CHAPTER 1 INTRODUCTION

Definition of Surveying The art and science of making and analyzing measurements(Distances,angles,levels, coordinates) using different surveying tools and instruments in both vertical and horizontal planes for specific part of earth surface and processing of the measurements to represent them in a map of specific scale. It also include the opposite activity: stakeout of locations from the map to the earth surface. Surveying measurements may be on ground ,under ground ,in the see, from space.

The figure of the earth and its relation to survey measurements Sine the earth surface is the reference of all surveying measurements so the figure and size of the earth is very necessary. The first trial to measure the earth size was by the ancient Greeks by Eratosthenes as follow: Sun Distance between Aswan and Alexandria = 5000 stadia (each stadia = 185.2 m.) Tan α = s/h = 7.2 º ( α = β ) 5000 stadia / L(earth perimeter) = β / 360 L = 46300 km. = 2 π R R = 7368.874 km. This value is larger than the accurate satellite measured value by only 16% Pole α h s Aswan Alexandria well R R β

Reference of measurements The earth surface is irregular and can not be represented by mathematical model. The geoid is the mean see level which result if there is no topography above mean see level and also this surface is irregular . If the earth has a uniform density and topographic variation does not exist the geoid will have the shape of ellipsoid.

For (areas < 50 km²) The reference is a plane. The reference for measurements depend on the the area to be surveyed as follow : For (areas < 50 km²) The reference is a plane. For ( 50 km² < areas < 500 km² ) the reference is a sphere of radius (R = 6372.2 km.) For (areas > 500 km²) The reference is Ellipsoid with the following properties : NP a : semi major axis b : semi minor axis=a(1-f)=a(1-e²)½ f (flattening) = ( a – b ) / a ( the value of f is between 0 “ sphere ” and 1 “ flat surface ”. e : eccentricity (e² = (a² - b²)/a² = 1-(b²/a²) =2f –f²) b a SP

Types of Surveying A- size of the area to be surveyed: 1- Geodetic surveying(Surveys of countries , small scale < 1:100,000 . 2- plane surveying . B- Methods of Surveying : 1- Ground Surveying . 2- Photogrammetric Surveying .

C-Purpose of the Surveying : 1- Cadastral or Land Surveying : To Produce Cadastral maps. 2- Topographical Surveying : To Produce Topographic Topographic maps. 3- Hydrographic Surveying : Surveys of water bodies. 4- Route Surveying : surveying works necessary for highways,railroads,canals,….. . 5- Construction Surveying . 6- Mine Surveying .

A- Units of Length Measurement : Units of Measurement A- Units of Length Measurement : English system 1Mile = 1760 yards (yd) = 5280 ft 1 Yard = 3 foot (ft) 1 Foot = 12 inches (in) Metric System 1 Kilometer (km) = 1000 meter (m). 1 Meter (m) = 10 decimeter (dcm). 1 Dicemeter(dcm) = 10 cm. 1 centimeter = 10 mm. 1 link = 0.20 m. 1 inch = 2.54 cm. & 1 ft = 0.3048 m & 1 km = 0.6214 mile 1 mile = (5280*12*2.54)/(100*1000 ) = 1.609344 km

B- Units of Area Measurement English System In² , ft ² , yd ² , mile² (ft² , acre) 1 acre = 43560 ft² Metric System 1 donum = 1000 m² 1 hectare (ha) = 10000 m² 1 km ² = 100 ha 1 fadan = 4200 m² C- Units of Volume Measurement Metric system 1 m³ = 1000000 cm³ 1 m³ = 1000 liters 1 liter = 1000 cm³ English System 1 ft³ = 1728 in³ 1 yd = 27 ft³

D-Units of Angle Measurement The sexagesimal System 1 degree(º) = 60 minutes (') 1 minutes (') = 60 seconds (") example : 200 º 20' 30" The Decimal or centesimal system (gradian system) 1 g = 100 c 1 c = 100 cc example : 140 g 30 c 25 cc 1 g = 0.90 º The radian System L θ in radian = L / R Angle in radian for complete circle = 2πR /R = 2π = 360 º = 400 g This leads that ( π = 180 º = 200 g ) R θ R

