Presentation on theme: "Plane Surveying Traverse, Electronic Distance Measurement and Curves"— Presentation transcript:
1 Plane Surveying Traverse, Electronic Distance Measurement and Curves Civil Engineering StudentsYear (1)Second semester – Phase II Dr. Kamal M. Ahmed
2 IntroductionTopics in Phase II: Angles and Directions, Traverse, EDM, Total Stations, Curves, and Introduction to Recent and supporting technologiesIntroduction of the InstructorBackground, honors, research interests, teaching, etc.Method of teaching:what to expect and not to expect, what is allowed.Language used.Lecture slides: NO DISTRIBUTION WITHOUT PERMITBreaks
10 Angles and Directions 1- Angles: Horizontal and Vertical Angles Horizontal Angle: The angle between the projections of the line of sight on a horizontal plane.Vertical Angle: The angle between the line of sight and a horizontal plane.Kinds of Horizontal AnglesAngles to the Right: clockwise, from the rear to the forward station, Polygons are labeled counterclockwise.Interior (measured on the inside of a closed polygon), and Exterior Angles (outside of a closed polygon).
11 Angles to the Left: counterclockwise, from the rear to the forward station. Polygons are labeled clockwise.Right (clockwise) and Left (counterclockwise) Polygons
12 Distinguish between angles, directions, and readings. Direction of a line is the horizontal angle between the line and an arbitrary chosen reference line called a meridian.We will use north or south as a meridian ““مرجعTypes of meridians:Magnetic: defined by a magnetic needle “ ” ابرةGeodetic “ جيوديسى” meridian: connects the mean positions of the north and south poles “ اقطا ب”.Astronomic الفلكى : instantaneous لحظى , the line that connects the north and south poles “ اقطا ب” at that instant. Obtained by astronomical observations.Grid شبكى : lines parallel to a central meridianDistinguish between angles, directions, and readings.
13 Angles and Azimuth الزوايا والانحرافات Horizontal angle measuredclockwise from a meridian (north) to the line, at the beginning of the lineThe line AB starts at A, the line BA starts at B.Back-azimuth “الانحراف الخلفى “ is measured at the end of the line.88
15 Azimuth and Bearing المختصر الانحراف و الانحراف Bearing (reduced azimuth): acute “حادة “ horizontal angle, less than 90°, measured from the north or the south direction to the line. Quadrant is shown by the letter N or S before and the letter E or W after the angle. For example: N30W is in the fourth quad “ الربع الرابع“.Azimuth and bearing: which quadrant “ اى ربع “ ?99
17 Departures and Latitudes المركبات السينية و الصادية
18 Azimuth EquationsHow to know which quadrant from the signs of departure and latitude?For example, what is the azimuth if the departure was (- 20 m) and the latitude was (+20 m) ?The following are important equations to memorize and understandAzimuth of a line (BC)=Azimuth of the previous line AB+180°+angle BAssuming internal angles in a counterclockwise polygon1010
19 NCNBNNNBAACAzimuth of a line such as BC = Azimuth of AB ± The angle B +180°
20 Easting and Northing α N P (E ,N) L E In many parts of the world, a slightly different form of notation is used.instead of (x,y) we use E,N (Easting, Northing) .In Egypt, the Easting comes first, for example: (100, 200) means that easting is 100In the US, Northing might be mentioned first.It is a good practice to check internationally produced coordinate files before using them.
21 Polar Coordinates u u N +P ( r , ) r E The polar coordinate system describes a point by (angle, distance) instead of (X, Y)We do not directly measure (X, Y in the fieldIn the field, we measure some form of polar coordinates: angle and distance to each point, then convert them to (X, Y)
23 Example (1)Calculate the reduced azimuth of the lines AB and AC, then calculate the reduced azimuth (bearing) of the lines AD and AELineAzimuthReduced Azimuth (bearing)AB120° 40’AC310° 30’ADS 85 ° 10’ WA EN 85 ° 10’ W
24 Example (1)-Answer Line Azimuth Reduced Azimuth (bearing) AB 120° 40’ S 59° 20’ EAC310° 30’N 49° 30’ WAD256° 10’S 85° 10’ WA E274° 50’N 85° 10’ W
25 Example (2) Compute the azimuth of the line : - AB if Ea = 520m, Na = 250m, Eb = 630m, and Nb = 420m- AC if Ec = 720m, Nc = 130m- AD if Ed = 400m, Nd = 100m- AE if Ee = 320m, Ne = 370m
26 Note: The angle computed using a calculator is the reduced azimuth (bearing), from 0 to 90, from north or south, clock or anti-clockwise directions. You Must convert it to the azimuth α , from 0 to 360, measured clockwise from North.Assume that the azimuth of the line AB is (αAB ),the bearing is B = tan-1 (ΔE/ ΔN)If we neglect the sign of B as given by the calculator, then,1st Quadrant : αAB = B ,2nd Quadrant: αAB = 180 – B,3rd Quadrant: αAB = B,4th Quadrant: αAB = B
27 - For the line (ab): calculate ΔEab = Eb – Ea and ΔNab = Nb – Na- If both Δ E, Δ N are - ve, (3rd Quadrant)αab = = 210- If bearing from calculator is – 30 & Δ E is – ve& ΔN is +veαab = = 330 (4th Quadrant)- If bearing from calculator is – 30& ΔE is + ve& ΔN is – ve,αab = = 150 (2nd Quadrant)- If bearing from calculator is 30 , you have to notice if both ΔE, ΔN are + ve or – ve,If both ΔE, ΔN are + ve, (1st Quadrant)αab = 30otherwise, if both ΔE, ΔN are –ve, (3rd Quad.)αab = = 210
29 Example (3)The coordinates of points A, B, and C in meters are (120.10, ), (214.12, ), and (144.42, 82.17) respectively. Calculate:The departure and the latitude of the lines AB and BCThe azimuth of the lines AB and BC.The internal angle ABCThe line AD is in the same direction as the line AB, but 20m longer. Use the azimuth equations to compute the departure and latitude of the line AD.
31 d) AZAD:The line AD will have the same direction (AZIMUTH) as AB = 54° 04’ 18”LAD = (94.02)2 + (68.13)2 = mCalculate departure = ΔE = L sin (AZ) = 94.02mlatitude = ΔN= L cos (AZ)= 68.13m
32 Example (4)120ECBA1159011010530DIn the right polygon ABCDEA, if the azimuth of the side CD = 30° and the internal angles are as shown in the figure, compute the azimuth of all the sides and check your answer.
33 Example (4) - Answer120ECBA1159011010530DCHECK : Bearing of CD = Bearing of BC + Angle C + 180= = 30 (subtracted from 360), O. K.Bearing of DE = Bearing of CD + Angle D + 180= = 320Bearing of EA = Bearing of DE + Angle E + 180= = 245 (subtracted from 360)Bearing of AB = Bearing of EA + Angle A + 180= = 180 (subtracted from 360)Bearing of BC = Bearing of AB + Angle B + 180= = 120 (subtracted from 360)