## Sunday, June 17, 2012

### Braking distance and time on level road and on an incline

Off topic entry. Nothing to do with electronics. But I've been obsessed with (or should I say possessed by) this topic for the past week or so. Just want to have all the equations and derivations in one place so I can refresh my memory if I need to in the future.

Let
vi = initial velocity, m/s
vf = final velocity, m/s
v = average velocity, m/s
a = deceleration rate, m/s2
t = time to stop the vehicle, s
d = braking distance, m
μ = coefficient of friction, unitless
Ff = force of friction, kg·m/s2
Fn = normal force (force perpendicular to surface), kg·m/s2
Fg = force due gravity pulling the vehicle down a slope, kg·m/s2
m = mass, kg
g = gravitational acceleration, 9.8 m/s2
θ = angle of inclination (positive if vehicle going uphill, negative if going downhill)

Final velocity is given by the equation:

$v_{f}=v_{i}+-at$

If vf  = 0 then

$t=\frac{v_{i}}{a}$

Distance is given by:

$d=\overline{v}t$

Average velocity is:

$\overline{v}=\frac{v_{i}+v_{f}}{2}$

Given  vf  = 0 and substituting we have:

$d=\frac{v_{i}}{2}\frac{v_{i}}{a}=\frac{v_{i}^{2}}{2a}$

Alternatively we can derive the distance formula using kinetic energy and work equations:

$\textup{kinetic energy}= \frac{1}{2}mv_{i}^{2}$

$F=ma$

$\textup{work} = Fd = mad$

Equating work and energy we have:

$mad= \frac{1}{2}mv_{i}^{2}$

Solving for d we obtain:

$d=\frac{v_{i}^{2}}{2a}$

The force of friction is given by the following formula:

$F_{f}=\mu F_{n}$

If the vehicle is on a level surface then normal force is just its weight. Therefore,

$F_{f}=\mu mg$

Since

$F=ma$

then

$ma =\mu mg$

Therefore,

$a =\mu g$

Substituting the above into the distance and time equations we get:

$d=\frac{v_{i}^{2}}{2\mu g}=\frac{v_{i}t}{2}$

$t=\frac{v_{i}}{ug}=\frac{2d}{v_{i}}$

Interesting to note that we can (at least in principle) determine the coefficient of friction by noting the velocity when the brakes are fully applied and time to bring the vehicle to a dead stop.

Now let's tackle the situation when the vehicle is braking on a slope.

When the vehicle is on an incline then a portion of the gravitational force acts to pull the vehicle down the slope. This force we denote as Fg and is derived via basic vector analysis. Refer to the free-body diagram. On the left of the diagram is an object on an incline with an angle of θ. Its weight is labeled as mg. Normal force Fn is the component of mg that is perpendicular to the incline. Fg is the component of mg that is parallel to the incline. To the right I've drawn a vector addition of the forces. Since this is a right triangle we can use the pertinent trigonometric functions: Sin θ = opposite side / hypotenuse. Cos θ = adjacent side / hypotenuse. Since the hypotenuse = mg we obtain the following:

$F_{n}=mg\cos \theta$

$F_{g}=mg\sin \theta$

An aside: Since the above are the legs of the right triangle it follows from the Pythagorean theorem that

$\textup{weight} = mg = \sqrt{ \left (mg\sin \theta \right )^{2} + \left (mg\cos \theta \right )^2}$

When on a slope with the brakes engaged so that the vehicle is at a standstill Ff is the force that keeps the vehicle from slipping and sliding down and acts in the opposite direction as Fg which is pulling it down the incline. The net force holding the vehicle in place is therefore Ff - Fg. It stands to reason that when Fg becomes >= Ff then the vehicle--even with the brake pads fully gripping the brake discs and drums--will begin to slip and slide. We can calculate the angle of inclination at which this will occur:

$F_{f}=\mu F_{n}=F_{g}$

$\mu F_{n}=F_{g}$

$\mu mg\cos \theta = mg\sin \theta$

$\mu = \frac{\sin \theta}{\cos \theta } = \tan \theta$

$\theta = \arctan \mu$

Bear in mind that in this case we are using the static coefficient of friction rather than kinetic (or sliding) coefficient of friction since the vehicle is at rest with respect to the road.

