Chapter 2 - Measurement of Horizontal Distances Illustrative Problem - Taping PDF

Chapter 2: Measurement of Horizontal Distances The accurate determination of the distance between two points on any surface is one of the basic operations in plane surveying. If the points are at different elevations, the distance is the horizontal length between plumb lines at the point...

Chapter 2: Measurement of Horizontal Distances The accurate determination of the distance between two points on any surface is one of the basic operations in plane surveying. If the points are at different elevations, the distance is the horizontal length between plumb lines at the points. There are several methods of determining distances. In surveying, the commonly employed methods include pacing, taping, tachymetric, graphical, mathematical, mechanical, photogrammetric, and electronic distance measurement. Pacing Pace is defined as the length of a step in walking. It may be measured from heel to heel or from toe to toe. Pacing. In surveying, pacing means moving with measured steps; and if steps are counted, distances can be determined if the length of a step is known. Pacing To pace a distance, it is necessary to first determine the length of one’s pace. This is referred to as the pace factor. Methods: Determine the average length of an individual’s normal step, Adjust one’s pace to some predetermined length such as 1 m. Length of pace varies with different persons. A stride is equivalent to two paces or a double step. DISTANCE BY PACING Problem: A 45-m course, AB, on level ground was paced by a surveyor for the purpose of determining his pace factor. The number of paces for each trial taken are shown in theTRIAL accompanying LINE tabulations. TAPED NO. OF DISTANCE PACES 1 AB 50 2 BA 53 3 AB 51 45.00 4 BA 53 5 AB 52 6 BA 53 DISTANCE BY PACING Requirements: Determine his pace factor. If the surveyor then took 771, 770, 768, 770, 772, and 769 paces in walking an unknown distance CD, what is the length of the line? Assuming that the taped length of line CD is 667.0, determine the relative precision of the measurement performed. DISTANCE BY PACING Solutions: A. Determine the pace B. Determining Unknown Distance L =factor 45 m (length of line AB) n = 6 2 (no. of n1 = 6 (no. of trials) trials) Sum2 = 771 + 770 + 768 + 770 Sum1 = 50 + 53 + 51 + 53 + 52 + 53 + 772 + 769= 4620 paces = M1312 paces/ n = 312/6 = 52 = Sum M2 = Sum2 / n2 = 4620/6 = 1 1 paces 770 paces PF = L / M1 PF = PD / M2 PF = 45.00m/52 PD = PF (M2) paces PF = 0.865 m/pace PD = 0.865 m/pace x 770 paces PD = 666.1 m (paced length of line CD) DISTANCE BY PACING Solutions: C. Relative Precision RP = (TD – PD) / TD (taped distance) = 667.0 TD m RP = (667.0 – 666.1) / PD (paced distance) = 666.1 m 667 RP = 0.9 / 667 RP = RP = 1 / 741 DISTANCE BY PACING In five trials of walking along a 90-m course on fairly level ground, Stefano, a pacer, counted 51, 52.5, 51.5, 52.5 and 51.5 strides respectively. He then started walking an unknown distance XY in four trials which were recorded as follows: 88.5, 89, 88, and 87 strides. Determine the following: 1. Pace factor of Stefano. 