3. COMPOUND VERTICAL CURVE Conpound vertical parabolic curves frequently occur i

Question

3. COMPOUND VERTICAL CURVE Conpound vertical parabolic curves frequently occur in rolling terrain. A compound vertical curve is one in whieh two parabolic curves of opposite concavity are joined· i.e., a sag is followed by a crest or vice versa (see Figure 5). In this problen of highway design ve are given the following information about a compound vertical curve There are three grade lines vith grades in order from left to right (1) of -2%, 3%, and -1%. (1) The point where the sag vertical curve meets the -2 gra line is denoted by A; B is the point where the sag meets the 3x grade 1ine; Cis a point on the crest vhere the sag is joined smoothly to the crest: and D is the point- where the crest meets te-1% grade line. (111) The horizontal distances between these points are: A to B 800 feet; A to C, 600 feet; and B to D, 900 feet. (iv) The elevation for A is 500 feet and it is 499.25 feet for c 200 900 FEET 600 FEET 3% -2% BLEVATION OF A: 500 FEET ELEVATION OF C: 499.25 EBT COMPOUND VERTICAL CURVE FIGURE a. Deternine the equation of the sag vertical parabolic curve. b. Deternine the equation of the crest vertical parabolic curve. c. Where should the drain be located on the sag vertical eurve d· What is the elevation of D? e. Make a chart which lists the elevations of points on the compound vertical curve starting at point A and proceeding in steps of 1 station to point D.

Explanation / Answer

a. equation of parabolic vertical curve

y=ax^2+bx+c

where a=(R2-R1)/2L, R2=final roadway grade in fraction, R1=initial roadway grade in fraction, L=length of curve

b=R1, c=elevation of Point A

a=(0.03+0.02)/2/800=0.00003125, b=-0.02, c=500

so y=0.00003125x^2-0.02x+500....(1)

(b)equation of crest vertical parabolic curve.

to find this equation first calculate slope at C by differentiating eq. 1 w.r.t. x

dy/dx=6.25x10^(-5)x-0.02

(dy/dx)x=600=6.25x10^(-5)*600-0.02=0.0175

y=ax^2+bx+c

a=(-0.01-0.0175)/2/1100=-0.0000125

b=0.0175

c=499.25

y=-0.0000125x^2+0.0175x+499.25 ....(2)

(c)drain can be provided in sag curve where grade of 0.3% is achieved in sag curve

differentiate eq. 1 with respect to x

dy/dx=6.25x10^(-5)x-0.02

put dy/dx=0.003

0.003=6.25x10^(-5)x-0.02

x=368ft.

d)elevation of D

put x=1100 in eq. (2)

yD=503.375 ft.

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