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 commonSpiral Curve: a curve whose radius decreases uniformly

from infinity at the tangent to that of the curve it meets.

Horizontal Curves

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 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.

Types of circular Curves

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 gradually in a spiral curve, inversely proportion to the radius.

When to Use WhatSimple circular curves are the most common type. Usually if there is only one condition such as the radius, or length, start or end point.Spirals are used at highway exits, sometimes, and all the times in railroad curves.The rest of curves are used when the designer has to.

Circular Curves NotationsDefinitions: Point of intersection (vertex) PI, back and forward tangents.Point of Curvature PC, beginning of the curvePoint of Tangency PT, end of the Curve.Tangent Distance T: Distance from PC, or PT to PILong Chord LC: the line connecting PC and PTLength of the Curve L: distance for PC to PT on the curveExternal 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 tangentIntersection Angle I: the change of direction of the two tangents,

equal to the central angle subtended by the curve.

Circular Curves FormulasRemember that : R is radius, perpendicular to the tangents at PC, and PT

D is the curve degree. We will use partial cord = 20m

L = 20 I

Dm, where D and I in same units.

L = R* I (I in radians)

R = 57.3 * 20

D(m)

T = R tan( I2

LC = 2R sin( I2

))

E = R[ 1cos (I/2) - 1] M = R(1 - cos

I

2 )E = T tan( I

4) M = E cos I2

D/360 = 20 / 2 R

L / I = 20 / D

ExampleCalculate the elements of a circular (simple) horizontal curve

of which the degree is 5 , and the intersection angle is 160.

Answer:L = 20 (I/D) = 20 *(160/ 5) = 640 mR = 57.3 *20 / D = 57.3 *20 / 5 = 229.2 mT = R tan (I/2) = 229.2 * tan (160 /2) = 1299.85mSimilarly, LC = 2R sin (I/2) = 2 ( 229.2) sin (80) = Then, Calculate M and E

Circular Curve

Road designers study the topography of the area and select the route of the centerline of the road. At this point, the C.L is a series of connected straight lines,

At this stage we can determine the Intersection angle (I).Then the designer selects the radius of the curves and the surveyors setout the curve on the ground. To do that:GIVEN: I and R, or I and DRequired: - curve elements such as T, E, M, etc

- prepare the information needed to mark the curve on the ground (stake it out)

Concept of stations• Station: a distance marking technique in which a

(station) in the metric system equals 1000m.

• Stations are given in the format: (A +B) such as:11+213, A is the distance in kilometers and B is in meters. That means that this point is at a distance = 11213m from a certain zero. Simply erase the + sign

• Surveyors consider “stations” names and values. For example, the Distance between station 11+213 and station 13+ 412 = 13412 –11213 = 2199 m

• They may also say that we will start work at station 12+000, everybody knows where that is.

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)

Example

da= Sa D40 (degrees)

Where: da = DSa 20

or, da = Sa D

20

Theory; the angle between the tangent and a chord is equal to half the central angle subtended by the chord, so get da

Also, sin da = Ca

2Rfrom which Ca = 2R sin da

Ca = 2R sin da

In a curve whose I = 8° 24’, station of PC is 62+ 917.08, D = 2° 00’, calculate the necessary information to stake out points at stations 63+000, and 63+020.

Answer:

.. δa= Sa D/40 deg, and Ca = 2R sin δa

.. At station 63+000, Sa = 63000 – 62917.8 = 82.92 m

then, δ = (82.92) (2)/40 = 4.146 = 04° 08’ 46”

C= 2 (57.3 *20/2) sin(04° 08’ 46”) = 82.86 m

At station 63+020, Sa = 102.92m

Then δ = (102.92) (2)/40 = 5 8”20”

C = 2 (57.3 *20/2) sin(5 8”20” ) = 102.65 m

Circular Curve Layout by Coordinates with a Total Station

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 ProblemsPassing a curve through a certain point:- When?

-The problem: 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-P3- 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:

R2 = (XP + R tanI

2)

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• What is the problem?

• Stopping distance depends on: speed, perception and reaction time, coefficient of friction, and pavement

condition.

• Available sight distance = C = 8mRWhere m is the distance from the obstruction to the center of the road, along a radius.

Two solutions if C is less than the minimum safe sight distance:- Move the obstruction- Reduce the speed.

Example• If the minimum required sight distance on a curve

was 55 m, would a house at 15m from the center of the road allow that sight distance if the house is within a curved portion of the road, and the radius of the horizontal curve is 110m?

Answer: available sight distance = √ 8 m R = √ 8 *15 * 110 =

= 114.9 m

Spiral Curves

• Used to provide gradual transition in horizontal curvature,

and hence super elevation.

• 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) 20 m,

• 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 = S

• Then, M = ( Ls/2) (D/4) = (Ls D/8) = S /4

LP

LS)2(

Vertical Curves• In vertical planes, to provide smooth transitions

between grade lines of tangent sections.

