Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation

Faculty of Engineering and Technology

Structural Department - Year 1

Dr Ahmed Ragheb

(Planimetric Surveying 1)

SCM 221

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4 1

Available Surfaces

Terrain

Geoid

Computational Surface

•Earth physical surface

•Non uniform shape

•No geometric characteristics No calculations

•All surveying measurements are taken on this surface

Only available tangible surface

•Imaginary non uniform surface

•No geometric characteristics No calculations

•Equipotential surface MSL surface used as datum for heights

•Direction of Gravity to the Geoid

•Vertical axis of any instrument takes gravity direction to the Geoid

•HGeoid

=0

•Imaginary Uniform Geometric surface Used for calculations

Plane (L<10km & A<50km ) Sphere (10km<L<20km & 50km <A<300km )

•Fitting Earth or part of the terrain as much as possible

2

2 2

2

Ellipsoid (L>20km & A>300km )

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4 2

Coordinate Systems

),,( hType of Coordinate Systems

1- Geographic or geodetic coordinate systems or

2- Plane coordinates

Polar (r,θ) and Cartesian (X,Y) or (E,N)

),,( ZYX

o

λ

ø

h

P

z

x

y

y

x

z

1- Geodetic Curvilinear Coordinates

Latitude ø, Longitude λ, height h

2- Geocentric Cartesian Coordinates

x, y, z

Meridian of point P

Equatorial plane

Perpendicular to Earth’s surface

True North at point p

λ=0

λ=90

NP

Ø = 0 – 90° N or S

λ = 0 – 360° or 180°E – 180°W

Geographic coordinate system

Part of sphere

Transformation is possible

between Curvilinear and

Cartesian coordinates

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation

3

Coordinate Systems

Why we use coordinate systems?

-No accumulation of errors

-Easy to use with computers

-Transformation of coordinates among different systems

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation Plane Coordinate System

4

Coordinates of point إحداثيات نقطة Components of a line مركبات الخط

Polar coordinate system

Cartesian coordinate system

r is the distance of the point from the origin

to the point P

θ is the angle between the polar axis and the

ray extending from the origin to point P

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation

5

Whole Circle Bearing

Azimuth of a line (Whole Circle Bearing WCB) انحراف الخط

هو الزاوية األفقية مه إتجاي الشمال إلى الخط مقاسة فى إتجاي عقارب الساعة

It is the angle measured from North direction to the line clockwise direction.

veN

veE

o

90360270

veN

veE

veN

veE

18090

veN

veE

270180

NB. If negative, then add 360, If then subtract 360 360

A

A

A

A

B

B

B

B

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation Forward & Backward Bearing

6

180

180

180

AB

AB

ABBA

if

if

298180118118

47180227227

BAAB

BAAB

Relation between Forward and Backward bearing العالقة بين االنحراف األمامي والخلفي

Ex:

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation Quadrant Bearing

7

900Q

Quadrant Bearing (Reduced Bearing RB) اإلنحراف المختصر

مع أو ضذ عقارب الساعة( شمال أو جىوب)هو الزاوية بيه الضلع واإلتجاي الزأسى

240If

60180240Q

WS 60

Ex:

S

E

N

W

NE NW

SW SE

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation Main Concept

8

CABABAC

Calculation of an angle between 2 bearings حساب زاوية من انحرافين

الزاوية دائماً فى إتجاي عقارب الساعة

ABABAB

ABABAB

LN

LE

cos

sin

ABABAB LEE sin

ABABAB LNN cos

Calculation of components of a line from length

and bearing

حساب مركبات الخط من الطول واالنحراف

If coordinates of A are given, then:

Angle always in clockwise direction

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation

ABAB

ABAB

NNN

EEE

22

ABABAB NEL

Calculation of length of a line from coordinates

حساب طول الخط من االحداثيات

ABAB Q

veN

veE

360

ABAB Q

veN

veE

180 ABAB Q

veN

veE

180

AB

AB

N

E

1-

AB tan Q

Calculation of bearing of a line from coordinates

حساب انحراف الخط من االحداثيات

بذون إشارات ثم وأخذ اإلشارات فى اإلعتبار أولً

ABAB Q

veN

veE

Calculation of Length and Bearing

9

First without signs, then take sign

into consideration

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation

Line AB was obstructed by a pond as shown in Figure.

The following were measured:

AC = L1 = 40.69m, CE = L2 = 30.53m,

ED = L3 = 29.38m and DB = L4 = 34.83m.

If the bearing of AB= 121° 38' 56" and

CE= N 81° 27' 14" E, calculate:

a) Length of AB in Yards.

b) Bearing of DE in Grad.

Solved Example

αAB

=121° 38' 56" Q

CE =N 81° 27' 14" E

αCE

= 81° 27' 14"

ECD = αCD

- α CE

= αAB

- αCE

= 121° 38' 56" - 81° 27' 14"

ECD = 40° 11' 42"

10

CDECDE

CE

ECD

ED

sin

53.30

421140sin

38.29

sinsin '''

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation

DEC = 180 – CDE – ECD = 97˚41' 15"

CD2 = CE

2 + ED

2 – 2.CE.ED. Cos DEC

CD = 45.11m

AB = AC+CD+DB = 40.69+45.11+34.83 = 120.63m

AB = 131.92 Yards

CDE = αDE

- αDC

αDE

= CDE + αDC

= CDE + αBA

αDE

= 42˚07' 03" + 121˚38' 56" + 180˚

αDE

= 343˚45' 59"

αDE

= 381g

96c 27

cc

CDE= 42° 07' 03"

11

Solved Example

Coordinate

Computation

Dr Ahmed Ragheb SCM 221 – Planimetric Surveying 1 Lecture 4

Coordinate

Computation

For the shown electric cables configuration, the following are given:

The coordinates of A (256.78m, 388.40m), the reduced bearing of AB = N75° 25'

40"E, the coordinates of C (310.25m, 505.32m) and the angle ADC = 65° 20' 50",

•Calculate coordinates of D.

•Given LDB

= 123m, calculate the coordinates of B.

•Check if electric tower E(410.25m,605.32m) lie on the path of the cable AB.

ΔEAC

= 310.25 – 256.78 = 53.47m, ΔNAC

= 505.32 – 388.40 = 116.92m

LAC

= 128.57m, QAC

= N24° 34' 32.25"E, αAC

= 24° 34' 32.25",

angle CAD = αAD

– αAC

= 75° 25' 40" - 24° 34' 32.25" = 50° 51' 7.75",

128.57 / sin65° 20' 50" = LCD

/ sin50° 51' 7.75",

LCD

= 109.71m,

65° 20' 50" = αDC

– αDA

,

αDA

= 75° 25' 40" + 180 = 255° 25' 40",

αDC

= 320° 46' 30",

αCD

= 320° 46' 30" – 180 = 140° 46' 30"

ED = E

C + L

CD × sinα

CD = 379.63m

ND = N

C + L

CD × cosα

CD = 420.33m

EB = E

D + L

DB × sinα

DB = 498.67m

NB = N

D + L

DB × cosα

DB = 451.28m

Solved Example

12