Survey Test 2

Errors which occur in the same direction and which finally tend to accumulate are said to be Cumulative errors. They are proportional to the length of the line.

Errors which occur in the both direction and which finally tend to compensate are said to be Compensating errors. They are proportional to the square root of the length of the line.

Systematic errors arise from the source that act in similar manner of source. They are cumulative in nature. Examples of systematic errors are Collimation in a level, Expansion of steel tape etc.

Random errors are all those discrepancies remaining after the mistakes and systematic errors are removed. It is mainly caused by limitations of observer and instruments and are random in nature.

Important Points:

Errors may arise from three sources:

1) Instrumental: Error may arise due to imperfection or faulty adjustment of the instrument with which measurement is being taken. For example, a tape may be too long or an angle measuring instrument may be out of adjustment. Such errors are known as instrumental errors.

2) Personal: Error may also arise due to want of perfection of human sight in observing and of touch in manipulating instruments. For example. an error may be there in taking the level reading or reading an angle on the circle of a theodolite. Such errors are known as personal errors.

3) Natural: Error may also be due to variations in natural phenomena such as temperature, humidity, gravity, wind, refraction and magnetic declination. If they not properly observed while taking measurements, the results will be incorrect. For example, a tape may be 20 metres at 20°C but its length will change if the field temperature is different.

But any error which cannot be classified in any of the above following are classified as random error and it cannot be rectified in surveying and detection of such error is noticed only after the completion if survey.

Concept:

The correct area is measured as follows,

\({\rm{Corrected\;Area}} = \frac{{{\rm{Measured\;Area}}}}{{{{\left( {{\rm{Shrinkage\;Factor}}} \right)}^2}}}\)

Where,

\({\rm{ShrinkageFactor = }}\frac{{{\rm{Shrunk}}\;{\rm{scale}}}}{{{\rm{Original}}\;{\rm{scale}}}}{\rm{ = }}\frac{{{\rm{Shrunk}}\;{\rm{length}}}}{{{\rm{Original}}\;{\rm{length}}}}\)

Calculation:

Given,

Original length = 5 cm

Shrunk length = 4.5 cm

Measured area = 400 m²

\({\rm{ShrinkageFactor = }}\frac{{{\rm{Shrunk}}\;{\rm{scale}}}}{{{\rm{Original}}\;{\rm{scale}}}}{\rm{ = }}\frac{{{\rm{Shrunk}}\;{\rm{length}}}}{{{\rm{Original}}\;{\rm{length}}}}\)

\({\rm{ShrinkageFactor = }}\frac{{{\rm{4.5}}\;{\rm{}}}}{{{\rm{5}}\;{\rm{}}}}=0.9\)

The shrinkage factor is 0.9

\({\rm{Corrected\;Area}} = \frac{{{\rm{Measured\;Area}}}}{{{{\left( {{\rm{Shrinkage\;Factor}}} \right)}^2}}}\)

\({\rm{Corrected\;Area}} = \frac{{{\rm{400}}}}{{{{\left( {{\rm{0.9}}} \right)}^2}}}=493.83 \ m^2 \)

The corrected area is 493.83 m²

Correction, \(C = \frac{{{W^2}l}}{{24{P^2}{n^2}}}\)

\(= \frac{{{{50}^2} \times 30}}{{24 \times {{300}^2} \times {1^2}}}\)

= 0.035 m

Concept:

Collimation error (e) is given by:

\({\rm{e}} = \frac{{ \left( {{{\rm{h}}_{\rm{a}}'} - {{\rm{h}}_{\rm{b}}'}} \right) - \left( {{\rm{h}}_{\rm{a}}^{\rm{}} - {{\rm{h}}_{\rm{b}}}{\rm{}}} \right)}}{2}\)

Where

h_a and h_b are the staff reading at station A and B respectively measured from instrument station A.

h_a’ and h_b’ are the staff reading at station A and B respectively measured from instrument station A.

Calculation:

\({\rm{e}} = \frac{{ \left( {1.770 - 1.850} \right) - \left( {1.870 - 1.970} \right)}}{2}\)

∴ e = 0.01 m

In the tangential method of tacheometry, the diaphragm of the tacheometer is not provided with Stadia hair.

The readings are taken by the Single horizontal hair.

In this method, the observations are made for vertical angles and staff intercepts are obtained with cross-wires only.

This method is quite similar to trigonometric leveling.

Given,

Area of different cross-sections are as follows: A₁ = 10 m², A₂ = 15 m², A₃ = 35 m²

Embankment interval = 40 m

Using the Prismoidal formula,

\(\text{V}=\frac{\text{h}}{3}\left( {{\text{A}}_{1}}+{{\text{A}}_{\text{n}}}+4\times {{\text{A}}_{\text{even}}}+2\times {{\text{A}}_{\text{odd}}} \right)\)

\(\text{V}=\frac{40}{3}\left( 10+35+4\times 15 \right)=1400\text{ }\!\!~\!\!\text{ }{{\text{m}}^{3}}\)

Flying height is the elevation of the exposure station above Mean Sea Level (MSL).

