Surveying Test 1

Concept:

The actual length of the chain is given by,

\({\rm{Actual\;length}} = \frac{{{\rm{Incorrect\;measurement}}}}{{{\rm{Actual\;measurement}}}} \times {\rm{Incorrect\;length\;of\;chain}}\)

Calculation:

\(\therefore \;{\rm{Actual\;length}} = \frac{{29.9}}{{30}} \times 300 = 299\;{\bf{m}}\)

Note:

Concept:

Error of a reading:

i) Addition

if c = a + b ⇒ Probable error (e c) \(= \pm \sqrt {{\rm{e}}_{\rm{x}}^2 + {\rm{e}}_{\rm{y}}^2} \)

ii) Subtraction

if c = a – b ⇒ Probable error (ec) \(= \pm \sqrt {{\rm{e}}_{\rm{x}}^2 + {\rm{e}}_{\rm{y}}^2}\)

iii) Multiplication

if c = a × b ⇒ Probable error (ec) \(= \pm {\rm{xy}}\sqrt {{{\left( {\frac{{{{\rm{e}}_{\rm{a}}}}}{{\rm{a}}}} \right)}^2} + {{\left( {\frac{{{{\rm{e}}_{\rm{b}}}}}{{\rm{b}}}} \right)}^2}} \)

iv) Division

if c = a/b ⇒ Probable error (ec) \(= \pm \frac{{\rm{a}}}{{\rm{b}}}\sqrt {{{\left( {\frac{{{{\rm{e}}_{\rm{a}}}}}{{\rm{a}}}} \right)}^2} + {{\left( {\frac{{{{\rm{e}}_{\rm{b}}}}}{{\rm{b}}}} \right)}^2}} \)

Calculation:

For the given situation:

c = 360° - (a + b)

This similar to standard function, where C is dependent on the added value of a and b

∴ c = 360° - (a + b)

\(\therefore {{\rm{e}}_{\rm{c}}} = \pm \sqrt {{\rm{e}}_{\rm{a}}^2 + {\rm{e}}_{\rm{b}}^2} \)

\(\therefore {{\rm{e}}_{\rm{c}}} = \pm \sqrt {{2^2} + {3^2}} \)

∴ e_c = ± 3.61^’’

Correction for temperature C_t = ∝ (T_m - T_o) L

Where T_m = Mean temperature during measurement

T_o = Temperature of standardization

So C_t might be +ve or –ve depending upon T_m and T_o

Correction for pull, \({C_p} = \frac{{P - {P_s}}}{{AE}}L\)

So, C_p might also be +ve or –ve

Sag correction, \({C_s} = - \frac{{{w^2}{l^3}}}{{24{p^2}}}\)

\(= - \frac{l}{{24}}.{\left( {\frac{W}{p}} \right)^2}\)

Where, W = wl

Sag correction is always negative

Mean sea level correction

\({C_h} = \pm \frac{{LH}}{R}\)

Sum of the individual weight = 4 + 3 + 2 + 1 = 10

Weighted arithmetic mean (W_m) is given by:

\(\left( {{\text{W}}_{\text{m}}} \right)=\frac{1}{10}\left[ \left( 35{}^\circ {{18}^{'}}{{18}^{''}} \right)\times 4+\left( 35{}^\circ {{18}^{'}}{{17}^{''}} \right)\times 3+\left( 35{}^\circ {{18}^{'}}{{20}^{''}} \right)\times 2+\left( 35{}^\circ {{18}^{'}}{7}' \right)\times 1 \right]\)

W_m = 35°18’17’’

For probable error:

Probable error of weighted mean \(=\pm \text{ }\!\!~\!\!\text{ }0.6745\text{ }\!\!~\!\!\text{ }\sqrt{\frac{\sum \text{w}{{\text{v}}^{2}}}{\sum \text{w}\left( \text{n}-1 \right)}}\)

\({{\text{e}}_{{{\text{w}}_{\text{m}}}}}=\pm 0.6745\text{ }\!\!~\!\!\text{ }\sqrt{\frac{115}{10\times \left( 4-1 \right)}}\)

Standard error of area is given by \({e_A} = \sqrt {{{\left( {{e_l}\frac{{dA}}{{dl}}} \right)}^2} + {{\left( {{e_b}\frac{{dA}}{{db}}} \right)}^2}} \)

∵ A = l × b

∵ \(\frac{{dA}}{{dl}} = b = 20\)

And \(\frac{{dA}}{{db}} = l = 60\)

Now, \({e_l} = {e_b} = \pm 1cm =\pm 0.01m\)

∴ \({e_A} = \pm 0.632\;{m^2}\)

Concept:

Error due to wrong measuring scale:

If a wrong measuring scale is used to measure the length of an already drawn line on the plan, the measured length will not be correct.

