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Palestine Polytechnic University

College of Engineering & Technology

Civil and Architectural Engineering Department

Surveying & Geomatics Engineering

ADVANCED PLANE SURVEYING

Lecture Notes Lecturer:

Eng. Ghadi Zakarneh

Hebron-Palestine

2007

Course Syllabus

COURSE: Advanced Plane surveying, 3 credit hours, CE223, summer semester 2006/2007. TIME and LOCATION: Sunday, Tuesday, Thursday 08:00-10:00,B303 INSTRUCTOR:

M.Sc. Ghadi Zakarneh Office: Wad Al-haria A404 Email: [email protected]

Eng.ghadi.googlepages.com Office Hour: Sunday, Tuesday 10:00-12:00 TEXT BOOK: Printed Lecture Notes by Lecturer REFERENCES:

James M. Anderson & Edward Michail ,”Surveying theory and practice”, 7th edition,1998.. Paul R. Wolf & Russell C. Brinkerl ,”Elementary Surveying”, 9th edition,. Andrew L.Harbin ,”Land Surveyor Reference Manual”, 2nd edition,1998.

ADDITIONAL MATERIALS or EQUIPMENT NEEDED FOR THE COURSE: Calculators, Theodelits, EDMs. COURSE DESCRIPTION: Theodolite concept and types. Theodolite adjustment. Angle measurements, bearings and azimuths by theodolite. Traverses. Traverse survey and computations. Campass rule. Introduction to coordinates systems. Rectangular coordinates and corrections. Coordinates by resection and intersection. Two dimensional conformal coordinate transformations. Two dimensional affine coordinate transformation. Three dimensional conformal coordinate transformations. Tacheometry and applications. INTENDED LEARNING OUTCOMES: [What are the Intended learning outcomes of this course?

Should be able to survey small lands and buildings using total stations and theodelites.

The student should be able to create traverses and their adjustment. The student should be able to determine the position of points using

resections and intersections. The student should be able to determine the elevations of points using

theodelites and tapes or EDMs.

mailto:[email protected]

Course Outline and Calendar:

Topics Weeks (or hours) Horizantal Angle measurement 2 week

Traverses 3 week Intersections 2 week Resections 2 weeks

Traverses missed data 2weeks Horizontal curves 1 week

Coordinate transformations 2 weeks

COURSE POLICIES: The University's and instructor's policy regarding the course:

Class attendance is important and mandatory. Unexcused absence is dealt through the grading. Reporting illnesses is wanted. Turning in late work is not accepted. Missing home works, tests or exams: needs formal proof for reasons.

TEACHING METHODS:

Lectures Assignments problem solving sessions discussions Lab and field work

ASSESSMENT MEASURES AND GRADING SYSTEM:

Written exams Assignments , quizzes, reports

Grading system: first exam 20%

Second exam 20% Final 40% Assignments 10% Class attendance 10%

TABLE OF CONTENTS

Chapter 1: Introduction.

Chapter 2:

Traverses Chapter 3:

Intersections Chapter 4:

Resection Chapter 5:

Traverse Missed Data Chapter 6:

Horizontal Curves Chapter 7:

Coordinates Transformations

Chapter 8: Survey Operations

Appendix: Stadia Units Conversions Trigonometric fuctions Previous Exams & Solutions

Advanced Plane Surveying Ch01: Introduction By: Eng.Ghadi Zakarneh

Ch01 Horizontal Angle

Measurement

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5

Palestine Polytechnic UniversitySurveying & Geomatics Eng.


HORIZONTAL ANGLE MEASUREMENT

1- Two Types of Theodolites

• Repeating - has both upper and lower motion • Directional - no lower motion

2- Theodolites or Transits can Measure

• horizontal angles • vertical angles • distances (stadia) • elevations

3- Basic components of the Theodelite

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4- Basic Components of an Angle

• reference or starting line (A) • direction of turn (B) • angular value (C) usually in DMS

AB

C

5- Setting up the theodelite

1. The tripod is centered approximately above the point. 2. The theodelite set above the tripod. 3. using the tripod legs the plummet is centered above the points.(moving the

legs of the tripod while looking in the eyepiece of the optical plummet) 4. The circular bubble is approximately catered by the three legs. 5. The longitudinal bubble is centered in two steps using the leveling screws

as shown below. 6. Slide the instrument carefully above the tripod while looking in the

eyepiece of the optical plummet to center the instrument above the point.

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8



6- Angle Measurement steps Follow these steps when turning each horizontal angle:

a. Allow one person at both the backsight point and foresight point to hold the plumb bob 1-3mm above the monument.

b. Set the horizontal circle to 0º 00’ 00” • For initial setup, align the upper motion mark and

intermediate plate mark. While looking through the reading sight, slowly adjust the plate until the view reads zero (see illustration below). Lock the upper motion and adjust by rotating the upper motion adjustment dial.

c. Sight the backsight using the LOWER MOTION • Lock the Lower motion and adjust until the vertical

crosshair splits the stringline of your rod person. d. Turn to the foresight using the UPPER MOTION

• Unlock the upper motion lock and advance to your foresight point. Use the lock and adjustment on the upper motion only.

e. Record horizontal angle to the nearest ¼ minute (15”) f. Sight the backsight using the LOWER MOTION

• Unlock the lower motion and turn back to your backsight point. Lock and adjust using only the lower motion. You will notice your reading will not change during this step.

g. Turn to the foresight again using the UPPER MOTION, using the same instructions as before to turn to the foresight point.

h. Record second horizontal angle, which should be double the first angle.

i. Divide the second reading by two (2) and compare with the first reading (should be within 30” of the first reading)

Face Right (Direct): Taking the angle measurement when the telescope in the normal case. Face Left (Reverse): Taking the angle when the telescope in the reverse case.(rotation the vertical angle by 180 degrees, and moving the telescope in the horizontal plane to point again), the difference between the direct and the reverse angle measurement is 180 degrees.

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7- Angles by repetition

1. The steps from a-i are done for n-times. 2. After the last measurement (the telescope is still in the foresight), reverse the

telescope and turn it back to the foresight, the difference between the last measurement and the new measurement is nearly 180 degrees.

