Horizontal Alignment:
Elements of Simple Circular Curves
INCI 4007
I. Cruzado
Element of a (Simple) Horizontal Curve
All equations so far
• Curve radius in feet and its relationship with degree of curvature D:
#$%&.#$(
!=
)
• Tangent of Curve (distance from PC to PI or PI to PT)
* = ! ∗ tan 0⁄%
• Length of Curve (arched distance from PC to PT) : 1 =
2∗3∗∆
5(6
• Long Chord (straight distance from PC to PT) 17 = 2! ∗ sin 0⁄%
• External distance ; = * ∗ tan 0⁄<
• Middle ordinate = = ! ∗ (1 − cos 0⁄% )
Example # 1
• Determine all the elements of a 870 ft – radius circular curve that has
an intersection angle of 25°25’ (R = 870; Δ = 25°25’)
• Elements:
•
•
•
•
•
•
Degree of Curvature
Tangent
Curve Length
Long Chord
External Distance
Middle Ordinate
D = 6°35’9”
T = 196.20 ft
L = 385.94 ft
LC = 382.78 ft
E = 21.85 ft
M = 21.31 ft
Example # 1 Extra - Answers
• The maximum speed at which a driver can safely maneuver the curve
• ! = 15%(' + )) =
• Vmax = 52.35 Mph
15 ∗ 870 ∗ (0.15 + 0.06) = 52.35 34ℎ
• The posted speed limit that you would suggest for the traffic sign.
• Vposted = 50 mph or 45 mph
Speed Limit cannot be higher than the maximum speed V from the equation!
Next
• PI, PC, and PT Stations
Stations Along Circular Curves
Starting with PI, PC and PT
INCI 4007
I. Cruzado
0+00
PC=???
2+00
1+00
Stations along highways
5+00
PT=???
Stations along highways
• Determine first PI station with first length given
• PC station is PI minus Tangent of Circular Curve
• PT station is PC plus Length of Circular Curve
• Move on to the next PI station and repeat.
Stations Along Highway
Given: “straight” distances
(highway tangents)
Stations Along Highway
1.
2.
3.
4.
Find PI
PC = PI - T
PT = PC + L
And repeat
1 Get PI with distance given
Go back to find PC
(minus tangent)
2
PI
PC
PT
3
Go forward along the curve to find PT
(plus length of curve)
4
And repeat!
Problem # 1
• Determine PI, PC, and PT stations for the curve below:
0+00
NOT TO SCALE
T
75°
R = 650FT
PT
ft
8
7
5. PC
7
10
PI
1. Start with the PI (with distance give):
PI = 10+75.78
2. Now find PC; you need the curve tangent
T = 498.76 ft
PC = 1075.78 – 498.76 = 577.02
PC = 5+77.02
3. From PC move to PT along the curve length
L = 850.85 ft
PT = 577.02 + 850.85 = 1427.87
PT = 14+27.87
Problem # 2
• Finish determining all PI, PC, and PT stations for the next two curves
65°
R = 800FT
NOT TO SCALE
R = 600 FT
ft
0+00
0
5.5
23
1
7
7.8
+2
14
PT
R = 650FT
95°
ft
3
4
.
2370
IMPORTANT
• Don’t assign PI stations based on previous PIs
• In other words: you have to “move” through the highway
PI1
.50
35
12
ft
8
.7
5
7
10
ft
ft
3
4
.
2370
PI2
Problem # 2 - Answers
• Curve 2
• PI = 21+64.60
• PC = 16+54.95
• PT = 25+62.52
• Curve 3
• PI = 44+23.29
• PC = 37+68.51
• PT = 47+63.34
Problem # 2 - check
• As you can see, you only need curve lengths (L’s) and tangent
distances (T’s) for each curve
• For the next curve: start with the PI
• You can determine the distance from PT of the first curve to the PI of the
second curve…
• … If you remember that there are two curve tangents
Practice Problem 1
• Two tangents of a route intersect at station 35+48, at an angle of
26°17’. Determine all elements of a 4°15’ curve (degree of curvature)
that will connect both tangents.
• Make sure that the curve built above is safe for 60 mph and a
superelevation of 6% (f = 0.13) (i.e. Is 60 mph a safe speed?).
Practice Problem 2
• Determine the radius for each of the three curves shown in the next
slide. Also determine the PI stations for each curve.
Stations for each curve:
STA PC1 = 65+48.7
STA PT1 = 68+11.6
STA PC2 = 71+46.9
STA PT2 = 72+79.9
1
STA PC3 = 77+62.7
STA PT3 = 78+73.5
2
Intersection angles for each
curve:
Δ1 = 63°15’
Δ2 = 48°20’
Δ3 = 42°45’
Disclaimer: the angles and
stations given are not the
real ones.
3
Next Class
• Location of Stations along Horizontal Curves