Surveying - traverse - for web.ppt - Memphis

CIVL 1112

Surveying - Traverse Calculations

Surveying - Traverse

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Introduction  Almost all surveying requires some calculations to reduce measurements into a more useful form for determining distance, earthwork volumes, land areas, etc.  A traverse is developed by measuring the distance and angles between points that found the boundary of a site  We will learn several different techniques to compute the area inside a traverse


Distance - Traverse Methods of Computing Area  A simple method that is useful for rough area estimates is a graphical method  In this method, the traverse is plotted to scale on graph paper, and the number of squares inside the traverse are counted

B

A C D

Distance - Traverse Methods of Computing Area B a A

Distance - Traverse Methods of Computing Area B

1 Area ABC  ac sin  2

b



a A

c

C

b



Area ABD 

1 ad sin  2

Area BCD 

1 bc sin  2

 C

d c D

Area ABCD  Area ABD  Area BCD

CIVL 1112


Distance - Traverse


Methods of Computing Area B

b

A

Area ABE 

c



Balancing Angles

C

a

 D

e

2/13

Area CDE 

d

1 ae sin  2

 Before the areas of a piece of land can be computed, it is necessary to have a closed traverse  The interior angles of a closed traverse should total:

1 cd sin  2

(n - 2)(180°) where n is the number of sides of the traverse

E

 To compute Area BCD more data is required



Balancing Angles

Balancing Angles A

Error of closure

B D

 A surveying heuristic is that the total angle should not vary from the correct value by more than the square root of the number of angles measured times the precision of the instrument  For example an eight-sided traverse using a 1’ transit, the maximum error is:

1' 8  2.83 '  3' C

Angle containing mistake



Balancing Angles

Latitudes and Departures

 If the angles do not close by a reasonable amount, mistakes in measuring have been made

 The closure of a traverse is checked by computing the latitudes and departures of each of it sides

 If an error of 1’ is made, the surveyor may correct one angle by 1’  If an error of 2’ is made, the surveyor may correct two angles by 1’ each  If an error of 3’ is made in a 12 sided traverse, the surveyor may correct each angle by 3’/12 or 15”

N

N

B

Latitude AB

Bearing  E

W Bearing 

A

W C

Departure AB Latitude CD

S

Departure CD

D S

E

CIVL 1112



3/13


Latitudes and Departures

Error of Closure

 The latitude of a line is its projection on the north–south meridian

 Consider the following statement:

N Latitude AB

E

W Bearing 

A

“If start at one corner of a closed traverse and walk its lines until you return to your starting point, you will have walked as far north as you walked south and as far east as you have walked west”

 The departure of a line is its projection on the east– west line

B

Departure AB

 A northeasterly bearing has: + latitude and + departure

 latitudes = 0

 Therefore

and

 departures = 0

S



Error of Closure

Error of Closure

 When latitudes are added together, the resulting error is called the error in latitudes (EL)

 If the measured bearings and distances are plotted on a sheet of paper, the figure will not close because of EL and ED

 The error resulting from adding departures together is called the error in departures (ED)

B

Error of closure

ED

Eclosure 

EL A

C

Latitudes and Departures - Example

Precision 

2

  ED 

2

Eclosure perimeter

 Typical precision: 1/5,000 for rural land, 1/7,500 for suburban land, and 1/10,000 for urban land

D


 EL 

Surveying - Traverse Latitudes and Departures - Example

A N N 42° 59’ E

S 6° 15’ W

234.58’

Departure AB

189.53’

B

W  (189.53 ft.)sin(615 ')  20.63 ft. A

W

E

E S 29° 38’ E 142.39’ 175.18’

N 12° 24’ W

S 6° 15’ W

Latitude AB

189.53 ft.

S  (189.53 ft.)cos(615 ')  188.40 ft.

175.18’

D N 81° 18’ W

C

B

S

CIVL 1112





Latitudes and Departures - Example Bearing

Side

N Departure BC

E  (175.18 ft.)sin(2938 ')  86.62 ft. B

W

4/13

AB BC CD DE EA

degree

m inutes

6 29 81 12 42

15 38 18 24 59

S S N N N

Length (ft.)