3. Turn to the backsight again using the UPPER MOTION. And register the angle measurements (it should be nearly 180 different from it,s corresponding direct measurement).

4. Turn to the foresight again using the LOWER MOTION. 5. Repeat 3 and 4 n-times. 6. Take the average of the differences of direct and reverse values of the angle.

Example: NOTE: we start direct up to down backsight to foresight. The reverse doun to up and foresight to backsight.

Direct Reverse Backsight foresight Difference Backsight foresight Difference00 00 00 45 12 30 4512 30 180 00 12 225 12 47 45 12 35 45 12 30 90 25 15 45 12 45 225 12 47 270 25 25 45 12 38 90 25 15 135 37 50 45 12 35 270 25 25 315 37 58 45 12 33 135 37 50 180 50 30 45 12 40 315 37 58 0 50 35(+360) 45 12 37

Average = 45 12 36.625 8- Zenith Angles in Direct and Reverse If the zenith angle is measured n-time direct and n-time reverse then:

Example:

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Advanced Plane Surveying Ch02: Traverses By: Eng.Ghadi Zakarneh

Ch02 Traverses

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TRAVERSES 1- A traverse Is a series of consecutive lines whose lengths and directions have been determined from field measurements. Traversing is the act of establishing traverse stations. 2- Types of traverse

1. Open traverse

2. closed traverse (link traverse)

3. closed loop traverse

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2- Deflection Angle Traverse

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MN

LMVMN

RVUVM

RUTUV

LRTTU

RSRRT

AzgivenChechAzAzAzAzAzAzAzAzAzAz

5

4

3

2

1

ααααα

−=+=+=−=+=

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15



3- Interior Angle Traverse

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The Azimuth calculation is done by taking anticlockwise direction:

000597112 ′′′+= oPAzAz

22123 α+= AzAz 1802332 ±= AzAz

33234 α+= AzAz 1803443 ±= AzAz Other method is to add the corrections to the calculated azimuth as described in the table below:

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4- Angles to Right Traverse

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1α+= YZYA AzAz 180±= YAAY AzAz

2α+= AYAR AzAz 180±= ARRA AzAz

3α+= RARS AzAz 180±= RSSR AzAz

Other method is to add the corrections to the calculated azimuth as described in the table below:

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5- Traverse Computations

ijADXiDepXiXj sin.. +=+=

ijADXiLatYiYj cos.. +=+=

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This also called: Relative accuracy, or Ratio of Misclosure (RoM)

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Example-1:

Solution:

21

5'1 ′′=−=cpa

Azimuth Calculations

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Relative accuracy= RoM = 0.36 / 1762.29 = 1 / 4894

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Example-2:

Relative accuracy= RoM = 0.28 / 1769.76 = 1 / 6320

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6 - Traverse Adjustment (Compass Rule) The errors in the traverse (linear misclosure) can be reduced according to compass rule:

• coordinates corrections

• latitude and Departure corrections

• Adjusted azimuth and distance Using the adjusted coordinates , the azimuth and the distance can be calculated again, these value are the adjusted values:

CYYXX

Aij

ijij +⎟

⎟⎠

⎞⎜⎜⎝

⎛

−

−= −1tan

Where ;

The adjusted distance:

( ) ( )22

ijijij YYXXd −+−=

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Example: For the last example,

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7- Traverse Accuracy standards

• Federal Geodetic Control Subcommittee (FGCS 1984)

• Palestine Land survey accuracy standards

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Problems

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Name: Traverse Adjustment Form Date: Page ____ of ____

Preliminary Corrections to Adjusted Coordinates Adjusted

Station Direction Distance Latitude Departure Latitude Departure Latitude Departure X Y Direction Distance

33


Name: Traverse Adjustment Form Date: Page: ____ of _____ Readings Preliminary Preliminary Corrections Corrected Adjusted

St D Az Dep. Lat. Easting (X) Northing(Y) Xδ Yδ Easting Northing D Az

ΣD = dX= dY= Misclosure Error =

___________1RoM = Traverse Class=____________ □Accepted / □Rejected

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Advanced Plane Surveying Ch03: Intersections By: Eng.Ghadi Zakarneh

Ch03 Intersections

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35



INTERSECTIONS

A- Two angles Intersections

We measure the angles from a line between two known points to the new point. Then we can calculate the coordinates of the new points.

• To find point C : <A = <CAB clockwise <B = <ABC clockwise

• To find point C’ : <A = <BAC clockwise <B = <CAB clockwise

Calculation procedure:

1. Calculate the Azimuth of AB

CYYXXA

AB

ABAB +⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

= −1tan Or CNNEEA

AB

ABAB +⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

= −1tan

Where ;

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2. Calculate the length of AB

( ) ( )22ABAB YYXXAB −+−=

3. Calculate the angle <C=180-(<A + <B) 4. Calculate the Azimuth of AC and BC.

AAzAz ABAc <−= 180mABBA AzAz =

BAzAz BABc <+=

To find Azimuth AC’ and BC’ AAzAz ABAc <+='

180mABBA AzAz = BAzAz BABc <−=

5. Calculate the lengths of the lines AC and BC using sine Rule.

BC

ABAC sinsin

= and AC

ABBC sinsin

=

6. Calculate the coordinates of the C from A.

ACAAC AACXDepXX sin.. +=+=

ACAAC AACYLatYY cos.. +=+= Or ACAAC AACEDepEE sin.. +=+=

ACAAC AACNLatNN cos.. +=+=

7. Check by calculating the coordinates of the point C from B. BCBBC ABCXDepXX sin.. +=+=

BCBBC ABCYLatYY cos.. +=+= Or BCBBC ABCEDepEE sin.. +=+=

BCBBC ABCNLatNN cos.. +=+=

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Example: For the figure above find the coordinates of the point C. Calculate the Azimuth of AB

823583tan 1 ′′′=+⎟⎟⎠

⎞⎜⎜⎝

⎛−−

= − oCNNEEA

AB

ABAB

Calculate the length of AB

( ) ( ) mYYXXAB ABAB 12.33622 =−+−=

Calculate the angle <C=180-(<A+ <B) 814493 ′′′= oC

Calculate the Azimuth of AC and BC.

