Fundamentals of Transportation/Sight Distance

Sight Distance is a length of road surface which a particular driver can see with an acceptable level of clarity. Sight distance plays an important role in geometric highway design because it establishes an acceptable design speed, based on a driver's ability to visually identify and stop for a particular, unforeseen roadway hazard or pass a slower vehicle without being in conflict with opposing traffic. As velocities on a roadway are increased, the design must be catered to allowing additional viewing distances to allow for adequate time to stop. The two types of sight distance are (1) stopping sight distance and (2) passing sight distance.

Derivations

Stopping Sight Distance

Forces acting on a vehicle that is braking

Stopping Sight Distance (SSD) is the viewable distance required for a driver to see so that he or she can make a complete stop in the event of an unforeseen hazard. SSD is made up of two components: (1) Braking Distance and (2) Perception-Reaction Time.

For highway design, analysis of braking is simplified by assuming that deceleration is caused by the resisting force of friction against skidding tires. This is applicable to both an uphill or a downhill situation. A vehicle can be modeled as an object with mass $m$ sliding on a surface inclined at angle $\theta$ .

While the force of gravity pulls the vehicle down, the force of friction resists that movement. The forces acting this vehicle can be simplified to:

$F=W\left({\sin \left(\theta \right)-f\cos \left(\theta \right)}\right)\,\!$

where

$W=mg\,\!$ = object’s weight,
$f\,\!$ = coefficient of friction.

Using Newton’s second law we can conclude then that the acceleration ( $a$ ) of the object is

$a=g\left({\sin \left(\theta \right)-f\cos \left(\theta \right)}\right)\,\!$

Using our basic equations to solve for braking distance ( $d_{b}\,\!$ ) in terms of initial speed ( $v_{i}\,\!$ ) and ending speed ( $v_{e}\,\!$ ) gives

$d_{b}={\frac {v_{i}^{2}-v_{e}^{2}}{-2a}}\,\!$

and substituting for the acceleration yields

$d_{b}={\frac {v_{i}^{2}-v_{e}^{2}}{2g\left({f\cos \left(\theta \right)-\sin \left(\theta \right)}\right)}}\,\!$

For angles commonly encountered on roads, $\cos(\theta )\approx 1$ and $\sin(\theta )\approx \tan(\theta )=G$ , where $G$ is called the road’s grade. This gives

$d_{b}={\frac {v_{i}^{2}-v_{e}^{2}}{2g\left({f\pm G}\right)}}\,\!$

Using simply the braking formula assumes that a driver reacts instantaneously to a hazard. However, there is an inherent delay between the time a driver identifies a hazard and when he or she mentally determines an appropriate reaction. This amount of time is called perception-reaction time. For a vehicle in motion, this inherent delay translates to a distance covered in the meanwhile. This extra distance must be accounted for.

For a vehicle traveling at a constant rate, distance $d_{r}$ covered by a specific velocity $v$ and a certain perception-reaction time $t_{r}$ can be computed using simple dynamics:

$d_{r}=\left(vt_{r}\right)\,\!$

Finally, combining these two elements together and incorporating unit conversion, the AASHTO stopping sight distance formula is produced. The unit conversions convert the problem to metric, with $v_{i}$ in kilometers per hour and $d_{s}$ in meters.

$d_{s}=d_{r}+d_{b}=0.278t_{r}v_{i}+{\frac {(0.278v_{i})^{2}}{19.6\left({f\pm G}\right)}}\,\!$

A Note on Sign Conventions

We said $d_{b}={\frac {v_{i}^{2}-v_{e}^{2}}{2g\left({f\pm G}\right)}}\,\!$

Use: $(f-G)\,\!$ if going downhill and $(f+G)\,\!$ if going uphill, where G is the absolute value of the grade

Passing Sight Distance

Passing Sight Distance (PSD) is the minimum sight distance that is required on a highway, generally a two-lane, two-directional one, that will allow a driver to pass another vehicle without colliding with a vehicle in the opposing lane. This distance also allows the driver to abort the passing maneuver if desired. AASHTO defines PSD as having three main distance components: (1) Distance traveled during perception-reaction time and acceleration into the opposing lane, (2) Distance required to pass in the opposing lane, (3) Distance necessary to clear the slower vehicle.