Types of scales Scale of Surveys A- Numeric Scale Dimensionless Ratio Ex. : 1:5000 & 1:500 Unit equivalents Ex. : 1 in = 200 ft & 1 cm = 100 m Dimensionless representative fraction Ex. : 1/1000 & 1/2500 B- Linear or planning Scale 10 8 6 4 2 10 20 30 40 50 m. Each 1 cm = 10 m ( 1:1000 )

1:100 & 1:1,000 & 1:10,000 & 1:100,000 & 1:1000,000 Common Scales 1:100 & 1:1,000 & 1:10,000 & 1:100,000 & 1:1000,000 1:50 & 1: 500 & 1: 50,000 & 1 : 50,000 & 1:500,000 1:25 & 1: 250 & 1: 25,000 & 1: 25,000 & 1:250,000 1:200 & 1: 2,000& 1: 20,000 & 1:200,000 Architectural works : 1/50 , 1/100 , 1/200 Civil works (site plans) : 1/500,1/1000,1/1250,1/2500 Town Surveys/highways surveys : 1/1250,1/2000,1/2500,1/5000,1/10000 Mapping : 1/25000,1/5000,1/100000,1/200000,1/1000000

Basic Geometric Principles of Surveying Working from Known control points of high degree of accuracy ( ) To establish new control points with less accuracy ( ) The location of unknown points can be established relative to known line in one of the following process 1 2 3 A B C A B C A B C θ2 d1 θ1 θ d d2

Exercises Ex1: If a distance was measured on a map of unknown scale and equal to 10 cm , and the same distance was measured on another map of scale 1/2500 and its equal to 5cm,what is the scale of unknown map? solution distance on a map : distance on the ground 1 : 2500 5 cm. : ? distance on the ground = 5*2500 =12500 cm. =125m. distance on the map : distance on the ground 10 cm. : 125 m 10 cm. : 12500cm. 1 : 1250 unknown scale = 1: 1250 Distance of unknown scale is twice the distance of the known map 5cm 10cm (enlarge tow times), IF we enlarge the known Scale(1:2500) tow times this will produce (1:1250)

Ex2: A line with length of 100m was rotated by 10" about one of its ends, what is the distance moved by the another end ? Solution Θ in radian = arc length (L) / Radius (R) Π = 180 º ? = 10 " Θ in radian =( 10"/ 60*60 )º * (π /180º) =0.00004848 Arc length (L)=Θ in radian*R=.00004848*100=0.004848m=4.848mm Ex3: what is the angle opposite to circular arc of length 24 mm, and radius of 50m? solution Angle in radian = L / R = 24 mm /( 50 *1000 mm) = 0.00048 angle in sexagesimal system : Angle in radian 180º : π ? : 0.00048 Angle in sexagesimal = (0.00048*180) / π = 0.0275º *60*60 = 99" = 1' 39“

Ex4: If an angle in sexagesimal System is equal to 51º24',what is the value of angle in both of gradian and radian systems ? Solution Π = 180º ? = 51º 24 ' Angle in radian = ( 51.40º * π ) / 180 º = 0.897 90º = 100 g 51.4 = ? Angle in gradian =( 51.40 * 100 )/ 90 = 57.11 g Ex5: If an area of a land is equal to 100 hectare ,what is the the area of land in acres? What is the relaelation between acre and hectare ? solution 1 acres = 43560 ft² = (43560 *12 ² * 2.54²)cm² /(100²) m² = 4046.8564 m² = 4046.8564/10000 = 0.4047 hectare Each one acre = 0.4047 hectare The area of land in acres = 100 / 0.4047 = 247.1 acres

