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by | Nov 30, 2023 | questions

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Question 1

Reduce the following level sheet showing all checks. Compute the adjusted reduced levels for all points.

BS INT FS Rise Fall RL Remarks 2.567           PSM 162 (RL = 37.855) 1.984   3.984       TBM A 0.355   0.768       TBM B 3.031   3.804       TBM C 0.783   2.547       TBM D   2.541         Natural Surface   2.895         Natural Surface 3.651   0.784       TBM E 3.981   2.189       TBM F     2.283        PSM 162 (RL = 37.855)              

Question 2

Calculate the index error for an automatic level based on the tabulated measurements below.
Two Peg Test At To Distance Reading Mid.Pt 501 30 2.136 Mid.Pt 502 30 1.826 502 502 1 1.625 502 501 60 1.959

Question 3

Line Bearing Distance ?E  ?N  E   N   Deg Min Sec metres         A     1000.000  2000.000  A-B 57 56 40 111.505 B-C 135 20 30 84.256 C-D 179 55 50 47.269 D-E 255 15 10 78.988 E-A   311 21  20  103.11                   
From the closed traverse information above, compute coordinates of Stations B, C, D and E.

Question 4

(a) 2.5764o to Degrees Minutes Seconds
(b) 204o 39’43” to Degrees Decimals

Question 5

What is the height difference between A and B given the following measurements:
– the height of the instrument (hi) at A is 1.272m and the height of the target at B (ht) is 1.333
– zenith angle (z) and slope distance (s) is 82o 52’36” and 127.025m respectively.

Question 6

The reduced level of 901 is to be determined by trig heighting from instrument station 1. Given the height difference from 1 to 728 is -0.421m and 1 to 901 is +1.527m. If the RL of 728 is 26.505m calculate the RL of 901.

Question 7

Given the bearing and direction of one line calculate the bearings of the remaining lines.

Question 8

Calculate the mean of the measured angle from the following observed bearings:
At Station To Face…


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