Measurement of Horizontal Distances

the distance between two points means the horizontal distance

Methods of Linear Measurements

Pacing

Mechanical Devices

Taping*

Tachymetric

Photogrammetric*

Electronic Distance Measurement (EDM)*

Distance by Pacing

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 Pace (Toe to Toe)

1 Pace (Heel to Heel)

1 Stride (2 Paces or Double Step)

1 Stride (2 Paces or Double Step)

Distance by Mechanical Devices

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 & Measuring Wheel converts the number of revolutions

of a wheel of known circumference to a distance.

Distance by Photogrammetric

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 EDM

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 Tachymetric

Based on the optical geometry of the instruments employed; indirect method of measurement

1. Stadia Method ◦ Factors:

Refinement with which the instrument was manufactured

Skill of observer

Length of measurement

Effects of refraction

Distance by Tachymetric

1. Stadia Method

CKsD

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

α α/2

D (Horizontal Distance)

S/2

S/2

S

Theodolite or Transit

2-m Long Subtense Bar

Left Target Mark

Right Target Mark

D

STan

2/)2/(

)2/(

2/

Tan

SD

)2/()2/(

1

Cot

TanD

Since S=2.00m

D = horizontal distance α = angle subtended by the targets

Illustrative Problem

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.

Distance by Taping

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. Steel Tape b. Meter c. Marking Pins d. Clamp Handles e. Range Pole f. Plumb bobs

Taping Corrections

• Incorrect Tape Length

• Slope

• Temperature

• Pull (Tension)

• Sag

• Alignment

• Wind

Incorrect Tape Length

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.

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 (construction surveys)

A 100m 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 (construction surveys)

A 100m 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

Incorrect Tape Length

TL = actual length of tape Cl = total correction to be applied to

the measured length or length to be laid out

CL = corrected length of the line to be

measured or laid out ML = measured length or length to be

laid out NL = nominal length of tape

l

l

CMLCL

NL

MLCorrC

NLTLCorr

Incorrect Tape Length

Illustrative Problem

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.

Due to Slope

s

hCh

2

2

Gentle Slope (s < 20%)

3

42

82 s

h

s

hCh

Steep Slope (20% < s < 30%)

)cos1( sCh

Very Steep Slope (s > 30%)

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

hCsd

Illustrative Problem

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.

Due to Temperature

)( 12 TTkLCt

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

For steel: k = 0.0000116/°C Standard Temperature = 20 °C

A temperature higher or lower than the standard temperature causes a change in length

If T2 > T1, +Ct (too long); otherwise, - Ct

Illustrative Problem

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

Due to Pull / Tension

p

p

CLL

AE

LPPC

'

)( 12

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)

If Pm > Ps, too long; otherwise, too short

Illustrative Problem

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.

Due to Sag

• 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

2

2

2

32

24

24

P

LWC

P

LwC

s

s

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)

Illustrative Problem

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