Latitude

Departure

189.53 175.18 197.78 142.39 234.58 939.46

-188.403 -152.268 29.916 139.068 171.607 -0.079

-20.634 86.617 -195.504 -30.576 159.933 -0.163

W E W W E

E 175.18 ft.

Latitude BC

S 29° 38’ E

S  (175.18 ft.)cos(2938 ')  152.27 ft.

C S



Latitudes and Departures - Example Bearing

Side AB BC CD DE EA

Eclosure 

S S N N N

 EL 

2

degree

m inutes

6 29 81 12 42

15 38 18 24 59

  ED  

Length (ft.)

Latitude

Departure

189.53 175.18 197.78 142.39 234.58 939.46

-188.403 -152.268 29.916 139.068 171.607 -0.079

-20.634 86.617 -195.504 -30.576 159.933 -0.163

A S 77° 10’ E

W E W W E

 0.079 

2

Group Example Problem 1

2

  0.163   0.182 ft.

0.182 ft. Eclosure  Precision   939.46 ft. perimeter

N 29° 16’ E

651.2 ft.

B 660.5 ft.

S 38° 43’ W

2

1 5,176

D

491.0 ft.

826.7 ft.

N 64° 09’ W C


Surveying - Traverse Balancing Latitudes and Departures


 Balancing the latitudes and departures of a traverse attempts to obtain more probable values for the locations of the corners of the traverse Bearing

Side AB BC CD DE

S S N N

Length (ft.)

degree

m inutes

77 38 64 29

10 43 9 16

E W W E

651.2 826.7 491.0 660.5

Latitude

Departure

 A popular method for balancing errors is called the compass or the Bowditch rule  The “Bowditch rule” as devised by Nathaniel Bowditch, surveyor, navigator and mathematician, as a proposed solution to the problem of compass traverse adjustment, which was posed in the American journal The Analyst in 1807.

CIVL 1112


5/13



Balancing Latitudes and Departures

Balancing Latitudes and Departures A

 The compass method assumes: N 42° 59’ E

1) angles and distances have same error 2) errors are accidental

S 6° 15’ W

234.58 ft.

189.53 ft.

B

 The rule states: E

S 29° 38’ E

“The error in latitude (departure) of a line is to the total error in latitude (departure) as the length of the line is the perimeter of the traverse”

142.39 ft. 175.18 ft.

N 12° 24’ W

175.18 ft.

D N 81° 18’ W


C




 Recall the results of our example problem

 Recall the results of our example problem

Bearing

Side AB BC CD DE EA

S S N N N

Length (ft)

degree

m inutes

6 29 81 12 42

15 38 18 24 59

W E W W E

Latitude

Departure

Bearing

Side

189.53 175.18 197.78 142.39 234.58

AB BC CD DE EA

S S N N N

degree

m inutes

6 29 81 12 42

15 38 18 24 59

W E W W E

Length (ft)

Latitude

Departure

189.53 175.18 197.78 142.39 234.58 939.46

-188.403 -152.268 29.916 139.068 171.607 -0.079

-20.634 86.617 -195.504 -30.576 159.933 -0.163





N

N Latitude AB

Departure AB

S  (189.53 ft.)cos(6 15 ')  188.40 ft. 

A

W

E

Correction in Lat AB LAB  EL perimeter

S 6° 15’ W 189.53 ft.

B

W  (189.53 ft.)sin(615 ')  20.63 ft. A

W

Correction in Lat AB 

EL  LAB 

189.53 ft.

B

Correction in Lat AB 

939.46 ft.

Correction in Dep AB LAB  ED perimeter

S 6° 15’ W

Correction in Dep AB 

perimeter

S

0.079 ft. 189.53 ft.

E



 0.016 ft.

ED  LAB  perimeter

S

Correction in Dep AB 

0.163 ft. 189.53 ft. 939.46 ft.



 0.033 ft.

CIVL 1112


6/13





N

N Latitude BC

Departure BC

S  (175.18 ft.)cos(29 38 ')  152.27 ft. 

B

W

E

Correction in LatBC LBC  EL perimeter

175.18 ft.