648250 ′′′=<−= oAAzAz ABAc 8235263180 ′′′== omABBA AzAz 8244316 ′′′=<+= oBAzAz BABc

Calculate the lengths of the lines AC and BC using sine Rule.

mBC

ABAC 48.268sinsin

==

mAC

ABBC 48.185sinsin

==

Calculate the coordinates of the C from A.

mAACEDepEE ACAAC 77.1751sin.. =+=+= mAACNLatNN ACAAC 03.1427cos.. =+=+=

Check by calculating the coordinates of the point C from B.

BCBBC ABCEDepEE sin.. +=+=


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B- Two Distance intersections

Two distances are measured from two known points to the new point.


1. Calculate the Azimuth of AB

CYYXXA

AB

ABAB +⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

= −1tan Or CNNEEA

AB

ABAB +⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

= −1tan

Where ;

2. Calculate the length of AB

( ) ( )22ABAB YYXXAB −+−=

3. Calculate the angle A using Cosine Rule.

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4. Calculate Angle <B using Sine Rule

BCAACB sinsin =

5. Calculate the angle <C=180-(<A + <B) 6. Calculate the Azimuth of AC and BC.

AAzAz ABAc <−= 180mABBA AzAz =

BAzAz BABc <+= To find Azimuth AC’ and BC’

AAzAz ABAc <+=' 180mABBA AzAz =

BAzAz BABc <−=

7. Calculate the coordinates of the C from A.

ACAAC AACXDepXX sin.. +=+=

ACAAC AACYLatYY cos.. +=+= Or ACAAC AACEDepEE sin.. +=+=

ACAAC AACNLatNN cos.. +=+=

8. Check by calculating the coordinates of the point C from B. BCBBC ABCXDepXX sin.. +=+=

BCBBC ABCYLatYY cos.. +=+= Or BCBBC ABCEDepEE sin.. +=+=


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Example: For the figure above find the coordinates of the point C.

Calculate the Azimuth of AB

640077tan 1 ′′′=+⎟⎟⎠

⎞⎜⎜⎝

⎛−−

= − oCNNEEA

AB

ABAB

Calculate the length of AB

( ) ( ) mYYXXAB ABAB 86.33522 =−+−=

Calculate the angle A using Cosine Rule.

108340 ′′′= oA

Calculate Angle <B using Sine Rule

254329sinsin ′′′== o

BCAACB

Calculate the Azimuth of AC and BC.

542236 ′′′=<−= oAAzAz ABAc 6400257180 ′′′=+= o

ABBA AzAz 8353286 ′′′=<+= oBAzAz BABc

Calculate the coordinates of the C from A.

mAACEDepEE ACAAC 95.729sin.. =+=+= mAACNLatNN ACAAC 01.1362cos.. =+=+=

Check by calculating the coordinates of the point C from B.

BCBBC ABCEDepEE sin.. +=+=


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C- finding the elevation of inaccessible point

To find the elevation of a point that can not be accessed like point P in the above figure, the following measurements are done:

• The vertical angles and or and 1v 2v 1z 2z• The horizontal angles A and B • The horizontal distance AB


• Calculate the horizontal angle <P=180-(<A + <B) • Calculate the lengths of the Horizontal lines AP and BP using sine Rule.

BP

ABAP sinsin

= and AP

ABBP sinsin

=

• Calculate the Elevation of the P from A.

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)cot(...

)tan(...

1

1

zAPhiLRLR

vAPhiLRLR

AAP

AAP

A

A

++=

++=

• Calculate the Elevation of the P from B.

)cot(...

)tan(...

2

2

zBPhiLRLR

vBPhiLRLR

BBP

BBP

B

B

++=

++=

• The elevation of P is:

2..

. BA

B

PPP

LRLRLR

+=

Example: Stations A and B have elevations of 298.65 ft, and 301.53 ft, respectively, and the instrument heights at A and B are hiA = 5.55 ft, and hiB = 5.48 ft. The other field measurements are:

What is the elevation of the chimney stack P?

• Calculate the horizontal angle <P=180-(<A + <B) 030296 ′′′°= • Calculate the lengths of the Horizontal lines AP and BP using sine Rule.

ftBP

ABAP 233.87sinsin

==

ftAP

ABBP 730.95sinsin

==

• Calculate the Elevation of the P from A.

ftvAPhiLRLR AAPA791.316)tan(... 1 =++=

• Calculate the Elevation of the P from B.

ftvBPhiLRLR BBPB795.316)tan(... 2 =++=

• The elevation of P is:

ftLRLR

LR BA

B

PPP 793.316

2..

. =+

=

H.W : find the zenith angle and use it to find the elevation of P?

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Notes about Coordinates Systems In some countries the coordinates systems is define through using X as the easting coordinates , and Y as the northing coordinates. As we normally deal. Then:

ACAACC AACXDepXEX sin.. +=+==

ACAACC AACYLatYNY cos.. +=+==

CNNEEC

YYXXA

AB

AB

AB

ABAB +⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

=+⎟⎟⎠

⎞⎜⎜⎝

⎛−−

= −− 11 tantan

this is used in the USA and other countries. In other countries the System is the opposite through using X as the northing coordinates, and Y as the easting coordinates. Then you can arrange your solution through:

• Interchange the names of the coordinates, use X as Y and Y as X. in the results interchange the names again.

• The other way is to calculate the coordinates and azimuths through:

ABAABB AABYDepYEY sin.. +=+==

ABAABB AABXLatXNX cos.. +=+==

CNNEEC

XXYYA

AB

AB

AB

ABAB +⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

=+⎟⎟⎠

⎞⎜⎜⎝

⎛−−

= −− 11 tantan

This system is used in Palestine, Britain, Germany and other countries.

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D- Points Stake out (layout)by angles to right to fix points on the ground which are not identified, but they have known coordinates the following procedure id followed, to fix pint C on the ground using the EDM on point B and zero direction at point A, where points A and B have known coordinates and well identified on the ground, while the point C has to be identified on the ground:

1. calculate the length of the line BC

( ) ( )22CBCB NNEEBC −+−=

2. calculate the azimuth of the line BA

CNNEEAz

BA

BABA +⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

= −1tan

3. calculate the azimuth of the line BC

CNNEE

AzBC

BCBC +⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

= −1tan

4. calculate the clockwise angle ABC BABC AzAzABC −=<

If 0<< ABC then 360+<=< ABCABC 5. Set the EDM above the point B. 6. Set the instrument zero to A.