Flight line is the line which represents the track of an aircraft on an existing map.

The principal point is the point where the perpendicular projected through the center of the lens intersects the photo image.

Nadir point is the point where a plumb line dropped from the front nodal point pierces the photograph. It is also called a photo nadir and a photo plumb point.

Hence, correct option is ‘3’.

Important Terms:

Isocenter: It is the point on the photo that falls on a line half- way between the principal point and the Nadir point.

Perspective Centre: The real or imaginary point of the origin of bundle of perspective rays.

Exposure or air station: Point in the air occupied by front nodal point of the camera lens at the instant of exposure.

Concept:

\({\rm{Length\;of\;the\;long\;chord\;}}\left( {{{\rm{L}}_{\rm{c}}}} \right) = 2{\rm{R}}\sin \left( {\frac{{\rm{\Delta }}}{2}} \right)\)

\({\rm{Length\;of\;mid\;ordinate}}\left( {{{\rm{L}}_{\rm{m}}}} \right) = {\rm{R}} \times \left( {1 - \cos \frac{{\rm{\Delta }}}{2}} \right)\)

Where

Δ = Angle of deflection or Deviation angle

R = Radius of the curve

Calculation:

Δ = 60°

∴ Deflection angle = 180° - 150° = 30°

\({\rm{Length\;of\;the\;long\;chord\;}}\left( {{{\rm{L}}_{\rm{c}}}} \right) = 2 \times 600 \times \sin \left( {\frac{{60}}{2}} \right)\)

∴ Lc = 600 m

\({\rm{Length\;of\;mid\;ordinate\;}}\left( {{{\rm{L}}_{\rm{m}}}} \right) = 600 \times \left( {1 - \cos \left( {\frac{{60}}{2}} \right)} \right)\)

∴ Lm = 80.38 m

For a triangle: The sum of all the angles should be 180°.

∴ ∠A + ∠B + ∠C = 180°

But, ∠A + ∠B + ∠C = 49°52’30’’ + 60°30’55’’ + 66°36’35’’ = 177°0’0’’

∴ Error = 180° - 170° = 3°

The error of 3° will be distributed based upon the weight of each angle measured.

∴ The corrected measurement of angle c is 67°9’19’’.

The correction per chain length = Temperature correction + Pull correction

Temperature correction, C_T = L ∝ (T_m – T_o)       

= 30 × 12 × 10 – 6 × (35 – 20)

= 0.0054 m (+ve)

Pull correction, \({C_{pull}} = \frac{{\left( {{P_m} - {P_0}} \right)L}}{{AE}}\)

\( = \frac{{\left( {15 - 10} \right) \times 30}}{{0.03 \times 2 \times {{10}^6}}}\)

= 0.0025 m (+ve)

Total correction, e = + 0.0054 + 0.0025

= 0.0079 m

∴ L = L + e

= 30 + 0.0079

= 30.0079 m

∴ True distance \( = \frac{{{\rm{L'}}}}{{\rm{L}}} \times {\rm{measured\;distance\;}}\) 

\(= \frac{{30.0079}}{{30}} \times 600\)

Reciprocal levelling: It eliminates the error due to collimation and error due to curvature of Earth completely, but as the refraction depends upon the atmosphere which may change every minute.

Trigonometric Levelling: In this method, the difference in elevation is determined from the observed vertical angles and measured distances. It is an indirect method of contouring.

Hypsometer: It is used to determine the altitude of station depends on the temperature at which water boils varies with the atmospheric pressure. Therefore, the method of levelling, which is used to estimate the height of mountain by observing the temperature at which water boil.

Profile levelling: It is the method of determining the level of a ground surface along a predetermined line which may be the centre-line of a road, canal, railways or pipeline. The predetermined line may be a single straight line or a series of connected straight lines. The method is also known as longitudinal levelling or sectioning. 

