Lecture 3 survey method on distance calculation.pdf
- 2. CHAIN SURVEYING ➢ The principle of chain surveying is triangulation ➢ In triangulation, the area to be surveyed is divided into number of small triangles which should be well conditioned ➢ In chain surveying the sides of the triangles are directly measured on the field using chain or tape, and no angular measurements are taken ➢ It is usually recommended when the area to be surveyed is small and ground surface is more or less leveled
- 3. WELL-CONDITIONED AND ILL-CONDITIONED TRIANGLES 60O 60O 60O 25O 75O 90O All anlges are between 30o and 120o An angle is less than 30o or more than 120o WELL-CONDITIONED ILL-CONDITIONED
- 4. TESTING OF A CHAIN ➢ Due to continuous use, a chain may be elongated or shortened ➢ So, the chain should be tested and adjusted accordingly ➢ If full adjustment is not possible, then the amount of shortening (known as “TOO SHORT”) and elongation (known as “TOO LONG”) should be noted clearly for necessary correction applicable to the chain ➢ For testing the chain, a test gauge is established on a level platform with the help of a standard steel tape
- 5. ➢ The steel tape is standardized at 20 oC and under a tension of 8 kg ➢ The test gauge consists of two pegs having nails at the top and fixed on level platform a required distance apart (20 or 30 m) ➢ The incorrect chain is fully stretched by pulling it under normal tension (8 kg) along the test gauge ➢ If the length of the chain does not tally with the standard length, then an attempt should be made to rectify the error ➢ Finally, the amount of elongation or shortening should be recorded
- 6. TEST GAUGE 20 m or 30 m Allowable error for 20 m chain = 5 mm Allowable error for 30 m chain = 8 mm
- 7. ADJUSTMENT OF CHAIN ➢ When the chain is too long, it is adjusted by; 1. Closing up the opened joints of the rings 2. Hammering the elongated rings 3. Replacing some old rings by new rings ➢ When the chain is too short, it is adjusted by; 1. Straightening the bent links 2. Opening the joints of the rings 3. Replacing the old rings by some larger rings
- 8. SCALE FOR MAPS ➢ Scales are generally represented in terms of Representative Factor (R.F.) or scale factor ➢ If 1 cm on map represents 10 m on ground then R.F. is given as 10 𝑚 = 1 𝑐𝑚 10000 𝑚𝑚 = 10 𝑚𝑚 𝑅. 𝐹. = 10 10000 = 1 1000 ➢ A map having R.F. of less than “1 is to 500” is considered to be a large-scale map, otherwise its is considered to be a small-scale map (Read as “1 is to 1000” or “1 ratio 1000”)
- 9. CHAINING ON LEVEL GROUND ➢ To chain the line, the leader moves forward by dragging the chain and by taking with him a ranging rod and arrows. ➢ The follower stands at the starting station by holding the other end of the chain ➢ When the chain is fully extended, Leader holds the ranging rod and the ranging process is done with the help of directions given by the follower ➢ Then the follower holds the zero end of the chain by touching the station point
- 10. ➢ The leader stretches the chain and finally place it on the line. Once done he inserts an arrow in the ground at the end of the chain ➢ Now the leader moves forward by dragging the chain with arrows and the ranging rod and the whole process is repeated again but this time there should be a surveyor on the starting station to conduct the ranging operation
- 12. TAPE CORRECTIONS 1. TEMPERATURE CORRECTION ➢ This correction is necessary because the length of the tape or chain may be increased or decrease due to rise or fall of temperature, respectively, during measurement ➢ The sign of correction may be positive or negative ➢ Coefficient of thermal expansion for the steel tape can be assumed to be 11 x 10-6 per degree centigrade, if not known