7. Read and angle with value of <ABC.

8. Fix the EDM direction.

9. Ask the reflector person to above nearer and further until the distance BC

is measured accurately in the fixed direction.

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Example: The given points below in graph are the centers of columns for a new building, the coordinates of the columns 60-72 are listed below. Prepare the calculations to mark the columns centers on the ground. The instrument is to be setup on the station 73 and the instrument (EDM) is to be zeroed at 74?

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Advanced Plane Surveying Ch04: Resection By: Eng.Ghadi Zakarneh

Ch04 Resection

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48



RESECTION

In this method, a theodelite is set up on a new point with unknown coordinates, and angles between three known points are measured to find the coordinates of the new point. 1- Derivation Using the figure below, O is the new point . A, B, and C are points with known coordinates. α and β are the observed angles. To solve the problem we need to calculate θ and γ .

We define R:

γθ +=R Using the Sine Rule in triangles OAC and OAB:

αθ

βγ

sinsin.

sinsin. cbAO ==

Arrange the above equation forθ :

βαγθ

sin.sin.sin.sin

cb

=

Because γθ −= R , using trigonometric functions for the sine of difference of two angles:

γγγθ sincoscos.sin)sin(sin RRR −=−=

βαγγγ

sin.sin.sin.sincoscos.sin

cbRR =−

Divide the above equations by γsin.sin R , then:

RcbR

sin.sin.sin.cotcotβαγ +=

γθ −= R

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2- Problem cases

1. The problem has no solution if the clockwise angle 180=++< βαBAC because . ∞⇒Rcot

2. The problem has basically three cases. Depending on the figure below. • Cases (a,b):

( )βαφγθ ++−=+= 360R

• Case ( c) : βαφγθ −−=+=R

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3- Calculation procedure

1. Determine the case of your solution, if: • bcase⇒>+ 180βα • acaseAzAzBAC AcAB ⇒<−=< 180 • ccaseAzAzBAC AcAB ⇒>−=< 180

2. Using the coordinates of points A,B and C ,calculate the distances : a,b, and c. 3. calculates the angle φ :

• case a : AcAB AzAzBAC −==<φ • case b: ABAC AzAz −=φ • case c: ABAC AzAz −=φ

4. compute R: • cases (a,b): ( )βαφ ++−= 360R • case ( c): βαφ −−=R

5. calculate γ :

RcbR

sin.sin.sin.cotcotβαγ +=

6. calculate θ γθ −= R

7. calculate AO:

αθ

βγ

sinsin.

sinsin. cbAO ==

8. compute <CAO and <BAO: ( )βγ +−=< 180CAO

)(180 αθ +−=< BAO 9. Calculate the line OC and OB, using sine Rule:

βsin)sin(. CAObOC <

=

αsin)sin(. BAOcOB <

=

10. calculate azimuth of CO and BO: • cases a,c: γ−= CACO AzAz θ+= BABO AzAz • case b: γ+= CACO AzAz θ−= BABO AzAz

11. calculate the coordinates of O: COCCO AOCXDepXX sin.. +=+=

COCCO AOCYLatYY cos.. +=+= 12. Check,

BOBBO AOBXDepXX sin.. +=+=

BOBBO AOBYLatYY cos.. +=+=

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Solution

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52



NOTE this solution depends of the figure: You follow the steps before to solve the problem.

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Problems:

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55


Advanced Plane Surveying Ch05: Traverse Missed Data By: Eng.Ghadi Zakarneh

Ch05 Traverse

Missed data

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56



TRAVERSE MISSED DATA When some measurements are missed in a traverse, you nee to complete these data, it is not always necessary to do field measurement again. Here are some examples that show how to find the values of the missed data: 1- Two Distances are missed Depending ing the figure below, for a closed polygon we can solve the problem as below:

In a traverse, we start at a known point (X,Y) and we end at other known point, then: firstlast XXAzDDep −==∑ ∑ sin.

∑ ∑ −== firstlast YYAzDLat cos.

NOTE: in close loop traverse the first point is also the last point. Then we have for the points with known distance and azimuths :

( ) ( ) ( )( ) ( ) ( )∑ ∑

∑ ∑Δ=−−=−−

Δ=−−=−−

YYYAzDYYLat

XXXAzDXXDep

firstlastfirstlast

firstlastfirstlast

cos.

sin. .1

For a closed loop traverse:

∑ ∑∑ ∑

Δ==

Δ==

YAzDLat

XAzDDep

cos.

sin. .1a

Then if the unknown distance are D1 and D2, we have :

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.a .b

By eq.a we get:

.2

.3a

Or .3b Example: For the following closed loop traverse find the missed data

Using eq.1a:

Using eq.3:

Using eq.2

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2- A Distance of a line and a direction of other line are missed Suppose the following:

• the unknowns are and , 1Az 2D• and 2sin AzM = 2cos AzN = .

Using eq.1 or eq.1a:

( ) ( ) ( )( ) ( ) ( )∑ ∑

∑ ∑Δ=−−=−−

Δ=−−=−−

YYYAzDYYLat

XXXAzDXXDep

firstlastfirstlast

firstlastfirstlast

cos.

sin. .1

∑ ∑∑ ∑

Δ==

Δ==

YAzDLat

XAzDDep

cos.

sin. .1a

We have: Arrange the above equations for and : 1sin Az 1cos Az

.4a

.4b Squaring the above equations:

.c

.d The summation of eq.c and eq.d:

But , then: 1cossin 22 =+ αα

.e

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In the above equation we have 2nd order equation with unknown : 2D

a

acbbD2

42

2−±−

=

But , then a=1, and : 1cossin 22 =+ αα

242

2cbbD −±−

= .5a

Or

.5b

The value of , we calculate the sine and cosine of the azimuth using eq.4a and eq.4b: 1Az

.4a

.4b To find the quarters by finding or using the following graph: Aztan

cAzAzAz +⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

1

111 cos

sintan .6

**** Keep the algebraic signs of when you add the value of c.1tan −

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Example: For the following closed loop traverse find the missed data

Using eq.1:

Using eq.5b:

Using eq.4a and eq.4b:

=-0.6723955146

=-0.7401920491 The sine and the cosine are negatives; this means the line is in the 3rd quarter. Then using eq.6 we get:

8051222180805142180cossintan

1

111 ′′′=+′′′=+⎟⎟

⎠

⎞⎜⎜⎝

⎛= − oo

AzAzAz

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To check calculate the traverse through the summation of latitudes and departure: ( )( )∑

∑−=

−=

firstlast

firstlast

YYLat

XXDep

Here it is a closed loop traverse, the first point is the last one, then:

∑∑

=

=

0

0

Lat

Dep

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3- Two azimuths are unknown Suppose the following:

• the unknowns are and , 1Az 2Az• and 2sin AzM = 2cos AzN = .

