Sight Distance
Introduction
INCI 4007
I. Cruzado
This week
1. Provide adequate sight distance (to make it “safer”)
Sight Distance along Horizontal Curves
• Sight distance (S): the ability to “look forwards”
• You must be able to see an obstacle and safely bring the car to a stop
Sight Distance along Horizontal Curves
• Sight distance (S): the ability to “look forwards”
• You must be able to see an obstacle and safely bring the car to a stop
Sight Distance along Horizontal Curves
• Along horizontal curves:
BAD
GOOD
Sight Distance along Horizontal Curves
• Along horizontal curves:
BAD
GOOD
Sight Distance along Horizontal Curves
“How much” sight distance?
• You must be able to see an obstacle and safely bring the car to a stop.
• Compare it to the Stopping Sight Distance (SSD)
$$
-.
!!" = %& ' ∗ )*+ + %&/
SSD in feet
V = speed in mph
tpr = perception-reaction time in seconds (2.5 seconds)
f = coefficient of longitudinal friction
Sight Distance along Horizontal Curves
!!" =
$$
%&
' ∗ )*+ +
perception
reaction distance
-.
%&/
braking
distance
RULE: Sight distance ≥ Stopping Sight Distance (S ≥ SSD)
SSD Example
• You are traveling at 55 mph and the longitudinal friction coefficient is
0.28. If you see an obstacle, how much distance you travel before
coming to a complete stop?
• !!" =
$$
%&
' ∗ )*+ +
-.
%&/
=
$$
%&
55 ∗ 2.5 +
33.
%&∗&.45
= 201.67 + 360.12 = ;<=. >? @A
perception
reaction distance
braking
distance
Stopping
Sight
Distance
Sight Distance along Horizontal Curves
$$
-.
!!" = %& ' ∗ )*+ + %&/
perception
reaction distance
braking
distance
RULE: Sight distance ≥ Stopping Sight Distance (S ≥ SSD)
SSD
325
402
486
S
330
410
490
Sight Distance along Horizontal Curves
• S = Sight Distance
• MO = clear distance from obstacle to center of the (closest) lane
Sight Distance along Horizontal Curves
• S = Sight Distance
• MO = clear distance from obstacle to center of the (closest) lane
2. Sight Distance along Horizontal Curves
Case 1: S < L
• The sight distance is shorter
than the length of the curve
• Approximate MO :
$%
!" =
8'
• More exact value:
!" = ' 1 − cos Where - =
.
%//
0
• MO = ?
• R = 870 ft; S = 300 ft; L = 386 ft
2. Sight Distance along Horizontal Curves
Case 1: S < L
• The sight distance is shorter
than the length of the curve
• Approximate MO :
$%
!" =
8'
• More exact value:
!" = ' 1 − cos Where - =
.
%//
0
• MO = ?
• R = 870 ft; S = 300 ft; L = 386 ft
• First equation:
• !" =
.1
23
=
4//1
2∗26/
= 12.93 ;<
• Second equation:
• 0=
=6%>.=62
=
26/
4//
6.58
%//
6.58°
• -=
= 9.88°
• !" = 870 1 − cos 9.88 =
12.90;<
M is a minimum! So 13 ft is a good number.
MO is a minimum value;
To provide at least 300 ft of sight
distance, the obstacle cannot be
less than 12.9 feet.
If MO is less than 12.9 feet, S is less than 300 ft
If MO is more than 12.9 feet, S is more than 300 ft
MO is a minimum value;
To provide at least 300 ft of sight
distance you, obstacle cannot be
less than 12.9 feet.
If MO is less than 12.9 feet, S is less than 300 ft
If MO is more than 12.9 feet, S is more than 300 ft
Next
Case 1: S < L
• The sight distance is shorter
than the length of the curve
Case 2: S > L
• The sight distance is longer than
the length of the curve
Sight Distance along Horizontal Curves
• !" = ?
• R = 870 ft; S = 500 ft; L = 386 ft
Case 2: S > L
• The sight distance is longer than
the length of the curve
• Only one equation and only one
that considers the length of the
curve:
$(2' − $)
!" =
8+
2. Sight Distance along Horizontal Curves
• !" = ?
• R = 870 ft; S = 500 ft; L = 386 ft
$(2' − $)
!" =
8+
386(2 ∗ 500 − 386)
!" =
8 ∗ 870
!" = 34.05 45
Case 2: S > L
• The sight distance is longer than
the length of the curve
• Only one equation and only one
that considers the length of the
curve:
$(2' − $)
!" =
8+
Insights
• Based on design speed, choose an S
• S >>> SSD
• You have S and L – choose a case
• Then calculate !"
