Question

The grade line is being developed for a six-lane freeway where a rising 2.5% grade meets a 1.5% falling grade at station 100 00. The elevation of the PVI of the two grades is 471.00 ft.a. Using a vertical curve length of 1400 ft, compute the middle ordinate of the equal-tangent vertical curve.b. Calculate elevations of the vertical curve for each station starting at the beginning of the curve (i.e., at stations 93+00, 94+00, etc.)

Answer 1

We are given:

Elevation of PVI = 471.00ft

Vertical curve = 1400ft

Rising grade, G1=2.5℅

Falling grade, G2= 1.5℅

at a 100 00 station

a) Let's find PVC and PVT elevation,

•PVC elevation = PVI- g1

471.00 - (2.5/100)(700) = 453.5ft

•PVT elevation = PVI +g2

= 471 +(1.5/100)700 = 460.5ft

For mid-ordinate, we have:

L/8 * (G2 + G1)

= (1400/8)[(1.5+2.5)/100] = 7ft

Mid-ordinate= 7ft

b) •station PVC = PVI - L/2

= (100+00) - (14/2 +00)

= 93 + 00

• Station PVT = PVI + L/2

= (100+00)+(14/2 + 00)

= 107 + 00

Let's find the elevation at PVI stations before,

PVI = elevation of PVI - g1(distance between PVI and station)

= 471 - 2.5℅ * [(100+00) - (94+00)]

= 471 -2.5℅*(6+00)

= 471 -2.5℅ *600ft

= 456ft

Let's now find elevation of station after PVI,

Elevation = PVI-g2

= 471-1.5℅[(101+00)-(100-00)]

= 471-1.5℅(1+00)

= 471-1.5℅*(100) = 469.5ft

After calculating, we have .....Station....... Elevation

PVC=> 93+00. 453.5

94+00. 456

95+00. 458.5

96+00. 461

97+00 463.5

98+00. 466

99+00. 466.85

PVI=> 100+00. 471

101+00. 469.5

102+00. 468

103+00. 466.5

104+00. 465

105+00. 463.5

106+00. 463

PVI=> 107+00. 460.5

(100+00) => 100chains

If 1 chain = 100ft

Then, (101+00) = 110 chains

= (101+00) = 110*100ft