Survey Test 1

Concept:

A representative fraction (R.F.) indicates the ratio between the number of units on the map to the number of units on the ground or a representative fraction (RF) is the ratio of distance on the map to distance on the ground.

\({\rm{Representative}}\;{\rm{fraction}}\;\left( {{\rm{R}}{\rm{.F}}{\rm{.}}} \right){\rm{ = }}\frac{{{\rm{Distance}}\;{\rm{on}}\;{\rm{the}}\;{\rm{map}}}}{{{\rm{Distance}}\;{\rm{on}}\;{\rm{the}}\;{\rm{ground}}}}\;\)

Calculation:

Given,

Distance on map = 90 mm

Distance on the ground = 100 mm

Reduction factor = \(\frac{{90}}{{100}} = 0.9\)

Revised scale = original scale × reduction factor

Revised scale \(= \frac{1}{{1000}} \times 0.9 = \frac{1}{{1111}}\)

Reduction factor in terms of area is given as

\({\left( {{\rm{RF}}} \right)^{\rm{2}}}{\rm{ = }}\left( {\frac{{{\rm{Area}}\;{\rm{on}}\;{\rm{the}}\;{\rm{map}}}}{{{\rm{Area}}\;{\rm{on}}\;{\rm{the}}\;{\rm{ground}}}}} \right)\)

Given.

Length of line measured = l = 428 m

For Case 1: The angle of slope between the points is 8° 

θ = Angle of slope = 8° 

∴ The horizontal distance = l × cos θ = 428 × cos 8° = 423.83 m

For Case 2: The Slope is 1 in 4

tan θ = (1/4) = 0.25

θ = 14.036° 

∴ The horizontal distance = l × cos θ = 428 × cos 14.036° = 415.22 m

Hence For Case 1: Horizontal distance = 423.83 m & For Case 2: Horizontal distance = 415.22 m

BS = 1.050 m

RL = 135.15 m

∴ HI of instrument = RL + BS

= 135.15 + 1.050

= 136.2 m

FS = -2.300 m (as inverted)

∴ RL of roof slab = HI – FS

= 136.2 – (-2.300)

= 138.5 m

From an old map:

True Bearing = Magnetic bearing – θ_W

T.B = 280° 45’ – 2° 15’

T.B = 278° 30’

Now At Present suppose declination is still to the west, Then

T.B = M.B – θ_W

278° 30’ = 272° 45’ – θ_W

θ_W = – 5° 45’

So,

Declination is present is to the east

In Direct Method of Contouring, the reduced level of different selected points is located and joined to form a contour. This method is accurate but very tedious and slow.

Square method or grid method of contouring – In this method we draw a grid point on the ground and measure the reduce level of all these points. It is generally used in plain terrain.

Tachometric method of contouring – This method is used for angular surveying.

This method is best suitable for hilly terrains as the number of stations which can be commanded by a tachometer is far more than those by a level and thus the number of instrument settings are considerably reduced.

Subtense Tacheometry:

Subtense procedures involve a method in which no graduated staff is used. Instead, a bar of fixed length, called a subtense bar, is positioned at one end of the line and a theodolite at the other.

The angle subtended by the bar is measured using a theodolite and knowing the length of the bar, the distance can be calculated since it is proportional to the subtense angle.

Subtense bar: 

A horizontal bar used in the subtense system of surveying by tacheometry. It is held at a distant point and its distance is calculated from its known length and the angle that it subtends at the observer's eye.

Sources of error in Subtense Tacheometry:

1. Incorrect length of bar

2. Incorrect setting of the bar

3. Non-horizontality and poor centering of bar

4. Errors in the measured subtense angle

5. The error due to refraction in subtense tacheometry is very less.

Applications of Subtense Tacheometry

It is suitable for measuring traverses of low accuracy.

It is adopted for flat areas, where only horizontal angles and distances are required.

