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Unsymmetrical Parabolic Curves

An unsymmetrical parabolic curve is a vertical curve in which the lengths of the curve on either side of the point of vertical intersection (PVI) are not equal. Unlike symmetrical curves—where the PVI lies at the midpoint—an unsymmetrical curve has different distances from the PVI to the beginning (PC) and to the end (PT) of the curve. These curves are commonly used when site constraints, existing structures, drainage requirements, or design limitations prevent equal curve lengths on both sides. Although the curve remains parabolic and the rate of change of grade is still constant, the vertex does not lie at the midpoint of the curve.

Let:

• $g_1$ = initial grade
• $g_2$ = final grade
• $g_3$ = grade at highest/lowest point (HP)
• $L_1$ = horizontal distance from PC to PVI
• $L_2$ = horizontal distance from PVI to PT
• $L = L_1 + L_2$ = total length of curve
• $H$ = maximum offset from PVI to curve

1. Maximum Offset (Unsymmetrical Curve)

2. Location of Highest/Lowest Point (HP)

Grade at HP:

Interpretation:

• $g_3 = 0$ → HP is at PVI
• $g_3 < 0$ → HP is left of PVI
• $g_3 > 0$ → HP is right of PVI

3. Relationship of Intermediate Grade

Consider algebraic signs of grades.

4. Station to Highest/Lowest Point

If $g_3 < 0$:

If $g_3 > 0$:

Problem: Unsymmetrical Parabolic Curve — Summit Curve

An unsymmetrical parabolic curve has a forward tangent of -8% and a back tangent of +5%. The length of curve on the left side of the curve is 40m long while that of the right side is 60m long. The PC is at station 6+780 and has an elevation of 110m.
a. Determine the elevation of the highest point.
b. Determine the elevation at 6+840.

Step 1: Compute Maximum Offset $H$

$$ H = \frac{(g_1 - g_2)L_2 L_1}{2L} $$ $$ H = \frac{(0.05 - (-0.08))(60)(40)}{2(100)} $$ $$ H = 1.56 \text{ m} $$

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Step 2: Locate the Highest Point (HP)

$$ g_3 = g_1 - \frac{2H}{L_1} = 0.05 - \frac{2(1.56)}{40} = -0.028 $$

OR

$$ g_3 = g_2 + \frac{2H}{L_2} = -0.08 + \frac{2(1.56)}{60} = -0.028 $$

Since $g_3 < 0$, the highest point is located to the left of the PVI.

$$ S_1 = \frac{g_1 L_1^2}{2H} = \frac{0.05(40)^2}{2(1.56)} = 25.64 \text{ m} $$

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Step 3: Compute Offset at HP using SPP

$$ \frac{y_1}{(25.64)^2} = \frac{1.56}{40^2} $$ $$ y_1 = 0.64 \text{ m} $$

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Step 4: Elevation of the High Point (Answer to Part a)

Using: $\text{El. HP} = \text{Elev. PC} + g_1 S_1 - y_1$

$$ \text{El. HP} = 110 + 0.05(25.64) - 0.64 $$ $$ = 110.642 \text{ m} $$

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Step 5: Elevation at Station 6+840 (Answer to Part b)

First compute elevation of PVT:

$$ \text{El. PVT} = 110 + 0.05(40) - 0.08(60) $$ $$ = 107.2 \text{ m} $$

Compute offset using SPP:

$$ \frac{y_2^2}{40^2} = \frac{1.56}{60^2} $$ $$ y_2 = 0.693 \text{ m} $$

Using: Elevation = El. PVT + $g_2(40) - y_2$

$$ \text{El. } 6+840 = 107.2 + 0.08(40) - 0.693 $$ $$ = 109.707 \text{ m} $$

For the grade diagram in unsymmetrical parabolic curves, we do not connect the back tangent grade and the forward tangent grade using a straight line. Instead, we divide the diagrams into two. First, we plot the ordinate of the back tangent (+5% or +0.05), then we plot the ordinate of PVI (g₃=-2.8% or -0.028), and we connect these two points with a straight line. Next, we plot the ordinate of the forward tangent (-8% or -0.08), and connect this to the ordinate of PVI using another straight line. The area under the diagram still defines the difference in elevation between two points, just like in symmetrical parabolic curves. Here, we compute the elevation of the highest point and the elevation at 6+840 using the grade diagram and verify that we still obtain the appropriate values. The small discrepancy is only due to rounding off error.