engi 3703 surveying and geomatics - faculty of …sitotaw/site/fall2007_files/lecture11.pdf · engi...

ENGI 3703Surveying and Geomatics

TopicInstructor: Prof. Ken Snelgrove

Lect 11 - Oct 10/07 Slide 1 of 10

Traverse Calculations

Traversing Computations: (Chapter 10) Closed traverses like closed level loops offer a opportunity

to check ones work and make adjustments because extra measurements have been make to form a

loop. Adjustment steps include:

1. Balance Interior Angles [Lecture 10]

2. Determine Azimuths (preliminary) [Lecture 10]

3. Calculate latitudes and departures and adjust misclosure

4. Compute rectangular coordinates (x,y)

5. Calculate new length and azimuths coordinates

A

BC

D

E

101o24’00”

149o13’00”

080o58’30”

116o19’00”

092o04’30”

Example Traverse

634.347sum

051o22’00”122.402AB

332o46’12”106.710EA

244o50’54”176.488DE

181o10’06”112.252CD

082o08’48”116.495BC

Preliminary

Azimuths

Length

(m)

Course

Measured Distances and Prelim. Azimuths





Calculate Departures and Latitudes: (Section 10.4 and 10.5) We have dealt the angle component of

traverses and now we must consider the distances of each course within the traverse. We will

separate these distances into two rectangular components:1. Latitudes - the N-S component or Y with North (+) and South (-).

2. Departures - the E-W component or X with East (+) and West (-).

To calculate components we make use ofour previously calculate azimuths ( ) as:

departure = Lsin

latitude = Lcos

The signs of the departures and latitudes are

essentially looked after by use of the

azimuths. If bearings are used then signmanagement is required after component Y

and X are calculated.

We can use the lats and deps to determine in

traverse adjustment since the sum of each

must equal zero if the survey is perfect.





A

BC

D

E

Example Lats and Deps: (Section 10.5) To calculate the latitude (lats) and departures (deps) we

use the azimuths calculated from either the clockwise or counterclockwise approach. In the example

below we start at Point B and work around in a clockwise direction.

deps=0.147

95.615

-48.827

-159.754

-2.289

115.402

Departure (m)

lats=-0.017634.347sum

76.420051o22’00”122.402AB

94.884332o46’12”106.710EA

-75.010244o50’54”176.488DE

-112.229181o10’06”112.252CD

15.918082o08’48”116.495BC

Latitude (m)Preliminary

Azimuths

Length

(m)

Course

Dep BC (+)

Lat B

C (

+)

Lats and Deps, Clockwise Approach (Start at B)

Lat C

D (-)

Lat D

E (-)

Dep CD (-)

Dep DE (-)Dep EA (-)

Lat E

A (

+)

Lat A

B (

+)

Dep AB (+)





A

B

C

D

E

B’

Linear Misclosure and Relative Precision: (Section 10.6) Once calculated we can must determine

the sum of the lats and deps to determine both the magnitude and direction of the misclosure.

linear misclosure = departure misclosure( )2

+ latitude misclosure( )2

E = deps( )2

+ lats( )2

relative precision =linear misclosure

traverse length=E

P

tan Bearing Angle( ) =deps

lats

Example

E = 0.147( )2

+ 0.017( )2

E = 0.148

relative precision =E

P=

0.148

634.147=

1

4285

1

4300

tan Bearing Angle( ) =0.1470.017

B'B Error Bearing = 83.4033 = S83o2 4 11. 7 E

B'B Azimuth =180 83o2 4 11. 7 = 96o3 5 48. 3

B

B’

E=0.148

deps=0.147

lats=-0.017





Adjust Misclosure: (Section 10.7) Now that we know the bearing and magnitude of the closure error

we can determine how to deal with it. One method is via the Compass Adjustment Rule. Here we

simply pro-rate deps and lats errors onto the sides of the traverse based on their length relative to

the traverse perimeter. Short legs get little error, long legs get a lot as:

