# Problem 03 - Symmetrical Parabolic Curve

**Problem**

A grade line *AB* having a slope of +5% intersect another grade line *BC* having a slope of –3% at *B*. The elevations of points *A*, *B* and *C* are 95 m, 100 m and 97 m respectively. Determine the elevation of the summit of the 100 m parabolic vertical curve to connect the grade lines.

A. 98.32 m

B. 99.06 m

C. 97.32 m

D. 96.86 m

**Solution**

*A*and

*B*= (100 - 95)/0.05 = 100 m

Horizontal distance between

*B*and

*C*= (100 - 97)/0.03 = 100 m

The figure above placed the parabolic curve at the middle-half:

$\text{Elev } PC = \text{Elev } A + 50(0.05)$

$\text{Elev } PC = 95 + 2.5$

$\text{Elev } PC = 97.5 \, \text{ m}$

Distance from PC to the summit:

$\dfrac{S_1}{0.05} = \dfrac{100}{0.08}$

$S_1 = 62.5 \, \text{ m}$

Elevation of the summit:

$\text{Elevation of the summit} = \text{Elev } PC + \frac{1}{2}S_1(0.05)$

$\text{Elevation of the summit} = 97.5 + \frac{1}{2}(62.5)(0.05)$

$\text{Elevation of the summit} = 99.0625 \text{ m}$ [ B ] *answer*