CHAPTER 4

Coordinate Geometry and Traverse Surveying

Horizontal and vertical anglesVertical angles: A vertical angles is measured in a vertical plane in

two ways depending on the reference from which the angle is measured:

1- Elevation or depression angle: using the horizontal plane as a reference ( the value of the angle is between -90º to 90º )

a- when the point being sighted on is above the horizontal plane, the angle is called an angle of elevation with positive value.

b- when the point being sighted on is below the horizontal plane, the angle is called an angle of depression with negative value.

2- Zenith angle : using the overhead extension of the plumb line as a reference line, its value ranges from 0º to 180º.

Horizontal Angles

In the horizontal plane, direction of all lines of a survey are referenced to the meridian ( true or magnetic ).

True meridian (at any point): the great circle that passes through that point and the geographic north and south poles of the earth.

Magnetic meridian : a direction that the magnetic needle takes when allowed to come to rest in the earth’s magnetic needle.

There two ways to determine the direction in surveying: Azimuth and Bearing.

1 -Azimuth : is the clockwise horizontal angle that the line makes with

the north end of the reference meridian, and its value ranges from

0º to 360º. And called true azimuth if the true meridian is used as the

reference and called magnetic azimuth if the magnetic meridian is used

2 -Bearing : is the acute angle that the line makes with the meridian, it

expressed as north or south and how many degrees to the east or west

and its value ranges from 0º to 90º and called true bearing if the true

meridian is used and magnetic bearing if the magnetic meridian is used.

Declination : The angle between the magnetic and true meridian is called the magnetic declination and expressed as the angular distance east or west of the

true meridian , and its value is about 3º .

Back Bearing and Back Azimuth: Back bearing or Back azimuth of a line going from A to B is the azimuth or Bearing of the same line going from B to A .

Coordinate Geometry:

1 -The Inverse Problem:

If the X and Y coordinate of two points are known ,the horizontal distance and the azimuth of the line joining them can be computed as following:

d i j = √ (x j – x i )² + ( y j – y i )²

α i j = tan -1 (( x j – x I ) / ( y j – y i )) + c

C = 0º if ∆x is positive and ∆y is positive ( 1st quadrant) .

C = 180º if ∆x is positive and ∆y is negative ( 2nd quadrant).

C = 180º if ∆x is negative and ∆y is negative ( 3rd quadrant).

C = 360º if ∆x is negative and ∆y is positive ( 4th quadrant).

x

y

1st quadrant

2nd quadrant

3rd quadrant

4th quadrant

α i j

i( x i ,y i )

j( x j ,y j )

Example:

Given the following horizontal coordinates for points 1 & 2

X1 = 1437.21 m. Y1 = 2681.46 m.

X2 = 1169.72 m. Y2 = 2004.53 m.

Compute the horizontal distance (d12) and azimuth ( α12 )

Solution

x2 – x1 = 1169.72 – 1437.21 = -267.49 m

y2 – y1 = 2004.53 – 2681.46 = -676.93 m

d 12 = √ (-267.49)² + (-676.93)² = 727.86 m.

α 12 = tan -1 (-267.49 / -676.93 )

= 21º 33' 42" + 180º = 201º 33' 42" ( third quadrant )

2 -Location BY angle and distance:

βαij

αik

i

k

j

x

yi & j are two points of known coordinates ,

the horizontal coordinate of a new point such As k can be determined by measuring the horizontal

Angle β and the distance dik αik=αij + β (if it is larger than 360º then subtract 360º)

xk = xi + dik sinαik

yk = yi + dik cosik

dik

Example

Given The following information:

X i = 3632.11 m. Yi = 1469.27 m

X j = 4987.22 m. Yj = 2073.91 m

β = 141º 27' 33" dik = 1423.55 m

compute the horizontal coordinates of point k

solution

αij = tan -1 (4987.22 – 3632.11) / (2073.91 – 1469.27) = 65º 57' 14 "

αik = αij + β = 65º 57' 14" + 141º 27' 33" = 207º 24' 47“

Xk = 3632.11 + 1423.55 sin ( 207º 24' 47" )

Yk = 1469.27 + 1423.55 cos ( 207º 24' 47" )

3 -Intersection By Angles:y

x

i

j

kβαij αik

dik

αjkØ djk

The coordinate of a new point (k) can be determineby measuring horizontal angles ( β & Ø) from twopoints of known coordinates ( i & j )

)d ik / sin Ø ) = ( d jk / sinβ) = ( d ij / sin (180-Ø-β)(Xk = Xi + dik sin α ik

Yk = Yi + dik cos α ikOr:

Xk = Xj + djk sin α jkYk = Yj+ djk sin α jk

Example

In the previous figure:

Xi = 5329.41 ft Yi=4672.66 ftXj = 6321.75 ft Yj= 5188.24 ft

β = 31º 26' 30" Ø = 42º 33 ' 41 "

