4.4 Route Surveying: Horizontal and Vertical Curves

4.4 Route Surveying: Horizontal and Vertical Curves

Highway, railway, and canal routes consist of straight tangents joined by smooth geometric curves. Horizontal curves transition changes in direction on a horizontal plane, while vertical curves transition changes in longitudinal gradient on a vertical plane. Mastery of both curve geometries is vital for route design and the PRC CE Licensure Examination.

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1. Simple Circular Horizontal Curves

A simple circular curve is a single circular arc connecting two straight tangents intersecting at a Point of Intersection (PI).

Primary Geometric Elements

• Point of Intersection (PI or Vertex V): The point where the back tangent and forward tangent intersect.
• Point of Curvature (PC): The point where the alignment transitions from the back tangent into the circular arc.
• Point of Tangency (PT): The point where the circular arc ends and transitions into the forward tangent.
• Intersection Angle ($I$ or $\Delta$): The deflection angle between the tangents, identical to the central angle subtended by the arc.
• Radius of Curvature ($R$): The radius of the circular arc.
• Degree of Curve ($D$): The measure of curvature sharpness. In metric Philippine practice, the 20-meter Arc Definition defines $D$ as the central angle subtended by a $20.00\text{-m}$ arc: $\frac{D}{20} = \frac{360^\circ}{2 \pi R} \implies D = \frac{1145.916}{R} \quad (\text{degrees})$ $R = \frac{1145.916}{D} \quad (\text{meters})$ (Note: Traditional 100-ft English practice uses $D = 5729.58 / R$.)

Mathematical Formulas for Simple Curves

$\text{Tangent Distance: } T = R \tan\left( \frac{I}{2} \right)$

$\text{External Distance: } E = R \left[ \sec\left( \frac{I}{2} \right) - 1 \right] = R \left[ \frac{1}{\cos(I/2)} - 1 \right] = T \tan\left( \frac{I}{4} \right)$

$\text{Middle Ordinate: } M = R \left[ 1 - \cos\left( \frac{I}{2} \right) \right] = E \cos\left( \frac{I}{2} \right)$

$\text{Long Chord: } C = 2 R \sin\left( \frac{I}{2} \right)$

$\text{Length of Curve: } L = R \cdot I_{\text{radians}} = \frac{\pi R I^\circ}{180^\circ} = \frac{20 I}{D}$

Stationing of Simple Curves

In route surveying, stationing measures actual ground travel distance along the centerline. Because the alignment travels along the arc rather than the tangents: $\text{Station PC} = \text{Station PI} - T$ $\text{Station PT} = \text{Station PC} + L$

Curve Layout by Deflection Angles

To stake intermediate points at station interval $d$ along the curve with an instrument set up at the PC: $\text{Deflection Angle } \delta = \frac{d}{20} \times \left( \frac{D}{2} \right)$ $\text{Chord Length } c = 2 R \sin \delta$

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2. Compound and Reverse Horizontal Curves

1. Compound Curves

A compound curve consists of two or more consecutive circular arcs of different radii curving in the same direction, joining at a common Point of Compound Curvature (PCC).

• The total intersection angle is the sum of component central angles: $I = I_1 + I_2$.
• At the PCC, the curves share a single common tangent ($V_1 V_2$): $T_{\text{common}} = t_1 + t_2 = R_1 \tan\left( \frac{I_1}{2} \right) + R_2 \tan\left( \frac{I_2}{2} \right)$
• In the vertex triangle formed by the PI and the two tangent points $V_1$ and $V_2$, the Law of Sines is used to solve for overall tangents $T_1$ (back tangent) and $T_2$ (forward tangent): $\frac{V_1 \text{PI}}{\sin I_2} = \frac{V_2 \text{PI}}{\sin I_1} = \frac{T_{\text{common}}}{\sin(180^\circ - I)}$ $T_1 = t_1 + V_1 \text{PI}, \quad T_2 = t_2 + V_2 \text{PI}$

2. Reverse Curves

A reverse curve consists of two consecutive circular arcs turning in opposite directions, joining at a Point of Reverse Curvature (PRC). While avoided in high-speed highway design due to instantaneous reversal of superelevation, reverse curves are widely applied in railway yards, mountainous access roads, and canal alignments.

Reverse Curve Connecting Parallel Tangents

When two parallel tangents separated by perpendicular offset distance $p$ are connected by a reverse curve: $p = R_1 (1 - \cos I_1) + R_2 (1 - \cos I_2)$ If both arcs have equal radii ($R_1 = R_2 = R$) and equal central angles ($I_1 = I_2 = I$): $p = 2 R (1 - \cos I)$ $R = \frac{p}{2 (1 - \cos I)}$ $\text{Total Length } L_{\text{total}} = 2 L = 2 \left( \frac{\pi R I^\circ}{180^\circ} \right)$

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3. Vertical Parabolic Curves

Vertical curves provide gradual transitions between intersecting roadway gradients ($g_1$ and $g_2$, expressed as percentages or decimals). The parabolic profile provides a constant rate of change of grade, guaranteeing uniform vertical acceleration for passenger comfort and clear sight distances.

