Representation Of Scale, Surveying Errors And Corrections ~ Civil Engineer Knowledge

SURVEYING (Part 2)

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REPRESENTATION OF SCALE

1. ENGINEER'S SCALE - This scale is a tool for measuring distances and transferring measurements at a fixed ratio of length. Example- 1 cm = 50 m, 1 cm = 180 m etc.

ENGINEER'S SCALE

2. REPRESENTATION FRACTION - A scale in which figures representing units (as centimeters, inches, or feet). Example- 1 cm = 50 m represent in Representation Factor = 1: 5000 or 1/5000.

3. GRAPHICAL SCALE - This is a scale which used to convert distances on the map to actual ground distances.

SURVEYING ERRORS

1. ERRORS DUE TO SHRINKAGE OF MAP-

Shrinkage Ratio=Shrinkage length/ Actual length.
Correction Length=Measured length/ shrinkage ratio.
Corrected Area=(Measured area/ Shrinkage ratio^2)

2. ERRORS DUE TO WRONG SCALE-

Corrected Length= ( Wrong Scale/Corrected Scale)* Measured Length.
Corrected Area= ( Wrong Scale/Corrected Scale)^2 * Measured Area.

SURVEYING CORRECTION

1. CORRECTION FOR STANDARDIZATION-

L= Actual length.

L'= Designated Length.

l = Measured Length.

When, L< L' Then, correction= negative(+ve) and Error = positive(-ve)

Correction Length = ( L- L')* (l/L)

2. CORRECTION FOR SLOPE-

C=(-d^2/ 2s) d= Slope Distance
Correction is always -Ve

CORRECTION DUE TO SLOPE

3. CORRECTION FOR PULL-

P>P' Correction +ve. P = Applied pull.
P<P' Correction -ve. P' = Standard pull.

C' = ( P- P')* ( L/AE) A = Cross section area of tape.

L = Measured Length.

E = Modulus of Elasticity.

4. CORRECTION FOR TEMPARTURE-

$C_{t}=C\cdot L(T_{m}-T_{s})$ $T_{m}$ > $T_{s}$ So, correction is +ve.

T_{m}

T_{s}

So, correction is -ve.

Where,

$C_{t}$ is the correction to be applied to the tape due to temperature.
C is the coefficient of thermal expansion of the metal that forms the tape.
L is the length of the tape or length of the line measured.
$T_{m}$ is the observed temperature of the tape at the time of measurement.
$T_{s}$ is the standard temperature, when the tape is at the correct length, often 20 °C.

5. CORRECTION FOR SAG-

$C_s = \frac {{\omega}^2 L^3}{24P^2}$

Where,

$C_{s}$ is the correction applied to the tape due to sag; meters.
$\omega$ is the weight of the tape per unit length; newtons per meter.
L is the length between the two ends of the catenary; meters.
P is the tension or pull applied to the tape; newtons.

IMPORTANT POINTS

Scale of Map = ( Distance between two points on Map)/ ( corresponding distance between those two point on ground)
When, Correction is + Ve then, Error will be -Ve.

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