It is suitable for topographic surveying because the relative location of points in the horizontal plane and the elevation of the points can be determined simultaneously.

The following table gives information regarding the Global Positioning System.

Hence the minimum number of satellite required to determine the position precisely is 4 

Concept:                                            

Scale of the photograph is given by,

\(Scale = \frac{f}{{H - h}}\)

f = focal length of camera

H = Average height of photograph from MSL

h = Average height of terrain from MSL

Calculation:

 Given, f = 15 cm; H = 1200 m; h = 80 m

∴ \({\bf{Scale}} = \frac{{0.15}}{{1200 - 80}} = \frac{1}{{7467}}\)

Remote Sensing: Remote sensing is an art and science of obtaining information about an object without physical contact with that object, typically from aircraft or satellites.

All old type of surveying cannot be completely replaced by remote sensing system because for preliminary survey or small scale work, old survey method like chaining, compass, triangulation etc are more economical than that of remote sensing.

Data is generally is collected by high level sensors, camera, scanners, radars etc. The collected data has to be analyzed at ground by using highly sophisticated techniques using high speed computers. 

Concept: Probable error of the mean [E_m]:

E_m = \(\pm 0.6745\sqrt {\frac{{\sum {{\rm{V}}^2}}}{{{\rm{n}}\left( {{\rm{n}} - 1} \right)}}} = \frac{{{{\rm{E}}_{\rm{s}}}}}{{\sqrt {\rm{n}} }}\)

v → difference between single observation and mean of series

E_s = Probable error of single observation

Calculations:

E_m = \(\pm 0.6745\sqrt {\frac{{0.065669}}{{4\left( {4 - 1} \right)}}}\)

E_m = ± 0.0499 metre

Concept:

If a chain is too short, the measured distance will be more, the error will be positive, and the correction will be negative.

\(True\;length\;of\;line\;\left( l \right)\)

\(= Measured\;length\;of\;line\;\left( {l'} \right) \times \frac{{Incorrect\;length\;of\;the\;chain\;\left( {L'} \right)}}{{True\;length\;of\;the\;chain\;\left( L \right)}}\) 

Calculations:

With 20 m chain:

Measured length of line = 1500 m

True length of chain = 20 m

Incorrect length of chain = 20 – 0.05 = 19.95 m

Therefore, True length of line =

\(= \;Measured\;length\;of\;line\;\left( {l'} \right) \times \frac{{Incorrect\;length\;of\;the\;chain\;\left( {L'} \right)}}{{True\;length\;of\;the\;chain\;\left( L \right)}}\)

\(= 1500 \times \frac{{19.95}}{{20}}\)

= 1496.25 m

Therefore, True length of line = 1496.25 m

With 30 m chain:

Measured length of line = 1476 m

True length of chain = 30 m

True length of line = 1496.25 m

\(1496.25 = \;1476 \times \frac{{Incorrect\;length\;of\;the\;chain\;\left( {L'} \right)}}{{30}}\)

\(Incorrect\;length\;of\;the\;chain\;\left( {L'} \right) = \frac{{1496.25 \times 30}}{{1476}}\)

\(Incorrect\;length\;of\;the\;chain\;\left( {L'} \right) = 30.41\;m\)

Therefore, the error in 30 m chain is (30.41 – 30) = 0.41 m

Hence, the 30 m chain is 41 cm too long.

Concept:

A station is free from local attraction when the difference between FB and BB is ± 180°.

Calculation:

It can be noticed that station C and D are free from local attractions since B.B and F.B differ by 180°

∴ BB of Line BC is true

F.B of Line BC = 296°35’ – 180 = 116°35’

∴ True F.B of Line BC = 116°35’

Correction = 116°35’ – 115° 20’ = +1°15’

Now correct BB of Line AB = 254°20’ + 1°15’ = 255°35’

∴ Correct FB of Line AB = 255° 35’- 180 = 75° 35’.

