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Superelevation and Spiral Curves

Transportation Engineering I

Objectives1. Define superelevation runoff length and

methods of attainment (for simple and spiral curves)

2. Calculate spiral curve length

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Attainment of Superelevation -General

1. Must be done gradually over a distance without appreciable reduction in speed or safety and with comfort

2. Change in pavement slope should be consistent over a distance

3. Methods a. Rotate pavement about centerlineb. Rotate about inner edge of pavementc. Rotate about outside edge of pavement

Superelevation Transition Section

• Tangent Runout Section + Superelevation Runoff Section

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Tangent Runout Section

• Length of roadway needed to accomplish a change in outside-lane cross slope from normal cross slope rate to zero

For rotation about centerline

Superelevation Runoff Section

• Length of roadway needed to accomplish a change in outside-lane cross slope from 0 to full superelevation or vice versa

• For undivided highways with cross-section rotated about centerline

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Source: CalTrans Design Manual online, http://www.dot.ca.gov/hq/oppd/hdm/pdf/chp0200.pdf

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Same as point E of GB

Attainment Location -WHERE

1. Superelevation must be attained over a length that includes the tangent and the curve (why)

2. Typical: 66% on tangent and 33% on curve of length of runoff if no spiral

3. Normally used 70% and 30% if no spiral4. Super runoff is all attained in Spiral if used

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Minimum Length of Runofffor curve

• Lr based on drainage and aesthetics

• Rate of transition of edge line from NC to full superelevation traditionally taken at 0.5% ( 1 foot rise per 200 feet along the road)

• current recommendation varies from 0.35% at 80 mph to 0.80% for 15mph (with further adjustments for number of lanes)

Minimum Length of Tangent Runout

Lt = eNC x Lr

ed

where• eNC = normal cross slope rate (%)• ed = design superelevation rate• Lr = minimum length of superelevation

runoff (ft)(Result is the edge slope is same as for

Runoff segment)

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Length of Superelevation Runoff

α = multilane adjustment factorAdjusts for total width

Relative Gradient (G)• Maximum longitudinal slope• Depends on design speed, higher

speed = gentler slope. For example:• For 15 mph, G = 0.78%• For 80 mph, G = 0.35%• See table, next page

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Maximum Relative Gradient (G)

Multilane Adjustment

• Runout and runoff must be adjusted for multilane rotation.

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Length of Superelevation Runoff Example

• For a 4-lane divided highway with cross-section rotated about centerline, design superelevation rate = 4%. Design speed is 50 mph. What is the minimum length of superelevation runoff (ft)

Lr = 12eαG

Lr = 12eα = (12) (0.04) (1.5)G 0.5

Lr = 144 feet

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Tangent runout length Example continued

• Lt = (eNC / ed ) x Lr

as defined previously, if NC = 2%Tangent runout for the example is:

LT = 2% / 4% * 144’ = 72 feet

From previous example, speed = 50 mph, e = 4%From chart runoff = 144 feet, same as from calculation

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Spiral Curve Transitions

• Vehicles follow a transition path as they enter or leave a horizontal curve

• Combination of high speed and sharp curvature can result in lateral shifts in position and encroachment on adjoining lanes

Spirals

1. Advantagesa. Provides natural, easy to follow, path for

drivers (less encroachment, promotes more uniform speeds), lateral force increases and decreases gradually

b. Provides location for superelevation runoff (not part on tangent/curve)

c. Provides transition in width when horizontal curve is widened

d. Aesthetic

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Minimum Length of Spiral

Possible Equations:L = 3.15 V3

RCWhere:

L = minimum length of spiral (ft)V = speed (mph)R = curve radius (ft)C = rate of increase in centripetal acceleration (ft/s3) use 1-3 ft/s3 for highway)

Minimum Length of Spiral

Or L = (24pminR)1/2

Where:L = minimum length of spiral (ft)R = curve radius (ft)pmin = minimum lateral offset between the tangent and circular curve (0.66 feet)

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Maximum Length of Spiral

• Safety problems may occur when spiral curves are too long – drivers underestimate sharpness of approaching curve (driver expectancy)

Maximum Length of Spiral

L = (24pmaxR)1/2

Where:L = maximum length of spiral (ft)R = curve radius (ft)pmax = maximum lateral offset between the tangent and circular curve (3.3 feet)

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Length of Spiral

o AASHTO also provides recommended spiral length based on driver behavior rather than a specific equation.

o Superelevation runoff length is set equal to the spiral curve length when spirals are used.

o Design Note: For construction purposes, round your designs to a reasonable values; e.g. round

o Ls = 147 feet, to use Ls = 150 feet.

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SPIRAL TERMINOLOGY

Attainment of superelevation on spiral curves

See sketches that follow:Normal Crown (DOT – pt A)1. Tangent Runout (sometimes known as crown

runoff): removal of adverse crown (DOT – A to B) B = TS

2. Point of removal of crown (DOT – C) note A to B = B to C

3. Length of Runoff: length from adverse crown removed to full superelevated (DOT – B to D), D = SC

4. Fully superelevate remainder of curve and then reverse the process at the CS.

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With SpiralsSame as point E of GB

With Spirals

Tangent runout (A to B)

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With Spirals

Removal of crown

With Spirals

Transition of superelevation Full superelevation

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Transition Example

Given:• PI @ station 245+74.24• D = 4º (R = 1,432.4 ft)• ∆ = 55.417º• L = 1385.42 ft

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With no spiral …

• T = 752.30 ft• BC = PI – T = 238 +21.94

For: • Design Speed = 50 mph • superelevation = 0.04 • normal crown = 0.02

Runoff length was found to be 144’Tangent runout length = 0.02/ 0.04 * 144 = 72 ft.

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Where to start transition for superelevation?

Using 2/3 of Lr on tangent, 1/3 on curve for superelevation runoff:

Distance before BC = Lt + 2/3 Lr =72 +2/3 (144) = 168

Start removing crown at:BC station – 168’ = 238+21.94 - 1+68.00=

Station = 236+ 53.94

Location Example – with spiral

• Speed, e and NC as before and • ∆ = 55.417º• PI @ Station 245+74.24• R = 1,432.4’• Lr was 144’, so set Ls = 150’

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Location Example – with spiral

• Spiral angle Θs = Ls * D /200 = 3 degrees• P = 0.65 (calculated)• Ts = (R + p ) tan (delta /2) + k = 827.63 ft

• TS station = PI – Ts = 245+74.24 – 8 + 27.63= 237+46.61

Runoff length = length of spiral Tangent runout length = Lt = (eNC / ed ) x Lr

= 2% / 4% * 150’ = 75’Therefore: Transition from Normal crown

begins at (237+46.61) – (0+75.00) = 236+71.61

Location Example – with spiral

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• With spirals, the central angle for the circular curve is reduced by 2 * Θs Lc = ((delta – 2 * Θs) / D) * 100Lc = (55.417-2*3)/4)*100 = 1235.42 ft

• Total length of curves = Lc +2 * Ls = 1535.42Verify that this is exactly 1 spiral length longer than when spirals are not used (extra credit for who can tell me why)

Location Example – with spiral

Also note that the tangent length with a spiral should be longer than the non-spiraled curve by approximately ½ of the spiral length used. (good check – but why???)

Location Example – with spiral

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Notes