Measurement of
Horizontal Distances
the distance between two points means
the horizontal distance
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Pacing
Mechanical Devices
Taping*
Tachymetric
Photogrammetric*
Electronic Distance Measurement (EDM)*
Methods of Linear Measurements

Pacing
◦ Consists of counting the number of steps, or
paces, in a required distance
 Length of a pace varies with different persons
Pace Factor =
Distance
Average No. of Steps
1 Stride
(2 Paces or Double Step)
1 Pace
(Heel to Heel)
1 Pace
(Toe to Toe)
1 Stride
(2 Paces or Double Step)
Distance by Pacing

This method can often be used to advantage on
preliminary surveys where precise distances are
not necessary

For low precision surveys or for quick measurements
 Odometer
Wheel
& Measuring
 converts the number of revolutions
of a wheel of known circumference
to a distance.
Distance by Mechanical Devices

Photogrammetry
◦ Measurement of images on a photograph
◦ Photographs taken from an aircraft with the
axis of the camera pointed vertically towards
the terrain photographed
Distance by Photogrammetric

These devices send out a beam of light or high-frequency
microwaves from one end of a line to be measured, and
directs it toward the far end of the line.

A reflector or transmitter-receiver at the far end reflects
the light of microwave back to the instrument where they
are analyzed electronically to give the distance between
the two points.
Distance by EDM
Distance by EDM

Based on the optical geometry of the
instruments employed; indirect method of
measurement
Stadia Method
1.
◦
Factors:




Refinement with which the instrument was manufactured
Skill of observer
Length of measurement
Effects of refraction
Distance by Tachymetric
1.
Stadia Method
D  Ks  C
D = horizontal distance
K = stadia interval factor of
the instrument
s = difference between the
upper and lower stadia hair
reading
C = stadia constant
Distance by Tachymetric
2.
Subtense Bar Method
2-m Long Subtense Bar
Left Target Mark
Theodolite or
Transit
S/2
α/2
α
S
S/2
Right Target Mark
D
(Horizontal Distance)
Tan( / 2) 
D
S /2
D
S /2
Tan( / 2)
Since S=2.00m
D
1
 Cot ( / 2)
Tan( / 2)
D = horizontal distance
α = angle subtended by the targets
Distance by Tachymetric
1.
A stadia rod held at a distant point B is sighted
by an instrument set-up at A. The upper and
lower stadia hair readings were observed as
1.50m and 0.80m, respectively. If the stadia
interval factor is 100, and the instrument
constant is 0, determine the length on line AB.
2.
The following subtended angles were read on a
2m long subtense bar using a transit: 0°55’45”,
and 0°10’50”. Compute the horizontal distance
from the transit to each position of the bar.
Illustrative Problem


Most common method of measuring or laying out
horizontal distances
Consists of stretching a calibrated tape between
two points and reading the distance indicated on
the tape
a.
b.
c.
d.
e.
f.
Distance by Taping
Steel Tape
Meter
Marking Pins
Clamp Handles
Range Pole
Plumb bobs
•
•
•
•
•
•
•
Incorrect Tape Length
Slope
Temperature
Pull (Tension)
Sag
Alignment
Wind
Taping Corrections
A systematic error occurs when incorrect
length of a tape is used.
 The true length of a tape can be obtained
by comparing it with a standard tape or
distance.
 An error caused by incorrect length of a
tape occurs each time the tape is used.