But Observed Reading is 75°5’

Correction = 75°35’ – 75°5’ = + 30’

Now, this correction has to be applied to BB of Line EA

Correct bearing of EA = 125°5’ + 30’ = 125°35’

∴ Corrected FB of Line EA = 125°35’ + 180° = 305°35’

∴ Correction = 305° 35’ – 304°50’ = +45’

∴ Corrected BB of Line DE = 44°5’ + 45’ = 44°50’

Corrected Bearing of stations are

\({\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + \left( {{{\rm{A}}_2} + {{\rm{A}}_3} + \ldots {{\rm{A}}_{{\rm{n}} - 1}}} \right)} \right]\)

d = 5

A1 = 32

An = 15

\({\rm{V}} = 5\left[ {\frac{{32 + 15}}{2} + \left( {26 + 24 + 18} \right)} \right]\)

= 5 [23.5 + 68]

= 457.5 ha – m

Concept:

i) Mid-Ordinate Rule

Area of plot = h1 × d + h2 × d + … + hn × d = d (h1 + h2 + … hn)

∴ Area = common distance × sum of mid-ordinates

ii) Average-Ordinate Rule

\(\text{Area}=\frac{{{\text{O}}_{1}}+{{\text{O}}_{2}}+\ldots +{{\text{O}}_{\text{n}}}}{{{\text{O}}_{\text{n}+1}}}\times \text{l}\)

\(\text{i}.\text{e}.\text{ }\!\!~\!\!\text{ Area}=\frac{\text{Sum }\;\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ ordinates}}{\text{No}.\;\text{ }\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ ordinates}}\times \text{length }\;\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ base }\;\!\!~\!\!\text{ line}\)

(iii) Trapezoidal Rule

\(\text{Total }{area}=\frac{\text{d}}{2}\left\{ {{\text{O}}_{1}}+2{{\text{O}}_{1}}+2{{\text{O}}_{2}}+\ldots +2{{\text{O}}_{\text{n}-1}}+{{\text{O}}_{\text{n}}} \right\}\)

\(\text{Total }{ Area}=\frac{\text{Common }\;\!\!~\!\!\text{ distance}}{2}\times \{\left( 1\text{st }\!\!~\!\!\text{ ordinate}+\text{last }\!\!~\!\!\text{ ordinate} \right)+2\left( \text{sum }\!\!~\!\!\text{ of }\!\!~\!\!\text{ other }\!\!~\!\!\text{ ordinate} \right))\}\)

(iv) Simpson’s Rule

\(\text{Total }{ area}=\frac{\text{d}}{3}\left( {{\text{O}}_{1}}+4{{\text{O}}_{2}}+2{{\text{O}}_{3}}+4{{\text{O}}_{4}}+\ldots +{{\text{O}}_{\text{n}}} \right)\)

\(\text{Total }{ Area}=\frac{\text{Common }\;\!\!~\!\!\text{ Distance}}{3}\times \left\{ \left( 1\text{st }\!\!~\!\!\text{ ordinate}+\text{last }\!\!~\!\!\text{ ordinate} \right)+4\left( \text{sum }\!\!~\!\!\text{ of }\!\!~\!\!\text{ even }\!\!~\!\!\text{ ordinates} \right)+2\left( \text{sum }\!\!~\!\!\text{ of }\!\!~\!\!\text{ remaining }\!\!~\!\!\text{ odd }\!\!~\!\!\text{ ordinates} \right) \right\}\)

Calculation:

By Simpson’s rule

A = d/3 × [(O₀ + O_n) + 4 (O₁ + O₂ + … + O_n-1) + 2 (O₂ + O₄… + O_n-2)]

d = 10 m,

O₀ + O_n = 3.25 + 5.65 = 8.9 m,

4 (O₁ + O₃ + …. O_n-1) = 4 (5.60 + 6.65 + 6.20 + 4.20) = 90.60,

2 (O₂ + O₄ + …. O_n-2) = 2 (4.20 + 8.75 + 3.25) = 32.40

∴ \(\text{A}=\frac{10}{3}\) × (8.9 + 90.60 + 32.40) = 439.67 sq. metres = 4.3967 acres

