Tutorial work - 1 - Problem with solutions PDF

Preview

Summary

problem with solutions...

Description

UNIVERSITY OF WOLLONGONG Faculty of Engineering

CIVL272 Surveying Practice Problems 1 Angles and Bearings

1

Convert the following whole circle bearings into quadrant bearings: 214o 30'; 311o 45';

2

157o 30'; 278o 04'.

S 34o 15' E; S 42o 45' W; N 82o 45' W; S 64o 14' E;

N 79o 30' W; S 34o 30' W.

wcb of AP = 10o 00' 00" wcb of AP = 92o 30' 40" wcb of AP = 337o 15' 20" wcb of AP = 05o 11' 54"

, , , ,

wcb of PB = 300o 00' 00" wcb of PB = 11o 27' 46" wcb of PB = 100o 18' 42" wcb of PB = 128o 00' 41"

, , , ,

qb of PB = N 60o 00' 00" W qb of PB = N 13o 45' 46" E qb of PB = S 71o 39' 40" E qb of PB = S 65o 59' 10" E

Find the clockwise angle APB, if: (a) (b) (c) (d)

5

093o 30'; 244o 14';

Find the clockwise angle APB, if: (a) (b) (c) (d)

4

287o 45'; 078o 45';

Convert the following quadrant bearings into whole circle bearings: N 25o 30' E; S 18o 15' W;

3

027o 45'; 218o 30';

qb of AP = Due North qb of AP = N 89o 35' 36" E qb of AP = N 11o 39' 40" W qb of AP = N 08o 10' 10" E

Using a 1 second theodolite the following clockwise angles were measured in a closed traverse. What is the angular closing error and is this acceptable? 163o 27' 36"; 324o 18' 22"; 62o 39' 27"; 330o 19' 18"; 181o 09' 15"; 305o 58' 16"; 188o 02' 03"; 292o 53' 02" 131o 12' 50".

2 6

The internal clockwise angles of a close polygonal traverse are as shown in the table. Correct them and tabulate the whole circle bearings of all lines, given the wcb of AB is 170o 08' 34". The final calculation should be the check on the wcb of AB.

Angle

Observed Value '

o

Corrn.

"

Corrected Angle '

o

"

wcb o

170

7

" 34

AB

ABC

120

20

00

BC

BCD

86

00

40

CD

CDE

341

34

20

DE

DEF

60

22

00

EF

EFA

100

28

20

FA

FAB

11

14

10

AB

From the theodolite readings given below, determine the angles of the traverse ABCDE. Having obtained the angles, correct them to the nearest 10 seconds and then determine the whole circle bearing of BC if the bearing of AB is 45o 20' 40". Back Station

E A B C D

8

' 08

Line

Theodolite Station

Forward Station

A B C D E

B C D E A

Readings Back Station Forward Station 0o 00' 00" 264o 49' 40" 164o 29' 10" 43o 58' 30" 314o 18' 20" (R.I.C.S. Answer

264o 49 '40" 164o 29' 10" 43o 58' 30" 314o 18' 20" 179o 59' 10" 125o 00' 20")

Measurement of the interior anticlockwise angles of a closed traverse ABCDE have been made with a theodolite reading to 20 seconds of arc. Adjust the measurements and compute the quadrant bearings of the sides if the bearing of the line AB is N 43o 10' 20" E. Angle ABC BCD CDE DEA EAB

60o 21' 142o 36' 89o 51' 111o 50' 135o 20'

20" 20" 40" 40" 40"

R.I.C.S. Answer. AB N 43o 10' 20" BC S 17o 10' 52" CD S 20o 12' 56" DE N 69o 38' 36" EA N 01o 29' 08"

E E W W W

Q6

Solution

Angle

Observed Value o

'

Corrn.

Corrected Angle o

"

'

"

wcb

Line

o

'

"

170

08

34

AB

ABC

120

20

00

+5"

120

20

05

110

28

39

BC

BCD CDE

86 341

00 34

40 20

+5" +5"

86 341

00 34

45 25

16 178

29 03

24 49

CD DE

DEF

60

22

00

+5"

60

22

05

58

25

54

EF

EFA

100

28

20

+5"

100

28

25

338

54

19

FA

FAB

11

14

10

+5"

11

14

15

170

08

34

AB

Sum

719

59

30

+30”

720

00

00

(2n+4).90 720

00

00

check

Error '

"

wcbAB

170

08

34

ABC

120

20

05

290 -180

28

39

wcbBC

110

28

39

BCD

86

00

45

196

29

24

-180 wcbCD

16

29

24

CDE

341 358

34 03

25 49

03 22 25

49 05 54

25 28 54

54 25 19

338 11

54 14

19 15

350 -180

08

34

-180 178 60 238 -180 wcbEF EFA

58 100 158 +180

wcbFA FAB

check

-30" o

wcbDE DEF

check

wcbAB

170

08 check

34

Q8 solution. o

Sum Measured Angles = 540 00' 40" Sum Internal Angles

wcbAB

43

10

20

ABC

-60

21

12

-17 +360

10

08

342

49

08

o

= 540 00' 00"

Misclosure = +40" Correction/angles = -40/5 = -8" Thus:

Angle EAB

Corrected Angle o ' " 135 20 32

ABC

60

21

12

BCD

142

36

12

CDE

89

51

32

DEA

111

50

32

sum

540 00 (check)

00

As angles are anticlockwise treat them as negative with respect to whole circle bearings. When the sum of whole circle bearing minus the angle is negative, add o o 360 before subtracting or adding 180 as per the normal process. Alternatively, if the value of the wcb is smaller than the o value of the angle, add 360 to the wcb first:

-180 wcbBC

162

49

08

BCD

-142

36

12

20 +180

12

56

200 -89

12 51

56 32

110

21

24

wcbCD CDE

+180 wcbDE

290

21

24

DEA

-111

50

32

178 +180

30

52

wcbEA

358

30

52

EAB

-135

20

32

223

10

20

10

20

-180 wcbAB

43

(check)

Solution requires answers as quadrant bearings, thus: o

qbAB = N 43 o qbBC = S 17 o qbCD = S 20 o qbDE = N 69 o qbEA = N 1

10' 10' 12' 38' 29'

20" E 52" E 56" W 36" W 08" W...