Surveying Test 2

Concept:

We know that for levelling work, following equation must be satisfied:

\(\sum BS - \;\sum FS = Last\;RL - First\;RL\)

Where

BS stands for Back Sight and FS stands for Fore Sight

RL stands for Reduced Level

Calculation:

Given, Summation of all back sights i.e. ΣBS = 7.475 m

Summation of all Fore sights i.e. ΣFS = 7.395 m

First Reduced Level = 100.00 m

Using above Equation

7.475 – 7.395 = Last RL – 100.00

⇒ Last RL = 100.08 m

Note: Last RL is more than that of First RL, this means that ground slope is upward or ground is rising from initial benchmark.

The permanent adjustments of a theodolite are so arranged so that the next adjustment do not creates any disturbance in the results obtained from previous adjustments. If the properly arranged sequence is not followed it may result in the error in the previous adjustment and hence the given order is followed:

1) Plate level Test: It is done to make the plate bubbles central to their run when the vertical axis is truly vertical.

2) Cross Hair Ring Test: It is done to make the vertical crosshairs lie in a plane perpendicular to the horizontal axis.

3) Spire Test: It is done to make the horizontal axis perpendicular to the vertical axis. 

4) Vertical Arc Test: It is done to make the vertical circle indicate zero when the line of sight is perpendicular to the vertical axis.

Important Point:

Some of the Temporary adjustments of Theodolite are:

1)Setting up

2) Centring

3) Levelling

4)Focussing the eyepiece

5)Focussing the objective

For a particular line, if the difference between fore bearing and back bearing is found exactly to be 180^o, the lines/measurement associated with this station is said to be free from local attraction.

From the observation, the difference between F.B and B.B for line DE is 180^o.

For Line DE: F.B – B.B = 242°45’ – 62°45’ = 180°

∴ Station D and E is said to be free from local attraction.

Error for the fore bearing of CD = 180^o – (200^o30’ – 22^o15’) = 1^o45’0’’

∴ Corrected fore bearing of CD = 22^o15’ – 1^o45’ = 20^o30’0’’

∴ Corrected back bearing of BC = 222^o30’ – 1^o45’ = 220^o45’0’’

Error for fore bearing of BC = 180^o – (220^o45’ – 39^o30’) = - 1^o15’

∴ Corrected fore bearing of BC = 39^o50’ + 1^o15’ = 40^o45’0’’

Concept:

When staff is held vertically downward, then Height of instrument is

H I = Reduce Level (RL) + Staff Reading (SR)

When Staff is held inverted then

HI = RL - SR

Height of Ceiling above floor = RL of Ceiling – RL of Floor

Calculation:

RL of Floor = 74.50 m

SR when staff held upright position is 1.815 m

Hence, HI = 74.5 + 1.815 = 76.315 m

SR when Staff held inverted = 2.870 m

In this case, HI = RL of Ceiling – SR = R L of Ceiling - 2.870

In both cases, height of instrument will not change as such location of instrument is not changed.

∴ 76.315 m = RL of ceiling – 2.870

⇒ RL of ceiling = 79.185 m

Now,

Height of Ceiling above floor = RL of Ceiling – RL of Floor

Or

Height of ceiling above floor = 79.185 – 74.50 = 4.685 m

Concept:

The sensitivity of Bubble tube is given by the following expression

\(\alpha = \frac{S}{{nD}} = \frac{l}{R}\)

Where

α is the sensitivity of bubble tube i.e. angle of rotation in radian due to one division of movement of bubble.

n is the divisions moved by bubble.

l is the length of one division.

S is staff intercept and it is given as

S = S₂ – S₁

S₁ is the staffs reading when bubble is exactly at center or actual staff reading.

S₂ is the staff reading when bubble moves some n divisions.

R is the radius of curvature of Bubble tube.

Further,

1 sec = 1/206265 radian

Calculation:

Given, α = 25 secs/division

l = 2 mm

n = 2 divisions

D = 200 m

S₂ = 4.54 m

Using

\({\rm{\alpha \;}}\left( {{\rm{in\;sec}}} \right) = \frac{{\rm{S}}}{{{\rm{nD}}}}\)

\({\rm{\alpha \;}}\left( {{\rm{in\;radian}}} \right) = \frac{{\rm{S}}}{{{\rm{nD}}}} \times 206265\)

\(25 = \frac{{\rm{S}}}{{2\; \times \;200}} \times 206265\)

⇒ S = 0.048 m

Now

S = S₂ – S₁

0.048 = 4.54 – S₁

⇒ S₁ = 4.492 m

Concept:

Reciprocal levelling is adopted to accurately determine the level difference between two points which are separated by obstacles like a river, ponds, lakes, etc.

It eliminates the following errors:

i) error in instrument adjustments i.e error due to collimation

ii) the combined effect of Earth's curvature and the refraction of the atmosphere

iii) variation in the average refraction

Difference of levels,

\(h = \frac{{\left( {{a_1} - {b_1}} \right) + \left( {a - b} \right)}}{2}\)

Calculation:

Let the error is ‘e’ and the line of sight is inclined upwards and A is higher than B

When instrument near station A:

ΔhAB = (1.40 - e) - 1.25

∴ ΔhAB = 0.15 - e      ---(i)

When instrument near station B:

ΔhAB = 0.6 - (0.8 - e) = e - 0.2      ---(ii)

But ΔhAB will remain same, thus equating (i) and (ii) we get,

0.15 - e = e - 0.2

⇒ 2e = 0.15 + 0.2 ⇒ e = 0.175

Putting the value of e in equation i or ii,

∴ ΔhAB = e - 0.2 = 0.175 - 0.2 = -0.025

Here, ΔhAB is negative and hence, the assumption made is incorrect and B is higher than A.

∴ R.L of B = R.L of A + |ΔhAB|

Concept:

To find the missing values of the table, the latitude and departure of all the lines for a closed traverse should be calculated and the summation of the all the latitude and departures must equals to zero.

∴ ∑ L = 0, and ∑ D = 0

Latitude of a line = ℓ cos θ, and Departure of a line = ℓ sin θ

Calculation:

Latitude of line PQ = l cos θ = 300 × cos 110° = - 102.60 m

Departure of line PQ = l sin θ = 300 sin 110° = +281.90

Similarly, the latitude and departure of all the lines IS calculated below:

Convert the quadrantal bearing system of bearing angle into whole circle bearing system

∑ L = - 102.60 – 100 + 0 + 450 cos y = 0

∑ D = 281.90 – 173.20 – x + 450 sin y = 0

450 cos y = 202.60                         …… (i)

- x + 450 sin y = -108.70                  …… (ii)

Solving equation, (i) and (ii)

\(\therefore \cos {\rm{y}} = \frac{{ 202.60}}{{450}}\)

∴ y = 63.24°

- x + 450 sin 116.75° = -108.70

For the given problem:

As the Reduced Level decreases from station 1 to station 2:

⇒ Indicates fall and thus reading at F.S will be more than that of B.S

Fall = 105.450 - 104.290 = 1.160

∴ F.S = B.S + Fall = 1.785 + 1.160 = 2.945 m

At station 2, reading of B.S > I.S at a station

⇒ As the reading reduces, it indicates a rise in the level.

Rise = 2.130 - 1.905 = 0.225

∴ R.L of station 3 = R.L of station 2 + Rise = 104.290 + 0.225 = 104.515

At station 3, I.S at station 3 > I.S at station 4

⇒ As the reading reduces, it indicates rise in the level.

Rise = 1.905 - 1.030 = 0.875

∴ R.L of station 4 = R.L of station 3 + Rise = 104.515 + 0.875 = 105.390