S 29° 38’ E

Correction in LatBC  C

S

Correction in LatBC 

EL  LBC 

E  (175.18 ft.)sin(2938 ')  86.62 ft. B

W

E

S 29° 38’ E

939.46 ft.

ED  LBC 

Correction in DepBC 

perimeter

0.079 ft. 175.18 ft.

Correction in DepBC LBC  perimeter ED

175.18 ft.

perimeter

C S

  0.015 ft.

Correction in DepBC 

0.163 ft. 175.18 ft.

  0.030 ft.

939.46 ft.





Length (ft.) Latitude 189.53 175.18 197.78 142.39 234.58 939.46

-188.403 -152.268 29.916 139.068 171.607 -0.079

Departure -20.634 86.617 -195.504 -30.576 159.933 -0.163

Corrections Latitude Departure 0.016 0.015

Balanced Latitude Departure

0.033 0.030


-188.403 -152.268 29.916 139.068 171.607 -0.079

Departure -20.634 86.617 -195.504 -30.576 159.933 -0.163

Corrections Latitude Departure 0.016 0.015

0.033 0.030

Balanced Latitude Departure -188.388 -152.253

Corrected latitudes and departures

Corrections computed on previous slides






-188.403 -152.268 29.916 139.068 171.607 -0.079

Departure -20.634 86.617 -195.504 -30.576 159.933 -0.163

Corrections Latitude Departure 0.016 0.015 0.017 0.012 0.020

-20.601 86.648

0.033 0.030 0.034 0.025 0.041

Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627 0.000

No error in corrected latitudes and departures

-20.601 86.648 -195.470 -30.551 159.974 0.000

Combining the latitude and departure calculations with corrections gives: Side

Corrections Length (ft.) Latitude Departure Latitude Departure

Bearing


degree m inutes

AB BC CD DE EA

S S N N N

6 29 81 12 42

15 38 18 24 59

W E W W E

189.53 175.18 197.78 142.39 234.58 939.46

-188.403 -152.268 29.916 139.068 171.607 -0.079

-20.634 86.617 -195.504 -30.576 159.933 -0.163

0.016 0.015 0.017 0.012 0.020

0.033 0.030 0.034 0.025 0.041

-188.388 -152.253 29.933 139.080 171.627 0.000

-20.601 86.648 -195.470 -30.551 159.974 0.000

CIVL 1112






Balance the latitudes and departures for the following traverse.

In the survey of your assign site in Project #3, you will have to balance data collected in the following form:

Corrections Balanced Length (ft) Latitude Departure Latitude Departure Latitude Departure 450.00 -285.00 -164.46 0.54

339.00 259.50 -599.22 -0.72

B

N 69° 53’ E

A N

600.0 450.0 750.0 1800.0

7/13

51° 23’

713.93 ft. 105° 39’ 606.06 ft.

781.18 ft.

78° 11’

C

124° 47’ 391.27 ft.

D


Surveying - Traverse Calculating Traverse Area

Group Example Problem 3 In the survey of your assign site in Project #3, you will have to balance data collected in the following form: Side

Corrections Length (ft.) Latitude Departure Latitude Departure

Bearing


degree m inutes

AB N BC CD DA

69

53

E

713.93 606.06 391.27 781.18

Eclosure =

Precision =

 The meridian distance of a line is the east–west distance from the midpoint of the line to the reference meridian  The meridian distance is positive (+) to the east and negative (-) to the west

ft. 1

Surveying - Traverse Calculating Traverse Area N

Surveying - Traverse Calculating Traverse Area

A N 42° 59’ E

S 6° 15’ W

234.58 ft.

189.53 ft.

B E S 29° 38’ E 142.39 ft.

Reference Meridian

 The best-known procedure for calculating land areas is the double meridian distance (DMD) method

175.18 ft.

N 12° 24’ W 175.18 ft.

