# Horizontal by Lq8sz3CE

VIEWS: 15 PAGES: 28

• pg 1
```									Readings: 24-1
to 24-8,
Figures: 24-1 to
24-6, 24-12, and
24-16
Examples: 24-1,
24-2,
Required:
Figures: 24-1 to 24-6, 24-12, and 24-16
Examples: 24-1 and 24-2.
Introduction
•Use of curves, horizontal and vertical.
•Types of horizontal curves: Circular and spiral.
We will cover circular curves only, spiral curves are
given for future reference.

Definitions:
Horizontal Curves: curves used in horizontal planes to
connect two straight tangent sections.
Simple Curve: circular arc connecting two tangents. The
most common
Spiral Curve: a curve whose radius decreases uniformly
from infinity at the tangent to that of the curve it meets.
Compound Curve: a curve which is composed of two or
more circular arcs of different radii tangent to each other,
with centers on the same side of the alignment.
Broken-Back Curve: the combination of short length of
tangent (less than 100 ft) connecting two circular arcs that
have centers on the same side.
Reverse Curve: Two circular arcs tangent to each other,
with their centers on opposite sides of the alignment.
Easement Curves: curves used to lessen the effect of the sudden change
in curvature at the junction of either a tangent and a curve, or of two
curves.

Super elevation: a difference of elevation between the edges of the
cross section, to overcome the effect of centrifugal force. Changes

When to Use What
Simple circular curves are the most common type.
Spirals are used at highway exits, sometimes, and all the times in

The rest of curves are used when the designer has to.
Degree of Circular Curve
Curves are identified either by:
Degree of Curve (D): central angle subtended by a 100ft:
circular arc (in arc definition), or chord in chord definition.

The arc definition is commonly used in highways, chord definition

In arc definition, D = 5729.58          degrees, R in ft.
R
50
In chord definition, D = 2 sin-1   (R)
Pay attention to units, we will use ft for length, how about angles?
Circular Curves Notations
Definitions:
Point of intersection (vertex) PI, back and forward tangents.
Point of Curvature PC, beginning of the curve
Point of Tangency PT, end of the Curve.
Tangent Distance T: Distance from PC, or PT to PI
Long Chord LC: the line connecting PC and PT
Length of the Curve L: distance for PC to PT:
measured along the curve, arc definition.
measured along the 100 chords, chord definition
External Distance E: The length from PI to curve midpoint.
Middle ordinate M: the radial distance between the midpoints of the
long chord and curve.
POC: any point on the curve.
POT: any point on tangent
Intersection Angle I: the change of direction of the two tangents,
equal to the central angle subtended by the curve.
Circular Curves Formulas
Remember that : R is radius, perpendicular to the tangents at PC, and PT
D is the curve degree, use ARC definition.

I
L = 100 I            ft, D and I in same units. =           (sta) = R I I in rad
D                                            D
5729.58
R=                    (ft)
D
I
T = R tan(       )                     LC = 2R sin( I )
2                                      2
1
[
E = R cos (I/2) - 1      ]                          (
M = R 1 - cos I
2       )
()
E = T tan 4
I
M = E cos I
2
Circular Curve Stationing
Route Survey are usually staked out as a series of tangents having
continuous stations. The station of the PI is determined.
Stations of PC, and PT are computed from PI, R is given by designers:
I
Compute T = R tan(          )
2
Station of PC = station of PI - T
Station of PT = station of PC + L
Station equations at PT: the route considering the curve is shorter
than it was computed considering the tangents.
= (station of PI + T) - (station of PC + L)
This amount should be subtracted from stations of all the points
after PT.
Circular Curves Layout by Deflection
Angles with a Total Station or an EDM
All stations will be positioned from PC.
Compute the chord length and the deflection angle from the direction
PC-PI as follows: (see fig 25-6)

da=
Sa D
(degrees)          Ca = 2R sin da
200
Where:
da = D                  or, da = Sa D
Sa 100                            100
