At a simple level, speeding in a car attempts to minimize the time required to travel a distance by utilizing the basic relationship: $$d=st$$

So for a given distance, time should be inversely proportional to speed: the faster you go the less time it takes.

My question is, on a practical level, does this actually help you get to your destination much faster? Say you are travelling, on average, 5 mph (sorry for non SI units, as an American driving in kph just seems wrong) over the speed limit, which is often considered 'safe' for avoiding a traffic ticket. Does this shave seconds off your commute time? Minutes?

I believe that traffic can be modeled using fluid dynamics (haven't actually seen those models myself just have been told this is the case) so, how does this come to in to play while speeding? It seems to me that it would depend on where you speed rather than how fast, i.e. speeding to avoiding getting stuck at a traffic light.

Any insights or calculations would be greatly appreciated, Thanks!

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    $\begingroup$ How is this related to fluid dynamics? Traffic flow is modeled similar to fluid dynamics when there are many many cars, driving on roads which fork, merge, turn, etc. If you're just looking to know how much time you can save by going, say, 5 mph above the speed limit, the answer is a trivial extension of the equation you've mentioned in your question. $\endgroup$
    – pho
    Commented Jul 9, 2014 at 5:24
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    $\begingroup$ @PranavHosangadi I assume that's exactly how the question is related to fluid dynamics - that is, jkeuhlen wants to know whether the effect of speeding (when possible) by a few MPH is worthwhile when you take traffic flow into account. jkeuhlen, can you confirm this? (Otherwise, it would be an incredibly lazy question!) I do think some clarifying edits are in order, though; for example, what sort of traffic flow should be considered? Highway traffic or city traffic? Rush hour? etc. $\endgroup$
    – David Z
    Commented Jul 9, 2014 at 5:26
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    $\begingroup$ I'm with @PranavHosangadi - I don't see fluid dynamics here. If you're capable of speeding, then there is no traffic flow issue. If there is obstructing traffic, then you can't choose your speed, so asking what speed to choose doesn't make any sense. $\endgroup$
    – user10851
    Commented Jul 9, 2014 at 8:59
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    $\begingroup$ In some cities, the traffic lights are timed so that when you drive down the main roads slightly below speed limit, they will always be green. Driving faster is pointless, because you will just spend the time you gained waiting at traffic lights. $\endgroup$
    – Philipp
    Commented Jul 9, 2014 at 13:59
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    $\begingroup$ Just an FYI: For Canada, it's Km/h, not kph. $\endgroup$ Commented Jul 9, 2014 at 15:33

5 Answers 5


Alright, let's start with your direct question. Since $d = vt$ the time it takes to travel a certain distance is inversely proportional to your speed $ t \propto v^{-1} $, and so the fractional change in time is proportional to the negative fractional change in your speed.

$$ \frac{dt}{t} = - \frac{dv}{v} $$ So, if we consider typical a typical highway speed of 65 mph and a 5 mph difference, this is a fractional change of about 8%. So going 5 mph faster on the highway will shave 8% off your travel time. So if you had an hour drive, it'll shave off 5 minutes.


But. Let's try to consider the added cost of going faster. If you go faster, the wind drag is higher, so your car needs more power to maintain speed. More power means more energy, more energy means more fuel, more fuel means more money.

If we consider just the contribution of wind resistance, we know that $ F \propto v^2 $ for cars, and power is $ P = F v $ so the power consumed by drag goes as $ P \propto v^3 $. Energy consumption is $ E = P t $, so if we consider a drive of fixed length, since $ t \propto v^{-1}$ we have for the contribution of air drag, $ E \propto v^2 $. Now, the energy you get from fuel is proportional to the number of gallons you buy, and the cost scales as the number of gallons so $ \text{fuel cost} \propto v^2 $. So the fractional change in the fuel consumed due to wind drag is: $$ \frac{ d(\text{fuel cost}_{\text{drag}} ) }{ \text{ fuel cost}_{\text{drag}} } = 2 \frac{ dv }{ v } $$ So, for the same 8% increase in speed, you pay an additional 16% in fuel costs due to the loss to air drag.