• if !" increases, S increases; if !" decreases…
• !" depends on S and S depends on !"
What happens if you have #$ , need to compute S,
but you don’t know if S> L or S<L?
Example: both equations
A curve with a 820 ft radius has a length of 700 ft. There is an
obstruction 15 ft from the centerline of the nearest lane.
1. What is the sight distance along the curve?
• Assuming S < L
• Assuming S > L
Solve for S with !" = 15, R = 820, L = 700
2. What should be the posted speed limit to ensure an appropriate
sight distance? (Assume f = 0.30)
Example: both equations
A curve with a 820 ft radius has a length of 700 ft. There is an
obstruction 15 ft from the centerline of the nearest lane.
1. What is the sight distance along the curve?
• Assuming S < L S = 313.7 ft
• Assuming S > L S = 420.3 ft
2. What should be the posted speed limit to ensure an appropriate
sight distance? (Assume f = 0.30)
Example: both equations
A curve with a 820 ft radius has a length of 700 ft. There is an
obstruction 15 ft from the centerline of the nearest lane.
1. What is the sight distance along the curve?
• Assuming S < L S = 313.7 ft
Both results say “S < L”,
so I choose the result
from the case S<L
• Assuming S > L S = 420.3 ft
2. What should be the posted speed limit to ensure an appropriate
sight distance? (Assume f = 0.30)
Example: both equations
A curve with a 820 ft radius has a length of 700 ft. There is an
obstruction 15 ft from the centerline of the nearest lane.
1. What is the sight distance along the curve?
• Assuming S < L S = 313.7 ft
• Assuming S > L S = 420.3 ft
2. What should be the posted speed limit to ensure an appropriate
sight distance? (Assume f = 0.30)
Example: both equations
A curve with a 820 ft radius has a length of 700 ft. There is an
obstruction 15 ft from the centerline of the nearest lane.
1. What is the sight distance along the curve?
• Assuming S < L S = 313.7 ft
• Assuming S > L S = 420.3 ft
2. What should be the posted speed limit to ensure an appropriate
sight distance? (Assume f = 0.30)
• SSD Equation, solve for V = 39.14 mph – posted speed limit 35 MPH
Problem to Practice
• What is the stopping sight distance for a highway with posted speed
limit of 45 mph (assume f = 0.21)?
• Along the same highway there is a curve with 900 ft radius
(intersection angle 65°); construction panels will be placed 6 feet
from the edge of the highway. If the highway has two lanes – each
one 11-ft wide – and shoulder on both sides (2-ft wide), determine if
you have safety problems due to the panels.
• Find out the three solutions (one of them is not realistic) for the
problem above.
EXTRA: Reverse Curves
and how they help you for Exam # 1
Two simple
curves in
opposite
directions
with no
space
between
them.
R
R
PC2
PC1
PT1
R
PT2
R
R
PC2
PC1
PT1
R
Straight distance between curves
PT2
R1>>
PC1
PT1=PC2
PT2
R2>>
R1>>
PC1
PRC
PT2
Deltas Δ’s remained the same
Tangents (T’s), radii (R’s), and Curve Lengths (L’s):
INCREASED!
PRC = point of reverse curve
R2>>
Degree of Curvature - SAME
5729.578
!=
)
Elements of a (Simple) Curve - SAME
How does this help for the exam?
• You have additional practice problems
Example
Given:
R1 = 750 ft
Δ1 = 40°15’
R2 = 800 ft
Δ1 = 35°30’
R
R
PC2
PC1
Sta PC1 35+92
Distance 550 feet between curves
PT1
Design a reverse curve by distributing evenly
(half and half) the straight distance between them.
What are the new PC, PRC, and PT stations?
R
PT2
Example: Steps (plus do the drawing)
• Start by finding the (current) horizontal curve tangents T1 and T2.
• Find the PI station for the first curve (it will remain the same after the
changes)
• Distribute the straight section: add half of it to each tangent and you
will have the tangents for the new curve.
• Intersection angles remained the same, so you can calculate the new
radius for each curve with each (increased) new tangent.
• Calculate length of curve (L) for each curve
• Determine the stations
Pause
• Try to do it yourself
Example: Results
• The (current) tangents T1= 274.83’ and T2=256.08’
• PI station for the first curve: PC1 + T1 = 38+66.83
• Distribute the straight section: 550/2 = 275 for each tangent; that
means T1new = 549.83’ and T2new = 531.08’
• Same deltas and new tangents: R1new = 1500.46’ and R2new = 1659.10’
• Length of curve (L): L1new = 1054.06’ and L2new =1027.96’
• Stations: PC = 33+17; PRC = 43+71.06; PT = 53+99.03
Good Luck!