These are sample problems about simple curves. Elementary Surveying for Engineering Students
Simple Curve Lecture 1Full description
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Surveying manual
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my first lecture on surveying
cot curve notesFull description
SOLVING A SIMPLE CURVE Problem 2 Find the length of curve and the station at PT if the degree of curve is 5° and the central angle is 72°30’. Solve also the radius, tangent distance, external distance and middle ordinate. Use 20 meter full station. Solution:
PI
I = 72°30’
E T
C
PC
M
T
R
R
Solve for the length of curve,
C=
20 I D
C=
20(72°30' ) 5°
C = 290m
Solving for station at PT, Sta @ PT = sta at PC + C
= (1 + 040) + 290 = 1+ 330
Solving for R,
R=
10 D sin 2
R=
10 5 sin 2
PT
LC
R = 229.256
Solving for T, T = R tan
I 2
T = 229.256 tan
72°30' 2
T = 168.097 m
Solving for the long chord, C = 2 R sin
I 2
C = 2(229.256)sin
72°30' 2
C = 271.123m
Solving for the external distance,
I ⎞ ⎛ E = R⎜ sec − 1⎟ 2 ⎠ ⎝
72°30' ⎞ ⎛ E = 229.256⎜ sec ⎟ 2 ⎠ ⎝
E = 55.024m
Solving for the middle ordinate,
I⎞ ⎛ M = R⎜1 − cos ⎟ 2⎠ ⎝
72°30' ⎞ ⎛ M = 229.256⎜1 − cos ⎟ 2 ⎠ ⎝
M = 44.374m
Problem 3 A 5° curve intersects a property line CD, at point D. The back tangent intersects the property line at point C which is 105.720 m from the PC which is at station 0+612.690. The angle that the property line CD makes with the back tangent is 110°50’. (a) Determine the distance CD. PI (b) Determine the stationing at CD. Solution:
C 105.720 m 110°50’
ß
D θ
α PT
PC
x
Ф
O
Solve for R:
Solving for x:
R=
1145.916 D
x = 180° − 45°30'−51°43'
R=
1145.916 5
x = 82°47'
R = 229.183m
Solving for CD:
Solving for Ф:
CD 229.183 = sin 82°47' sin 45°30'
tan φ =
105.720 229.183
CD = 318.776m
φ = 24°40'
Solving for C:
Solving for α:
C = R (φ + x)
α = 180° − 90° − 24°40'
⎛ π ⎞ C = (229.183)(107°27')⎜ ⎟ ⎝ 180° ⎠
α = 65°20'
C = 429.8m
Solving for ß:
Solving for Station @ CD:
β = 110°50'−65°20'
sta @ CD = Sta @ PC + C
β = 45°30'
sta @ CD = (0 + 612.690) + 429.8
Solving for OC:
sta @ CD = 1 + 042.490
OC =
105.720 sin 24°40'
OC = 254.241m
Considering ΔOCD: OC 229.183 = sin θ sin 45°30'
θ = 57°43'
Instructions: 1. The class will be divided into 10 groups with 5 members each. 2. Each group will read, discuss and understand the lecture notes especially the sample problems. 3. Try to answer problems 4 and 5. Write the group’s solution in a 1 whole yellow paper. 4. Write down the things which the group doesn’t understand in a ¼ yellow paper. 5. Submit the papers after the class period at the CE Department. Remember to list all the names of the group members. For next week: Tuesday – continue with the lecture discussion Thursday – Pre Midterm Examination Coverage: Unit 1. Route Surveying Unit 2. Railroad and Highway Curves Unit 3. Horizontal Curves -Geometry of Simple Curves - Methods of Laying Out Simple Curves Type of Exam: Objective Type, Problem Solving, Essay
Problem 4 It is required to layout a simple curve by deflection angles. The curve is to connect two tangents with an intersection angle of 32° and a radius of 240 m. Compute the deflection angles to each 20 m full stations on the curve. If the transit is set up at the P.C. which is at station 5+767.2. What is the stationing of the P.T.? Problem 5 Two tangents intersect at V at station 12+705.84. The angle of intersection is 34°. The stationing of P.C. is 12+628. Without changing the location of P.C. and the sharpness of the curve, it is desired to shorten the length of curve by 50 meters. Determine the stationing of the new point of intersection (P.I.) and new point of tangency (P.T.).
Problem 4 Two tangents intersect at V at station 12+705.84. The angle of intersection is 34°. The stationing of P.C. is 12+628. Without changing the location of P.C. and the sharpness of the curve, it is desired to shorten the length of curve by 50 meters. Determine the stationing of the new point of intersection (P.I.) and new point of tangency (P.T.). Solution:
V or PI
T
I = 34°
PIN PTN
TN
L1
50m
PC θ
α R
34°
Solve for Tangent Distance: T = stationV − stationPC T = 12705.84 − 12628 T = 77.84m