Converting the Quadrantal Bearing into whole circle Bearing

Taking sum of all the latitudes

∑L = 0

150 cos 45° + x cos 155° + 75 cos 195° + 225 cos y = 0

33.621 – 0.906x + 225 cos y = 0

0.906x – 225 cos y = 33.621

0.906x – 33.621 = 225 cos y – (i)

Taking sum of all the departures

∑D = 0

150 sin 45° + x sin 155° + 75 sin 195° + 225 sin y = 0

88.65 + 0.423x + 225 sin y = 0

0.423x + 225 sin y = - 88.65

0.423x + 88.65 = -225 sin y – (ii)

From (i) and (ii)

Squaring and adding equation (i) and (ii)

(0.906x – 33.621)² + (0.423x + 88.65)² = (225)² cos² y + (225)² sin² y

(0.906x)² + (33.621)² – 2 × 0.906x × 33.621 + (0.423x)² + (88.65)² + 2 × 0.423x × 88.65 = 225²

0.999x² - 14.08x – 41635.81 = 0

x = 21.13, – 197.22

x ≠ - 197.22 m

So, x = 21.13 m

0.423x + 88.65 = - 225 sin y

0.423 × 21.13 + 88.65 = - 225 sin y

- 0.4337 = sin y

y = 205.70° (WCB)

y = S 25.70° W

For closed traverse ∑ Latitude = 0;  ∑ Departure =0.

∑ Latitude,

\(\begin{array}{l} 200\cos 0^\circ + 1000\cos \;45^\circ + 907\cos 180^\circ + {l_{SP}}\cos {\theta _{SP}} = 0\\ {l_{SP}}\;cos\;{\theta _{SP}} + 0.107 = 0 \end{array}\)

∑ Departure,

\(\begin{array}{l} 200\sin 0^\circ + 1000\sin 45^\circ + 907\sin 180^\circ + {l_{SP}}\;sin\;{\theta _{SP}} = 0\\ {l_{SP}}\;sin\;{\theta _{SP}} + 707.1 = 0 \end{array}\)

Now,

\(\begin{array}{l} {\left( {{l_{SP}}\;cos\;\theta } \right)^2} + {\left( {{l_{SP}}\;sin\;\theta } \right)^2} = {\left( { - 0.107} \right)^2} + {\left( { - 707.1} \right)^2}\\ {l_{SP}} = \;707\;m\\ \frac{{{l_{SP}}\sin \theta }}{{{l_{SP}}\cos \theta }} = \frac{{ - 707.1}}{{ - 0.107}} \end{array}\)

\(\theta\) = 270°

Concept:

Horizontal distance = k × s + c

RL of staff station = RL of line of collimation (or) Height of Instrument (HI) – Staff reading

HI = RL of instrument station + Height of trunnion axis

Calculation

Given:

Reading of tachometer is 0.85, 1.3 and 1.75.

k = 100, c = 0.2

RL of instrument station is 102.67 m and height of trunnion axis is 1.5 m.

s = 1.75 – 0.85 = 0.9

Horizontal distance = k × s + c = 100 × 0.9 + 0.2 = 90.2 m

RL of line of collimation = RL of instrument station + Height of trunnion axis = 102.67 m + 1.5 m = 104.17 m

RL of staff station = RL of line of collimation – Staff reading = 104.17 – 1.3 = 102.87 m

∴ Option a), c) and d) are correct options.

Concept:
Ground coordinate of A = X_A = \(\frac{{{\rm{H}} - {\rm{\;}}{{\rm{h}}_{\rm{A}}}}}{{\rm{f}}} \times {{\rm{x}}_{\rm{a}}}\) and Y_A = \(\frac{{{\rm{H}} - {\rm{\;}}{{\rm{h}}_{\rm{A}}}}}{{\rm{f}}} \times {{\rm{y}}_{\rm{a}}}\)

Ground coordinate of B = X_B = \(\frac{{{\rm{H}} - {\rm{\;}}{{\rm{h}}_{\rm{B}}}}}{{\rm{f}}} \times {{\rm{x}}_{\rm{b}}}\) and Y_B = \(\frac{{{\rm{H}} - {\rm{\;}}{{\rm{h}}_{\rm{B}}}}}{{\rm{f}}} \times {{\rm{y}}_{\rm{b}}}\)

x_a, y_a, x_b, y_b are the photo coordinates,

H is the flying height and f is the focal length.

Horizontal length of the line AB in the ground = \(\sqrt {{{\left( {{{\rm{X}}_{\rm{B}}} - {{\rm{X}}_{\rm{A}}}} \right)}^2} + {\rm{\;}}{{\left( {{{\rm{Y}}_{\rm{B}}} - {{\rm{Y}}_{\rm{A}}}} \right)}^2}} \)

Calculation

Given:

x_a = +50 mm, y_a = +40 mm, x_b = - 60 mm, y_b = - 90 mm

The elevations of point A and B are 300 m and 400 m

i.e., h_A and h_B are 300 m and 400 m

focal length (f) = 150 mm

X_A = 400 m

In order to find the ground level coordinates, we first need to determine the value of flying height which is not given directly in the question. But we know the value of ground level coordinate of A using which the flying height can be determined.