Corr deps AB

deps=

AB

P

Corr lats AB

lats=

AB

P

Corr deps AB =AB deps

P= AB kdeps Corr lats AB =

AB lats

P= AB klats

where: AB - is the measured distance AB

Corr deps AB - is the departure correction to apply to AB

Corr lats AB - is the latitude correction to apply to AB

P - is the perimeter of the traverse

lats & deps are the latitude and departure error

klats & kdeps are constants for each traverse = lats/P and deps/P, respectively

Note: the negative sign is used since these corrections are added





0.000

76.423

94.887

-75.005

-112.226

15.921

Corrected

Latitudes

0.000

95.587

-48.852

-159.795

-2.315

115.375

Corrected

Departures

0.017

0.003

0.003

0.005

0.003

0.003

Corr Lats

-0.147

-0.028

-0.025

-0.041

-0.026

-0.027

Corr Deps

deps

=0.147

95.615

-48.827

-159.754

-2.289

115.402

Departure

(m)

lats

=-0.017

634.347sum

76.420051o22’00”122.402AB

94.884332o46’12”106.710EA

-75.010244o50’54”176.488DE

-112.229181o10’06”112.252CD

15.918082o08’48”116.495BC

Latitude

(m)

Preliminary

Azimuths

Length

(m)

Course

Corrected Latitudes and Departures

Example: (cont)

kdeps =deps

P k lats =

lats

P

=0.147

634.347= 231.7 10 6 =

0.017

634.347= 26.80 10 6

Corr Dep BC = 231.7 10 6 116.495( ) Corr Dep BC = 26.80 10 6 116.495( )

= 0.027 = 0.003

Note: the sum of both corrected departures and latitudes must equal zero when complete.





Calculate Rectangular Coordinates: (Section 10.7) Prior to determining the revised azimuths and

distances of each traverse segment it is convenient first compute X, Y coodinates. By convention

we often make these positive quantities by assuming a large starting value. These are referred to as

Eastings and Northings. Coordinates are calculated as:

XB = XA + departure AB

YB = YA + latitude AB

Example (cont)

1000.000 (check)1000.000 (check)B

923.577

828.690

903.695

1015.921

1000.000 (assumed)

Northing

904.413

953.265

1113.060

1115.375

1000.000 (assumed)

Easting

A

E

D

C

B

76.423

94.887

-75.005

-112.226

15.921

Corrected

Latitudes

95.587

-48.852

-159.795

-2.315

115.375

Corrected

Departures

AB

EA

DE

CD

BC

StationCourse





A

B

C

D

E

X

Y

1000.0N

1000.0E

923.557N

904.413E

828.690N

953.265E

903.695N

1113.060E

1015.921N

1115.375E

Adjusted Coordinates:

0.0N

0.0E

1000 m

1000 m





Adjusted Distances and Azimuths: (Section 10.10) [skip 10.9] We must often layout an adjusted

traverse in the field. To do this we might use eastings and northings as input for a total station or

distances and azimuths of each course. Departures and coordinates can be used to calculate these:

From Latitudes and Departures

tan bearing AB( ) =dep ABlat AB

length AB = dep AB( )2

+ lat AB( )2

Since dep AB = X

and lat AB = Y

From Coordinates

tan bearing AB( ) =X

Y=XB XA

YB YA

length AB = X( )2

+ Y( )2

= XB XA( )2

+ YB YA( )2

Note: because tan-1 only produce angles 0 -> 90 (and -90 -> 0) we must

determine their quadrant from the signs of the dep and lat to make a final

azimuth conversion

NE

dep(+)

lat(+)

SE

dep(+)

lat(-)

SW

dep(-)

lat(-)

NW

dep(-)

lat(+)





51°21’25.9”

332°45’31.0”

244°51’19.8”

181°10’54.1”

82°08’35.5”

Azimuth

122.382

106.724

176.522

112.250

116.468

Course

Length

N51.3572E

N27.2414W

S64.8555W

S1.1817W

N82.1432E

Bearing

1000.0001000.000B

923.577

828.690

903.695

1015.921

1000.000

Northing

904.413

953.265

1113.060

1115.375

1000.000

Easting

A

E

D

C

B

76.423

94.887

-75.005

-112.226

15.921

Corrected

Latitudes

95.587

-48.852

-159.795

-2.315

115.375

Corrected

Departure

AB

EA

DE

CD

BC

StationCourse

Example: (cont)

tan bearing EA( ) =X

Y=XA XE

YA YE=

904.413 953.265

923.577 828.690=

48.852

94.887= 0.514844

bearing EA = tan 1 0.514844( ) = N27.2414W

Az EA = 360 bearing EA = 360 - 27.2414 = 332o4 5 31. 0

length EA = XA XE( )2

+ YA YE( )2

= 904.413 953.265( )2

+ 923.577 828.690( )2

=106.724

EA from Coordinates

engi 3703 surveying and geomatics - faculty of …sitotaw/site/fall2007_files/lecture11.pdf · engi...

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