Introduction to Transportation Engineering https://en.wikipedia.org/wiki/Atlantic_Ocean_Road#/media/File:Atlanterhavsveien.jpg Instructor Dr. Norman Garrick Tutorial Instructor Hamed Ahangari April 2016

Horizontal Alignment Spiral Curve

Circle Curve Lc Ls Ls Spiral Curve Δc Δs Δs

Spiral Curves k = 100 D/ Ls Δ = Δc + 2 Δs Δs = Ls D / 200 http://www.nh.gov/dot/cadd/msv8/spiral.htm

Circle Curve Spiral Curve 1- Ltotal = LC + 2 Ls = Lcircle + Ls 2- Δ = Δc + 2 Δs 3- Δs = Ls D / 200 4- Δc = Lc D / 100 5- k = 100 D/ Ls

Example 4 The central angle for a curve is 40 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

Solution Given: R=2000 , Δ=40, k=3 D = 5730/2000. Therefore D = 2.86 L = 100(Δ)/D = 100(40)/2.86 = 1398 ft. Part i) Ls k = 100 D/ Ls,  3=100*2.86/Ls  Ls= 95 ft

Part iii) Spiral Central Angle Part ii) Total Length Total Length of curve = length with no spiral + Ls Total Length= 1398+95=1493 ft Part iii) Spiral Central Angle Δs = Ls D / 200  Δs = 95*2.86/200= 1.35  Δs =1.35  Δc= Δ- 2*Δs=40-2*1.35  Δc = 37.30 degree

Example 5 In a spiral curve the Δc is 33 degree. If the degree of curvature is 6 and the rate of increase in degree of curvature is 4. determine the length of each spiral leg, the intersection angel, Radius total length of curve

Solution Given: Δc =33, D=6, k=4 i) Ls ? k = 100 D/ Ls,  4=100*6/Ls  Ls= 150 ft ii) Δ ? Δ = Δc + 2.Δs Δs = Ls D / 200  Δs = 150*6 / 200  Δs = 4..5 Δ = 33+ 4.5*2= 42

iii) Radius R=5730/D  R =5730/6= 955 ft iv) total length of curve Ltotal = LC + 2 Ls = Lcircle + Ls Method a) Ltotal =Lcircle + Ls L = 100(Δ)/D = 100*(42)/6 = 700 ft. Ltotal =700 + 150 =850 ft Method b) Ltotal = LC + 2 Ls Δc = Lc D / 100  Lc = Δc *100 / D =33*100/6=550 Ltotal =550 + 2*150 =850 ft

Vertical Curve

Horizontal Alignment Vertical Alignment Crest Curve G2 G3 G1 Sag Curve

Design of Vertical Curves

Parabolic Curve BVC G1 G2 EVC PI L/2 L/2 L 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 curve y = 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)

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

PI EVC BVC 2.5% -1.5%

Solution 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

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

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=0 X=0.025/0.00008= 312.5 y(312.5) = 203.75+0.025*312.5-0.00004*(312.5)^2  y(312.5)= 207.66

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 r ii) station of BVC and EVC iii) elevation of point at a distance, L/4, from BVC iv) station of turning point v) elevation of turning point vi) elevation of mid-point of the curve vii) grade at the mid-point of the curve

Solution i) K and r K = L / |A| A=G2-G1, A=-4-(1)=-5 K=300/5=60 -4% i) K and r K = L / |A| A=G2-G1, A=-4-(1)=-5 K=300/5=60 r=(g2-g1)/L r=(-0.01-(-0.04))/300=0.03/300 r=0.0001 -1%

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 + g1x + 1/2 rx2, 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’

v) elevation of turning point iv) station of turning point at turning point: dy/dx=0 dy=dx=- 0.04+0.0001x=0 x(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)

vi) elevation of mid-point x=150’ Y(150)= 206-0.04*(150)+0.00005.(150)^2 y (150)=201.12’ vii) grade at the mid-point of the 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