Problem 1

Given

L = 300 G1 = -3.20%

G2 = 1.80%

PI = 3030

ElevPVI = 465.92

1. 1 3.2*300465.92 470.72

2 2*100

G LElevBVC ElevPVI ft

2. 2 1.8*300465.92 468.62

2 2*100

G LElevEVC ElevPVI ft

3. Lowest Point

1min

3.2*300192

1.4

G LX ft

A

2

1G(min) ElevBVC

100 200

x AxElev

L

23.2 (1.8 3.2)(min) 470.72 467.65

100 200

x xElev ft

L

Problem 2

Given

G1 -6%

G2 5%

PI 11+00

ElevPVI 43

Distance d =

1100 1000 100d StaPVI StaCulvert ft

3 53 3 56ElevE ElevCulvert ft ft

1

2

6*10043 49

100 100

5*10043 38

100 100

G dElevF ElevPVI ft

G dElevG ElevPVI ft

56 38 18

56 49 7

EG ft ft ft

EF ft ft ft

Length of curve

1811

72 * 2*100* 862.73

181 1

7

EG

EFL d ftEG

EF

Problem 3

Given

L 1200

G1 -4.10%

G2 6.20%

StaPVI 36+50

ElevPVI 100

1

100*2

4.1*1200100 124.6

200

G LElevBVC ElevPVI

ElevBVC

2

1

1

GElevBVC

100 200

G 4.1 (6.2 4.1)*x 4.5

100 100 100 100*1200 100

(4.5 4.1)*12001001.942

10.3

x

x

x AxElev

L

dElev Ax

dx L

x ft

StaX = StaPVI +

2

1

2

G(min) ElevBVC

100 200

4.1 10.3*(1001.94)(min) 124.6 126.6

100 200*1200

x

x

x AxElev

L

Elev ft

Problem 4

Given

StaPVI

10+500

ElevPVI

350

ElevGrat

358.3

a. Minimum continuos curve length

b. Given L = 1900ft then

/ 2 10500 1900 / 2 9550StaBVC StaPVI L

StaBVC = 95 + 50

/ 2 10500 1900 / 2 114 50StaEVC StaPVI L

c. Elevation of lowest points

1min

4*19001266.67

6

G LX ft

A

2

1

2

G(min) ElevBVC

100 200

4*1266.67 6*(1266.67)(min) 138 112.67

100 200*1200

x

x

x AxElev

L

Elev ft

d. Elev at Sta 106+40

Distance = StaX – StaBVC = 10640 – 9550 =

2

1

2

GElevBVC

100 200

4*1090 6*(1090)138 113.16

100 200*1900

x

x

x AxElev

L

Elev ft

Problem 5

L

1400

G1

4.50%

G2

-3.00%

PI

4200

ElevPVI

436.56

/ 2 4200 1400 / 2 35 00StaBVC StaPVI L