Ex6: A map of scale 1: 2500 ,and have 20cm. X 30cm Ex6: A map of scale 1: 2500 ,and have 20cm. X 30cm. Dimensions , what is its new dimensions if it was enlarged to scale 1:625 ? Solution Length on map : length on ground 1 : 2500 20cm : ? 30cm : ? Dimension of the area covered by the map on the ground = (20*2500) & (30*2500) =50000cm X 75000 cm. 1 : 625 ? : 50000 cm ? : 75000 cm New dimension of the map sheet = (50000/625)cm & (75000/625)cm = 80 cm X 120 cm Note: when we move from scale 1:2500 to scale 1:625,this means that we Enlarged the map 4 times so if we multiply the dimension of the first map in 4 we will have the new dimensions ( 4*20cm =80cm & 4*30cm = 120cm )

CHAPTER 2 CHAIN SURVEYING Tape Measurements

Mapping Details using chain surveying In chain surveying all ground features (natural or industrial) are located and mapping by measuring lengths using tapes to a selected lines called chain lines .This done by one of two ways: 1- Ties method : 2- Offsets method: Pole Building Pole Building Chain lines offsets Ties Ground Details may be located by the method of ties or offsets or a combination of both. The two methods depend on measuring horizontal distances between points and setting Right angles using some of surveying equipments.

Chain Surveying Equipments 1- Equipments used for the measurements of lines : A- The chain B- Tapes: synthetic, glass fiber, coated steel or plain steel of lengths: 10m, 20m, 30m, 50m . , tapes are more accurate than chain. C- Invar Tapes: accurate tape made of steel (65%) & Nickel (35%) but they are very expensive . 2- Equipments used for making right angles : A- The cross staff: B- Site square: consists of two telescopes, their lines of sight at 90º .

C- The optical square: There are two types: mirrors & prism 3- Other equipments: a- Ranging Rods: 2m, 2.5m, 3m long & painted red and white each 0.5m with a pointed steel shoe. b- Arrows: 40cm long & 3- 4 mm in diameter. c- Pegs: wooden (square or circular cross-section) & steel d- Plumb Bob: a metallic cone object used to erect vertical lines. e- Clinometer & Abney level: used to measure the inclination angle of slope lines.

Processes in Chain Surveying 1- Ranging and measurement of lines: a- Level ground. b- Uniformly Sloping Ground. c- Uneven Ground (non-uniformly sloping ground): using stepping process. 2- Setting out right angles: A- Dropping a perpendicular from a point to a line:

B- Setting out a perpendicular to a line from a specific point: 1- Using the optical square. 2- The equilateral triangle method. 3- Pythagoras theorem method.

Plan of Mapping details using chain surveying 1-Reconnaissance: The surveyor should visit the area to be mapped in order to: a- Notice the shape of the area, the existing details and draws a reasonable sketch of the area showing all details such as roads, buildings, fences, electric poles, ….etc. in addition to the approximate north direction. b- Choose the most suitable location for the survey station which form the chain lines, the following points should be taken into consideration when choosing survey stations: 1- chain lines should form well-conditioned triangles with internal angles is between 30º and 120º . 2- chain lines should be chosen as close as possible to the mapping details. 3- It should be possible to see at least two other survey stations from each station. 4- The number of chain lines should be kept to a minimum, but enough to locate all details. 5- Survey station should be chosen in away such that check lines is available. 6- Survey Station should be chosen on firm ,easy to reach grounds.

2- Booking the measurements 20cm. 1.50cm 15cm

3- Plotting the details: a- Choose the appropriate scale b- If you are drawing the map manually, using a pencil to draw the chain lines on a transparent paper, in order to center the drawing on the final map sheet. c- Try to make the north direction pointing towards the top of the sheet. d- Offsets and ties are plotted systematically in the same order in which they were measured and booked. e- After finishing the drawing by pencil, the plan is taken to the site and checked. Accuracy of Measurements: A good draughtsman can plot a length to within 0.20 mm. (0.20mm on a map of scale 1:500 represents 10cm on the ground). (0.20mm on a map of scale 1:100 represents 2cm on the ground). As a result ,measuring a line to 1 cm accuracy is practically sufficient for such scales.

