Trav Assgn DATA 2-1 - Assessment PDF

Title Trav Assgn DATA 2-1 - Assessment
Course Mechanical Engineering
Institution Macquarie University
Pages 3
File Size 244.6 KB
File Type PDF
Total Downloads 102
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Summary

Assessment...


Description

TRAVERSE CLOSE ASSIGNMENT A closed traverse was run from A to B, then C and back to A, with radiations made at B and C to Y and X respectively. The diagram below shows the reduced observations made at A and B and the field notes recorded when the theodolite was set up at C. When the distances were observed at C, the tape was placed on evenly sloping ground and the appropriate vertical angle observed. Reduce the field notes made at C, in the spaces on the booking sheet. (5 Marks)

HORIZONTAL ANGLES TARGET Face Left B A X

00-00-00 55-29-10 204-15-30

Face Right

Mean

179-59-40 235-30-30 24-16-50

-(00 -00- 10) 55-29-50 204-16-10

Station C Reduced Mean 00-00-00 55-30-00 204-16-20

VERTICAL ANGLES TARGET A X

Face FL FR FL FR

Observed Vertical Circle 86-02-50 273-59-10 92-11-30 267-50-10

Vertical Angle -(3-57-10) -(3-59-10) 2-11-30 2-9-50

Mean Vertical Angle -(3-58-10) 2-10-40

Observed Distance 29.285 10.958

Calculate the angular misclose, angular adjustment and the adjusted bearings of all the traverse lines and radiations. (6 marks) Angular Misclose of traverse : __________ Given data Angle ABC = 63-29-30 · Angle CAB = 60-58-30 · Angle BCA = 55-30-00 [Reduced observations] Sum of interior angle = 179-58-00 But theoretical sum of triangle inside angles must be 180 degree. Angular misclose = (180-00-00) – (179-58-00) = 00-02-00

Angular adjustment : _______________

Since angle ABC and CAB were already given and angle BCA was measured now, the error caused at measuring angle BCA only. Hence the above error shall be adjusted in angle BCA only. Hence adjusted angle BCA = (55-30-00) + (00-02-00) = 55-32-00 Given Data, · Angle ABY = 186-15-00 · Angle BCX = 204-16-20 [Reduced observations] · Azimuth of AB = 26-26-40 · Angle ABC = 63-29-30 · Angle BCA = 55-32-00 [Adjusted]

Bearing of B-C : _26ᵒ26’40” + 180ᵒ + 63ᵒ29’30” = 269ᵒ56’10” Bearing of C-A :_ 269ᵒ56’10”+ 55ᵒ32’00”- 180ᵒ = 145ᵒ28’10” Bearing of B-Y: _ 26ᵒ26’40” - 180ᵒ +186ᵒ15’00” = 32ᵒ41’40” Bearing of C-X: _269ᵒ56’10” + 204ᵒ16’20” - 180ᵒ = 294ᵒ 12’30” Traverse Close (4 marks) Complete the traverse table to find the linear misclose, proportional accuracy of the traverse, the coordinates of all points and the bearing and distance of the line X-Y. Horizontal distance CA = 29.285 x cos (-3ᵒ58’10”) = 29.215

LINE A-B B-C C-A

Adjusted Bearing

Horiz. Dist

 E (+)

26-26-40 269-56-10 145-28-10 Misclose total

26.915 28.54 29.215

11.986

E W (-)

 N N (+) S(-) 24.099

-28.54

-0.032 -24.068

16.56

CO-ORD E 200.000 211.986 183.446 200.006 0.006

INATES N 300.000 324.099 324.067 299.999 0.001

PT. A B C A

84.67

Linear Misclose = √((𝑚𝑖𝑠𝑐𝑙𝑜𝑠𝑒 𝑖𝑛 𝐸)2 + (𝑚𝑖𝑠𝑐𝑙𝑜𝑠𝑒 𝑖𝑛 𝑁)2 ) 2 2 √(0.006 + 0.001 ) = 0.00608m

Proportional Accuracy =

𝐿𝑖𝑛𝑒𝑎𝑟 𝑀𝑖𝑠𝑐𝑙𝑜𝑠𝑒 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡𝑟𝑎𝑣𝑒𝑟𝑠𝑒

=

0.00608 84.67

= 76/1058375 = 7.1808 x10-5

Radiations (2 marks) Transcribe the coordinates you have calculated for C and B above, to these tables so you can calculate the coordinates of the radiated points. Calculated so far Bearing of BY = 32ᵒ41’40” Bearing of CX = 294ᵒ 12’30” Horizontal distance B-Y = 6.75m Horizontal distance C-X = 10.958 cos (2ᵒ10’40) = 10.95m

LINE

Adjusted Bearing

Horiz. Dist

C- X

294-12-30

10.95

LINE

Adjusted Bearing

Horiz. Dist

 E (+)

B-Y

32-41-40

6.75

3.646

 E (+)

E W (-)

 N (+)

-9.987

4.49

E W (-)

 N (+)

N S(-)

CO-ORD E 183.446 173.459

INATES N 324.067 328.557

PT. C X

N S(-)

CO-ORD INATES E N 211.986 324.099 215.632 329.78

PT. B Y

5.681

Missing Line Calculation (3 marks) Transcribe the coordinates calculated from the radiation tables to this table to find the missing bearing and distance XY. Calculated so far, Co ordinates of X (E,N) = (173.459m , 328.557m) Co ordinates of Y (E,N) = (215.632m , 329.78m) E = Departure = E coordinate of end point - E Coordinates of start point N =Latitude= N coordinate of end point - N Coordinates of start point Horizontal distance =√Δ𝐸2 + Δ𝑁 2 Bearing = tan-1 (E/N)

LINE

XY

Bearing

88-20-22.63 (88ᵒ20’23”)

Horiz. Dist

 E (+)

42.191

42.173

E W (-)

 N (+)

1.223

N S(-)

CO-ORD E 173.459 215.632

INATES N 328.557 329.78

PT. X Y...


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