Simple Curve PDF
Simple Curve PDF
Simple Curve PDF
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 T
C
PC M
LC PT
R R
Sta @ PT = sta at PC + C
= (1 + 040) + 290
= 1+ 330
Solving for R,
10 10
R= R= R = 229.256
D 5
sin sin
2 2
Solving for T,
I 72°30' T = 168.097 m
T = R tan T = 229.256 tan
2 2
I 72°30' C = 271.123m
C = 2 R sin C = 2(229.256)sin
2 2
Solving for the external distance,
⎛ I ⎞ ⎛ 72°30' ⎞ E = 55.024m
E = R⎜ sec − 1⎟ E = 229.256⎜ sec ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
⎛ I⎞ ⎛ 72°30' ⎞
M = R⎜1 − cos ⎟ M = 229.256⎜1 − cos ⎟ M = 44.374m
⎝ 2⎠ ⎝ 2 ⎠
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.
(b) Determine the stationing at CD. PI
Solution:
C
ß
105.720 m 110°50’ D
θ
α
PC PT
O
Solve for R: Solving for x:
1145.916
R= x = 180° − 45°30'−51°43'
D
1145.916
R= x = 82°47'
5
CD 229.183
Solving for Ф: =
sin 82°47' sin 45°30'
105.720
tan φ = CD = 318.776m
229.183
Solving for α: C = R (φ + x)
⎛ π ⎞
α = 180° − 90° − 24°40' C = (229.183)(107°27')⎜ ⎟
⎝ 180° ⎠
α = 65°20' C = 429.8m
105.720
OC =
sin 24°40'
OC = 254.241m
Considering ΔOCD:
OC 229.183
=
sin θ sin 45°30'
θ = 57°43'
Instructions:
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:
T V or PI
I = 34°
PIN
PTN
TN
L1
50m PT
PC
θ
α
R 34°
T = stationV − stationPC
T = 12705.84 − 12628
T = 77.84m
T 77.84
R= R= R = 254.60m
I 34°
tan tan
2 2
Solve for θ:
L = Rθ 50 = ( 254.60 ) θ θ = 11°15'
Solve for α:
α = I −θ α = 34° − 11°15' α = 22°45 '