D N 81° 18’ W

C

 The most westerly and easterly points of a traverse may be found using the departures of the traverse  Begin by establishing a arbitrary reference line and using the departure values of each point in the traverse to determine the far westerly point

CIVL 1112




Calculating Traverse Area Length (ft.) Latitude 189.53 175.18 197.78 142.39 234.58 939.46

Departure

-188.403 -152.268 29.916 139.068 171.607 -0.079

-20.634 86.617 -195.504 -30.576 159.933 -0.163

-20.601 -195.470

D -30.551

E

D

Calculating Traverse Area

Corrections Latitude Departure 0.016 0.015 0.017 0.012 0.020

B

0.033 0.030 0.034 0.025 0.041

N

Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627 0.000

-20.601 86.648 -195.470 -30.551 159.974 0.000

Reference Meridian

86.648

E

142.39 ft.

S 29° 38’ E 175.18 ft.

D N 81° 18’ W

C


DMD Calculations

DMD Calculations The meridian distance of line EA is:

A

N

Meridian distance of line AB

N

B

 The meridian distance of line AB is equal to:

A

the meridian distance of EA + ½ the departure of line EA + ½ departure of AB

B

 The DMD of line AB is twice the meridian distance of line AB

A

E

E C

189.53 ft.

N 12° 24’ W


D

S 6° 15’ W

234.58 ft.

E C C

Point E is the farthest to the west

A

Reference Meridian

N 42° 59’ E

175.18 ft.

159.974

N

A

B

A

B

8/13

E

DMD of line EA is the departure of line

Surveying - Traverse DMD Calculations

N

Meridian distance of line AB

A

B E

Surveying - Traverse DMD Calculations

The DMD of any side is equal to the DMD of the last side plus the departure of the last side plus the departure of the present side

Side AB BC CD DE EA

Balanced Latitude Departure -188.388 -152.253 29.933 139.080 171.627

-20.601 86.648 -195.470 -30.551 159.974

The DMD of line AB is departure of line AB

DMD -20.601

CIVL 1112




DMD Calculations Side AB BC CD DE EA

DMD Calculations


-20.601 + 86.648 + -195.470 -30.551 159.974

Side DMD -20.601 45.447

The DMD of line BC is DMD of line AB + departure of line AB + the departure of line BC


AB BC CD DE EA

DMD -20.601 -20.601 45.447 86.648 -195.470 + -63.375 -30.551 + -289.397 159.974


AB BC CD DE EA


DMD -20.601 -20.601 45.447 86.648 -63.375 -195.470 -30.551 + -289.397 159.974 + -159.974

The DMD of line EA is DMD of line DE + departure of line DE + the departure of line EA

Traverse Area - Double Area  The sum of the products of each points DMD and latitude equal twice the area, or the double area


DMD -20.601 45.447 -63.375


DMD Calculations

AB BC CD DE EA

-20.601 86.648 + -195.470 + -30.551 159.974

The DMD of line CD is DMD of line BC + departure of line BC + the departure of line CD

Side

The DMD of line DE is DMD of line CD + departure of line CD + the departure of line DE

Side

-188.388 -152.253 29.933 139.080 171.627

DMD Calculations


AB BC CD DE EA



DMD Calculations Side

9/13

-20.601 86.648 -195.470 -30.551 159.974

DMD -20.601 45.447 -63.375 -289.397 -159.974

Notice that the DMD values can be positive or negative

Side AB BC CD DE EA


-20.601 86.648 -195.470 -30.551 159.974

DMD Double Areas -20.601 3,881 45.447 -63.375 -289.397 -159.974

 The double area for line AB equals DMD of line AB times the latitude of line AB

CIVL 1112




Traverse Area - Double Area


 The sum of the products of each points DMD and latitude equal twice the area, or the double area Side AB BC CD DE EA


-20.601 86.648 -195.470 -30.551 159.974

DMD Double Areas -20.601 3,881 -6,919 45.447 -63.375 -289.397 -159.974


Balanced Latitude Departure -20.601 86.648 -195.470 -30.551 159.974



1 acre = 43,560 ft.2

-20.601 86.648 -195.470 -30.551 159.974

 The sum of the products of each points DMD and latitude equal twice the area, or the double area

AB BC CD DE EA

-188.388 -152.253 29.933 139.080 171.627

-20.601 86.648 -195.470 -30.551 159.974

DMD Double Areas -20.601 3,881 -6,919 45.447 -1,897 -63.375 -40,249 -289.397 -27,456 -159.974