2. Length of line XY. 3. Percentage of error in the measurement if the taped length of XY is 150.5 meters. DISTANCE BY PACING B. Determining Unknown Distance n =4 (no. of trials) 2 Solutions: Sum2 = 88.5 + 89 + 88 + 87 A. Determine the pace = 352.5 strides or 705 L =factor 90 m (length of course) paces M2 = Sum2 / n2 = 705/4 = n1 = 5 (no. of trials 176.25 paces PD = PF (M2) taken) Sum1 = 51 + 52.5 + 51.5 + 52.5 PD = 0.869 m/pace x 176.25 +51.5 paces PD = 153.2 m (paced length = 259 strides or 518 paces M1 = Sum1 / n1 = 518/5 = 103.6 C. of line XY) Determining Percentage paces PF = L / M1 of Error TD (taped distance) = PF = 90.00m/ 103.6 PD 150.50 (paced m distance) = paces 153.20 m PF = 0.869 m/pace PE = X 100% PE = EXERCISE (15 PTS) A line 100 m long was paced by a surveyor for four times with the following data: 142, 145, 145.5, and 146. Another line was paced four times again with the following results: 893, 893.5, 891, and 895.5 1. Determine the pace factor. 2. Determine the average number of paces for the new line. 3. Determine the distance of the new line. DISTANCE BY PACING Solutions: A. Determine the pace B. Determining Unknown Distance L =factor 100 m (length of line) n = 4 2 (no. of n1 = 4 (no. of trials) trials) Sum2 = 893 + 893.5 + 891 + Sum1 = 142 + 145 + 145.5 + 146= 895.5 578.5 M = paces Sum / n = 578.5/4 = M2 = Sum 2 / npaces = 3573 2 = 3573/4 = 1 1 1 144.625 paces 893.25 PF = L / M1 PF = PD / M2 PF = 100.00m/ 144.625 PD = PF (M2) paces PF = 0.691 m/pace PD = 0.691 m/pace x 893.25 paces PD = 617.632 m (paced length of the new line DISTANCE BY TAPING The use of graduated tape is probably the most common method of measuring of laying out horizontal distances. Taping consists of stretching a calibrated tape between two points and reading the distance indicated on the tape. Taping is a form of direct measurement which is widely used in engineering as well as non- engineering activities. MEASURING TAPES STEEL TAPES MEASURING TAPES INVAR TAPES MEASURING TAPES 1. Steel Tape- surveyor’s or engineer’s tape, made of ribbon of steel 0.5 to 1.0 cm in width, and weighs 0.8 to 1.5 kg per 30 meters. Lengths of 10, 20, 30, 50, 100 m are available. The 30- m tape is most common. It is designed for most conventional measurements in surveying and construction work. 2. Metallic Tape 3. Non-Metallic 4. Invar Tape- a special tape made of an alloy of nickel (35%) and steel (65%) with a very low coefficient of thermal expansion which makes it less affected by temperature changes. 5. Lovar Tape 6. Fiberglass Tape 7. Wires 8. Builder’s Tape 9. Phosphor-Bronze Tape PROCEDURES IN TAPING Aligning the Tape Stretching the Tape Plumbing Marking Full Tape Lengths Tallying Taped Measurements Measuring Fractional Lengths BREAKING THE TAPE A standard practice of measuring shorter distances which are accumulated to total a full tape length. This procedure is always done especially when the course is sloping or has an uneven terrain surface.  CORRECTIONS IN TAPING  SLOPE TAPING where d = horizontal distance between the two points, s = measured slope length between the points, and α is the angle of inclination from d = s cos α the horizontal. If the difference in elevation, d= h, is known, the horizontal distance is computed using the following expression SLOPE TAPING A measurement is made along a line that is inclined by a vertical angle of 15ᵒ25’. The slope measurement is 756.52 m. What is the corresponding horizontal distance? Given Solution: : Inclination Angle (α) = s 15ᵒ25’ Horizontal