Incorrect Tape Length
Incorrect Tape Length
When measuring,
•If the tape is long, add the correction.
•If the tape is short, subtract the correction.
Incorrect Tape Length
Measured Distance
A
100m
B
AB is measured using 2 tape lengths
But, the tape length is actually 50.02m (Tape is too long)
So AB is actually: 2(50.02) = 100.04m
Must add a correction of 2(0.02) = 0.04m
Incorrect Tape Length
Measured Distance
A
100m
B
AB is measured using 2 tape lengths
But, the tape length is actually 49.98m (Tape is too short)
So AB is actually: 2(49.98) = 99.96m
Must subtract a correction of 2(0.02) = 0.04m
Incorrect Tape Length
When laying out,
•If the tape is long, subtract the correction.
•If the tape is short, add the correction.
Incorrect Tape Length
Layout Distance
A
100m
(construction surveys)
B
The distance between A and B must be 100m.
But, the tape length is actually 50.02m (Tape is too long)
2 tape applications: 2(50.02) = 100.04m
Must subtract a correction of 2(0.02) = 0.04m
Incorrect Tape Length
Layout Distance
A
100m
(construction surveys)
B
The distance between A and B must be 100m.
But, the tape length is actually 49.98m (Tape is too short)
2 tape applications: 2(49.98) = 99.96m
Must add a correction of 2(0.02) = 0.04m
Corr  TL  NL
ML
Cl  Corr 
NL
TL = actual length of tape
CL  ML  Cl
CL = corrected length of the line to be
measured or laid out
Cl = total correction to be applied to
the measured length or length to
be laid out
ML = measured length or length to be
laid out
NL = nominal length of tape
Incorrect Tape Length
1.
A rectangular lot was measured using a 50-m
steel tape which was found to be 0.025m too
short. If the recorded length and width of the
lot are 180.455m and 127.062m, respectively,
determine the following:
a. Actual dimension of the lot.
b. Error in area introduced due to the
erroneous length of tape.
Illustrative Problem
Gentle Slope
(s < 20%)
Ch
d  s  Ch
h2

2s
Steep Slope
(20% < s < 30%)
Ch
h2
h4


2s
8s 3
Very Steep Slope
(s > 30%)
Ch  s(1  cos  )
Due to Slope
s = measured slope distance
between points A and B
h = difference in elevation between A
and B
d = equivalent horizontal distance
AC
Ch = slope correction
2.
Slope distance AB and BC measures 300.50m
and 650.01m, respectively. The differences in
elevation are 15.00m for point A and B, and
20.05m for point B and C. using the
approximate slope correction formula for gentle
slopes, determine the horizontal length of line
ABC. Assume that line AB has a rising slope
and BC a falling slope.
Illustrative Problem
Ct  kL(T2  T1 )
For steel:
k = 0.0000116/°C
Standard Temperature = 20 °C
k= coefficient of linear expansion or
the amount of change on length per
unit length per degree change in
temperature
L = length of the tape or length of
line measured
T2 = observed temperature of the
tape at the time of measurement
T1 = temperature at which the tape
was standardized
A temperature higher or lower than the standard
temperature causes a change in length
If T2 > T1, +Ct (too long);
otherwise, - Ct
Due to Temperature
3.
A steel tape with a coefficient of linear
expansion of 0.0000116/°C is known to be 50m
long at 20°C. The tape was used to measure a
line which was found to be 656.29m long when
the temperature was 40°C. Determine the
following:
a. Temperature correction per tape length
b. Temperature correction for the measured line
c. Correct length of the line
Illustrative Problem
Cp
( P2  P1 ) L

AE
L'  L  C p
If Pm > Ps, too long;
otherwise, too short
Cp= total elongation in tape due to
pull or the correction due to
incorrect pull applied on the
tape (m)
P2 = pull applied to tape the during
measurement (kg)
P1 = standard pull (kg)
L = measured length of line
A = cross-sectional area of the tape
(sq. cm)
E = modulus of elasticity (kg/cm2)
L’= corrected length of the measured
line (m)
Due to Pull / Tension
4.
A heavy 50-m tape having a cross-sectional
area of 0.05cm2 has been standardized at a
tension of 6.0kg. If E=2.10x106 kg/cm2,
determine the elongation of the tape if a pull of
15kg is applied.
Illustrative Problem
• A steel tape not supported along its entire length
sags in the form of a catenary curve
• Because of sag the horizontal distance is less
than the graduated distance between tape ends
• Sag can be reduced by applying great tension,
but not eliminated unless the tape is supported
throughout
Due to Sag
w 2 L3
Cs 
24 P 2
W 2L
Cs 
24 P 2
Cs= correction due to sag or the
difference between the tape
reading and the horizontal
distance between supports (m)
w = weight tape per unit length
(kg/m)
W = total weight of tape between
supports (kg)
L = interval between supports or the
unsupported length of tape (m)
P = tension or pull applied on the
tape (kg)
Due to Sag
5.
A 50-m steel tape weighing 0.035kg/m is
constantly supported at mid-length and its end
points, and is used to measure a line AB with a
steady pull of 6.5kg. If the measured length of
AB is 1200.00m, determine the following:
a. Correction due to sag between supports and
for the whole tape length
b. Total sag correction for the whole length
measured
c. Correct length of line AB
Illustrative Problem