Symmetrical Vertical Curves

A symmetrical vertical curve has equal horizontal tangent lengths on either side of the Point of Vertical Intersection (PVI), meaning the curve extends $L/2$ before and $L/2$ after the PVI:

• PVC (Point of Vertical Curvature): $\text{Station PVC} = \text{Station PVI} - \frac{L}{2}$
• PVT (Point of Vertical Tangency): $\text{Station PVT} = \text{Station PVI} + \frac{L}{2} = \text{Station PVC} + L$
• Rate of Change of Grade ($r$): $r = \frac{g_2 - g_1}{L}$ (Where $g_1, g_2$ are algebraic tangent grades; e.g., $+4\% = +0.04$, $-2\% = -0.02$.)

General Parabolic Elevation Equation

The elevation $y$ of any station located at horizontal distance $x$ from the PVC ($0 \le x \le L$) is: $y(x) = \text{Elev}_{\text{PVC}} + g_1 x + \frac{r}{2} x^2 = \text{Elev}_{\text{PVC}} + g_1 x + \left( \frac{g_2 - g_1}{2 L} \right) x^2$

Tangent Offset Method

The vertical offset $y'$ from the tangent line to the curve at distance $x$ from PVC is: $y'(x) = a x^2 = \left( \frac{g_2 - g_1}{2 L} \right) x^2$ At the PVI ($x = L/2$), the external offset is: $E_v = \frac{(g_2 - g_1) L}{8} = \frac{A L}{800} \quad (\text{with } A = |g_2 - g_1| \text{ in } \%$

Location of the Summit (Crest) or Invert (Sag)

The highest point on a crest curve (or lowest point on a sag curve) occurs where the instantaneous slope is zero ($\frac{dy}{dx} = 0$): $\frac{dy}{dx} = g_1 + r x = 0 \implies x_0 = -\frac{g_1}{r} = \frac{-g_1 L}{g_2 - g_1}$ When grades have opposite signs, this simplifies using absolute magnitudes: $x_0 = \frac{|g_1| L}{|g_1| + |g_2|}$ $\text{Station of High/Low Point} = \text{Station PVC} + x_0$ $\text{Elevation of High/Low Point} = \text{Elev}_{\text{PVC}} + g_1 x_0 + \frac{r}{2} x_0^2$

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4. Comprehensive Worked Examples

Worked Example 1: Simple Horizontal Circular Curve

Problem: A proposed provincial bypass alignment has an intersection angle $I = 42^\circ 00'$ and a degree of curve $D = 5^\circ 00'$ based on the metric 20-meter arc definition ($D = 1145.916 / R$). The PI is located at Station $14 + 280.00$ with an elevation of $85.40\text{ m}$. Calculate:

1. Radius $R$, Tangent distance $T$, and Length of curve $L$.
2. External distance $E$ and Middle ordinate $M$.
3. Centerline stationing of the PC and PT.

Solution:

1. Radius, Tangent, and Curve Length: $R = \frac{1145.916}{D} = \frac{1145.916}{5.00} = 229.183\text{ m}$ $T = R \tan\left( \frac{I}{2} \right) = 229.183 \times \tan(21^\circ) = 229.183 \times 0.383864 = 87.975\text{ m}$ $L = \frac{20 I}{D} = \frac{20 \times 42.00}{5.00} = 168.000\text{ m}$ (Verification: $R \cdot I_{\text{rad}} = 229.183 \times \frac{42 \pi}{180} = 229.183 \times 0.733038 = 168.000\text{ m}$)

2. External Distance and Middle Ordinate: $E = R \left[ \frac{1}{\cos(21^\circ)} - 1 \right] = 229.183 \times \left[ \frac{1}{0.933580} - 1 \right] = 229.183 \times 0.071145 = 16.305\text{ m}$ $M = R [1 - \cos(21^\circ)] = 229.183 \times (1 - 0.933580) = 229.183 \times 0.066420 = 15.222\text{ m}$

3. Stationing: $\text{Station PC} = \text{Station PI} - T = (14 + 280.00) - 87.98 = 14 + 192.02$ $\text{Station PT} = \text{Station PC} + L = (14 + 192.02) + 168.00 = 14 + 360.02$

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Worked Example 2: Symmetrical Vertical Crest Curve Summit

Problem: A vertical crest parabolic curve connects an ascending grade $g_1 = +4.0\%$ with a descending grade $g_2 = -3.0\%$. The horizontal length of the curve is $L = 280.00\text{ m}$ (14 full 20-m stations). The PVI is situated at Station $6 + 450.00$ at elevation $128.500\text{ m}$. Determine:

1. Stationing and elevations of the PVC and PVT.
2. The location (stationing) and elevation of the highest point (summit) on the curve.
3. The elevation of Station $6 + 400.00$ on the curve.