Using eq.1 or eq.1a:

( ) ( ) ( )( ) ( ) ( )∑ ∑

∑ ∑Δ=−−=−−

Δ=−−=−−

YYYAzDYYLat

XXXAzDXXDep

firstlastfirstlast

firstlastfirstlast

cos.

sin. .1

∑ ∑∑ ∑

Δ==

Δ==

YAzDLat

XAzDDep

cos.

sin. .1a

We have:

.f

.g

Squaring the above two equations:

By The sum of the two equations, and , we: 1cossin 22 =+ αα

Rearranging:

Divide by :

If : Then:

5 - 8

63



But: We get:

But:

Then ,

.7 Using eq.f , eq.g , and and 2sin AzM = 2cos AzN = :

cAzAzAz +⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

1

111 cos

sintan .8

****NOTE: for this solution the quarter of the must be known. Unless the solution can not be correct.

2Az

Example: If the 4th line is in the 3rd quarter, what are value of the missed data :

h.w ? use eq.1 use eq.7 use eq.8

5 - 9

64



Problems

5 - 10

65


Advanced Plane Surveying Ch06: Horiz. Circular Curves By: Eng.Ghadi Zakarneh

Ch06 Horizontal

Circular Curves

6 -1

66



HORIZANTAL CIRCULAR CURVES

Many features, like roads, consist of a series of straight-line segments brought together by horizontal and vertical curves. A horizontal curve is an arc placed between two lines called tangents. Types of horizontal Curves:

• Simple Curve. • Compound Curve. • Broken-Back Curve. • Reverse Curve.

Other type of horizontal curves is used; it is called the transition or spiral curves. These curves are used to move from straight line to a circular curve and to apply the superelevation. The figure below shows the methods of applying transition curves.

6 -2

67



1- Degree of the Curve (D) An important relationship is the association between the degree of curve and the radius. The degree of curve is the angle at the center of the circle that lays out 100’. The radius is the distance from the center of the circle to the curve itself. There are two methods of defining this relationship:

• Arc method; most common approach. • Chord method.

• Arc Method

ft .1a

In metric units :

DR 1746= meter .1b

• Chord Method

( )2sin'50

DR = ft 1.c

)2sin(

24.15D

R = meter 1.d

6 -3

68



2- Elements of the Circular curve

The elements of the horizontal circular curve are shown in the figure below:

6 -4

69


Advanced Plane Surveying Ch06: Horiz. Circular Curves By: Eng.Ghadi Zakarneh Tangent distance:

2. The long chord:

.3 The middle ordinate:

.4

The external distance:

.5 the length of curve: in radians .6a

in degrees .6b

6 -5

70


Advanced Plane Surveying Ch06: Horiz. Circular Curves By: Eng.Ghadi Zakarneh EXAMPLE: Calculate the elements of the horizontal curve, while D=3º40`00``, ∆=24º20`00``, ST. (PI) =5+062.48. Solution:

28.47637.1746==

DR

T= R*tan )2

(Δ =102.685 m.

Lc=2R*sin )2

(Δ =200.75 m.

M=R-R*cos )2

(Δ =10.97 m.

E= .943.10)

2cos(

mRR=−

Δ

L= .17.202180

mR=

ΔΟ

π

ST. (PC) =ST (PI) – T= 4+959.795 m. ST (PT) =ST (PC) +L= 5+ 161.965 m.

6 -6

71



3- Circular curve setting out (layout) 3.1- small chords and deflection angles (using thedelite and tape )

The principle is that the angle from the PC where the theodelite is setup to a point on a circular curve (δ) is one-half of the central angle (d) subtended.

d1/21 =δ .7

π

π 180180

.RsddRs =⇒= .8

( )2sin..2 dRc = .9

Using the above principle the circular curve is divided to sub chords ( c ) , these chords are normally smaller than R/20 t, this make the chords nearly equal to arcs.

scRc ≈⇒≤20

6 -7

72



• The first selection of the arc ( s1 ) is done to make the station to be rounded to regular number. Then the deflection angle is:

π18011

Rsd =

And the the chord length ( c1 ) is : ( )2

1sin..21 dRc =

• the internal arcs ( s ) are selected to have rounded number of stations, then the deflection angle is:

π180

Rsd =

And the the chord length ( c1 ) is : ( )2sin..2 dRc =

• for the last arc the length is fixed to make ( s1 + n.s + s2 =L ), then the deflection

angle is:

π18022

Rsd =

And the the chord length ( c1 ) is : ( )2

2sin..22 dRc =

• The deflection angles to points on the circular curve are cumulative as follows:

6 -8

73



M

M

)2/(52/12/2/2/2/2/14)2/(32/12/2/2/2/14

)2/(22/12/2/2/132/2/12

2/11

ddddddddddddd

ddddddd

d

+=++++=+=+++=

+=++=+=

=

δδδδδ

6 -9

74


Advanced Plane Surveying Ch06: Horiz. Circular Curves By: Eng.Ghadi Zakarneh EXAMPLE:- Find the chords and deflections by the way of (deflection with small chords) while R=476.28 m, L=202.17 m, ∆=24º20`00``, ST (PC) = 4+959.97 m, ST (PT) = 5+161.965 m , use S=20 m ? Solution:-

ST (1) = 4+980 m. S1=ST (1) – ST (PC) = 20.21 m.