 The double area for line EA equals DMD of line EA times the latitude of line EA

 The sum of the products of each points DMD and latitude equal twice the area, or the double area Side

DMD Double Areas -20.601 3,881 45.447 -6,919 -63.375 -1,897 -289.397 -40,249 -159.974 -27,456 2 Area = -72,641

Area =




-188.388 -152.253 29.933 139.080 171.627

DMD Double Areas -20.601 3,881 -6,919 45.447 -1,897 -63.375 -289.397 -159.974



AB BC CD DE EA

-20.601 86.648 -195.470 -30.551 159.974

 The double area for line CD equals DMD of line CD times the latitude of line CD

Side DMD Double Areas -20.601 3,881 -6,919 45.447 -1,897 -63.375 -40,249 -289.397 -159.974

 The double area for line DE equals DMD of line DE times the latitude of line DE

Side

-188.388 -152.253 29.933 139.080 171.627



-188.388 -152.253 29.933 139.080 171.627

AB BC CD DE EA




AB BC CD DE EA

 The sum of the products of each points DMD and latitude equal twice the area, or the double area Side

 The double area for line BC equals DMD of line BC times the latitude of line BC

Side

10/13

36,320 ft.2 0.834 acre

AB BC CD DE EA


1 acre = 43,560 ft.2

-20.601 86.648 -195.470 -30.551 159.974

DMD Double Areas -20.601 3,881 45.447 -6,919 -63.375 -1,897 -289.397 -40,249 -159.974 -27,456 2 Area = -72,641

Area =

36,320 ft.2 0.834 acre

CIVL 1112


Surveying - Traverse Traverse Area - Double Area

11/13

Surveying - Traverse Traverse Area - Double Area

 The word "acre" is derived from Old English æcer (originally meaning "open field", cognate to Swedish "åker", German acker, Latin ager and Greek αγρος (agros).

 The word "acre" is derived from Old English æcer (originally meaning "open field", cognate to Swedish "åker", German acker, Latin ager and Greek αγρος (agros).

 The acre was selected as approximately the amount of land tillable by one man behind an ox in one day.

 A long narrow strip of land is more efficient to plough than a square plot, since the plough does not have to be turned so often.

 This explains one definition as the area of a rectangle with sides of length one chain (66 ft.) and one furlong (ten chains or 660 ft.).

 The word "furlong" itself derives from the fact that it is one furrow long.

Surveying - Traverse Traverse Area - Double Area  The word "acre" is derived from Old English æcer (originally meaning "open field", cognate to Swedish "åker", German acker, Latin ager and Greek αγρος (agros).

Surveying - Traverse Traverse Area – Example 4  Find the area enclosed by the following traverse

Side

Balanced Latitude Departure DMD

AB BC CD DE EA

600.0 100.0 0.0 -400.0 -300.0

Double Areas

200.0 400.0 100.0 -300.0 -400.0 2 Area =

1 acre = 43,560 ft.2

Surveying - Traverse DPD Calculations

Area =

ft. 2 acre

Surveying - Traverse Rectangular Coordinates

 The same procedure used for DMD can be used the double parallel distances (DPD) are multiplied by the balanced departures

 Rectangular coordinates are the convenient method available for describing the horizontal position of survey points

 The parallel distance of a line is the distance from the midpoint of the line to the reference parallel or east–west line

 With the application of computers, rectangular coordinates are used frequently in engineering projects  In the US, the x–axis corresponds to the east–west direction and the y–axis to the north–south direction

CIVL 1112




Rectangular Coordinates Example


In this example, the length of AB is 300 ft. and bearing is shown in the figure below. Determine the coordinates of point B y

Latitude AB =300 ft. cos(4230’) = 221.183 ft.

B

In this example, it is assumed that the coordinates of points A and B are know and we want to calculate the latitude and departure for line AB y

A

Latitude AB = -400 ft.

x

Departure AB = x B – x A Departure AB = 220 ft.

x B = 200 + 202.667 = 402.667 ft.