X_A = \(\frac{{{\rm{H}} - {{\rm{h}}_{\rm{A}}}}}{{\rm{f}}} \times {{\rm{x}}_{\rm{a}}} \Rightarrow H = \frac{{{\rm{f}} \times {{\rm{X}}_{\rm{A}}}}}{{{{\rm{x}}_{\rm{a}}}}} + {{\rm{h}}_{{\rm{A}}}} = \frac{{150 \times 400}}{{50}} + 300 = 1500{\rm{\ m}}\)

∴ The flying height is 1500m.

Now the other ground level coordinates can be determined.

\({Y_A} = \frac{{{\rm{H}} - {{\rm{h}}_{\rm{A}}}}}{{\rm{f}}} \times {{\rm{y}}_{\rm{a}}} = \frac{{1500 - 300}}{{150}} \times 40 = 320\;{\bf{m}}\)

\({X_B} = \frac{{{\rm{H}} - {{\rm{h}}_{\rm{B}}}}}{{\rm{f}}} \times {{\rm{x}}_{\rm{b}}} = \frac{{1500 - 400}}{{150}} \times - 60 = - 440\;{\bf{m}}\)

\({Y_B} = \frac{{{\rm{H}} - {{\rm{h}}_{\rm{B}}}}}{{\rm{f}}} \times {{\rm{y}}_{\rm{b}}} = \frac{{1500 - 400}}{{150}} \times - 90 = - 660\;{\bf{m}}\)

Horizontal length of the line AB in the ground = \(\sqrt {{{\left( {{{\rm{X}}_{\rm{B}}} - {{\rm{X}}_{\rm{A}}}} \right)}^2} + {{\left( {{{\rm{Y}}_{\rm{B}}} - {{\rm{Y}}_{\rm{A}}}} \right)}^2}}=\sqrt {{{\left( { - 440 - 400} \right)}^2} + {{\left( { - 660 - 320} \right)}^2}} \) = 1290.7 m

∴ Option a), b) and c) are correct.

Concept:

The number of photographs required is N = \(\left[ {\frac{{{{\rm{L}}_1}}}{{\left( {1 - {{\rm{L}}_{\rm{o}}}} \right) \times \frac{{\rm{l}}}{{\rm{s}}}}} + 1{\rm{\;}}} \right] \times \left[ {\frac{{{{\rm{W}}_1}}}{{\left( {1 - {{\rm{S}}_{\rm{o}}}} \right) \times \frac{{\rm{w}}}{{\rm{s}}}}} + 1{\rm{\;}}} \right]\),

One extra photograph is always considered to cover the area at the ends.

where,

L₁ = Length of ground to be covered

W₁ = Width of ground to be covered

l = Length of the photograph in the direction of flight

w = Width of the photograph

s = Scale of the photograph

L_o = Longitudinal overlap

S₀ = Side overlap

Calculation:

Given:

L₁ = 31 km, W₁ = 10 km,

l = 250 mm, w = 250 mm,

L_o = 0.5, S_o = 0.3, s = \(\frac{1}{{10000}}\)

Ground length covered by each photograph = L = \(\left( {1 - {{\rm{L}}_{\rm{o}}}} \right) \times \frac{1}{{\rm{s}}} \times {\rm{l}}\)

Time interval between exposure is = \(\frac{{\rm{L}}}{{\rm{V}}}\)

The number of photographs required is

N = \(\left[ {\frac{{{{\rm{L}}_1}}}{{\left( {1 - {{\rm{L}}_{\rm{o}}}} \right) \times {\rm{\;}}\frac{{\rm{l}}}{{\rm{s}}}}} + 1} \right] \times \left[ {\frac{{{{\rm{W}}_1}}}{{\left( {1 - {{\rm{S}}_{\rm{o}}}} \right) \times {\rm{\;}}\frac{{\rm{w}}}{{\rm{s}}}}} + 1} \right]\)

\(N=\left[ {\frac{{31000}}{{\left( {1 - 0.5} \right) \times \frac{{0.25}}{{\left( {\frac{1}{{10000}}} \right)}}}} + 1} \right] \times \left[ {\frac{{10000}}{{\left( {1 - 0.3} \right) \times \frac{{0.25}}{{\left( {\frac{1}{{10000}}} \right)}}}} + 1} \right]{\rm{\;}}\)

N = [ 25.8] × [ 6.7] = 26 × 7 = 182

The number of photographs required = 182

Length covered by each photograph

= \(\left( {1 - {{\rm{L}}_{\rm{o}}}} \right) \times \frac{1}{{\rm{s}}} \times {\rm{l}} = \left( {1 - 0.5} \right) \times \frac{1}{{\left( {\frac{1}{{10000}}} \right)}} \times 0.25 = 1250{\rm{\ m}} = 1.25{\rm{\;km}}\)

Ground speed of aircraft = 240 kmph

The time interval between exposure = \(\frac{{1.25}}{{240}} \times 3600\sec = 18.75{\rm{\;sec}}\)

∴ Option c) and d) are incorrect options.