44.5621121144.5252218011 ′′′°==⇒′′′°==

dRsd δ

π

( ) mmdRc 21.20209.2021sin..21 ≈==

S=20

57.102112

94.21422180 ′′′°==⇒′′′°==d

Rsd δ

π

( ) mmdRc 20999.192sin..2 ≈==

S2=21.96

61.1591122223.3083218022 ′′′°==⇒′′′°==

dRsd δ

π

( ) mmdRc 96.21958.2122sin..22 ≈==

Station S C d/2 δ (PC) =

4+959.97 0 0 0 0

4+980 20.21 20.209 44.56211 ′′′° 44.56211 ′′′° 5+000 20 20 57.10211 ′′′° 91.7522 ′′′° 5+020 20 20 57.10211 ′′′° 49.17733 ′′′° 5+040 20 20 57.10211 ′′′° 96.28944 ′′′° 5+060 20 20 57.10211 ′′′° 44.39106 ′′′° 5+080 20 20 57.10211 ′′′° 91.50317 ′′′° 5+100 20 20 57.10211 ′′′° 49.0628 ′′′° 5+120 20 20 57.10211 ′′′° 96.11839 ′′′° 5+140 20 20 57.10211 ′′′° 44.220510 ′′′°

5+161.96 21.96 21.96 61.15911 ′′′° 6.37912 ′′′° 2/6.37912 Δ≈′′′°=∑

6 -10

75



3.2- long chords and deflection angles(using Total Stations) the idea is the same basic idea as in the previous method the following relations are used, the total station is setup on point PC , and distance are measured from PC to the point on the curve. As shown in the figure below.

The basic relations are as describe below:

• The arc ( S ) is the distance from PC to the point on the curve.

• The deflection angle from the tangent is :

π180iδ

21

2 RSdi

i ==

RCi sin..2

• The chord that represents the distance which has to be read with the Total station is:

( ) mmi 21.20209.20 ≈=δ =

6 -11

76


Advanced Plane Surveying Ch06: Horiz. Circular Curves By: Eng.Ghadi Zakarneh Example :- Prepare the calculation for Setting out the curve with (deflection angle and long chord) , where L=202.17 m, R =476.28 m, ∆=24º20`, T=102.68 m.

Station S C δ

(pc) 4+959.79 0 0 0 4+980 20.21 20.22 1º12`58.48`` 5+000 40.21 40.21 2º25`11.4`` 5+020 60.21 60.21 3º37`24.46`` 5+040 80.21 80.15 4º49`37.40`` 5+060 100.21 100.07 6º1`50.43`` 5+080 120.21 119.952 7º14`00`` 5+100 140.21 139.78 8º26`16.14`` 5+120 160.21 159.537 9º38`29.39`` 5+140 180.21 179.228 10º50`42.35``

5+161.965 202.175 200.765 12º10`1.08``

6 -12

77



3.3- Setting out the curve using offset from the tangent (x, y) coordinate from the tangent

Using the calculation of the above method the xy-coordinates are calculated for a point are calculated:

δcos.ii Cx = δsin.ii Cy =

Example: Prepare the calculation for Setting out the curve with (deflection angle and long chord) , where L=202.17 m, R =476.28 m, ∆=24º20`, T=102.68 m.

Station S C δ x y

(pc) 4+959.79

0 0 0 0 0

4+980 20.21 20.22 1º12`58.48`` 20.21 0.429 5+000 40.21 40.21 2º25`11.4`` 40.17 1.698 5+020 60.21 60.21 3º37`24.46`` 60.08 3.804 5+040 80.21 80.15 4º49`37.40`` 79.86 6.744 5+060 100.21 100.07 6º1`50.43`` 99.52 10.5135+080 120.21 119.952 7º14`00`` 118.997 5+100 140.21 139.78 8º26`16.14`` 5+120 160.21 159.537 9º38`29.39`` 5+140 180.21 179.228 10º50`42.35``

5+161.965 202.175 200.765 12º10`1.08`` 6 -13

78



3.4- setting out the horizontal curve by offset from the long chord

Using the calculation of the of the deflections and chords in section 3.4 the xy-coordinates shown in the figure below are calculated for a point as follows:

δθ −Δ= 2

θcos.ii Cx = θsin.ii Cy =

3.5- setting out the curve by radial distances from the tangent to the curve centre

Using the calculation of the deflections and chords in section 3.4 the xy-coordinates shown in the figure below are calculated for a point as follows:

⎟⎠⎞

⎜⎝⎛=⇒= −

Rx

ddRx ii

1tan)tan(.

RxRyi −+= 22

6 -14

79


Advanced Plane Surveying Ch06: Horiz. Circular Curves By: Eng.Ghadi Zakarneh Problems 6.1 For a highway a curve is to be designed with degree=2°, and deflection angle is 75°,

the station of the of PI= 2+500, calculate the curve elements using chord definition, and use small chords method to prepare setting out table?

6.2 For a highway a curve is to be designed with degree=3°, and deflection angle is 75°,

the station of the of PI= 2+500, calculate the curve elements using arc chord definition, and use small chords method to prepare setting out table?

6.3 For problem 6.2, using long chords method to prepare setting out table? 6.4 For problem 6.2, using offsets from tangent method to prepare setting out table? 6.5 For problem 6.2, using offsets from the long chord method to prepare setting out

table? 6.6 For problem 6.2 , using radial distances from the tangent to the curve centre method

to prepare setting out table?

6 -15

80


Advanced Plane Surveying Ch07: Coord. Transformations By: Eng.Ghadi Zakarneh

Ch07 Coordinates

Transformations

7 -1

81



COORDINATES TRANSFORMATIONS

When dealing with map, we have maps with different scales, orientations and coordinates origins. Then coordinates are used to move the coordinates from one map system to the other. 1- 2D conformal coordinates transformations In this type of coordinate’s transformations as shown in the figure below, we have three steps; scale change, rotation and two translations:

Step 1: Scale Change

• The lengths of lines ab and AB are unequal; hence the scales of the two coordinate systems are unequal.

• The scale of the XY system is made equal to that of the EN system by multiplying each X and Y coordinate by a scale factor s. The scaled coordinates are designated as X’ and Y’.

• By use of the two control points, the scale factor is calculated in relation to the two lengths AB and ab as:

Step 2: Rotation

• If the scaled X’Y’ coordinate system is superimposed over the EN system , so that line AB in both systems coincide, the result is as shown in Figure below.