Correct length \( = \frac{{R.F.\;of\;the\;wrong\;scale}}{{R.F.\;of\;the\;correct\;scale}} \times \) Measured length

As the area is the product of two distances, hence

Correct area \( = {\left[ {\frac{{R.F.\;of\;the\;wrong\;scale}}{{R.F.\;of\;the\;correct\;scale}}} \right]^2} \times \) Measured area

Calculation:

R.F of the wrong scale \(= \frac{1}{{50}}\) 

R.F of the correct scale \(= \frac{1}{{100}}\)

∴ Correct length \( = \frac{{R.F\;of\;wrong\;scale}}{{R.F\;of\;correct\;scale}} \times Measured\;Length = \frac{{\left( {\frac{1}{{50}}} \right)}}{{\left( {\frac{1}{{100}}} \right)}} \times 100\) 

= 200 m

Correct Area \( = {\left[ {\frac{{R.F\;of\;wrong\;scale}}{{R.F\;of\;correct\;scale}}} \right]^2} \times Measured\;Area\; = {\left[ {\frac{{\left( {\frac{1}{{50}}} \right)}}{{\frac{1}{{100}}}}} \right]^2} \times 100\) 

= 400 m2

Concept:

1. Correction for temperature:

C_b = L ∝ (T_x - T_o)

Where

C_t = Temperature correction

L = Length of tape/chain

T_m = Measurement temperature

T_o = Standardization temperature

2. Correction for pull (C_p)

\({{\rm{C}}_{\rm{p}}} = \frac{{\left( {{\rm{P}} - {{\rm{P}}_{\rm{o}}}} \right) \times {\rm{L}}}}{{{\rm{AE}}}}\)

P = Pull at the measurement

P_o = Pull at standardization

A = Area of the tape

E = Youngs modulus of tape material

3. Correction for sag (C_s)

\({{\rm{C}}_{\rm{s}}} = \frac{{{{\rm{W}}^2}{\rm{l}}}}{{24{{\rm{n}}^2}{{\rm{P}}^2}}}\)

W = Weight of chain tape

n = number of suspensions

Calculation:

1) Correction of temperature (C_f)

C_t = 20 × 9.4 × 10^-6 × (95 - 77)

C_t = (+) 3.384 × 10^-3 m

2) Correction for pull (C_p)

\({{\rm{C}}_{\rm{p}}} = \frac{{\left( {75 - 50} \right) \times 20}}{{10 \times {{10}^{ - 6}} \times 128 \times 10}} = \left( + \right){\rm{\;}}0.39 \times {10^{ - 3}}{\rm{\;m}}\)

3) Correction for sag (C_o)

\({\rm{W}} = 89.6 \times {10^3} \times 20 \times {10^{ - 6}} \times \frac{{20}}{5} = 7.168{\rm{\;N}}\)

\({{\rm{C}}_{\rm{S}}} = \frac{{{{7.168}^2} \times 20}}{{24 \times {5^2} \times {{75}^2}}} = \left( - \right){\rm{\;}}0.304 \times {10^{ - 3}}{\rm{\;m}}\)

Combined correction (C_c) is given as:

C_c = C_t + C_p + C_s

C_c = (3.384 + 0.39 - 0.304) × 10^-3

C_c = 3.47 × 10^-3 m

C_c = 0.00347 m

∴ The length of tape available = 30 + 0.00347 m

Concept: Least count = \(\frac{S}{n}\)

n → no. of divisions on vernier scale

s → smallest division of main scale

Calculation: s = 20’ = 20 minutes

L.C = 5” = \(\frac{{5'}}{{60}} = \frac{1}{{12}}\) minutes

\(\frac{1}{{12}} = \frac{{20}}{n}\)

n = 20 × 12

n = 240

Concept:

Actual area on Ground = (Shrunk scale)² × Present area on Drawing

Shrunk scale = shrunk factor × original scale

Shrunk factor = \(\frac{{Shrunk\;length}}{{Original\;length}}\)

Calculations:

Shrunk factor = \(\frac{{19.1}}{{20}}\) = 0.955

Original scale = \(\frac{{1{\rm{cm}}}}{{10{\rm{m}}}} = \frac{1}{{1000}}\)

Shrunk scale = 0.955 × \(\frac{1}{{1000}} = \frac{1}{{1047.12}}\)

i.e. 1cm = 10.4712 metres

Actual area = (10.4712)² × 135.6

= 14868 m²