Chaining Obstacles 1-Vision obscured, Chaining possible: 2- Vision Possible, Chaining Prevented: A- Measurements can be made around the obstacle: AB = AC + EF + DB AB = AG + KL + HB

B- The width of obstacles is greater than tape length: Case a: HG = FD .FG /ED = CG × FG / (EC – FG) Case b: The unknown distance FH Is equal to JG which can easily measured Case c: EG = CE × EF / ED

3-Both Chaining and Vision prevented: GD = FC × GA / FA HE = FC × HA / FA ED is the missing part of AB Length AB = AD + DE + EB DE = HJ MN = 2 CD

ERRORS IN CHAIN SURVEYING 1- Blunders (Mistakes). 2- Systematic errors: a- Temperature correction : Ct = 0.0000116 ( T1 – T0 ) L 0.0000116 is the coefficient of thermal expansion of steel per 1º C T1 = is the field temperature. T0 = is the temperature under which the tape is calibrated L = is the length of the line b- Sag correction: Cs = -W²L / 24 P² or = - w² L³ / 24 P² W = the total weight of the section of tape located between supports. w = weight per meter of tape L = the interval between supports P = the tension on the tape Ex : calculate the sag correction for : a- A 100 ft steel tape weighting 2 ib and supported at the ends only with a 12 Ib pull b- A 30 m steel tape weighting 0.0112 kg/m and supported at 0, 15, 30m points under a tension of 5 kg. Solution a- Cs = -W²L / 24 P² = - (2² × 100 ) / ( 24 × 12² ) = -0.116 ft b- Cs = - 2 ( w² L³ / 24 P²) = -2 (( 0.0112² × 15² ) / (24 × 5² )) = -0.001 m.

Cp = the elongation of the tape of length L in m. c- Tension correction: Cp = (P1 – P0 ) × L / AE Cp = the elongation of the tape of length L in m. P1= the applied tension P0= the calibration tension A = the cross-sectional area of the tape in cm² E = the modulus of elasticity of the tape material (for steel E= 29,000,000 ib/in² ). d- Length correction: Cl = ( la – l0 ) L / l0 Ex: A line is measured with a tape believed to be 100 ft long which gives a length of 705.76 ft . On checking ,the tape is found to measure 100.02 ft . What is correct length of the line ? Solution la = 100.02 ft , l0 = 100 ft , L = 705.76 ft Cl = ( 100.02 – 100 ) × 705.76 / 100 = + 0.14 ft correct length = 705.76 + 0.14 ft = 705.9 ft.

CHAPTER 3 LEVELING

Definitions Definitions An elevation of a point : The vertical distance between the point and the reference level surface ( datum ) ,the most commonly used datum is the mean sea level (MSL ) . Leveling : The process by which the elevation of a point above or bellow the MSL or the elevation difference between points is determined, There are several ways to do that: A- Chain surveying: by measuring the slope distance and angle of inclination B- Barometric leveling: by measuring the atmospheric pressure using barometer or an altimeter. C- Trigonometric Leveling: by measuring the horizontal or slope distance between points and the vertical angle ( an elevation or zenith ). D- Differential Leveling: by using an instrument called a level . Bench Mark ( BM) : well marked points whose elevation has been accurately measured. Level line: a line perpendicular to the direction of gravity at all points.

Basic Principle of A level A level consists essentially of : a- telescope for sighting. b- a leveling device to maintain the line of sight a horizontal as Bubble Tube, circular bubble, and some levels use a bubble consists of two separate halves which can be seen through an eyepiece near the telescope. To use the level instrument in making leveling there is a need for : a- Tripod: A three-legged stand used to support a level or other surveying equipments. b- level rods ( staff ): used to measure the vertical distance between the points and is usually 3 – 5 m long ( telescopic staff, folding staff, one piece staff ). There three kinds of levels : The dumpy level: the line of sight ( collimation ) is fixed at right angles to the vertical axis of rotation of the instrument, ( the spirit level attached to the telescope) should also be perpendicular to the vertical axis and parallel to the line of collimation. The tilting level: the vertical axis is made approximately vertical using circular bubble, and the telescope is leveled horizontal using the telescope spirit level. The Automatic level: use a system of self-leveling compensators within the optical system of the telescope to bring the line of sight to the horizontal plane.

