Problem 02 - Symmetrical Parabolic Curve

From the grade diagram:
$\dfrac{S_1}{0.06} = \dfrac{160}{0.06 + 0.02}$

$S_1 = 120 \, \text{ m}$
 

 

Horizontal distance from the lowest point to point Q:
$s_Q = S_1 - 40 = 120 - 40$

$s_Q = 80 \, \text{ m}$
 

Grade at point Q by ratio and proportion of triangles:
$\dfrac{g_Q}{s_Q} = \dfrac{0.06}{S_1}$

$g_Q = \dfrac{0.06s_Q}{S_1} = \dfrac{0.06(80)}{120}$

$g_Q = 0.04$
 

Elevation of PC:
$\text{Elev } PC = \text{Elev } V + 80(0.06)$

$\text{Elev } PC = 14.375 + 80(0.06)$

$\text{Elev } PC = 19.175 \, \text{ m}$
 

Difference in elevation between PC and Q:
$DE_{PC-Q} = \text{shaded area in the grade diagram}$

$DE_{PC-Q} = \frac{1}{2}(g_Q + 0.06)(40)$

$DE_{PC-Q} = \frac{1}{2}(0.04 + 0.06)(40)$

$DE_{PC-Q} = 2 \, \text{ m}$
 

Elevation of the first quarter point Q:
$\text{Elev } Q = \text{Elev } PC - DE_{PC-Q}$

$\text{Elev } Q = 19.175 - 2$

$\text{Elev } Q = 17.175 \, \text{ m}$           [ C ]   answer