Question

he vertical curve is 600-ft long, equal-tangent, and connects an initial grade of 1.2% and a final grade of -1.08%. determine what is the offset at the pvt.

Answer 1

To find the offset at the PVT (point of vertical tangency) of a 600-ft long equal-tangent vertical curve connecting an initial grade of 1.2% and a final grade of -1.08%, you can use the following formula:

`\rm \: Offset \: at \: PVT = ( (L)/(2) ) * ((E1 + E2))/(100)`

Where:

L = Length of the curve = 600 ft

E1 = Initial grade = 1.2%

E2 = Final grade = -1.08%

Substituting the given values into the formula, we get:

`\rm \: Offset \: at \: PVT = (600)/(2) * ((1.2 - 1.08))/(100)`

`\rm \: Offset \: at \: PVT = 300 * 0.012`

`\rm \: Offset \: at \: PVT = 3.6 \: feet`

Therefore, the offset at the PVT of this vertical curve is 3.6 feet