Measuring Elevation Difference using a level The difference in elevation between A & B = the vertical distance AC = RA -RB = 2.56 -0.93 = 1.63 m. ”positive value means Rise” B 0.93 2.56 Line of collimation A C Second reading smaller than first reading represents a rise A 2.97 C 0.64 The difference in elevation between A & B = the vertical distance BC = RA -RB = 0.64 -2.97 = -2.33 m. ”Negative value means Fall” Line of collimation B Second reading greater than first reading represents a fall

Calculation of Unknown Points B A 2.97 C 0.64 Line of collimation B 0.93 2.56 Line of collimation A C A- The height of Instrument (collimation ) method: Height of instrument (HI) = Elevation of A + staff reading at A Elevation of B ( RL ) = HI – staff reading at B HI = 520.43 + 2.56 = 522.99 m AMSL RLB = 522.99 – 0.93 = 522.06 AMSL HI = 520.43 + 0.64 = 521.07 m AMSL RLB = 521.07 – 2.97 = 518.10 m AMSL B- The Rise and Fall method: Elevation Difference = first reading at A – second reading at B Elevation of B ( RL ) = elevation of A + Rise or Fall ELev. Diff. = 2.56 – 0.93 = 1.63 m. (Rise) RLB = 520.43 + 1.63 = 522.06 m. AMSL ELev. Diff. = 0.64 – 2.97 = -2.33 m. (Fall) RLB = 520.43 + (-2.33) = 518.10 m. AMSL ∆HAB = HB – HA = RA - RB

Procedure in Differential Level Definitions: 1- Backsight ( BS ): The first reading taken at every instrument station . 2- Foresight ( FS ): The last reading taken at every instrument station . 3- Intermediate sight ( IS ): Any reading taken at an instrument station between BS & FS 4- Turning point ( TP ): A point at which both BS & FS are taken before moving the staff General Procedure: The main purpose is to provide RL for a large number of points, such as the center line of highway, or an area to produce a contour plan, this will be illustrated through the following example of production longitudinal section (profile ) of a road.

Booking the work

Calculations A- The Height of Instrument ( HI ) method: Checks: 1- # of BS reading = # of FS reading = 2 2- ∑BS - ∑FS= RL last point – RL first point(3.450 – 2.231 = 99.979 – 98.760=1.219) 3- ( # of readings = 9 ) = (# of points) + (# of TP) = 8 + 1 4- Sum of all RL excluding the first reduced level = sum of the HI for each setup multiplied by the number of IS and FS readings taken from that setup – sum of IS - sum of FS 689.741= (99.423×4 + 100.625×3 = 699.567) – (∑IS=7.595)–(∑FS =2.231) =689.741

B- The rise and Fall method: Checks: 1- # of BS = # of FS = 2 2-( # of readings = 9 ) = ( # of points = 8 ) + ( # of TP = 1 ) 3- (∑ BS - ∑ FS) = (3.450 – 2.231 ) = 1.219 = (∑ R - ∑ F) = ( 3.079 – 1.860) = 1.219 = ( RL last - RL1st ) = ( 99.979 – 98.760 ) = 1.219

General Notes 1- The accuracy of RL depend on both calculations & correct measurements( staff readings) To ensure accurate elevations of the level points, the field work should start at BM and close at another known BM, or should go back and close at the starting point . 2- If the purpose of leveling is to find out the elevation difference between two points, no intermediate sights are necessary, only BS & FS readings are made. 3- The BS & FS distances should be approximately equal to avoid errors. 4- IF possible staff readings should be made to the nearest mm at TP, and to the nearest cm at other points. 5- TP points should be chosen on firm ground. On soft ground, a special triangular base is used. 6- IF the point whose elevation is to be calculated lie above the level of the sight line like a bridge or a ceiling or top of wall or column, then the staff reading is recorded in negative. 7- If the BM point is the last point, then use the equation ( ∑BS - ∑FS= RL last–RL1st ) to calculate the RL of first point But if the BM lie in the middle of level chain, then calculate the RL of last point ,and after that use the previous equation to calculate the RL of first point and complete the calculations in the normal way.

