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Spiral Curves

Spiral curves are used to aid the abrupt change in curvature and superelevation that occurs between a tangent and a circular curve. The spiral curve gradually changes the curvature and superelevation of the road, and is therefore also called a transition curve or easement curve.

Elements of the Spiral Curve

Summary of Formulas

1. Spiral angle at any point

$$ S = \frac{L^2}{2R_c L_c} \times \frac{180}{\pi} $$

2. Spiral angle at $S.C.$ (Arc Basis, Metric System)

$$ S_c = \frac{D_c L_c}{40} $$

3. Spiral angle at $S.C.$ (in terms of radius)

$$ S_c = \frac{L_c}{2R_c} \times \frac{180}{\pi} $$

4. Offset from the tangent at $S.C.$

$$ x_c = \frac{L_c^2}{6R_c} $$

5. Offset from the tangent at any point

$$ x = \frac{x_c L^3}{L_c^3} $$

6. Deflection angle at $S.C.$

$$ i_c = \frac{S_c}{3} $$

7. Deflection angle at any point

$$ i = \frac{S}{3} $$

8. Distance along the tangent from $T.S.$ to $S.C.$

$$ y_c = L_c - \frac{L_c^3}{40R_c^2} $$

9. Distance along the tangent to any point

$$ y = L - \frac{L^5}{40R_c^2 L_c^2} $$

10. Spiral tangent distance

$$ T_s = \frac{L_c}{2} + \left(R_c + \frac{x_c}{4}\right)\tan\frac{I}{2} $$

11. External distance

$$ E_s = \left(R_c + \frac{x_c}{4}\right)\sec\frac{I}{2} - R_c $$

12. Central angle of the circular curve

$$ I_c = I - 2S_c $$

13. Length of throw

$$ P = \frac{x_c}{4} = \frac{L_c^2}{24R_c} $$

14. Superelevation ($K$ in kph)

$$ e = \frac{0.0079K^2}{R_c} $$

15. Superelevation ($K$ in kph, considering that 75% of $K$ counteracts the superelevation)

$$ e = \frac{0.004K^2}{R_c} $$

16. Desirable length of spiral

$$ L_c = \frac{0.036K^3}{R_c} $$

Degree of curve (arc basis) — used to convert between $D_c$ and $R_c$:

$$ R_c = \frac{1145.916}{D_c} $$

Problem 1: Superelevation, Deflection Angle, and External Distance

Two tangents having azimuths of 240° and 282° are connected by an 80 m spiral curve with a 6° circular curve. The width of the roadway is 10 m. If the design velocity is 60 kph, determine the following:
a. Super-elevation at quarter points.
b. Deflection angle at the end point S.C.
c. External distance.

Given: $I = 282^\circ - 240^\circ = 42^\circ$, $L_c = 80 \text{ m}$, $D_c = 6^\circ$, roadway width $= 10 \text{ m}$, $K = 60 \text{ kph}$.

a. Super-elevation at quarter points

$$ R_c = \frac{1145.916}{D_c} = \frac{1145.916}{6} = 190.986 \text{ m} $$ $$ e = \frac{0.0079K^2}{R_c} = \frac{0.0079(60)^2}{190.986} = 0.149 \text{ m/m (width of roadway)} $$

This is the superelevation at S.C. Distributing it across the spiral at the quarter points:

$$ e_4 = 0.149(10)\left(\frac{80}{80}\right) = \boxed{1.49 \text{ m}} $$ $$ e_3 = 0.149(10)\left(\frac{60}{80}\right) = \boxed{1.118 \text{ m}} $$ $$ e_2 = 0.149(10)\left(\frac{40}{80}\right) = \boxed{0.745 \text{ m}} $$ $$ e_1 = 0.149(10)\left(\frac{20}{80}\right) = \boxed{0.373 \text{ m}} $$

b. Deflection angle at the end point S.C.

$$ S_c = \frac{D_c L_c}{40} = \frac{6(80)}{40} = 12^\circ $$ $$ i_c = \frac{S_c}{3} = \frac{12^\circ}{3} = \boxed{4^\circ} $$

c. External distance

$$ E_s = \left(R_c + \frac{x_c}{4}\right)\sec\frac{I}{2} - R_c = \left(R_c + \frac{L_c^2}{24R_c}\right)\frac{1}{\cos\frac{I}{2}} - R_c $$ $$ E_s = \left[190.986 + \frac{(80)^2}{24(190.986)}\right]\frac{1}{\cos\frac{42^\circ}{2}} - 190.986 $$ $$ \boxed{E_s = 15.083 \text{ m}} $$

Problem 2: Replacing a Simple Curve with Spiral Ends

A simple curve having a radius of 280 m connects two tangents intersecting at an angle of 58°. It is to be replaced by another curve having 80 m spirals at its ends such that the point of tangency shall be the same. Determine the following:
a. Radius of the new circular curve.
b. Central angle of the circular curve.
c. Distance that the curve will move nearer the vertex.
d. Offset from the tangent at the end point of spiral.
e. Distance along the tangent at the midpoint of the spiral.