Solution:

1. PVC and PVT Elements: $\frac{L}{2} = \frac{280.00}{2} = 140.00\text{ m}$ $\text{Station PVC} = (6 + 450.00) - 140.00 = 6 + 310.00$ $\text{Station PVT} = (6 + 450.00) + 140.00 = 6 + 590.00$ $\text{Elev PVC} = \text{Elev PVI} - g_1 \left( \frac{L}{2} \right) = 128.500 - (0.04)(140.00) = 128.500 - 5.600 = 122.900\text{ m}$ $\text{Elev PVT} = \text{Elev PVI} + g_2 \left( \frac{L}{2} \right) = 128.500 + (-0.03)(140.00) = 128.500 - 4.200 = 124.300\text{ m}$

2. Summit Location and Elevation: $r = \frac{g_2 - g_1}{L} = \frac{-0.03 - (+0.04)}{280.00} = \frac{-0.07}{280.00} = -0.00025\text{ m}^{-1}$ $x_0 = \frac{|g_1| L}{|g_1| + |g_2|} = \frac{4.0 \times 280.00}{4.0 + 3.0} = \frac{1120}{7.0} = 160.00\text{ m from PVC}$ $\text{Station Summit} = (6 + 310.00) + 160.00 = 6 + 470.00$ $\text{Elev Summit} = \text{Elev PVC} + g_1 x_0 + \frac{r}{2} x_0^2$ $\text{Elev Summit} = 122.900 + (0.04)(160.00) + \frac{-0.00025}{2} (160.00)^2$ $\text{Elev Summit} = 122.900 + 6.400 - 0.000125 \times 25,600 = 122.900 + 6.400 - 3.200 = 126.100\text{ m}$

3. Elevation of Station 6+400.00: Distance from PVC: $x = (6 + 400.00) - (6 + 310.00) = 90.00\text{ m}$. $y(90) = 122.900 + (0.04)(90.00) + \frac{-0.00025}{2} (90.00)^2$ $y(90) = 122.900 + 3.600 - 0.000125 \times 8,100 = 122.900 + 3.600 - 1.0125 = 125.488\text{ m}$

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5. Construction Layout and Survey Control

The table of specifications asks candidates to "explain layout techniques and field observations" and to "develop sound construction control methodologies." Layout means setting out the design on the ground. Control means keeping the work within tolerance as it rises.

Horizontal and vertical control

• Control points. A network of fixed points with known coordinates and elevations is established and protected beyond the work area. Benchmarks are placed where traffic and excavation will not disturb them.
• Total station stakeout. The instrument occupies a control point, backsights another, and turns computed angles and distances to each design point. Stakeout lists are prepared from design coordinates.
• GNSS (RTK) stakeout. Real-time kinematic receivers give centimeter-level positions in open areas. They are less reliable under trees or beside tall buildings.

Setting out a building

1. Stake the main corners from the control network, then check the diagonals. A rectangle is square when both diagonals are equal.
2. Erect batter boards (offset boards) outside the excavation limits. Mark grid lines on them with nails, and stretch strings between nails to recover lines after digging.
3. Transfer elevations with a level to a datum mark on each batter board.
4. On upper floors, transfer grid lines vertically with a plumb laser or by total station from offset lines, rather than by measuring up from the floor below.

3-4-5 check. To square a corner with a tape, measure 3 m along one line and 4 m along the other. The hypotenuse should read 5 m. For larger corners use multiples, such as 6-8-10 or 9-12-15.

Highway and earthwork staking

• Slope stakes mark where cut or fill slopes meet natural ground. They are marked with the cut or fill height and the offset from centerline.
• Grade stakes and blue tops give finished subgrade elevations at stations and offsets.
• Reference stakes are offset from the centerline so the line can be recovered after clearing.

Construction control and tolerances

• As-built checks. After each lift or pour, survey the actual positions and elevations and compare them with the design. Approve the next stage only when the work is within the specified tolerances.
• Settlement and deformation monitoring. Repeated leveling of settlement plates, survey pins on structures, and inclinometers near deep excavations detects movement early.
• Redundancy. Close every level loop on a benchmark and every traverse on a known point. A layout that cannot be checked is not controlled.

6. Board Exam Traps & Practical Review Tips

• Arc vs. Chord Definition: In Philippine CELE problems, always assume the 20-meter Arc Definition ($D = 1145.916 / R$) unless chord definition is explicitly specified. For chord definition, $R = \frac{10}{\sin(D/2)}$.
• Station PT Trap: Never compute $\text{Sta PT} = \text{Sta PI} + T$. Always add curve length $L$ to the PC station: $\text{Sta PT} = \text{Sta PC} + L$.
• Sign of Grades: Pay rigorous attention to signs in vertical curve formulas. An ascending grade is positive ($+$), and a descending grade is negative ($-$). In the rate of change formula $r = \frac{g_2 - g_1}{L}$, a crest curve produces a negative $r$, while a sag curve produces a positive $r$.
• Summit/Invert Boundary: The computed high or low point distance $x_0$ must fall within the physical curve limits ($0 \le x_0 \le L$). If $x_0 < 0$ or $x_0 > L$, the curve does not turn, and the maximum elevation occurs at one of the tangent terminals.