25 Apr 2019 - tsp

The following calculations try to answer the basic question at what speed a street or a highway can - under optimal conditions - provide the highest throughput (i.e. cars per second). Despite the personal impression that higher speed always leads to higher throughput the calculations will show the opposite.

Please note that this is not the *only* evidence that shows that some
speed limits may make sense - there are way more important ones like
the model Wramborg uses to estimate fatality rates during car crashes
at various speeds.
These models show that at a speed limit of about $30 \frac{\text{km}}{\text{h}}$
the fatality rate is about $10\%$ whereas for a speed of about $50 \frac{\text{km}}{\text{h}}$
it already reaches a fatality rate of more than $80\%$. These figures are
way more relevant for speed limits in inner city regions / residential areas
than the througput estimations which are more relevant to highways.

The calculations are going to be really simplified to illustrate the idea and give a basic ballpark figure about the topic. The assumptions are:

- Traffic is flowing ideal. No acceleration and deceleration (this would in fact reduce the carrying capacity of a street by some orders of magnitude but would not modify the speed at which the highest throughput is reached). Acceleration and deceleration would also lead to shockwaves through traffic and may lead to traffic jams (without any other cause than for example someone changing lanes. Maybe there will be a blog post on that later on)
- All cars go at the same (maximum) speed.
- All cars have equal length of about $4.5$ meters
- All tires are providing a maximum deceleration of about $3.5g$ which is a reasonable assumption for an emergency brake on many cars - but of course not on all and not under all weather conditions (maybe more on that in a followup post too)
- Reaction time is assumed to be optimal (but it will cancel out for the speed optimum anyways as shown later) to be about $0.25$ seconds for visual stimulus.
- Assumes a 100 percent utilization (maybe an update will follow that shows the change in maximum speed vs. car frequency for various utilization factors)

The basic idea is to first calculate the space a single car occupies including it’s own length, the reaction space and the emergency braking distance. Some people may now argue that one doesn’t have to drive with that distance to the next car (this may be a bad idea in case of some unlikely accidents like loosing cargo, etc. - but it wouldn’t affect the optimum speed value but only the maximum optimal car througput as shown later)

The total space occupied is calculated as

$l_{occupied} = l_{car} + l_{react} + l_{stop}$

The car length $l_{car}$ is assumed to be constant $4.5$ meters. Reaction time is assumed to be constant $0.25$ seconds:

$ \begin{align} l_{car} = 4.5 \\ l_{react} = \frac{v_{max}}{t_{react}} = \frac{v_{max}}{0.25} \end{align} $

The stopping distance is calculated by basic accelerated/decelerated movement with a constant deceleration of $a=-3.5g \approx 3.5 * \pi^2 \approx 35 \frac{\text{m}}{\text{s}^2}$.

$ \begin{align} v(t) = v_{max} - 35*t \\ v(t_{stop}) = v_{max} - 35*t_{stop} = 0 \\ \to v_{max} = 35*t_{stop} \\ \to t_{stop} = \frac{v_{max}}{35} \\ s(t) = -35 * \frac{t^2}{2} + v_{max}*t \\ l_{stop} = s(t_{stop}) \end{align} $

Adding all factors up leads to the occupied space

$ \begin{align} l_{occupied} = 4.5 + v_{max} * t_{react} - \frac{35}{2}*t_{stop}^2 + v_{max}*t_{stop} \\ = 4.5 + v_{max} * t_{react} - \frac{35}{2} * \left(\frac{v_{max}}{35}\right)^2 + v_{max} * \frac{v_max}{35} \\ = 4.5 + v_{max} * t_{react} - \frac{1}{70} v_{max}^2 + \frac{1}{35} v_{max}^2 \\ = 4.5 + v_{max} * t_{react} + \frac{1}{70} v_{max}^2 \end{align} $

To calculate car frequency one can use

$ \begin{align} f_{car} = \frac{v_{max}}{l_{occupied}} \\ = \frac{v_{max}}{4.5 + v_{max} * t_{react} + \frac{1}{70} v_{max}^2} \end{align} $

This is illustrated in the following graphs for speeds between $0$ to $250 \frac{\text{km}}{\text{h}}$ (i.e. from $0$ to $69.4 \frac{\text{m}}{\text{s}}$):

As expected we see an optimum around 60-80 kilometers per hour. To locate the estimated maximum we search for maxima as usual:

$

\begin{align}
\frac{\partial f_{car}}{\partial v_{max}} = \frac{(4.5 + v_{max}*t_{react} + \frac{1}{70}*v_{max}^2) - v_{max} * (t_{react} + \frac{2}{70} * v_{max})}{(4.5 + v_{max} * t_{react} + \frac{1}{70} * v_{max}^2)^2} \\
\frac{\partial f_{car}}{\partial v_{max}} = 0 \\
\to 4.5 + v_{max}*t_{react} + \frac{1}{70} * v_{max}^2 = v_{max} * t_{react} + \frac{2}{70} * v_{max}^2 \\
\to v_{max}^2 = 70 * 4.5 \\
\to v_{max} = \sqrt{315} = 17.748 \frac{\text{m}}{\text{s}} = 63.89 \frac{\text{km}}{\text{h}}
\end{align}
$

As expected (by anyone who ever has done some more extensive calculations) an optimal speed in the range between 60 and 80 kilometers per hour leads to the highest frequency of cars on a single lane.

This article is tagged: Physics, Traffic, School math

Dipl.-Ing. Thomas Spielauer, Wien (webcomplains389t48957@tspi.at)

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