a. Radius of the new circular curve

Since the point of tangency will be the same, the tangent distance $T$ of the simple curve equals the tangent distance $T_s$ of the spiral curve.

$$ T = R\tan\frac{I}{2} = 280\tan\frac{58^\circ}{2} = 155.207 \text{ m} = T_s $$ $$ T_s = \frac{L_c}{2} + \left(R_c + \frac{L_c^2}{24R_c}\right)\tan\frac{I}{2} $$ $$ 155.207 = \frac{80}{2} + \left(R_c + \frac{80^2}{24R_c}\right)\tan\frac{58^\circ}{2} $$ $$ \boxed{R_c = 206.548 \text{ m}} $$

b. Central angle of the circular curve

$$ S_c = \frac{L_c}{2R_c}\times\frac{180}{\pi} = \frac{80}{2(206.548)}\times\frac{180}{\pi} = 11.096^\circ $$ $$ I_c = I - 2S_c = 58^\circ - 2(11.096^\circ) = \boxed{35.808^\circ} $$

c. Distance that the curve will move nearer the vertex

The distance from the vertex to the curve is the external distance, so the distance the curve moves nearer the vertex is the difference between the external distance $E$ of the simple curve and the external distance $E_s$ of the spiral curve.

$$ E = R\left(\sec\frac{I}{2} - 1\right) = 280\left(\sec\frac{58^\circ}{2} - 1\right) = 40.139 \text{ m} $$ $$ E_s = \left[206.548 + \frac{(80)^2}{24(206.548)}\right]\frac{1}{\cos\frac{58^\circ}{2}} - 206.548 = 31.086 \text{ m} $$ $$ d = E - E_s = 40.139 - 31.086 = \boxed{9.053 \text{ m}} $$

d. Offset from the tangent at the end point of spiral

$$ x_c = \frac{L_c^2}{6R_c} = \frac{80^2}{6(206.548)} = \boxed{5.614 \text{ m}} $$

e. Distance along the tangent at the midpoint of the spiral

At the midpoint, $L = \dfrac{L_c}{2} = 40 \text{ m}$.

$$ y = L - \frac{L^5}{40R_c^2 L_c^2} = 40 - \frac{40^5}{40(206.548)^2(80)^2} = \boxed{39.991 \text{ m}} $$

Problem 3: Spiral Angle, Deflection Angle, and Offset

A spiral 80 m long connects a tangent with a 6°30′ circular curve. If the stationing of the T.S. is 10+000, determine the following:
a. The spiral angle at the first quarter point.
b. The deflection angle at the end point of spiral.
c. Offset from the tangent at the second quarter point of the spiral.

Given: $L_c = 80 \text{ m}$, $D_c = 6^\circ 30' = 6.5^\circ$.

$$ R_c = \frac{1145.916}{D_c} = \frac{1145.916}{6.5} = 176.295 \text{ m} $$

a. The spiral angle at the first quarter point ($L = 20 \text{ m}$)

$$ S = \frac{L^2}{2R_c L_c}\times\frac{180^\circ}{\pi} = \frac{(20)^2}{2(176.295)(80)}\times\frac{180^\circ}{\pi} = \boxed{0.812^\circ} $$

b. The deflection angle at the end point of spiral

$$ S_c = \frac{L_c}{2R_c}\times\frac{180^\circ}{\pi} = \frac{80}{2(176.295)}\times\frac{180^\circ}{\pi} = 13^\circ $$ $$ i_c = \frac{S_c}{3} = \frac{13^\circ}{3} = \boxed{4.333^\circ} $$

c. Offset from the tangent at the second quarter point ($L = 40 \text{ m}$)

$$ x_c = \frac{L_c^2}{6R_c} = \frac{(80)^2}{6(176.295)} = 6.05 \text{ m} $$ $$ x = \frac{x_c L^3}{L_c^3} = \frac{(6.05)(40)^3}{(80)^3} = \boxed{0.756 \text{ m}} $$

Problem 4: Degree of Curve, Length of Spiral, and Superelevation

The tangents of a spiral curve have azimuths of 226° and 221° respectively. The minimum length of the spiral is 40 m and a minimum super-elevation of 0.10 m/m width of roadway. The maximum velocity to pass over the curve is 70 kph. Assume the width of roadway to be 9 m. Compute the following:
a. Degree of simple curve.
b. Length of spiral at the end of the simple curve.
c. The super-elevation of the first 10 m from S.C. on the spiral.
USE: $e = \dfrac{0.004K^2}{R}$

a. Degree of simple curve

$$ e = \frac{0.004K^2}{R} \;\Rightarrow\; 0.10 = \frac{0.004(70)^2}{R_c} $$ $$ R_c = 196 \text{ m} $$ $$ D_c = \frac{1145.916}{R_c} = \frac{1145.916}{196} = \boxed{5.847^\circ} $$

b. Length of spiral at the end of the simple curve

$$ L_c = \frac{0.036K^3}{R} = \frac{0.036(70)^3}{196} = \boxed{63 \text{ m}} $$

c. Super-elevation of the first 10 m from S.C. on the spiral

Measuring 10 m from S.C. back along the spiral leaves $63 - 10 = 53 \text{ m}$ from the T.S., so the superelevation is proportioned by $\dfrac{53}{63}$.