2- (∑BS - ∑FS)=(-0.67-0.85)=-1.52 & (∑R - ∑F)=(3.38-4.90) = -1.52 Example: Using the data in the following figure, calculate the RL of points: A, B, C, D, using both of R & F and HI methods 1- # of BS = # of FS = 2 2- (∑BS - ∑FS)=(-0.67-0.85)=-1.52 & (∑R - ∑F)=(3.38-4.90) = -1.52 & (RL last – RL1st )= (498.48 – 500.00) = -1.52 3- RL excluding first RL = 501.13 + 503.38 + 498.48 = 1502.99 = (500.6 × 1 + 499.86 × 2 = 1500.32) – (∑FS=0.85) - (∑IS=-3.52) = 1502.99

Uses of Leveling 1- Longitudinal Sections: (center line of a railway, road, canal, sewer or water main) Levels are taken at: 1- every 20m, 50m or 100m depending on the topography (for earthworks computations, A spacing of 20m is common. 2- points at which gradient changes, and at streets intersections. Notes: Staff readings to 0.01 m accuracy are generally adequate Common scale for roads works are 1/1000 for horizontal ( Distances ) & 1/100 for vertical ( levels or elevations ). Points are connected by straight lines on the profile.

2- Cross-Sections: Some Engineering works require that cross sections be taken at right angles to the center line of a proposed or existing project such as a road. The width of these of sections are taken 15 m either side of the center for a normal road. A scale of 1/50 or 1/100 is used for both horizontal and vertical axes.

3- contouring: Definitions: A contour: is an imaginary line connecting points on the ground that have the same elevation. Contour Interval ( CI ): The vertical distance or elevation difference between two successive contours. Contour interval = 20m

Characteristics of Contours 1- Closely contours represent a steep slope, but spaced far contours represent a flat slope. 2- Contours of different values do not cross each other except in a cave, nor do they merge except in a vertically standing surface such as a wall. 3- A single contour can not split into two contours of the same value, and must be a closed circuit not necessary in limits of the contour plan. 4- Irregular contours represent a rough and uneven terrain. 5- contours are perpendicular to the direction of the steepest slope. 6- A hill or depression are represented by closed lines. Factors affecting choice of contour intervals: 1- Contour plan scale: The larger scale is the smaller contour interval. 2- The importance and purpose of a contour plan : for more details, a small contour interval is used. 3- Accuracy, time and cost : For higher accuracy, a smaller interval is used. 4- The topography of the ground : For steep ground a large contour interval is used, but for flat ground a small contour interval is used. 5- The area covered by the plan: For larger areas, a large contour interval is used.

Methods of Contouring 1- Griding : This method is suitable for flat terrain of small sites ,rectangles or squares of 10 to 20m side are set out on the ground 1 2 A 3 4 5 10-20m B C D E The reduced levels of corners are plotted as a grid on a plan with suitable scale, Then the required contour lines are plotted by a process called linear interpolation. X / 12 = ( 1 / 3 ) X= (1 / 3)×12 = 4m

2-Radiating Lines: Rays are set out on the ground from a central point, and levels are taken along these rays at measured distance from the center, and linear interpolation is used to give the contour lines. 3- Cross-Section Method : In this method, a cross-section are made on a line or a traverse inside the area, levels are taken at points where the topography changes, and linear interpolation is used to give the contour lines.