Theory; the angle between the tangent and a chord is equal to half the
central angle subtended by the chord, so get da
Ca
Also, sin da =          from which Ca = 2R sin da
2R
Example
In a curve whose I = 8° 24’, station of PC is
62+ 17.08, D = 2° 00’, calculate the necessary
information to stake out points at stations
63+00, 64+00, and at the PT.
.. δa= Sa D/200 deg, and Ca = 2R sin δa
.. At station 63+00, Sa = 6300 – 6217.08 = 82.92 ft
then, δ = (82.92) (2)/200 = 0.8292 = 00° 49’ 45”
C= 2 (5729.58/2) sin(00° 49’ 45”) = 82.92 ft
At station 64+00, Sa = 182.92 ft
Then δ = (182.92) (2)/200 = 1.8292 = 1° 49’ 45”
C = 2 (5729.58/2) sin(1° 49’ 45”) = 182.89 ft
At the PT: Sa = ?, and Ca = ?
Circular Curve Layout by
Coordinates with a Total Station
{this topic and all the following until “sight distance” is mentioned
for future reference and will not be covered.}

Given: Coordinates and station of PI, a point from which the curve
could be
observed, a direction (azimuth) from that point, AZPI-PC , and curve
info.
Required: coordinates of curve points (stations or parts of stations)
and the data to lay them out.
Solution: - from XPI, YPI, T, AZPI-PC, compute XPC, YPC
- compute the length of chords and the deflection angles.
- use the deflection angles and AZPI-PC, compute the azimuth
of each chord.
- knowing the azimuth and the length of each chord, compute
the coordinates of curve points.
- for each curve point, knowing it’s coordinates and the total
station point, compute the azimuth and the length of the
line connecting them.
- at the total station point, subtract the given direction from the
azimuth to each curve point, get the orientation angle.
Special Circular curve Problems
Passing a curve through a certain point:
- When?
-The problem: fig(24-15)
Given PI, point (P) that should be on the curve, and the tangents.
Required: R.
Solution:
1-Establish an arbitrary coordinate system, origin is at
PI, X axis is the line PC-PI. In that system we know the
coordinates of PI, PC.
In that system the coordinates of the origin O is:
Xo = -T = -R tan (I/2)                        Yo = -R
2- Measure the angle and the distance PI-P
3- Compute the coordinates of P:
Xp = - d cosθ Yp = - d sin θ
4- Substitute in the general equation of a circle:
R2 = (XP – Xo )2 +(YP – Yo)2

Solve the equation to compute R:

2
R2 =   (   XP + R tan
I
2
)   + (YP+ R)2
Intersection of a circular curve
and a straight line
• Form the line and the circle equations, solve them
simultaneously to get the intersection point.

Intersection of two Circular Curves
simultaneously solve the two circle equations.
Sight distance on Horizontal Curves
{Required topic}
• What is the problem?
• Stopping distance depends on: speed, perception and
reaction time, coefficient of friction, and pavement
condition.
• Available sight distance = C =                8mR
Where m is the distance from the obstruction to the center of

Two solutions if C is less than the minimum safe sight distance:
- Move the obstruction
- Reduce the speed.
Spiral Curves
{This topic will not be covered}
• Used to provide gradual transition in horizontal curvature,
and hence superelevation.
• Definitions:
–   Back and forward tangents.
–   Entrance and exit spirals. Geometrically identical.
–   TS, SC, CS, ST. What is in between?
–   SPI: the angle beteen the tangents at TS and SC.
–   Spiral Angle S: the angle between the two tangents.
–   Spiral Length LS: the arc length of the spiral.
Spiral Geometry
• Basic spiral properties:
– Radius changes uniformly from infinity at TS to the
radius of the circular curve at the SC. So, it’s degree of
curve DS changes uniformly from 0o to D at the SC.
• Average degree of curve is D/2.
• In circular curves, L = (I/D) 100 ft, or I = LD stations
• similarly, S= Ls (D/2)       S and D in deg, L in stations
– Spiral angles at any point is proportional to the square
of the distance Lp from TS to the point. P = (LP )2 S
L       S
• In Fig 25-15, M is the mid point of the spiral, Lp = Ls/2 but
M is not = (S /2).Since D changes uniformly, degree of the
curve = D/2 at M. But D changes uniformly, so the average
degree of curvature between TS and M is (D/2)/2 = D/4
• Then, M = ( Ls/2) (D/4) = (Ls D/8) = S /4

```
To top