$$ e = 0.10\,\frac{\text{m}}{\text{m}}(9 \text{ m})\left(\frac{53}{63}\right) = \boxed{0.757 \text{ m}} $$

Problem 5: Long & Short Tangents, External Distance, Throw, and Velocity

A spiral curve was laid out in a certain portion of the Manila-Cavite coastal road. It has a length of spiral of 80 m and an angle of intersection of the two tangents of 40°. If the degree of curve is 6°, determine the following elements of the spiral curve:
a. Length of the long and short tangent.
b. External distance.
c. Length of throw.
d. Maximum velocity of a car that could pass through the curve without skidding.

a. Length of the long and short tangent

$$ R_c = \frac{1145.916}{6} = 190.986 \text{ m} $$ $$ S_c = \frac{D_c L_c}{40} = \frac{6(80)}{40} = 12^\circ $$ $$ x_c = \frac{L_c^2}{6R_c} = \frac{(80)^2}{6(190.986)} = 5.585 \text{ m} $$

Short tangent ($ST$):

$$ \sin S_c = \frac{x_c}{ST} \;\Rightarrow\; ST = \frac{5.585}{\sin 12^\circ} = \boxed{26.862 \text{ m}} $$

Long tangent ($LT$):

$$ \tan S_c = \frac{x_c}{z} \;\Rightarrow\; z = \frac{5.585}{\tan 12^\circ} = 26.275 \text{ m} $$ $$ y_c = L_c - \frac{L_c^3}{40R_c^2} = 80 - \frac{(80)^3}{40(190.986)^2} = 79.649 \text{ m} $$ $$ LT = y_c - z = 79.649 - 26.275 = \boxed{53.374 \text{ m}} $$

b. External distance

$$ E_s = \left(R_c + \frac{L_c^2}{24R_c}\right)\sec\frac{I}{2} - R_c $$ $$ E_s = \left(190.986 + \frac{(80)^2}{24(190.986)}\right)\sec\frac{40^\circ}{2} - 190.986 = \boxed{13.743 \text{ m}} $$

c. Length of throw

$$ P = \frac{x_c}{4} = \frac{5.585}{4} = \boxed{1.396 \text{ m}} $$

d. Maximum velocity without skidding

$$ L_c = \frac{0.036K^3}{R} \;\Rightarrow\; 80 = \frac{0.036K^3}{190.986} $$ $$ \boxed{K = 75.15 \text{ kph}} $$
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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: t1189

MSTE - Highway Engineering / Curves, Earthworks, and Traffic Engineering / Gemini mapped Chapter 7 to 10

The angle of intersection of a circular curve is 42° 30'. If the external distance is 25.42 m, determine the length of the curve.

  1. 321.45 m
  2. 187.56 m
  3. 258.47 m
  4. 287.45 m

Solution pending in psadquestions/t1189.json.

Question Bank: t1233

MSTE - Highway Engineering / Curves, Earthworks, and Traffic Engineering / Gemini mapped Chapter 7 to 10

A highway intersection of 40° is to be connected by a 6-degree simple curve and 80-m long spiral on each end of the simple curve. Use arc basis.

What is the length of throw in meters?

  1. 5.6
  2. 1.8
  3. 1.4
  4. 2.6

What is the length of the short tangent in meters?

  1. 26.86
  2. 23.54
  3. 28.74
  4. 36.52

What is the total length of the curve in meters?

  1. 327.55
  2. 167.55
  3. 254.55
  4. 312.55

Solution pending in psadquestions/t1233.json.

Question Bank: t1236

MSTE - Highway Engineering / Curves, Earthworks, and Traffic Engineering / Gemini mapped Chapter 7 to 10

A central simple curve has a spiral curve 120 m long on two sides. For a design speed of 90 kph:

Find the desirable radius of central curve in meters.

  1. 236
  2. 275
  3. 258
  4. 219

Find the radius of the central curve if the rate of change of centripetal acceleration is limited to 0.5 m/s^3.

  1. 260.4 m
  2. 241.2
  3. 285.7
  4. 210.3

If the radius of the central curve is 180 m, what is the length of throw, in meters.

  1. 3.33
  2. 2.54
  3. 3.87
  4. 4.21

Solution pending in psadquestions/t1236.json.