Distance = α 756.52 m d Required: Horizontal Distance d = s Cos α d = 756.52 Cos 15ᵒ25’ D = 729.30 m SLOPE TAPING A horizontal distance of 325.75 m is to be established along a line that slopes at a vertical angle of 13ᵒ06’. What slope distance should be laid out? Given Solution: : Inclination Angle (α) = s 13ᵒ06’ Horizontal Distance = α 325.75 m d Required: d = s Cos α Slope Distance s = d / Cos α s = 325.75 / Cos 13ᵒ06’ s = 334.45 m SLOPE TAPING A line XYZ is measured on the slope in two segments. The first segment XY measures 824.45 m and the second segment YZ measures 1244.38m. If the difference of elevation between two points X and Y is 4.25 m and that between Y and Z is 6.47m, determine the horizontal length of the measured line. Given Required: : S = 824.45 m (segment a. Horizontal length of 1 XY) S2 = 1244.38 m segment XY length of b. Horizontal (segment h1 = 4.25 YZ) m (diff in elev. of point segment YZ length of line c. Horizontal X h2&Y) = 6.47 m (diff in elev. of point XYZ X &Y) Solution: Z s 2 =.38 m h2 = 6.47 44 m Y 12 D2 4.45 2 s1=8 h1 = 4.25 X m D1 m D a. Compute horizontal length of b. Compute horizontal length of segment XY (d1): segment YZ (d2): D1 = D2 = = D2 = D2 = D1 = b. Total Length (Horizontal) D1 = D1 + D2 = 824.44 +1244.36 = 2068.68 m MEASUREMENTS WITH The lengthTAPE of a line AB measured with a 50-m tape is 465.285 m. When the tape is compared with a standardized invar tape it is found to be 0.016 m too long in almost the same conditions of support, tension, and temperature that existed during measurement of the line. Determine the correct length of AB. Given : NL = 50 m (nominal or indicated length of tape used) ML = 465.285 m (measured length of line CorrAB) = 0.016 m (correction per tape length, tape too long) Required: Correct length of AB MEASUREMENTS WITH Solution: TAPE 𝐶1 𝐶𝑜𝑟𝑟 ; 𝐶 =𝐶𝑜𝑟𝑟 ( 𝑀𝐿 )¿ 0.016 ( 465.285 ) = 1 𝑀𝐿 𝑁𝐿 𝑁𝐿 50 ¿+ 0.149𝑚 (Total correction to be applied to measured CL length of line AB) CL (𝑎𝑑𝑑 𝑠𝑖𝑛𝑐𝑒𝑡𝑎𝑝𝑒𝑖𝑠𝑡𝑜𝑜 𝑙𝑜𝑛𝑔 ,𝑚𝑒𝑎𝑠𝑢𝑟𝑒) CL (𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡h𝑜𝑓 𝑙𝑖𝑛𝑒 𝐴𝐵 ) Ref: Jonas’ Class Notes https://www.youtube.com/watch?v=yP0Xi2vUcBM Correction due to Slope 2 a. Gentle Slope h 𝐶h= 2𝑠 2 4 b. Steep Slopes (between 20% h h 𝐶h= + and 30%) 2𝑠 8 𝑠3 c. Very Steep Slopes (greater 𝐶 h =𝑠(1 − cos ∅) than 30%) d. Equivalent Horizontal 𝑑=𝑠 − 𝐶 h 𝑵𝒐𝒕𝒆 Distance : 𝒎= 𝐭𝐚𝐧 𝜽 Sample Problem Slope distances AB and BC measures 330.49m and 660.97 m, respectively. The differences in elevation are 12.22 m for points A and B, and 10.85 m for points 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. Given : S = 330.49 m (slope length of AB) 1 S2 = 660.97 m (slope length of BC) h1 = 12.22 m (diff in elev. between h A2&=B) 10.85 m (diff in elev. between B & C) Required: Horizontal length of ABC Solution: B S2 S1 C d2 A d1 D = d1 + d2 d1 = s1 - = 330.49 - = 330.26 m (horizontal length of d2 = s2 -AB)= 660.97 - = 660.88 m (horizontal length of BC) D = d1 + d2 D = 330.26 + 660.88 = Ref: Jonas’ Class Notes https://www.youtube.com/watch?v=GfVbASs0SWg Correction due to Temperature The value of Ts is usually taken as 20°C. If the tape used is made of steel, the value of c is 0.0000116 / ᵒC. The resulting sign of Ct will be either positive or negative and is added algebraically to the length measured to obtain the correct length. As a