Horizontal Curve and Road Design Problems

Posted on Apr 5, 2025 in Civil Engineering, Transportation, and Urban Services

Problem 1: Circular Curve Design

A simple circular curve exists with a degree of curve D=12 and e=0.08. A structure is proposed on land on the inside of the curve. Assume the road is on level grade. Determine:

  1. The radius of the curve.
  2. The current maximum safe speed on the curve.
  3. The maximum distance allowable between the proposed structure and the centerline of the curve, such that the current maximum safe speed of the curve would not be reduced. Perception-reaction time is 2.5 seconds and fs=0.16

Solution:

  1. R=5729.6/12 = 477.467 ft
  2. 477.467= u^2/15(0.08+0.16), u=41.459 mph
  3. S=1.47×41.459×2.5 + 41.459^2 / 30[(11.2/32.2)-0] = 317.085 ft
    m= 477.467x(1-cos28.65×317.085/477.467)=26.085 ft

Problem 2: Vehicle Speed Estimation from Skid Marks

In an attempt to estimate the speed of a vehicle just before it hit a signal pole, a traffic engineer measured the length of the skid marks made by the vehicle and performed trial runs at the site to obtain an estimate of the coefficient of friction. Determine the estimated unknown velocity if the following data were obtained:

  • Length of skid marks: 585 ft, 590 ft, 580 ft, and 595 ft.
  • Speed of trial run: 30 mph
  • Distance traveled during trial run: 300 ft.
  • Examination of the vehicle just after the crash indicated that the speed of impact was 35 mph.

Solution:

Average length of skid marks= (585+590+580+595)/4= 587.5 ft
Uu=(587.5/300 x 30^2 +1225)^1/2 = (1762.5+1225)^1/2 = 54.66 mph

Problem 3: Horizontal Curve Design in Mountainous Terrain

A horizontal curve is to be designed for a two-lane road in mountainous terrain. The following data are known: intersection angle=40°, tangent length=436.76 ft, Station of PI=2700+10.65, and fs=0.12, e=0.08. Determine:

  1. Design speed
  2. Station of the PC
  3. Station of the PT

Solution:

  • R=436.76/Tan(40/2)= 1200 ft
  • V= √(1200x(15(0.08+0.12)))= 60 mph
  • Station of PC = (2700+10.65)-(4+36.76)= 2695+73.89
  • L of curve= 1200 x 40 x pi/180 = 837.76 ft
  • Station of PT= (2695+73.89)+(8+37.76)=2704+11.65
  • Deflection angle= delta= 180(100-73.89)/(pi x 1200) = 1.25°
  • L of chord= C1=2 x 1200 x sin(1.25/2)= 26.18 ft

Problem 4: Horizontal Curve Design (Revisited)

A horizontal curve is to be designed for a two-lane road (angle=40°, Tl=436.76, spi=2700+10.65, fs=0.12, e=0.08)

  1. Design speed: T=Rtan(delta / 2) = 1200, R=u^2/15(e+fs). u=√(15R(e+fs))= √(15(1200)(0.08+0.12))=60 mph
  2. Station of the PC: Pc station = PI station – T= (2700+10.65)-436.76= station 2695+73.89
  3. L= R*angle*pi/180 = 1200(40)pi/180 = 837.7, pt station=pt station + L=(2695+73.89)+837.7= station=2704+11.65

Problem 5: Stopping Distance on an Expressway Ramp

A motorist traveling at 65 mph on an expressway intends to leave the expressway using an exit ramp with a maximum speed of 35 mph. At what point on the expressway should the motorist step on her brakes in order to reduce her speed to the maximum allowable on the ramp just before entering the ramp if this section of the expressway has a downgrade of 3%?

Solution:

a/g= 11.2/32.2 = 0.35
Db= (65^2 – 35^2) / (30 * (0.35-0.03))= 299.47 ft

Problem 6: Horizontal Curve Improvement

An existing horizontal curve on a highway has a radius of 465 ft, which restricts the posted speed limit to only 61.5% of the design speed on this section of the highway. If the curve is to be improved, so that its posted speed will be the design speed of the highway, determine the minimum radius of the new curve. Assume that the rate of superelevation is 0.08 for both the existing curve and the new curve to be designed and fs=0.16 for 40 mph, 0.14 for 50 mph, 0.12 for 60 mph, and 0.10 for 70 mph.

Solution:

e= vmax^2/gr
0.08=vmax^2/(32.1737 x 465) = 34.6
vdesign= 34.6 x 0.615 = 56.26
e=vo^2/gr , 0.08=56.26^2/(32.1737 x R), R=1229.726 ft