The first distance component $d_{1}$ is defined as:

$d_{1}=1000t_{1}\left(u-m+{\frac {at_{1}}{2}}\right)\,\!$

where

$t_{1}\,\!$ = time for initial maneuver,
$a\,\!$ = acceleration (km/h/sec),
$u\,\!$ = average speed of passing vehicle (km/hr),
$m\,\!$ = difference in speeds of passing and impeder vehicles (km/hr).

The second distance component $d_{2}\,\!$ is defined as:

$d_{2}=\left(1000ut_{2}\right)\,\!$

where

$t_{2}\,\!$ = time passing vehicle is traveling in opposing lane,
$u\,\!$ = average speed of passing vehicle (km/hr).

The third distance component $d_{3}\,\!$ is more of a rule of thumb than a calculation. Lengths to complete this maneuver vary between 30 and 90 meters.

With these values, the total passing sight distance (PSD) can be calculated by simply taking the summation of all three distances.

$d_{p}=\left(d_{1}+d_{2}+d_{3}\right)\,\!$

Demonstrations

Examples

Example 1: Stopping Distance

TProblem

Problem:

A vehicle initially traveling at 66 km/h skids to a stop on a 3% downgrade, where the pavement surface provides a coefficient of friction equal to 0.3. How far does the vehicle travel before coming to a stop?

Example

Solution:

$d_{b}={\frac {\left({66*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left(0\right)^{2}}{2*\left({9.8}\right)*\left({0.3-0.03}\right)}}=63.5m\,\!$

Example 2: Coefficient of Friction

TProblem

Problem:

A vehicle initially traveling at 150 km/hr skids to a stop on a 3% downgrade, taking 200 m to do so. What is the coefficient of friction on this surface?

Example

Solution:

$d_{b}={\frac {\left({150*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left(0\right)^{2}}{2*\left({9.8}\right)*\left({f-0.03}\right)}}=200m\,\!$

$\left({f-0.03}\right)={\frac {\left({150*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left(0\right)^{2}}{2*\left({9.8}\right)*200}}\,\!$

$f=0.47\,\!$

Example 3: Grade

TProblem

Problem:

What should the grade be for the previous example if the coefficient of friction is 0.40?

Example

Solution:

$d_{b}={\frac {\left({150*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left(0\right)^{2}}{2*\left({9.8}\right)*\left({0.40-G}\right)}}=200m\,\!$

$\left({0.40-G}\right)={\frac {\left({150*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left(0\right)^{2}}{2*\left({9.8}\right)*200}}\,\!$

$G=0.44-0.40=0.04\,\!$

Thus the road needs to be a 4 percent uphill grade if the vehicles are going that speed on that surface and can stop that quickly.

Example 4: Crash Reconstruction

TProblem

Problem:

You are shown an crash scene with a vehicle and a light pole. The vehicle was estimated to hit the light pole at 50 km/hr. The skid marks are measured to be 210, 205, 190, and 195 meters. A trial run that is conducted to help measure the coefficient of friction reveals that a car traveling at 60 km/hr can stop in 100 meters under conditions present at the time of the accident. How fast was the vehicle traveling to begin with?

Example

Solution:

First, Average the Skid Marks.

$(210+205+190+195)/4=200\,\!$

Estimate the coefficient of friction.

$d_{b}={\frac {\left({60*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left(0\right)^{2}}{2*\left({9.8}\right)*\left({f-0}\right)}}=100m\,\!$

$f={\frac {\left({60*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left(0\right)^{2}}{2*\left({9.8}\right)*100}}=0.14\,\!$

Third, estimate the unknown velocity

$d_{b}={\frac {\left({v*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left({50*\left({\frac {1000}{3600}}\right)}\right)^{2}}{2*\left({9.8}\right)*\left({0.14-0}\right)}}=200m\,\!$

$\left({v*\left({\frac {1000}{3600}}\right)}\right)^{2}-\left({50*\left({\frac {1000}{3600}}\right)}\right)^{2}=200m*\left({2*\left({9.8}\right)*\left({0.14}\right)}\right)\,\!$