Measurement of Horizontal Distances. the distance between two points means the horizontal distance

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1 Measurement of Horizontal Distances the distance between two points means the horizontal distance

2 Pacing Mechanical Devices Taping * Tachymetric Photogrammetric * Electronic Distance Measurement (EDM) * Methods of Linear Measurements

3 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

4 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 Mechanical Devices

5 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.

6 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

7 Distance by EDM

8 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

9 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

10 2. Subtense Bar Method 2-m Long Subtense Bar Left Target Mark Theodolite or Transit α α/2 S/2 S S/2 Right Target Mark D (Horizontal Distance) Tan( / 2) S / 2 D S / 2 D Tan( / 2) Since S=2.00m D 1 Cot( / 2) Tan( / 2) D = horizontal distance α = angle subtended by the targets Distance by Tachymetric

11 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: , and Compute the horizontal distance from the transit to each position of the bar. Illustrative Problem

12 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 Distance by Taping

13 Incorrect Tape Length Slope Temperature Pull (Tension) Sag Alignment Wind Taping Corrections

14 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

15 Incorrect Tape Length When measuring, If the tape is long, add the correction. If the tape is short, subtract the correction.

16 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) = m Must add a correction of 2(0.02) = 0.04m

17 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

18 Incorrect Tape Length When laying out, If the tape is long, subtract the correction. If the tape is short, add the correction.

19 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) = m Must subtract a correction of 2(0.02) = 0.04m

20 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

21 Corr TL NL C l Corr CL ML ML NL C l TL = actual length of tape C l = 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 Incorrect Tape Length

22 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 m and m, 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%) d s C h C h 2 h C h 2s

B h = difference in elevation between A and B d =

23 Gentle Slope (s < 20%) d s C h C h 2 h C h 2s Steep Slope (20% < s < 30%) C h 2 h h 2s 8s Very Steep Slope (s > 30%) Due to Slope 4 3 s( 1 cos ) s = measured slope distance between points A and B h = difference in elevation between A and B d = equivalent horizontal distance AC C h = slope correction

24 2. Slope distance AB and BC measures m and m, 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

C t kl( T ) 2 T1 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 For steel: k =

25 C t kl( T ) 2 T1 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 For steel: k = / C Standard Temperature = 20 C T 2 = observed temperature of the tape at the time of measurement T 1 = temperature at which the tape was standardized A temperature higher or lower than the standard temperature causes a change in length If T 2 > T 1, +C t (too long); otherwise, - C t Due to Temperature

26 3. A steel tape with a coefficient of linear expansion of / C is known to be 50m long at 20 C. The tape was used to measure a line which was found to be m 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

27 C p ( P2 P1) L AE C p = total elongation in tape due to pull or the correction due to incorrect pull applied on the tape (m) P 2 = pull applied to tape the during measurement (kg) L' L C p P 1 = standard pull (kg) L = measured length of line If P m > P s, too long; otherwise, too short A = cross-sectional area of the tape (sq. cm) E = modulus of elasticity (kg/cm 2 ) L = corrected length of the measured line (m) Due to Pull / Tension

28 4. A heavy 50-m tape having a cross-sectional area of 0.05cm 2 has been standardized at a tension of 6.0kg. If E=2.10x10 6 kg/cm 2, determine the elongation of the tape if a pull of 15kg is applied. Illustrative Problem

29 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

30 C s 2 w L 24P 3 2 C s = 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) C s 2 W L 2 24P 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

31 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 m, 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

GEOMATICS ENGINEERING

GEOMATICS ENGINEERING CHAPTER 2 Direct and Indirect Distance Measurement Methods Distance Measurement Methods Equipment Classification Usage Fieldwork procedure Booking system Adjustment and plotting Introduction