When a vehicle is going downhill its braking distance will increase because a fraction of the gravitational force is acting to pull it down even when the brakes are applied. θ will be negative if going downhill, and so Fg will be negative. When going uphill part of gravitational force adds to the force of friction thereby decreasing braking distance.

The sum of forces acting on the vehicle parallel to its direction of motion is given by:

$F_{f}+F_{g}$

Since

$F_{f} =\mu F_{n}$

Therefore,

$F_{f} + F_{g} = \mu F_{n} + F_{g}$

$= \mu mg\cos \theta + mg\sin \theta = mg\left ( \mu \cos \theta + \sin \theta \right )$

Since

$F=ma$

$ma = mg\left ( \mu \cos \theta + \sin \theta \right )$

$a = g\left ( \mu \cos \theta + \sin \theta \right )$

Substituting into our distance and time equations we obtain:

$d = \frac {v_{i}^2} {2g\left ( \mu \cos \theta + \sin \theta \right )}$

$t = \frac {v_{i}} {g\left ( \mu \cos \theta + \sin \theta \right )}$

The above are exact equations in contrast to the approximate formula using road grade. The gradient is equal to the rise over run, or tangent of θ. The arctangent of the gradient = angle θ.

Reminder: when using the above equations θ should be negative when the vehicle is going downhill. Given -90° < θ < 0, sin θ is negative while cos θ is positive. Hence Fg is effectively subtracted from Ff. When going uphill and θ is positive then Ff is effectively increased by amount Fg. As anyone who's driven knows, it's harder to bring a vehicle to a stop when going downhill. And that it's easier to do so when going uphill than when on a level road.

Beware that the above distances and times do not take into account the driver's perception and reaction time--the time to perceive the need to stop and the time to actually to apply the brakes. This takes a minimum of around half a second for a driver who's alert and concentrating on the road, but may be much greater if the driver is drowsy, tired, intoxicated, on his cell phone, has poor eyesight, etc. or if visibility is poor among other possible factors beyond the driver's control.

The distance traveled by the vehicle during this time is simply

$v_{i}t_{perception+reaction}$

Here are a couple of online braking distance calculators which don't, however, have provisions for vehicles on an incline:
http://www.csgnetwork.com/stopdistcalc.html
http://forensicdynamics.com/stopping-distance-calculator

Tables of coefficients of friction:
http://www.engineeringtoolbox.com/friction-coefficients-d_778.html
http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm

Summary of the relevant data from the above tables:

tire material road type and condition kinetic coefficient of friction
rubber dry concrete0.6 - 0.85
rubber wet concrete 0.45 - 0.75
rubber dry asphalt 0.5 - 0.8
rubber wet asphalt 0.25 - 0.75

A point worth mentioning: While the wheels are turning static coefficient of friction is operative because at the point of contact the tire is not sliding along the road. This is desirable because the static coefficient is higher than the kinetic coefficient of friction thus leading to decreased braking distance. Anti-lock brakes (ABS) prevent the wheels from locking even when the driver slams the brake pedal, therefore preventing the tires from skidding and sliding. However, the speed of the tire is lower than the vehicle's and so some slip occurs between tire and road. The static coefficient of friction cannot therefore be applied to vehicles even if they're equipped with ABS. For safety reasons it's best to be conservative and use the kinetic coefficient.

The take home lessons from the above formulas:

Braking distance varies with the square of velocity. Thus, doubling one's speed quadruples the distance. Increasing one's speed from 40kph to 100kph increases the braking distance by (100/40)2 = 6.25 times. Braking time, on the other hand, increases linearly. And thus, since 100kph is 2.5 times 40kph, braking time at 100kph is 2.5x that at 40kph.

The worst braking scenario is when the coefficient of friction is low--when the roads are wet, muddy, or iced up--and one is going downhill, with braking time and distance becoming greater as the inclination increases.

Hence, from the standpoint of safety the best thing a driver can do is to keep speeds low and/or to keep one's distance from other vehicles (and pedestrians) as large as practically possible. One can also apply the two-second rule. Although as weather and road conditions worsen this should be increased to three or four seconds.