CIVIL ENGINEERING A03 SURVEYING AND MEASUREMENT TERM TEST, Fall 011 Page 1 of 6 FACULTY OF ENGINEERING McMaster University Hamilton, Ontario, Canada CIVIL ENGINEERING A03 SURVEYING AND MEAUREMENT TERM TEST (Day Class) Instructor: Dr. Dean Inglis DURATION OF TEST: 50 Minutes October 011 This examination paper includes 5 pages and 1 questions. You are responsible for ensuring that your copy of the paper is complete. Bring any discrepancy to the attention of your invigilator. SPECIAL INSTRUCTIONS: ONLY McMaster Standard Calculator (Casio fx-991) will be permitted. Attempt all questions. All answers should be in logical sequence with clear explanations. Assume a reasonable value for any parameter if you consider that the corresponding value is not given herein. Write all rough work and answers on the test sheets (do not detach sheets). The following formulae may be useful. El. A + HI ± V RR. B = El. B C T = 0.0000116( T 0) L C T = 0.000006 45( T 68) L W L C S = 4 P Y = H cos( α) X = H sin( α) C C lat dep AB = AB = AB lat P AB dep P n i = 1 α = i ( n ) 180 C P ( P P ) L S = E = lat + A E dep n A = = X ( Y 1 1 Y i i i i+ 1 ) + r) 0.0675 ( c = m K i = 1 i = n ( i 1) n ( i + 1) 1
CIVIL ENGINEERING A03 SURVEYING AND MEASUREMENT TERM TEST, Fall 011 Page of 6 Short Answer Questions: 1. Name four sources of systematic taping errors. Erroneous length, Slope, Temperature, Tension, Sag 4 marks. Name four common mistakes made in taping. Reading the tape incorrectly, recording numbers incorrectly, missing a tape length over long runs, mistaking the end point of the tape, making 1 ft / 1 m mistakes 4 marks 3. Name three distance measurement techniques. Pacing, Odometer, Stadia, Electronic Distance Measurement (EDM), Fibreglass Tape, Steel Tape 3 marks 4. True or false: A 3.63-percent slope is one that rises.08 m in 77.77 m. TRUE 1 mark 5. True or false: Random taping errors can occur when the tape is not held exactly horizontal each time the tape is used. Because the surveyor may hold the tape too high on some occasions and too low other times, these random errors will cancel out and not affect the accuracy of the survey. FALSE 1 mark 6. Definition: An azimuth is The direction of a line given by an angle measured CW from the North end of a meridian. marks 7. Definition: A turning point (TP) is A point temporarily used to transfer an elevation. marks 8. Definition: A vertical line is marks A line from the surface of the earth to the earth s center.
CIVIL ENGINEERING A03 SURVEYING AND MEASUREMENT TERM TEST, Fall 011 Page 3 of 6 Calculation Questions: 9. A pre-engineering baseline was run down a steep hill. Rather than measure horizontally downhill with the steel tape, the surveyor measures the vertical angle with a theodolite and the slope distance with a 00-ft steel tape. The vertical angle is -1 o 6 turned to a point on a plumbed range pole that is 4.88 ft above the ground. The slope distance from the theodolite to the point on the range pole is 148.61 ft. The theodolite s optical center is 4.66 ft above the upper baseline station at 1+71.5. a) If the elevation of the upper station is 318.71, what is the elevation of the lower station? b) What is the chainage of the lower station? part a) using El.A +HI +- V - RR.B = El.B 1 El.A = elevation of upper station 318.71 ft HI = height of instrument at El.A 4.66 ft deg min dec. deg. vertical angle -1 6-1.4333 S = slope distance 148.61 ft V = vertical distance = S*sin( vertical angle) -54.3048 ft RR.B = rod reading on plumbed range pole 4.88 ft El.B = elevation of lower station 64.19 ft part b) using H = S*cos(vertical angle) 1 H = horizontal distance 138.336 ft 0's ft < 0's ft upper station chainage 1 71.5 upper station chainage 171.5 ft H 138.336 ft lower station chainage = upper station chainage + H 19.58 ft lower station chainage 11 9.58 ft
CIVIL ENGINEERING A03 SURVEYING AND MEASUREMENT TERM TEST, Fall 011 Page 4 of 6. Station +04.67 must be marked in the field. If the steel tape to be used is only 99.98 ft (under standard conditions) and if the temperature is 87 o F at the time of measurement, how far from the existing station mark at 0+79.3 will the surveyor have to measure to locate the new station? reported tape length 0 ft actual tape length 99.98 ft erroneous tape length correction per tape = 99.98-0.00-0.0 ft ambient temperature 87 deg. F temperature correction per tape = 0.00000645*(87-68)*99.98 0.0153 ft total correction per tape = -0.0 + 0.0153-0.00775 ft to do a layout, apply the opposite of the correction for a reading layout correction per tape 0.007747 ft/tape total distance to layout 0's ft < 0's ft existing station 0 79.3 79.3 ft requested station 4.67 04.67 ft distance to layout 15.44 ft number of tape lengths = distance to layout / ideal tape length 1.544 ft/(0 ft) correction = layout correction * number tape lengths 0.009718 ft 5 actual layout = distance to layout + correction 15.45 ft 1
CIVIL ENGINEERING A03 SURVEYING AND MEASUREMENT TERM TEST, Fall 011 Page 5 of 6 11. Reduce the accompanying set of differential leveling notes by filling in the table below. Show the arithmetic check and the resulting error in closure if the elevation of BM 1 is known to be 54.19 ft. Station BS (ft) HI (ft) IS (ft) FS (ft) Elevation (ft) BM 0 8.7 40.48 TP 1 9.1.60 0+00 11.3 TP 7.33 4.66 +00 4.97 BM 1 3.88 Station BS (ft) HI (ft) IS (ft) FS (ft) Elevation (ft) BM 0 8.7 48.75 40.48 TP 1 9.1 55.36.6 46.15 0+00 11.3 44.06 TP 7.33 58.03 4.66 50.7 +00 4.97 53.06 BM 1 3.88 54.15 sum 4.81 11.14 8 arithmetic check = BM 0 + sum BS - sum FS = BM 1 = 40.48 + 4.81-11.14 54.15 check ok 1 error of closure = final elevation - known elevation known 54.19 ft final 54.15 ft error -0.04 ft 1
CIVIL ENGINEERING A03 SURVEYING AND MEASUREMENT TERM TEST, Fall 011 Page 6 of 6 1. A closed traverse ABCD has the following bearings: AB = N 60 o 50 E, BC = S 4 o 48 E, CD = S 7 o 33 W, DA = N 66 o 03 W. Compute the interior angles and show a geometric check of your work. deg min brg AB 60 50 brg BC 4 48 angle B = brg AB + brg BC angle B 98 angle B 3 38 brg CB 4 48 brg CD 7 33 angle C = 180 - brg CB - brg CD 180 0 angle C 131-81 angle C 19 39 brg DC 7 33 brg DA 66 3 angle D = brg DC + brg DA angle D 73 36 angle D 73 36 brg AB 60 50 brg AD 66 3 angle A = 180 - brg AB - brg AD 180 0 angle A 54-53 angle A 53 7 geometric check: sum = (n-)*180 n 4 angle to check against 360 deg angle deg min A 53 7 B 3 38 C 19 39 D 73 36 sum 358 check 360 0 ok! TOTAL MARKS = 59