• An auxiliary axis system E’N’ is constructed through the origin of the X’Y’ axis system parallel to the EN axes.

• It is necessary to rotate from the X’ Y’ system to the E’N’ system, or in other words, to calculate E’N’ coordinates for the unknown points from their X’ Y’ coordinates.

7 -2

82



• The E’N’ coordinates of point C may be calculated in terms of the clockwise angle θ by using the following equations:

Where,

βαθ +=

Or more general : ABab AzAz +=θ

Step 3: Translation • The final step in the coordinate transformation is a translation of the origin of the

E’N’ system to the origin of the EN system. • The translation factors required are and , which are illustrated in the above

figure. Final E and N ground coordinates for points C then are: ET NT

7 -3

83



7 -4

84



7 -5

85



Other method to write the equations: For the following coordinates transformations equation:

If we suppose θcos.sa = and θsin.sb = , then:

7 -6

86



Matrix Solution to find the parameters of coordinates transformations Suppose we have two points A and B, known in the source coordinates system XY and in the target coordinates system EN. We use these points to find the parameters of the coordinate’s transformation.

The matrix solution for any group of equations is:

LAX = Where,

A: is the design matrix, where each equation gives a row and each unknown gives a column. The elements of A are the factors multiplied by the unknowns in the equation. Each point gives two equations in E-direction and N-direction. So if we have n-points then the size of A is 2nX4. In the above group of equations:

4210011001

×

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=

nB

B

A

A

NE

XbYbYbXb

XaYaYaXa

NENE

TTba

A

X: the vector of unknowns the size is fixed: 14× .

14×

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

N

E

TTba

X

L: the vector of observations, it contains the observed EN-coordinates of the known points. If we have n-points the size of L is 12 ×n .

12 ×

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nB

B

A

A

NENE

L

The solution to find the unknown is as follows: ( ) LAAAX TT 1−

= For the previous example the calculated parameters can be used to calculate the EN-coordinates of the new point C.

***NOTE: to solve the matrix solution software is needed to multiply and inverse the matrices. e.g. EXCEL, MATLAB, Maple, MathCAD… etc.

7 -7

87



2- Affine coordinates Transformations In the affine coordinates transformations, to transform from xy-coordinates to XY-coordinates. As shown in the figure below, we have:

• Different scales in the x direction and y-direction. • Rotation angle. • The x-axis and y-axis are not orthogonal. • Two translations in the x and y directions.

Step 1: scaling in the x and y direction If we scale the xy-coordinates we get new coordinates system y’y’-coordinates:

Step 2: non-orthogonality correction

this step we find new coordinates system x”y”. so that x”-axis and y”-axis are orthogonal. In the figure below, ε is the non-orthogonality angle.

7 -8

88



Step 3: Rotation In this step we rotate the x”y”-coordinates X’Y’-coordinates, where X’Y’-coordinate are parallel to XY-coordinates. θ is the rotation angle between the y”-axis and the Y’-axis:

Step 4: Translation

In this step we add shift values to X’Y’-coordinates to get XY required coordinates:

Using the values of X’Y’, we get

This can be arranged as follows:

7 -9

89



In other way, we arrange the above equations as follows:

Using the trigonometric function relation (see the appendix):

Other form of the above equation is:

Where,

In the opposite direction we get:

7 -10

90



Matrix Solution to find the parameters of the affine coordinates transformations Suppose we have n-points, known in the source coordinates system xy and in the target coordinates system XY. We use these points to find the parameters of the co ordinate’s transformation. For each point we have two equations.

The matrix solution for any group of equations is:

LAX = Where,

A: is the design matrix, where each equation gives a row and each unknown gives a column. The elements of A are the factors multiplied by the unknowns in the equation. Each point gives two equations in X-direction and Y-direction. So if we have n-points then the size of A is 2nX6. If we have 4-points known in :

6244

44

33

33

22

22

11

11

4

4

3

3

2

2

1

1

210210

10000001100000011000000110000001

×⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nyxyx

yxyx

yxyx

yxyx

YXYXYXYX

bbbaaa

A

X: the vector of unknowns the size is fixed: 16× .

162

1

0

2

1

0

×⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

bbbaaa

X

L: the vector of observations, it contains the observed EN-coordinates of the known points. If we have n-points the the size of L is 12 ×n .

7 -11

91



124

4

3

3

2

2

1

1

×⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nYXYXYXYX

L

The solution to find the unknown is as follows: ( ) LAAAX TT 1−

=

Solution: The four common points A,B,C, and D form the following group of equations:

7 -12

92



The matrix form is:

( ) LAAAX TT 1−=

For the new points 1,2, and 3, the calibrated coordinates:

7 -13

93



3- 3D-conformal coordinates transformation In this type of coordinate’s transformations we transform the coordinates from xyz-coordinates system to other system XYZ-coordinates system. This type is commonly used in GPS and Photogrammetry.

The scale is not the same in both systems. And translations are needed. As shown in the figure above, the two system xyz-system and XYZ-system are not parallel. So we have three rotations show in the figure below:

• ω : Rotation about x-axis. • φ : rotation about y-axis. • : Rotation about z-axis. k

7 -14

94



Step 1: Rotation In this step we define a new coordinates system x’y’z’ that has the same scale of xyz-system, and it is parallel to XYZ-system.