Concept:

Scale of the photograph is given by,

\(S{\rm{cale}} = {\rm{\;}}\frac{{{\rm{Distance\;on\;photo\;}}}}{{{\rm{Distance\;on\;ground\;}}}} = \frac{{\rm{f}}}{{{\rm{H}} - {{\rm{h}}_{\rm{a}}}}}\)

The distance between images of top and bottom of the tower measures on photographs is called relief displacement (d) and it is given by:

\(d = \frac{{rh}}{{{\rm{H}} - {{\rm{h}}_{\rm{a}}}}}\)

ha = Average height from MSL, f = Focal length, H = Height from where the photograph is taken, r =  Distance of the Image of the top of the tower, h = Height of tower

Calculation:

Given, h_a = 250 m, f = 25 cm, The distance on ground = 300 m and distance on photo = 15 cm

We know that

\(S{\rm{cale}} = {\rm{\;}}\frac{{{\rm{Distance\;on\;photo\;}}}}{{{\rm{Distance\;on\;ground\;}}}} = \frac{{\rm{f}}}{{{\rm{H}} - {{\rm{h}}_{\rm{a}}}}}\)

\(\frac{{15{\rm{\;}}}}{{300}} = \frac{{25}}{{{\rm{H}} - 250}}\)

∴ H = 750 m

\(d = \frac{{rh}}{{{\rm{H}} - {{\rm{h}}_{\rm{a}}}}}\)

Where r = 10 cm

\(0.5 = \frac{{10\; \times \;h}}{{{\rm{\;}}750\; - \;250}}\)

∴  h = 25 m

Area of photogrammetric survey (A) = 200 sq km.

Photograph format = 250 × 250 mm

Scale \(\left( {\rm{S}} \right) = \frac{1}{{10000}}\)

Net Area covered in each photograph = (a) = \(\frac{{\left( {1 - 0.6} \right)\left( {1 - 0.3} \right)\left( {0.25 \times 0.25} \right)}}{{{{\left( {\frac{1}{{10000}}} \right)}^2}}}\) = 1750000 m²

∴ Net Area covered in each photograph = 1.75 km²

Number of photographs required are (N)\({\rm{\;}} = \frac{{\rm{A}}}{{\rm{a}}} = \frac{{200}}{{1.75}} = 114.28 \approx 115\)

Concept:

Length of the curve is given by:

\({{\rm{L}}_{\rm{c}}} = \frac{{2{\rm{\pi R}}}}{{360^\circ }} \times {\rm{\Delta }}\)

Where R → Radius of curveΔ → Deflection angle

Calculation:

\({{\rm{L}}_{\rm{C}}}_1 = \frac{{2{\rm{\pi }} \times 530^\circ }}{{360^\circ }} \times {{\rm{\Delta }}_1}\)

\({{\rm{L}}_{\rm{C}}}_1 = \frac{{2{\rm{\pi }} \times 530^\circ }}{{360^\circ }} \times 55\)

\({{\rm{L}}_{\rm{C}}}_1 = 508.76{\rm{m}}\)

Δ₂ = 1.60 × 55°

Δ₂ = 88°

\({{\rm{L}}_{\rm{C}}}_2 = \frac{{2{\rm{\pi }} \times 530^\circ }}{{360^\circ }} \times 88\)

\({{\rm{L}}_{\rm{C}}}_2 = 814.02{\rm{m}}\)

\({\rm{\Delta }}{{\rm{L}}_{\rm{C}}} = {{\rm{L}}_{\rm{C}}}_2 - {{\rm{L}}_{\rm{C}}}_1 = 814.02 - 508.76\)