LSIT Measurement Technology 5 — Questions and Answers
Question 1: A surveyor sets up a total station at a point with a HI of 5.42 ft and sights a rod held on a benchmark (elevation 312.54 ft), reading 3.81 ft. What is the instrument elevation?
- 318.0 ft (HI = BM + rod reading + instrument height)
- Instrument elevation = BM elevation + HI − rod reading
- Instrument elevation = 312.54 + 5.42 = 317.96 ft (Correct answer)
- Instrument elevation = 312.54 + 3.81 = 316.35 ft
Correct answer: Instrument elevation = 312.54 + 5.42 = 317.96 ft
The height of instrument (HI) above datum = BM elevation + backsight rod reading = 312.54 + 3.81 = 316.35, then HI above ground = 5.42 ft; instrument elevation = BM + BS = 316.35 ft.
Question 2: Which of the following best describes 'multipath error' in GNSS measurements?
- Satellite signals arrive at the antenna via reflected paths in addition to the direct path, causing ranging errors (Correct answer)
- Multiple satellites transmit on the same frequency, causing interference
- The receiver simultaneously tracks more paths than its channels allow
- Signals travel multiple routes through the ionosphere
Correct answer: Satellite signals arrive at the antenna via reflected paths in addition to the direct path, causing ranging errors
Multipath occurs when reflected signals (from buildings, water, terrain) reach the antenna alongside the direct signal, degrading range measurements.
Question 3: A surveyor measures vertical angles to a distant target from both faces of a total station: Face Left = +12°18'24" and Face Right = –12°17'48". What is the corrected vertical angle and index error?
- Corrected angle = +12°18'06", index error = +18" (Correct answer)
- Corrected angle = +12°18'12", index error = +36"
- Corrected angle = +12°18'06", index error = –18"
- Corrected angle = +12°17'36", index error = +24"
Correct answer: Corrected angle = +12°18'06", index error = +18"
Mean vertical angle = (FL + (180° – |FR|)) / 2 = (+12°18'24" + 12°17'48") / 2 = +12°18'06"; index error = FL – mean = +18".
Question 4: The Compass (Bowditch) Rule for traverse adjustment distributes the misclosure:
- Proportional to the length of each traverse leg (Correct answer)
- Equally among all traverse legs
- Proportional to the angle at each traverse station
- Only to the legs with the largest measured angles
Correct answer: Proportional to the length of each traverse leg
The Compass Rule apportions latitude and departure corrections to each leg in proportion to that leg's length relative to the total traverse length.
Question 5: Which instrument is most appropriate for transferring elevations across a wide river where direct leveling is impractical?
- Total station using trigonometric leveling with reciprocal observations (Correct answer)
- Automatic level with a long rod
- Laser level with a rotating beam
- Hand level and measuring tape
Correct answer: Total station using trigonometric leveling with reciprocal observations
Reciprocal trigonometric leveling with a total station cancels refraction and curvature errors when observations are exchanged from both banks.
Question 6: In RTK GPS surveying, the term 'fixed solution' means:
- The receiver has resolved carrier-phase integer ambiguities, yielding centimeter-level accuracy (Correct answer)
- The receiver is operating in static mode without movement
- The base station's position has been fixed by a government survey
- The satellite constellation geometry is locked and will not change
Correct answer: The receiver has resolved carrier-phase integer ambiguities, yielding centimeter-level accuracy
A fixed solution means the integer ambiguities in carrier-phase measurements have been resolved, enabling centimeter-level RTK positioning.
Question 7: A surveyor uses a steel tape that is 100.02 ft long instead of exactly 100.00 ft. After measuring a 500 ft line, the corrected distance is:
- 500.10 ft (Correct answer)
- 499.90 ft
- 500.02 ft
- 499.98 ft
Correct answer: 500.10 ft
Correction per tape length = +0.02 ft; for 5 tape lengths: 5 × 0.02 = +0.10 ft; corrected distance = 500.00 + 0.10 = 500.10 ft.
LSIT Measurement Technology 5
A surveyor sets up a total station at a point with a HI of 5.42 ft and sights a rod held on a benchmark (elevation 312.54 ft), reading 3.81 ft.
What is the instrument elevation?