QUESTION 1 The levelling data below relates to a flight of levels observed along the…

QUESTION 1 The levelling data below relates to a flight of levels observed along the…
GEOM 2088 Introduction to Surveying
1
QUESTION 1
The levelling data below relates to a flight of levels observed along the centreline of a road. A sight has
also been taken to the underside of a footbridge over the road at chainage 2190. The chainage of each
point in the table below is its horizontal distance from a survey mark measured along the road centreline.
BS IS FS RL Chainage (m) Remarks
1.830 139.350 2050 Road Surface
1.732 2.730 2150 Road Surface
-1.098 2190 U/side of bridge
0.816 2.632 2250 Road Surface
1.207 1.726 2350 Road Surface
2.107 2450 Road Surface
a) Compute the reduced level of the points using the rise and fall method and apply all the
usual checks. [Do NOT apply any adjustment methods. Show your answers in meters
correct to 3 decimal places]
b) What is the average grade of the road centreline between chainages 2050 and 2450?
[Write your answer in percentage grade correct to 1 decimal place]
c) A short section of the road, between chainage 2160 and 2240, is missing as a result of the
washaway and is to be rebuilt. The tallest vehicle that will pass over the rebuilt section stands
4.10 m above the road surface. If the road is rebuilt in the same place from which it appears to
have been washed away, will the tallest vehicle pass safely under the bridge, or will it hit?
Explain your answer.
(2 + 1.5 + 1.5 = 5 marks)
GEOM 2088 Introduction to Surveying
2
QUESTION 2
In the figures below (Figure 1 and Figure 2), calculate the bearings of all the sides in the
traverses, if the bearing of the line CB in Figure 1 is 305°40’02” and the bearing of the line FA
in Figure 2 is 331°20’40”. Provide your answers in sexagesimal system rounded to the nearest
1”. Note: figures are not to scale.
Figure 1 Figure 2
In Figure 1, calculate the bearings of: In Figure 2, calculate the bearings of:
AB = ? AB = ?
BC = ? BC = ?
CD = ? CD = ?
DA = ? DE = ?
EF = ?
(3 marks; 1/3 marks for each bearing)
A
B
C
D
85o
07’ 07’’
110o
30’ 45’’
72o
51’ 28’’
91o
30’ 40’’
A
B
C
D E
F
85o
15’ 10’’
103o
00’ 30’’
214o
31’ 27’’
248o
20’ 20’’
278o
13’ 43’’
167o
10’ 10’’
North
GEOM 2088 Introduction to Surveying
3
QUESTION 3
Using the “Radiation” technique, the bearings and horizontal distances are measured between
the station pairs A-B, B-C, C-D, D-E and E-A. From these stations, bearings and horizontal
distances are also measured to six fence posts 1, 2, 3, 4, 5, 6. All these measurements are shown
in the figure below. The coordinates of A are Easting: 1000.000m, Northing: 5000.000m.
95 18’ 30”
321
57’ 41”
215 43’ 00”
163
27’ 40”
60
35’ 20”
357.36
215.68
582.63
666.389
382.87
A
B
C
D
E
1
2
3
4
5
6
271 54’
31.35
135
27’
29.84
58 31’
10.42
2
01 10’
3.55
135 26’
5.76
158
55’
2.88
a) Calculate the coordinates (Easting and Northing) of the fence posts 1, 2, 3, 4, 5, and 6.
Provide your answers in meters correct to 3 decimal places.
b) Calculate the bearings and horizontal distances between the fence post pairs 1-2, 2-3,
3-4, 4-5, and 5-6. Provide your answer for bearings in sexagesimal system rounded to
the nearest 1”, and for distances in meters rounded to the nearest 1mm.
(3 + 2 = 5 marks)
North
GEOM 2088 Introduction to Surveying
4
QUESTION 4
A four-sided traverse is measured in the field, as illustrated below. Point A is a survey mark
with known coordinates of 1000.000E, 2500.000N.
From To Bearing Distance
A B 70º 30′ 00″ 78.934 m
B C ? 124.569 m
C D ? 298.012 m
D A ? 170.706 m
a) Determine the angular misclosure of the traverse ABCD. Provide your answer in
sexagesimal system rounded to the nearest 1”.
b) Determine the adjusted bearings of lines BC, CD, DA. Provide your answer in
sexagesimal system rounded to the nearest 1”.
c) Using the adjusted bearings and distances for the traverse ABCD, calculate the
following:
i. the linear misclosure (in meters correct to 3 decimal places)
ii. the precision in reciprocal form
d) remove the linear misclosure using Bowditch’s rule and calculate the adjusted
coordinates in meters correct to 3 decimal places.
(1 + 1.5 + (0.75 + 0.75) + 3 = 7 marks)
Internal Angle
A 83° 00′ 00?
B 228° 28′ 46?
C 13° 20′ 20?
D 35° 10′ 14?
Note: Not to scale
Attachments: Assignment-1-….pdfApr 28 2022 09:41 AM

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