• Almost always parabolic to provide constant rate of change of grade.

• Crest and sag curves.

Design Criteria

• Minimize cut and fill.

• Balance cut and fill.

• Maintain adequate drainage.

• Not to exceed max. Grade.

• Meet fixed elevations, other roads or bridges.

• Provide sufficient sight distance.

General Equation of a Vertical Parabolic Curve

• For a second order parabola:

Yp = a + bXp +cX2p

• What is the physical meaning of: a, b, c?

DefinitionsDefine: – BVC = VPC V= VPI EVC = VPT

– Percent grades g1, g2, r = rate of change of grade = (g2 - g1)/ L

– The curve length L: is it horizontal or curved length?

– What is an equal tangent vertical curve?

Equation of an Equal Tangent Vertical Parabolic Curve in Surveying Terminology

• For any point (p) on the curve, if E is elevation and X is distance from the beginning of the curve, then

Ep = EBVC + g1 Xp + g2 - g1 Xp2

2L

Vertical Curve Computation Using the Tangent offset Equation

• Select the grades, and hence find V’s

• The designer defines L, sight distance maybe?

• Compute the station of BVC, from the station of V and L, then compute the station of EVC, add L/2 to V?

• Note that you are not trying to locate the curve in the horizontal plane, just compute the elevations

• The problem:

– Given: g1, g2, station and elevation of V, and L

– Required : Elevation at certain distances

Example

g1 = +3.00%, g2 = -2%, V station is 46+700 and V elevation is 853.48 m, L = 200 m, compute the curve for stakeout at 20 m intervals.

Answer:

BVC station = (46+700) - (0+200 / 2) = 46 +600

EVC station = (46+600) + (0+200) = 46+800

Elevation of BVC = 853.48 – (0.03) (100) = 856.48

For each point, compute X and substitute in the equation below to compute Y:

Y = 856.48 + 0.03 (X) +{(-2-3}/2*200)} X2

For example, at station 46+ 620, X = 20m, E =

at station 46+710: X = 110m, E =

High or Low Points on a Curve• Why: sight distance, clearance, cover pipes, and

investigate drainage.

• At the highest or lowest point, the tangent is horizontal, the derivative of Y w.r.t x = 0.

• Deriving the general formula gives:

• X = g1 L/(g1 - g2) where: X is the distance in from BVC to the high or low point.

• Substitute in the tangent offset equation to get the elevation of that point.

• Example: compute the station and elevation of the highest point on the curve in the previous example.

• Answer: X = 3.00 *200/( 3+2.4) = 111.11 m

Plug X back into the equation get elevation = ??

Designing a Curve to Pass Through a Fixed Point

• Given: g1, g2 , VPI station and elevation, a point (P) elevation and station on the curve.

• Required: You need five values to design a curve: g1, g2 , VPI station and elevation, and curve length. The only missing value is the length of the curve.

• Solution:

– Substitute in the tangent offset formula, the only unknown is L:

Y = YBVC + g1 X + (r/2) X2

Solution: Substitute in the tangent offset formula, the only unknown is L: Y = YBVC + g1 X + (r/2) X2

Remember that: •YBVC = YV - g1 (L/2), only L is unknown•X is distance from BVC in stations, it is not the given station of P, to compute it, add or subtract the distance V-P to/from L/2.

X = (L/2) (V-P)stations

Since you are given stations of VPI and the unknown point, you should be able to know: to add or to subtract.

• r = (g2 - g1)/ L

Example :

..g1 = -4.00, g2 = +3.80 percent, V station is 52+00 and elevation is 1261.5, the curve passes by point P at station 53+50 and elevation 1271.2 ft.

Answer

1271.2 = {1261.5 +4.00 (L/2)} + { -4.00((L/2) +1.5)} + {(3.8+4.00)/2L)( L/2 +1.5)}

Sight Distance• SSD : is the sum of two distances: perception

reaction time, and the vehicle stopping distance.

• The length of the curve should provide enough SSD at design speed, and minimize cut and fill if possible.

• AASHTO design standards (1994): H1= 1.07m, driver’s eye height, and

• H2 = 0.15m, obstruction height.

• SSD given in tables.

Safe Length of Crest Curve To determine the safe length of a curve :

– compute the SSD, or use tables, according to design conditions.

– Compute (A): the absolute difference in grade.

– Apply one of the formulas, neglecting the sign of (A):

• Lm = 2 SSD - {200( H1- H2)2/A} Substituting

for H1 = 1.07m and H2 = 0.15m get Lm = 2 SSD - (404/A) if SSD>L

• Similarly, Lm = A (SSD2/404) if SSD < L

• Assume that SSD < L first, then check the answer.

Example

• What is the min length of a vertical curve that connects two slopes -1.5% and 2% if the required SSD is 120m?

• Answer:

• First eq. provides 124.6 m, does not fulfill the condition

• Second equation provides 124.8, ok