFrom triangle O1’A”F, h1 = D tan1 ----------------

(i)From triangle O2’A”F, h2 = (D + d) tan2 -------

(ii)Subtract Eq. (i) from Eq. (ii) to geth2 – h1 = (D + d) tan2 - D tan1 S = D tan2 – D tan1 + d tan2

= D (tan2 –tan1) + d tan2

or D =

But h1 = D tan1

or h1 =

Hence, R.L. of F = R.L. of B.M. + S1 + h1

C. Instrument axes at very different levels (Fig. 5 and 6)

If the difference in elevation (S2 – S1) 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:

Set up the instrument at O1 and measure the vertical angle at the point F (Fig. 5)

Transit the telescope and establish point O2, at a distance d from O1.

Shift the instrument to O2 and measure the vertical angle at the point F.

Observe the staff reading r on the staff at O1 (Fig. 6). Let S be the difference in level between the two axes at O1 and O2.

S = h2 – h1

We know that

D =

and h1 =Height of station O1 above the axis at O2 = h –

r= d tan

- r S = d tan - r + h’ Hence, R.L. of F = R.L. of B.M. + S1 + S + h1

= R.L. of B.M. + S1 + (d tan - r + h’) + h1

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 RPQ1 , 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.

In Fig. 7,AQ’ = horizontal line through AQ’ = vertical projection of QThus, AQQ’ is a vertical planeSimilarly, BQQ” is a vertical planeQ” = vertical projection of Q on a horizontal line through BPRQ1 = horizontal planeQ1 = vertical projection of QR = vertical projection of B on a horizontal plane passing

through P and = horizontal angles1 and 2 = vertical angles measured at A and B respectively.

From triangle AQQ’ , QQ’ = h1 = D tan 1 ---------- (1)

From triangle PRQ1 , PQ1R = 180 - ( + ) = - ( + )From the sine rule, = =

=

PQ1 = D1 = ------------------ (2)

and RQ1 = D2 = --------------------- (3)

Substituting the value of D in (1), we get

h1 = D1 tan 1 =

R.L. of Q = R.L. of B.M. + s + h1

As a check, h2 = D2 tan 2 =

If a reading on B.M. is taken from B, the R.L. of Q can be known by adding h2 to R.L. of B.