Fast ballpark calculation of maximum speed for maximum car througput (using simple high-school physics and math)

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:

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

Total length occupied by a single car (handdrawn)

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

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 $69.4 \frac{\text{m}}{\text{s}}$):

Car frequency in kilometers per hour

Car frequency in meters per second

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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