Civil Engineering May 2021 Review Innovations Surveying 2

ELEMENTS OF A SIMPLE CURVE REVERSE CURVE Situation 1: A simple curve have tangents AB and BC
Simple Curve – a circular arc, extending from one Problem 3: intersecting at a common point B. AB has an azimuth
tangent to the next. The perpendicular distance between two parallel of 1800 and BC has an azimuth of 2300. The stationing
tangents of the reverse curve is 35m. The azimuth of of the point of curvature at A is 10+140.26. If the
-Point of Curvature (PC) – point where the curve the back tangent is 2700 while the common tangent is degree of curve of the simple curve is 40, determine the
leaves the back tangent (first tangent). 3000. The first radius of the curve is 160m and the following:
-Point of Tangency (PT) – point where the curve joins stationing of PRC is 2+578. Determine the following: 3. Length of the long chord
the forward tangent (second tangent). a. Radius of the second curve. 4. Stationing of a point “X” on the curve on which a
-Intersection Angle (I) – angle of intersection of b. Stationing of PC. line passing through the center of the curve makes an
tangents. c. Stationing of PT. angle of 580 with line AB, intersects the curve at point
-Vertex or Point of Intersection (V or PI) – intersection SPIRAL CURVE “X”.
of back tangent and forward tangent. Problem 4: Answer: C = 242.14 m
-Tangent Distance (T) – distance from vertex to the PC A spiral curve having a length of 100 m is to be laid Sta. X = 10+300.26
or PT. out in a certain portion of road. The degree of the
-Chord Distance (C) – line connecting PC and PT. central curve is 6°. Situation 2: The common tangent BC of a reverse curve
-External Distance (E) – distance from the vertex to the a. Find the offset distance at the first quarter point is 280 m and has an azimuth of 312 0. AB is tangent of
curve. of spiral. the first curve whose azimuth is 2520 and a distance of
-Middle Ordinate (M) – line joining the middle of the b. Determine the spiral angle at the third quarter 320 m while CD is tangent of the second curve whose
curve and the middle of the chord. point of spiral. azimuth is 2180 and a distance of 260 m. If the radius
c. Compute the maximum speed of the car that of the second curve is 190 m, determine the following:
SIMPLE CURVE could pass through the spiral without skidding. 5. radius of the first curve
Problem 1: 6. length of the second curve
Two tangents adjacent to each other having bearings Problem 5: 7. total area of road from A to D using a width of 10m
N 65030’ E and S 85010’ E meet at station 11+157.98. If A spiral easement curve has a spiral angle at SC of 12° Answer: a. 132.07m; b. 311.72 m; c. 7500.20 m 2
the radius of the simple curve connecting these two and an offset distance at SC equal to 3.4m. Distance
tangents is 249.17m, determine the following: along tangent up to SC is 79.62m. What is the length Situation 3: An 80 simple curve connecting two
a. Tangent Distance of the short tangent, long tangent and length of throw? tangents that intersect at an angle of 88 0 is to be
b. Long Chord replaced by a symmetrical three centered compound
c. Middle Ordinate Problems for Practice: curve having a 60 end curves and 100 curve at the
d. External Distance 1. Two tangents AB and BC intersect at an angle of 240. center maintaining the same PC.
e. Degree of Curve A point P is located 21.03 m from point B and has a 8. Find the central angle of the 100 center curve.
f. Length of Curve perpendicular distance of 2.79 m from line AB. 9. Find the central angle of the 60 end curves.
g. Stationing of PT Calculate the radius of the simple curve connecting 10. Determine the stationing of the PT if PC is at
the two tangents and passing point P. 10+185.42
COMPOUND CURVE Answer: 285.33 m Answer: a. 51.460; b. 18.270; c. 10+410.14
Problem 2:
A compound curve has the following elements: 2. It is desired to lay out a curve at right angle. The 11. A spiral 80 m long connects a tangent with a 6.5 0
I1= 30° I2= 24° curve is compound with radius R1 equal to 310m and circular curve. Determine the deflection angle at the
D1= 4° D2= 5° radius R2 equal to 260m. If the distance from the vertex first quarter point.
If the stationing of the vertex is 4+ 620, to PC is 290 m, what is the distance from vertex to PT? Answer: 0.270
a. Determine the stationing of PC. Answer: 270 m
b. Find the stationing of PCC.
c. Determine the stationing of PT.
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Civil Engineering May 2021 Review Innovations Surveying 2

SIMPLE CURVE

I
a. T = R tan tangent distance
2
I
b. C = 2R sin long chord
2
1145 .916
c. D= degree of curve
R

d. L C = RI length of curve
180
 I 
e. E = R  sec − 1 external distance
 2 

 I
f. M = R1 − cos  middle ordinate
 2

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Civil Engineering May 2021 Review Innovations Surveying 2
L 2
180
= 
2RL S 
ELEMENTS PI
I 
LT = Long tangent i=
ST = Short tangent 3
R = Radius of simple curve
L = Length of spiral from TS to any point along the
spiral Es p=
(L ) S
2

Ls = Length of spiral Ts 24R

I = Angle of intersection Xc Ic
Ic = Angle of intersection of the simple curve
θs SC CS θs L3
p = Length of throw or the distance from tangent x=
that the circular curve has been offset 6RL S
X = Offset distance (right angle distance) from R I R
tangent to any point on the spiral
Ls L5
Xc = Offset distance (right angle distance) from y =L−
tangent to SC θs Ic θs 40R 2 (L S )
2

Es = External distance of the simple curve TS ST

θ = Spiral angle from tangent to any point on the P LS I
spiral TS = + (R + p) tan
θs = Spiral angle from tangent to SC 2 2
i = Deflection angle from TS to any point on the
I
E S = (R + p) sec
spiral, it is proportional to the square of its
distance x SC −R
is = Deflection angle from TS to SC 2
y = distance from TS along the tangent to any point
on the spiral
θ 0.036k 3
LS =
i R
0.0079 k 2
e=
TS R
D L
=
DC L S

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