4- Setting Out Levels: One of the basic applications of leveling is setting out sight rails which enable The excavation operators to cut out earth to an even gradient, and enable the pipe-layer to lay the pipes to this gradient using a boning rod looking The capital letter T with a sight bar across it, the convenient height above Invert would be 3.75 m . Sight rail A, RL= 30.02 + 3.75 =33.77m Distance AB = 60 m Fall = 60 ×( 1 / 100 ) =0.60 m Invert Level at B = 30.02 – 0.60=29.42 m Sight rail at B = 29.42 + 3.75 = 33.17 m If a level is setup nearby has a height Of collimation of 34.85 m, then the staff Reading at A = 34.85 – 33.77 = 1.08 m &Reading at B= 34.85 – 33.17 = 1.68 m

Classes and Accuracy of leveling: The allowable closure error in differential leveling is normally given in the form: ε = ± x √ K mm where: K Is the total leveled distance in kilometers. x = from 10 to 30 for ordinary leveling x = from 2 to 5 for precise leveling Errors in Differential Leveling: 1- Systematic errors: A- Inclination of line of sight due to the earth’s curvature & atmospheric refraction: BC = line of sight refraction from horizontal BD = error due to earth’s curvature CD = Actual error in the staff reading = BD-BC In triangle ABO: (BD + DO)² =AB² + AO² BD² + 2R.BD + R² = L² + R² BD ≈ L² / 2R ≈ 0.0786 L² ( R = 6365 km.) BD in meter & L in km. Refraction = BC ≈ BD/7 ≈0.0786L²/7 CD = 0.0786L² - 0.0786L²/7 = 0.0673L² Where CD is in meter & L is in km When L=1km CD ≈ 0.07m ≈7cm L ≈ 100m ≈0.10km CD ≈.001m ≈1mm

B-Maladjustment of the level collimation Errors: Due to maladjustment of the level, the line of sight will be actually inclined from the horizontal, this error can be completely eliminated by balancing the BS & FS distances ε 1 = L1.tanα & ε 2 = L2.tanα ∆ H = m – n =(a- ε1) – (b- ε2) = ( a – b ) – ( ε1 – ε2 ) =( a – b ) – tanα (L1 – L2) If L1 = L2 ( BS distance = FS) ∆ H = ( a – b ) 2-Random errors: the sources of random error are: - The staff not held plumb. - The bubble of the level not perfectly centered. - The incorrect reading of the staff. - The instability of turning points. - Wind 3- Blunders or Mistakes: - Not setting the staff on the same point for a FS and subsequent BS readings. - Recording or Booking of data, like reading 2.58m as 2.85 m or booking a FS in the BS column. - Misreading the staff especially when the marks on the staff are obscured by a tree or fence.

Example TO check a level for the existence of collimation error, the level was set up mid-way between A and B and the following two staff readings were taken:1.92m at A and 1.40m at B. The level was then moved to another position and the readings as in the following figure. Is there a collimation error ? If the answer is yes, then calculate the angle of inclination of the line of sight from the horizontal, as well as correct readings that should have been taken at A & B in the second setup if there was no collimation error? For setup 1: ∆H1 = (correct)=1.92–1.40=0.52m ∆H2 = 1.75 - 1.20 = 0.55 m ∆H1 ≠ ∆H2 there is a collimation error 0.52 =( 1.75 - 1.20 )–tanα(58–23) α = 0º 2’ 57” Correct reading at A (m) = 1.75 – 58 tanα = 1.70 m Correct reading at B (m) = 1.20 – 23 tanα = 1.18 m Check: ∆H = 1.70 – 1.18 = 0.52 m = ∆H1

Reciprocal leveling : 1- Setup the level at point C (first setup a), about 2 to 3 m. from A and take the readings a1 at A and b1 at B ( ∆H1 = a1 – b1 ) 2- Move the level to point D where ( AC = BD ) .Take the readings: a2 at A and b2 at B ( ∆H2 = a2 – b2 ) 3- ∆H =( ∆H1 + ∆H2 ) / 2 =( ( a1 – b1 ) + ( a2 – b2 ) ) / 2 ex: IF RLA = 917.34 m & a1 = 1.44m (BS) & b1 = 1.90m (FS) & a2 = 1.80m (BS) & b2 = 2.34m (FS), what is the elevation of point B? ∆H AB =((1.44 – 1.90) + (1.80 – 2.34)) / 2 = -0.50 m The elevation of B (HB) = HA + ∆H AB = 917.34 + (-0.50) = 916.84 m.