B



Rectangular Coordinates Example y

Consider our previous example, determine the x and y coordinates of all the points A

Side AB BC CD DE EA

B

E

D

-20.601 86.648 -195.470 -30.551 159.974

x

C

Surveying - Traverse Rectangular Coordinates Example y

A

E

 x coordinates E = 0 ft.

B D

A = E + 159.974 = 159.974 ft. x

C

B = A – 20.601 = 139.373 ft. Side AB BC CD DE EA

Balance d Latitude De parture -188.388 -152.253 29.933 139.080 171.627

-20.601 86.648 -195.470 -30.551 159.974

AB BC CD DE EA

D = C – 195.470 = 30.551 ft. E = D – 30.551 = 0 ft.

Rectangular Coordinates Example y

A (159.974, 340.640)

C = 0 ft. D = C + 29.933 ft. C

x B (139.373, 152.253)

E = D + 139.080 = 169.013 ft. Side

C = B + 86.648 = 226.021 ft.


 y coordinates

B D

A

E


x

Coordinates of Point B (320, -100)

y B = 300 + 221.183 = 521.183 ft.


y

Latitude AB = y B – y A

Coordinates of Point A (100, 300)

A

Departure AB =300 ft. sin(4230’) = 202.677 ft.

N 42 30’ E

Coordinates of Point A (200, 300)

12/13


-20.601 86.648 -195.470 -30.551 159.974

(0.0, 169.013)

A = E + 171.627 = 340.640 ft.

E

B = A –188.388 = 152.252 ft. (30.551, 29.933)

D

C = B –152.252 = 0 ft. C (226.020, 0.0)

x

CIVL 1112



Surveying - Traverse Area Computed by Coordinates

Group Example Problem 5 Compute the x and y coordinates from the following balanced. Side

Balanced Length (ft.) Latitude Departure Latitude Departure

Bearing

Coordinates x y

Points

degree m inutes

AB BC CD DE EA

S S N N N

6 29 81 12 42

15 38 18 24 59

W E W W E

189.53 175.18 197.78 142.39 234.58 939.46

-188.403 -152.268 29.916 139.068 171.607 -0.079

-20.634 86.617 -195.504 -30.576 159.933 -0.163

-188.388 -152.253 29.933 139.080 171.627 0.000

13/13

-20.601 86.648 -195.470 -30.551 159.974 0.000

A B C D E

100.000

100.000

The area of a traverse can be computed by taking each y coordinate multiplied by the difference in the two adjacent x coordinates (using a sign convention of + for next side and - for last side)



Area Computed by Coordinates

Area Computed by Coordinates

y

A (159.974, 340.640)

Twice the area equals: = 340.640(139.373 – 0.0)

 There is a simple variation of the coordinate method for area computation y

B (139.373, 152.253) (0.0, 169.013) E

A (159.974, 340.640)

+ 152.253(226.020 – 159.974)

x1 y1

+ 0.0(30.551 – 139.373)

x2 y2

x3 y3

x4 y4

x5 y5

B (139.373, 152.253) (30.551, 29.933)

+ 29.933(0.0 – 226.020)

D

(0.0, 169.013) E

Twice the area equals:

C (226.020, 0.0) x

+ 169.013(159.974 – 30.551) = 72,640.433 ft.2 Area = 0.853 acre

Area Computed by Coordinates  There is a simple variation of the coordinate method for area computation A (159.974, 340.640)

B (139.373, 152.253)

Twice the area equals: 159.974(152.253) + 139.373(0.0) + 226.020(29.933) + 30.551(169.013) + 0.0(340.640)

(0.0, 169.013) E

(30.551, 29.933)

- 340.640(139.373) – 152.253(226.020) - 0.0(30.551) – 29.933(0.0) – 169.013(159.974)

D

= x1y2 + x2y3 + x3y4 + x4y5 + x5y1

D C (226.020, 0.0) x

- x2y1 – x3y2 – x4y3 – x5y4 – x1y5

Area = 36,320.2 ft.2


y

(30.551, 29.933)

C (226.020, 0.0) x

= -72,640 ft.2

Area = 36,320 ft.2

End of Surveying - Traverse Any Questions?

x1 y1

Surveying - traverse - for web.ppt - Memphis

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