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Introduction to Transportation Engineering

Instructor Dr. Norman Garrick

Hamed Ahangari17th April 2014

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Horizontal Curve

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Elements Δ– Intersection angle

R

Δ R - Radius

Δ

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ΔΔ = Deflection Angle

R = Radius of Circular Curve

L = Length of Curvature (L = EC – BC)

BC = Beginning of Curve (PC) EC = End of Curve (PT)

PI = Point of Intersection

T = Tangent Length (T = PI – BC = EC - PI)

C = Chord Length

M = Middle Ordinate

E = External Distance

R R

BC(PC)

EC(PT)

PI

T TL

M

E

C

ΔΔ/2

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Basic Definitions

• BC/PC: Point of Curvature• BC = PI – T

– PI = Point of Intersection – T = Tangent

• EC/PT: Point of Tangency• EC = BC + L

– L = Length

BC(PC) EC

Δ

Δ

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Degree of Curvature• D used to describe curves

• D defines Radius

• Arc Method: – D/ Δ = 100/L (1) (360/D)=100/(2R)

R = 5730/D (2)

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Curve Calculations

Length L = 100.Δ/D (3)

Tangent T = R.tan(Δ /2) (4)

Chord C = 2R.sin(Δ /2) (5)

External Distance E = = R sec(Δ/2) – R (6)

Mid Ordinate M = R-R.cos(Δ /2)) (7)

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Example 1• A horizontal curve is designed with a 2000 ft.

radius. The tangent length is 500 ft. and the EC station is 30+00. What are the BC and PI stations?

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• Since we know R and T we can use T = R.tan(Δ /2) so Δ = 28.07 degrees

• D = 5730/R. Therefore D = 2.86

• L = 100(Δ)/D = 100(28.07)/2.86 = 980 ft.

• BC = EC – L = 3000 – 980 =2020~20+20

• PI = BC +T = 2020 + 500 = 2520~25+20

Solution

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Example 2A curve has external angle of 20.30’ degrees, a degree of curvature is 2°30’ and the PI is at 175+00. Calculate:

RadiusLength of Curve BC and EC ChordExternal Distance Mid Ordinate

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• Given: D = 2°30’, Δ=22.30’

• Part i) Radius:

• Part ii ) Length of Curve

• Part iii ) BC and EC

83.22915.2

5730

R

'87.45525.22tan38.2291

T

13.44171)87.554()00175( BC

00.9005.25.22100

L

13.44180)009()13.44171( EC

Solution

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• Part iv ) Chord

• Part v ) External Distance

• Part vi ) Mid Ordinate

'23.89425.22sin)83.2291(2

C

'90.44125.22sec(83.2291

E

'04.4425.22cos183.2291

M

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Example 3The central angle for a curve is 30 degrees - the radius of the circular curve selected for the location is 2000 ft.If a spiral with k=3 degrees is selected for use, determine the

i) length of each spiral leg, ii) total length of curve iii) Spiral Central Angle

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http://www.nh.gov/dot/cadd/msv8/spiral.htm

Spiral Curves

Δs = Ls D / 200

Δ = Δc + 2 Δs

k = 100 D/ Ls

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Given: R=2000 , Δ=30, k=3

• D = 5730/2000. Therefore D = 2.86• L = 100(Δ)/D = 100(30)/2.86 = 1047 ft.

Part i) Lsk = 100 D/ Ls, 3=100*2.86/Ls

Ls= 95 ft

Solution

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• Part ii) Total LengthTotal Length of curve = length with no spiral + LsTotal Length= 1047+95=1142 ft

Part iii) Spiral Central Angle

Δs = Ls D / 200 Δs = 95*2.86/200= 1.35

Δs =1.35

Δc= Δ- 2*Δs=30-2*1.35

Δc = 27.30 degree

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Vertical Curve

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Horizontal Alignment

Vertical Alignment

Crest Curve

Sag Curve

G1

G2 G3

Design of Vertical Curves

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Parabolic CurveBVC

EVC

L

G2

G1

L/2L/2

PI

Change in grade: A = G2 - G1 G is expressed as %

Rate of change of curvature: K = L / |A|Rate of change of grade: r = (g2 - g1) / L

Equation for determining the elevation at any point on the curvey = y0 + g1x + 1/2 rx2

where,y0 = elevation at the BVC g = grade expressed as a ratio

x = horizontal distance from BVC r = rate of change of grade (ratio)

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Example 1• The length of a tangent vertical curve equal to 500 m.

The initial and final grades are 2.5% and -1.5% respectively. The grades intersect at the station 10+400 and at an elevation of 210.00 m

• Determine:a)the station and the elevation of the BVC and EVC points

b) the elevation of the point on the curve 100 and 300 meters from the BVC point

c) the station and the elevation of the highest point on the curve

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BVCEVC

PI

2.5% -1.5%

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• Part a) the station and the elevation of the BVC and EVC points

– Station-BVC= 10400-250=10150~10+150– Station-EVC=10400+250=10650~10+650

– Elevation-BVC= 210-0.025*250= 203.75 m– Elevation-EVC= 210-0.015*250= 206.25 m

Solution

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• Part b) the elevation of the point on the curve 100 and 300 meters from the BVC point.

y = y0 + g1.x + 1/2 .r.x^2,y0= 203.75, g1= 0.025

– r=(g2-g1)/L– r=(-0.015-(0.025))/500=-0.04/500 r=-

0.00008

y = 203.75 + 0.025.x - 0.00004.x^2, y(100)=203.75+0.025*100-0.00004*100^2 y(100) = 205.85 y(300)= 207.65

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• Part c) the station and the elevation of the highest point on the curve– The highest point can be estimated by setting the first

derivative of the parabola as zero.

• Set dy/dx=0,dy/dx= 0.025-0.0008*x=0X=0.025/0.00008= 312.5

y(312.5) = 203.75+0.025*31.25-0.00004*(31.25)^2 y(312.5)= 207.65

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Example 2• For a vertical curve we know that G1=-4%, G2=-1%,

PI: Station 20+00, Elevation: 200’, L=300’• Determine:

i)K and rii) station of BVC and EVCiii) elevation of point at a distance, L/4, from BVCiv) station of turning pointv) elevation of turning pointvi) elevation of mid-point of each curvevii) grade at the mid-point of each curve

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• i) K and r

– K = L / |A|

– A=G2-G1, A=-4-(-1)=-3

K=300/3=100

– r=(g2-g1)/L

– r=(-0.01-(-0.04))/300=0.03/300 r=0.0001

Solution -4%

-1%

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• ii) station of BVC and EVC– L=300’, L/2=150’– BVC= 2000-150=1850’, ~ 18+50– EVC= 2000+150=2150’, ~ 21+50

• iii) elevation of point at a distance, L/4, from BVC– y = y0 + g1.x + 1/2 r.x^2,– r= -0.0001, g1=0.04– y0=200+150*0.04=206

y=206-0.04x+0.00005.x^2

L/4~ x=75 Y(75)=203.28’

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• iv) station of turning point– at turning point: dy/dx=0– dy=dx=- 0.04+0.0001x=0x(turn)=400’

• v) elevation of turning point– x=400’– Y (400) = 206-0.04*(150)+0.00005.(150)^2 Y(400)=198’

This point is after vertical curve (Turning point is not in the curve)

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• vi) elevation of mid-point – x=150’– Y(150)= 206-0.04*(150)+0.00005.(150)^2y (150)=201.12’

• vii) grade at the mid-point of each curve– Grade at every points: dy/dx= =- 0.04+0.0001x– if x= 150 Grade (150)=-0.04+0.0001*150 grade(150)= -0.025