general rule, to obtain more accurate measured values, taping should be undertaken when the temperature does not vary significantly from the temperature used during standardization Sample Problem 1.A steel tape with a coefficient of linear expansion of 0.0000116 / ᵒC is known to be 50m long at 20ᵒC. The tape was used to measure a line which was found to be 532.28 meters long when the temperature was 35ᵒC. Determine the following: a. Temperature correction per tape length b. Temperature correction for the measured line c. Correct length of the line Give n: Coefficient of thermal expansion: 0.0000116 / ᵒC Standard temperature = 20 ᵒC Temperature at instant = 35ᵒC Tape length = 50 m Length of the line = 532.28 m Required:a. Temperature correction per tape length b. Temperature correction for the measured line c. Correct length of the line Solution: a. Temperature correction per tape length Ct = (0.0000116 / ᵒC) (35 ᵒC - 20 ᵒC) (50m) = (0.0000116 / ᵒC) (15ᵒC) (50m) Ct = +0.0087 m (tape is too long) b. Temperature correction for the measured C ’ line = (0.0000116 / ᵒC) (35 ᵒC - 20 ᵒC) (532.28m) t = (0.0000116 / ᵒC) (15ᵒC) (532.28m) Ct’ = +0.0926 m (tape is too c. long)length of the line Correct L’ = L Ct’ = 532.28 + 0.0926 = Sample Problem 2. A steel tape, known to be of standard length at 20ᵒC, is used in laying out a runway 2500 m long. If its coefficient of linear expansion is 0.0000116/ᵒC, determine the temperature correction and the correct length to be laid out when the temperature is 42ᵒC. Give Coefficient of thermal expansion: 0.0000116 / ᵒC n: Standard temperature = 20 ᵒC Temperature at instant = 42ᵒC Length of Runway= 2500 m Requir Temperature correction | Correct length to laid out ed: Solution: a. Temperature correction for the runway to lay out Ct’ = (0.0000116 / ᵒC) (42 ᵒC - 20 ᵒC) (2500 m) = (0.0000116 / ᵒC) (15ᵒC) (532.28m) Ct’ = +0.638 m (tape is too long) b. Correct length of the runway to be laid out L’ = L Ct’ = 2500 - 0.638 = ,subtract since tape is too long 2,499.36 m  Correction due to Pull/Tension During calibration, a tape is subjected to a certain amount of standard pull or tension on its ends. If the pull is greater than that for which it was calibrated, the tape elongates and become too long. Correspondingly, it will stretch less than its standard length when an insufficient pull is applied thus making it too short. Errors due to variation of pull may be either random or systematic. It can be eliminated by using a spring balance, to measure and maintain the standard pull. Ref: Jonas’ Class Notes https://www.youtube.com/watch?v=b5pZZH_Czng Ref: https://www.youtube.com/watch?v=b5pZZH_Czng Jonas’ Class Notes Sample Problem 1.A heavy 50-m tape having a cross-sectional area of 0.05 cm2 has been standardized at a tension of 5.5 kg. If E = 2.10 x 106 kg/cm2, determine the elongation of the tape if a pull of 12 kg is applied. Given Pm = 12 kg Required Ps = 5.5 kg : Elongation of the L = 50 meters tape/correction A = 0.05 cm2 E = 2.10 x 106 kg/cm2  Solution: Elongation of the Tape ( 𝑃𝑚 − 𝑃𝑠 ) 𝐿 𝐶𝑝= 𝐴𝐸 ( 12 𝑘𝑔 − 5.5 𝑘𝑔 ) 50 𝑚 𝐶𝑝= 6 2 10 𝑘𝑔 ( 0.05 𝑐𝑚 )(2.0 𝑥 2 ) 𝑐𝑚 𝑪 𝒑=+ 𝟎.