Compute the horizontal coordinates X k & Yk

Solution

X j - Xi = 6321.75 – 5329.41 = 992.34 ftY j – Yi = 5188.24 – 4672.66 = 515.58 ftd ij = √ (922.34)² + (515.58)² = 1118.29 ftα ij = tan -1 ( 992.34 / 515.58 ) = 62º 32' 44"α ik = α ij + β = 62º 32' 44" + 31º 26' 30" = 93º 59' 14"

180 - β – Ø = 180 – 31º 26' 30" – 42º 33' 41" = 105º 59' 49 "d ik = 1118.29 sin ( 42º 33' 41" )/ sin (105º 59' 49" ) = 786.86 ftX k = 5329.41 + 786.86 sin (93º 59' 14" ) = 6114.37 ft

Y k = 4672.66 + 786.86 cos (93º 59' 14" ) = 4617.95 ft

4 -Intersection By distance:

i

j

kβαij αik

dik

αjkØ djk

The coordinate of a new point k can be determined by measuring distances )dik & djk (from two points of known coordinates i & j

d jk² = d ij² + d ik ² - 2 dij dik cos ββ = cos -1 ( d ij² + d ik ² - d jk² ) / 2 dij dik

x

y

5 -location by distance and offset:

y

x

i

jp

ko2o1

m

n

αij

If the point lie to the left of line ij, then the coordinates of point p calculated from the following equations:

Xp = Xi + dim sin αij + o1 sin (αij – 90º ) = Xi + dim sin αij - o1 cos αijYp = Yi + dim cos αij + o1 cos (αij – 90º ) = Yi + dim cos αij + o1 sin αij

If the point lie to the rigth of line ij, then the coordinates of point k calculated from the following equations:Xk = Xi + din sin αij + o2 sin (αij + 90º ) = Xi + din sin αij + o2 cos αij

Yp = Yi + din cos αij + o2 cos (αij + 90º ) =Yi + din cos αij - o2 sin αij

Resection:

As in the following figure, the horizontal position of a new point like P can be Determined by measuring the horizontal angles to three points of known coordinates like: A & B & C

A

P

CB

NM

c bӨ

R

Фβ

Let J = β + Ф then J = 360º – ( M+ N+ R ) &Let H = sin β / sin Ф

1 -compute αAB & αAC & b & c & R from the known coordinates of points: A , B ,C.

2 -compute J = 360º – ( M+ N+ R )3 -compute H = b sin M / c sin N

4 -compute Ф ( tan Ф = sin J / (H + cos J ))5 -compute Ө = 180º - N – Ф

6 -compute αAP = αAC + Ө 7 -compute AP = b sin Ф / sin N

8 -compute Xp & Yp

Xp = XA + AP sin αAP

Yp = YA + AP cos αAP

Traverse Surveying:

Def: Traverse is one of the most commonly used methods for determining the

relative positions of a number of survey points .

Purpose of the Traverse:

1 -property survey to establish boundaries.

2 -Location and construction layout surveys for highways, railways and other works.

3 -Ground control surveys for photogrammetric mapping.

Types of Traverse:

a- open Traverse: b- closed Traverse :

Computations and correction of errors:

A- Azimuth of a line:

1 -when ( α1 + Ө ) > 180º

α2 = Ө - ( 180º – α1) = Ө + α1 - 180º

2 -when ( α1 + Ө ) < 180º

α2 = Ө + 180º + α1 = Ө + α1 + 180º

B- Checks and correction of errors:

X last point – X first point = ∑ ∆ X all lines

Y last point – Y first point = ∑ ∆ y all lines

In order to meet the previous two conditions, the following corrections are performed:1 -Angle correction:

a- Closed loop traverse: For a closed traverse of n sides ,

sum of internal angles = (n – 2 ) × 180 º error = sum of measured angles – ((n – 2 ) × 180 º)

correction = - error / no of internal angles

b- For both loop and connecting closed traverse: If the azimuth of the last line

in the traverse is known, then the error εα = αc (calculated azimuth) - αn (known azimuth)

correction / angle = - εα / n the corrected azimuth αi = α’i ( initially computed azimuth)– i(εα / n)

2 -Position correction: IF the calculated and known coordinates of last point are:

) X c , Y c ) & ( X n , Y n( respectively, then

Closure error in x-direction(ε x ) = X c – X n

Closure error in y-direction(ε y ) = Y c – Y n

Closure error in the position of the last points = √ ε x² + ε y ²

Compass ( Bowditch ) Rule : used for position correction as follow:

Correction to departure of side ij( ∆x) = -(length of side ij / total length of traverse)(ε x )

Correction to departure of side ij( ∆y) = -(length of side ij / total length of traverse)(ε y )

Correction can be done directly to coordinates:

Cxi = - (Li / D) (ε x ) & Cyi = - (Li / D) (ε y )

Where:Li=the cumulative traverse distance up to station i &D=total length of the traverse

The corrected coordinates of station i ( x'i , y'i ) are: X'i = Xi + Cxi & Y'i = Yi + Cyi

Allowable error in Traverse surveying

the following figure:

X Y

x

y

x y

Preliminary coordinates

Corrected coordinates

Final results

X-coordinate

Y-coordinate

X XY Y

X Y

X X

Y Y