In derivation we derive starting from x’y’z’ to xyz:

1. Rotation about x’: in this step we transform x’y’z’ to x1y1z1 by rotating x’y’z’ by angle ω .

2. Rotation about y1: in this step we transform x1y1z1 to x2y2z2 by rotating x1y1z1 by angleφ .

7 -15

95



Substituting x1,y1,z1:

3. Rotation about z2: in this step we transform x2y2z2to x3y3z3 by rotating x2y2z2 by angle k .

Substituting x2,y2,z2:

Then:

7 -16

96



In simpler form for writing:

Where,

The above equations can be written in matrix form:

Where,

For the matrix M. it is orthogonal matrix, this mean:

TMM =−1

Then to find the coordinates in the opposite direction:

Step 2: Scaling and translating

In this step we scale thy x’y’z’-coordinates to the same scale factor of XYZ-coordinates, and we shift them the to get same origin for both system:

7 -17

97



Matrix Solution of 3D conformal coordinates transformations Equations of 3D conformal coordinate’s transformations are non linear equations; to

solve these equations need to be linearized using Taylor series. The linearized form of

those equations is:

xz

Py

y

Px

x

PPPPPPP dT

TXdT

TXdT

TXdXdXdXds

sXXX

00000000)( ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

+= κκ

φφ

ωω

xz

Py

y

Px

x

PPPPPPP dT

TY

dTTY

dTTY

dY

dY

dY

dss

YYY

00000000)( ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠

⎞⎜⎝

⎛∂∂

+⎟⎠

⎞⎜⎝

⎛∂∂

+= κκ

φφ

ωω

xz

PY

y

Px

x

PPPPPPP dT

TZ

dTTZ

dTTZ

dZ

dZ

dZ

dss

ZZZ

00000000)( ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠

⎞⎜⎝

⎛∂∂

+⎟⎠

⎞⎜⎝

⎛∂∂

+= κκ

φφ

ωω

The above equation can be written for three control points as follows:

[ ]017161514131211 )( PPzyx XXdTadTadTadadadadsa −=++++++ κφω[ ]027262524232221 )( PPzyx YYdTadTadTadadadadsa −=++++++ κφω[ ]037363534333231 )( PPzyx ZZdTadTadTadadadadsa −=++++++ κφω[ ]047464544434241 )( QQzyx XXdTadTadTadadadadsa −=++++++ κφω[ ]057565554535251 )( QQzyx YYdTadTadTadadadadsa −=++++++ κφω

[ ]067666564636261 )( QQzyx ZZdTadTadTadadadadsa −=++++++ κφω[ ]077767574737271 )( RRzyx XXdTadTadTadadadadsa −=++++++ κφω[ ]087868584838281 )( RRzyx YYdTadTadTadadadadsa −=++++++ κφω

[ ]097969594939291 ( )RRzyx ZZdTadTadTadadadadsa −=++++++ κφω

where,

)()()( 31211111 ppp zmymxmsXa ++=∂∂

=

012 =∂∂

=ωXa

[ ]szyxXa ppp )(cos)(sinsin))(cossin(13 φκφκφφ

++−=∂∂

=

115 =∂∂

=xT

Xa

016 =∂∂

=yT

Xa 017 =∂∂

=zT

Xa

7 -18

98



)()()( 32221221 ppp zmymxmsYa ++=∂∂

=

[ ]szmymxmYa ppp )()()( 33231322 −−−=∂∂

=ω

[ ]szyxYa ppp ))(sin(sin)(sincossin())(coscossin(23 φωκφωκφωφ

+−+−=∂∂

=

[ ]symxmYa pp )()( 122224 −=∂∂

=κ

025 =∂∂

=xT

Ya 126 =∂∂

=yT

Ya 027 =∂∂

=zT

Ya

)()()( 33231331 ppp zmymxmsZa ++=∂∂

=

[ ]szmymxmZa ppp )()()( 32221232 ++=∂∂

=ω

[ ]szyxZa ppp ))(sincos()(sincos(cos))(coscoscos(33 φωκφωκφωφ

−++−=∂∂

=

[ ]symxmZa pp )()( 132334 −=∂∂

=κ

035 =∂∂

=xT

Za 036 =∂∂

=yT

Za 137 =∂∂

=ZT

Za

In this solution we intial values of the unknowns (scale factor , 3 rotations, 3

translations). Using these value we calculate the coordinates of the common points:

, , using the 3D-conformal equations: 0)( PX 0)( PY 0)( PZ

The final estimate for those parameters is reached iteratively.

7 -19

99



AX=L

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

97969594939291

87868584838281

77767574737271

67666564636261

57565554535251

47464544434241

37363534333231

27262524232221

17161514131211

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

A

17×⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

z

y

x

dTdTdTdddds

X κφω

, ,

190

0

0

0

0

0

0

0

0

)()()()()()()()()(

×⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−−−−

=

RR

RR

RR

QQ

QQ

QQ

PP

PP

PP

ZZYYXXZZYYXXZZYYXX

L

19×

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

ZR

YR

XR

ZQ

YQ

XQ

Zp

Yp

Xp

VVVVVVVVV

V

The least squares solution for the above system is solved by equation(4-4):

X= (ATA)-1ATL

To solve these equations, initial values have to be assumed for each transformation

parameter. The following initial estimates are used:

1-Scal factor = 1.

2-Rotations: w = φ =0.

3-Tx, Ty, and Tz = 500 (initial are not so sensitive).

4-The rotation angle about the z-axis (k), is very sensitive. The initial value of k is

estimated as the difference between the azimuths of a line in the model

coordinates system and it's azimuth in the ground coordinates system as:

( ) ( ) elabgroundAB AzAzk mod−=

7 -20

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Problems

C-3. Solve C-1 using matrices?

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Advanced Plane Surveying Ch08: Survey Operations By: Eng.Ghadi Zakarneh

Ch08 Survey Operations

In Palestine

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معاملة افراز

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معاملة افراز مباني

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معاملة طابو

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معاملة تنظيم منطقة B

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معاملة تنظيم منطقة C

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معاملة تنظيم منطقة B

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معاملة ترخيص داخل حدود البلدية

Advanced Plane Surveying Appendix: Stadia By: Eng.Ghadi Zakarneh

Appendix

A - 1

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APPENDIX-A

STADIA 1- Stadia for Level Instrument

A - 2

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2- Stadia for theodelites

A - 3

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Unit conversions

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References 1. Andrew L.Harbin ,”Land Surveyor Reference Manual”, 2nd edition,1998. 2. James M. Anderson & Edward Michail ,”Surveying theory and practice”, 7th

edition,1998.

3. Najeh S. Tamim,” Surveying for Engineers”, 2nd edition, 2006.

4. Paul R. Wolf & Bon A. Dewitt,” Elements of Photogrammetry with application in GIS”, 3rd edition, 2000.

5. Paul R. Wolf & Charles D. Ghilani ,”Elementary Surveying: An Introduction to

Geomatics”; 11th edition,2006.

6. Paul R. Wolf & Russell C. Brinkerl ,”Elementary Surveying”, 9th edition,2002.

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