𝟎𝟎𝟑𝟐𝟓 𝒎 Sample Problem 2. A 30-m steel tape weighing 1.45 kg is of standard length under a pull of 5 kg, supported for full length. The tape was used in measuring a line 938.55 m long on smooth level ground under a steady pull of 10 kg. Assuming E = 2.0 x 106 kg/cm2 and the unit weight of steel to be 7.9 x 10-3 kg/cm3, determine the a. Cross-sectional area of the tape following: b. Correction for increase in tension c. Correct length of the line measured Given Pm = 10 kg Required Ps = 5kg : a. Cross-sectional area of the L = 30 meters tape D = 938.55 m b. Correction for increase in W = 1.45 kg tension γ = 7.9 x 10 kg/cm -3 3 2c. Correct length of the line E = 2.10 x 10 kg/cm 6 measured Solution a. Cross-sectional area of the tape 𝑊 1.45 𝑘𝑔 𝐴= 𝐴= 7.9 x 10−3 kg 𝑐𝑚 3 )(30 m ) 100𝑐𝑚 γ𝐿 (/ 𝑐𝑚 3 1𝑚 Solution: a. Cross-sectional area of the tape 𝑊 𝐴= γ𝐿 1.45 𝑘𝑔 𝐴= 7.9 𝑥 10 − 3 𝑘𝑔 100 𝑐𝑚 ( )(30 𝑚)( ) 𝑐𝑚 3 1𝑚 2 𝑨=+ 𝟎. 𝟎𝟔𝟏𝟐 𝑐𝑚 Solution: b. Correction for increase in tension ( 𝑃𝑚 − 𝑃𝑠 ) 𝐿 𝑀𝐿 𝐶𝑝= 𝐶 𝑝 1=𝐶 𝑝 ( ) 𝐴𝐸 𝑁𝐿 938.55 𝑚 ( 10 𝑘𝑔 − 5 𝑘𝑔 ) 30 𝑚 𝐶 𝑝 1=0.00122 ( ) 𝐶𝑝= 6 30 𝑚 2 10 𝑘𝑔 ( 0.0612 𝑐𝑚 ) (2.0 𝑥 2 ) 𝑪 𝒑 𝟏=𝟎. 𝟎𝟑𝟖 𝒎 𝑐𝑚 Tape is too long 𝑪 𝒑=+ 𝟎.𝟎𝟎𝟏𝟐𝟐 𝒎 (correction per tape length) Solution: c. Correct length of the line L’ = L Ct’ = 938.55 + 0.038 = 938.588 m / 938.59 m Exercise 1. A steel tape having a correct length at 22 ᵒC was used to measure a baseline and the recording readings gave the total of 856.815 m. If the average temperature during the measurement was 18 ᵒC, determine the correct length of the line. (5 points) 2. A steel tape is 30-m long under a pull of 6.0 kg when supported throughout. It has a cross- sectional area of 0.035 cm2 and is applied fully supported with a 12-kg pull to measure a line whose recorded length is 308.32 m. Determine the Solution #Prob1: a. Temperature correction of the baseline measured Ct’ = (0.0000116 / ᵒC) (18 ᵒC - 22 ᵒC) (856.815 m) = (0.0000116 / ᵒC) (-4ᵒC) (856.815m) Ct’ = -0.039756 m (tape is too short) b. Correct length of the baseline to measure L’ = L Ct’ = 856.815 - 0.039756 ,subtract since tape is too short = 856.775 m (measure)  Solution #Prob2: Elongation of the Tape ( 𝑃𝑚 − 𝑃𝑠 ) 𝐿 𝑀𝐿 𝐶𝑝= 𝐶 𝑝 1=𝐶 𝑝 ( ) 𝐴𝐸 𝑁𝐿 308.32 𝑚 ( 12 𝑘𝑔− 6 𝑘𝑔 ) 30 𝑚 𝐶 𝑝 1=0.002445 ( ) 𝐶𝑝= 30 𝑚 6 2 10 𝑘𝑔 ( 0.035𝑐𝑚 )(2.1 𝑥 2 ) 𝑪 𝒑 𝟏=𝟎. 𝟎𝟐𝟓𝟐 𝒎 𝑐𝑚 𝑪 𝒑=+ 𝟎.𝟎𝟎𝟐𝟒𝟒𝟓 𝒎 c. Correct length of the line L’ = L Ct’ = 308.32 + 0.0252 = 308.34 m Sample Problem 1. A 50-m steel tape weighs 0.04 kg/m and is supported at its end points and at the 8-m and 25-m marks. If a pull of 6 kg is applied, determine the following: Correction due to sag between the 0-m and 8-m marks, 8-m and 25-m marks, and the 25-m and 50-m marks. Correction due to sag for one tape length Correct distance between the ends of the tape Sample Problem Given L = 50 m (total length of the tape) (length of the 1st L1 = 8 m span) (length of the 2nd L2 = 17 m span) (length of the 3rd L3 = 25 m span) (pull applied on ends of P = 6 kg tape) (unit weight of tape) w = 0.04 kg/m  𝜔 2 𝐿23 𝐶 𝑠 2= Solution: 24 𝑃 2 2 3 ( 0.04) (17) A. Determining Correction 𝐶 𝑠 2= 2 Due to Sag for each Span 24 (6) 2 𝜔 𝐿 3 𝑪 𝑺 𝟐 =+𝟎. 𝟎𝟎𝟗𝟏 𝒎 1 𝐶 𝑠 1= 2 (Correction due to sag between 8- 24 𝑃 25m mark) 𝜔 𝐿2 3 3 2 ( 0.04) (8) 3 𝐶 𝑠 3= 2 𝐶 𝑠 1= 24 𝑃 2 24 (6) 2 ( 0.04) (25) 3 𝐶 𝑠 3= 𝑪 𝑺 𝟏 =+𝟎. 𝟎𝟎𝟎𝟗 𝒎 24 (6) 2 (Correction due to sag between 0-8m mark) 𝑪 𝑺 𝟑 =+𝟎. 𝟎𝟐𝟖𝟗 𝒎 (Correction due to sag between 25- Solution: c. Correct length of the line b. Determining Total Sag L’ = L Ct’ Correction for One Tape Length = 50 - 0.0389 + = 49.961 m + 0.0289 𝑪 𝑺=+ 𝟎.𝟎𝟑𝟖𝟗 𝒎