- 14. Where, Ct Correction for temperature (m) Tm Temperature during measurement (oC) To Temperature at which tape was standardized (oC) L Length of the tape (m) Coefficient of thermal expansion (oC-1) 𝐶𝑡 = 𝛼 𝑇𝑚 − 𝑇𝑜 𝐿
- 15. 2. PULL CORRECTION ➢ Due to the elastic properties of the tape material, when pull is applied, the strain will vary according to the variation of applied pull, and hence necessary correction should be applied ➢ The sign of correction may be positive or negative ➢ Modulus of elasticity of the tape may be assumed to be 2.1 x 106 kg/cm2, if not known
- 18. Where, Cp Pull Correction (m) Pm Pull applied during measurement (kg) Po Pull at which tape was standardized (kg) L Length of the tape (m) E Modulus of elasticity (kg/cm2) A Cross-sectional area of tape (cm2) 𝐶𝑝 = 𝑃𝑚 − 𝑃𝑜 𝐿 𝐴 𝐸
- 19. 3. SLOPE CORRECTION 𝐶ℎ = 𝐿 − 𝐿 cos 𝜃 Where, Ch Slope correction (m) h Vertical distance between two points (m) 𝐶ℎ = 𝐿 − 𝐿2 − ℎ2 𝐶ℎ = ℎ2 2𝐿 h L Ch (always negative) A B C (APPROXIMATE) (EXACT) (EXACT)
- 20. 4. SAG CORRECTION ➢ This correction is necessary when the measurement is taken with tape in suspension (always negative) 𝐶𝑠 = 𝐿 𝑤𝐿 2 24 𝑛2 𝑃𝑚 2 = 𝐿𝑊2 24 𝑛2 𝑃𝑚 2 Where, Cs Correction for sag (m) Pm Pull applied during measurement (kg) n number of spans w weight of tape per unit length (Kg/m) W Total weight of the tape (kg)
- 22. 5. NORMAL TENSION ➢ The tensions at which the effect of pull is neutralized by the effect of sag is known as normal tension ➢ At this the elongation due to pull is balanced by the shortening due to sag. Mathematically, 𝐶𝑠 = 𝐶𝑝 𝐿𝑊2 24 𝑛2 𝑃𝑚 2 = 𝑃𝑚 − 𝑃𝑜 𝐿 𝐴 𝐸
- 23. 𝑊2 24 𝑃𝑛 2 = 𝑃𝑛 − 𝑃𝑜 𝐴 𝐸 Let, Pn be the Normal tension or pull, then (Considering n = 1) 𝑊2 𝐴 𝐸 24 = 𝑃𝑛 2 𝑃𝑛 − 𝑃𝑜 By substituting the values of Po, W, A and E, we get the final equation of the following form, 𝑥𝑃𝑛 3 ± 𝑦𝑃𝑛 2 ± 𝐶 = 0 Pn is determined by trial and error method
- 24. CHAIN CORRECTIONS 1. CORRECTION FOR INCORRECT LENGTH 𝐿𝑡 = 𝐿′ 𝐿 𝐿𝑚 Where, L Standard/ true length of chain (m) Lm Measured length of the line (m) Lt True length of the line (m) L’ True length error (m) (Positive when chain is too long and negative when chain is too short)
- 25. 2. CORRECTION FOR INCORRECT AREA 𝐴𝑡 = 𝐿′ 𝐿 2 𝐴𝑚 Where, L Standard/ true length of chain (m) L’ True length error (m) Am Measured length of the line (m2) At True area (m2)
- 26. Numerical 1 A 20 m chain was used to measure the distance between two stations, which was found to be 2500 m. Then same distance when measured with a 30 m chain was found to be 2495 m. If the 20 m chain was 0.07 m too long, what was the error in the 30 m chain? 𝐻𝑖𝑛𝑡: 𝐿𝑡 = 𝐿′ 𝐿 𝐿𝑚 (Ans. 0.17 m)
- 27. Numerical 2 A line was measured by a 20 m chain which was accurate before starting the day’s work. After chaining 900 m, the chain was found to be 6 cm too long. After chaining a total distance of 1, 575 m, the chain was found to be 14 cm too long. Find the true distance of the line. 𝐻𝑖𝑛𝑡: 𝐿𝑡 = 𝐿′ 𝐿 𝐿𝑚 (Ans. 1,579.72 m)
- 28. Numerical 3 An old map was plotted to a scale of 40 m to 1 cm. Over the years, this map has been shrinking, and a line originally 20 cm long is only 19.5 cm long at present. Again the 20 m chain was 5 cm too long. If the present area of the map measured by planimeter is 125.50 cm2, find true area of the land surveyed. 𝐻𝑖𝑛𝑡: 𝐴𝑡 = 𝐿′ 𝐿 2 𝐴𝑚 (Ans. 21.23 hectares)
- 29. CONCLUDED