trigonometri levellingII

TRIGONOMETRICAL LEVELING

Objectives:
1. To determine the height of the building i.e. base accessible case and
2. To determine the height of the chimney i.e. base inaccessible case.

Instruments required:
1. Transit Theodolite
2. Ranging rods
3. Pegs and hammer
4. Tape
5. Plumb bob

Theory
Trigonometrical leveling:
Trigonometrical leveling is the process of determining the differences of elevations of stations from observed vertical angles and known distances, which are assumed to be either horizontal or geodetic lengths at mean sea level. The vertical angle may be measured by means of an accurate theodolite and the horizontal distances may either be measured (in the case of plane surveying) or computed (in case of geodetic observations).
In order to get the difference in elevation between the instrument station and the object under observation, we may consider different cases.
1. Base of the object accessible
Let us assume that the horizontal distance between the instrument and the object can be measured accurately. In figure,

Fig 3.1. Base accessible
Let, D = horizontal distance between P (instrument station) and Q (point to be observed i.e. tower)
V1 = QQ'
V2 = Q'Q1
H = height of the tower
s = reading of staff kept at B.M. with line of sight horizontal
α1 = angle of elevation from A to Q
α2 = angle of depression from A to Q1
From Δ AQQ',
V1 = D tanα1
V2 = D tanα2
 R.L. of Q = R.L. of instrument axis + V1
= R.L. of B.M. + s + V1
R.L. of Q = R.L. of B.M. + s + D tanα1
Also, H = V1 + V2
= D tanα1 + D tanα2
H = D (tanα1 + tanα2)

2. Base of the object inaccessible: instrument stations in the same vertical plane with the instrument axes at different level

Fig 3.2 Instrument axes at different level
In Δ OQP', V1 = D tanα1 .................(i)
In Δ ORP", V2 = (H + D) tanα2 .................(ii)
Where α1 = the angle of elevations from instrument stations A
α2 = the angle of elevations from instrument stations B and A is at higher level than B
D = distance between target tower OP and nearer instrument station A
H = distance between instrument stations A and B (measured directly)
Let, s1 = reading of staff kept at B.M. from instrument station A with line of sight horizontal
s2 = reading of staff kept at B.M. from instrument station B with line of sight horizontal
and s = s2 + s1
Then, from figure,
s = V2 − V1
= (H + D) tanα2 − D tanα1
= H tanα2 + D (tanα2 − tanα1)
 D = s − H tanα2tanα2 − tanα1 ....................................(iii)
From equation (i), we get,
V1 = s − H tanα2tanα2 − tanα1 × tanα1
= (s cotα2 − H) tanα1 tanα2tanα2 − tanα1 = (s cotα2 − H) sinα1 sinα2sin (α2 − α1)
Then,
R.L. of top of tower i.e., point O = R.L. of line of sight of staion A to B.M. + V1
= R.L. of B.M. + s1 + V1

3. Base of the object inaccessible: instrument stations are at different planes

Fig 1.4. Instrument stations are at different plane
From Δ ABT,
ATB = 180° − ( α + β) = π − ( α + β)
Also, in Δ ABT, applying sine rule,
BTsinα = ATsinβ = ABsin(π −(α +β)) = dsin(α +β)
 BT = d sinαsin(α +β) and
AT = d sinβsin(α +β)
From second figure,
VA = AT tanαA
= d sinβ tanαAsin(α +β)
 R.L. of T = R.L. of B.M. + sA + VA
As similar,
VB = BT tanαA
= d sinα tanαBsin(α +β)
 R.L. of T = R.L. of B.M. + sB + VB