Closure error Closure correction for ∆hi = (- ni / ∑ ni ) × ( ε ) Corrected ∆hi = measured ∆hi + closure correction for ∆hi ε (closure error) = h’ ( calculated elevation ) – h ( known elevation ) ni refers to the number of level setups & ∆hi refers to the elevation difference between consecutive stations.

CHAPTER 4 Coordinate Geometry and Traverse Surveying

Horizontal and vertical angles Vertical 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

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

d i j = √ (x j – x i )² + ( y j – y i )² 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). y j( x j ,y j ) α i j i( x i ,y i ) 4th quadrant 1st quadrant x 3rd quadrant 2nd quadrant

Given the following horizontal coordinates for points 1 & 2 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: y i & 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 j αik αij k β dik i x

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: The coordinate of a new point (k) can be determine by measuring horizontal angles ( β & Ø) from two points 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 ik Or : Xk = Xj + djk sin α jk Yk = Yj+ djk sin α jk y j αjk Ø djk αik αij β k i dik x

Example In the previous figure: Xi = 5329.41 ft Yi=4672.66 ft Xj = 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 ft Y j – Yi = 5188.24 – 4672.66 = 515.58 ft d ij = √ (922.34)² + (515.58)² = 1118.29 ft α 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 ft X 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: j αjk Ø djk αik αij β k i dik x 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 β β = cos -1 ( d ij² + d ik ² - d jk² ) / 2 dij dik

5- location by distance and offset: x j n p o2 o1 k αij m i y 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 αij Yp = 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 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 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 A P C B N M c b Ө R Ф β

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)

Allowable error in Traverse surveying 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:

Preliminary coordinates x y X Y x y

Corrected coordinates X-coordinate Y-coordinate Final results

X Y X X Y Y

X X Y Y

CHAPTER 5 AREAS AND VOLUMES

Mathematical formula for area calculation: Area= ½( b×h ) Area = ½ (a ×c ×sinB ) Area =√s(s-a)(s-b)(s-c) s=(a+b+c)/2 Area = a² Area = a×b Area=π r² Area = ½ ((a+b)/2)h Area=1/4 n.a² cot(180º/n) Area=π(r2² - r1²) Area=(∆ / 360).π.r² Area = ½ r²((π∆/180)-sin∆) Area= 2/3 .b .h Area= π.a.b

Areas by the method of coordinates: x y Y1x2 x1y2 Y2x3 x2y3 Y3x4 x3y4 Y4x5 x4y5 Y5x1 x5y1 x Area = ½ ( (y1x2 + y2x3+ y3x4 + y4x5 + y5x1) – (x1y2 + x2y3 + x3y4 + x4y5 + x5y1)) = ½ ( x1(y5-y2) + x2(y1-y3) + x3(y2-y4) + x4(y3- y5) + x5( y4-y1) )

x y x y

Areas of irregular figures: Trapezoidal rule:

Simpson’s one-third rule:

Calculation of volumes: All highways and railroad construction projects involve Earthworks( Cut & Fill ) There are two methods to calculate the volumes of cut and fill.

Average End Area Method (AEA):

Volume By Prismoidal Formula:

Volume From Contour Maps The figure

Volume From Spot Levels The following figure Volume (V) = (A / 4)(∑h1 + ∑2h2 + ∑3h3 + ∑4h4 ) where A is the area of one rectangle or square and h1,h2,h3,h4 are the corner heights common to one, two, three and four rectangles respectively. Solution