Naturally air drag isn't the only way we use up fuel to keep a car running, there are all kinds of losses in a car, from inefficiencies in the engine itself, to friction in the various components of the engine, etc. As a simple model, let's say that the power a car consumes is the sum of the air drag and a constant term independent of speed: $P \sim \alpha v^3 + \beta $ for some appropriate choices of $\alpha$ and $\beta$ This would mean that our energy consumption would still be $ E = P t $, so for a constant distance drive, we're talking $$ E \sim \alpha v^2 + \frac{\beta}{v} $$. We can test this model against data from a government study (figure from wikipedia:Fuel_economy_in_automobiles) where for our model, we have $$ \text{mpg} = \frac{ \alpha }{ v^2 + \frac{\beta}{v} } $$

Here I've shown the figure as well as an example fit of our model:

Observed mpg versus speed and fit

The fit is overlayed in red, and corresponds to $\alpha = 1.5 \times 10^5, \beta = 1.28 \times 10^5 $. Notice that our simple model does pretty well and corresponds to a car that has a highway mph of about 25 mpg. Notice that at high speeds we are seeing the scaling we expect due to air drag alone, for at high speed our operating costs are dominated by air resistance, but it was useful to create the simple model and do the fit in this case because the region of interest is in the overlap region.

Per hour

Now that we know how the efficiency of our car varies with speed, knowing the average price of gas of $\$3.752$/gallon (from wolfram alpha) we can compute the cost to operate an average car at a given speed:

Operating cost of vehicle versus speed

An in particular, we can compute the additional cost per hour per 5mph increase in speed as a function of speed:

Additional operating cost for a 5 mph increase as a function of speed

Per 10 miles

Here I've shown the operating costs as a function of the time spent driving, so as to give costs per hour, which I think is useful for longer drives and something people have a handle on from other areas of life.

If we instead want to look at it as function of the distance travelled, we can look at the car efficiency as the cost to travel 10 miles as a function of speed.

Operating cost per 10 miles

Or we can consider again the change created by a 5 mph increase in speed for a fixed travel distance.

Change in operating cost per 10 miles per 5 mph difference

Where here it becomes clear that for a fixed travel distance, as long as you are going less than 40 miles per hour (which for our model was the maximum fuel efficiency speed, and varies per car but the data seems to indicate is about 40 mph across the board), you can always justify speeding by 5 mph from purely economic terms, but at something like highway speeds, it costs you an additional 15 cents or so per 10 miles to go 5 over.

Traffic Lights

So, up till now we have considered the effectiveness of speeding from an economic perspective in the limit that we are travelling unimpeded down the road. As people have requested in the comments, let's try to figure out how effective speeding is in a more city type environment. This is a difficult problem to address, as traffic lights can have fairly complicated controllers. In particular, in some regions there are Green waves where the lights are designed to allow people travelling at the proper speed to pass unimpeded down long stretches of road. Obviously in this case, you would want to travel at the speed of the green wave and speeding wouldn't help you and would in fact hurt you.

But, sophisticated traffic light controllers are not all that common outside of rich large cities. So, let's try to adopt a spherical cow type approximation to traffic lights and assume that traffic lights are independent and just operating on some fixed cycle of green and red. $p$ will be the fraction of time the average traffic light is green, $\tau$ will be the length of a red light, $d$ will be the average distance between traffic lights. If the lights are all operating independently, the distribution of waiting times when we reach a light can be modeled as $$ P(t) = p \delta(t) + \frac{1}{\tau} ( 1 - p ) \quad 0 \leq t \leq \tau $$ or in words, with probability $p$ we don't have to wait at all, otherwise our waiting time will be uniform up to $\tau$. This distribution has mean and variance $$ \mu = \frac{\tau}{2} ( 1 - p ) $$ $$ \sigma^2 = \frac{\tau^2}{12} ( 1 - p) ( 3 p + 1 ) $$

Now, if we travel for $N$ blocks, we will have for the average time it takes $$ \langle t \rangle = N \left( \frac{d}{v} + \frac{\tau}{2} ( 1 - p ) \right) $$ $$ \sigma^2_t = N \frac{\tau^2}{12} ( 1- p) ( 3 p + 1 ) $$ where we have added in the travel time between the lights themselves.

So, for instance, with $d = 1/10$ mile between lights on average, $\tau = 30$ seconds, and $p = 0.65$ we get for an average city speed as function of target speed: Average city speed versus target speed

So, for a target speed of about 45 mph for a major road in a city, we get for an average speed something like 28 mph, which seems to agree moderately well with observations.

Now, as we have modeled it, if you speed you will get there faster, but what we should compare against is the intrinsic variability introduced by the traffic lights, and a case could be made that speeding 5 mph over is really only worth it if the gains you get in timing are larger than the natural variations in times you would have given the lights, otherwise you'll hardly notice the effect. So in particular, we can compare the fractional reduction in your travel time for going 5 mph over, versus the fractional change in your travel time due to the intrinsic variation due to the random light timings $(\sigma/\mu)$ for different number of blocks. We obtain:

Speeding versus light variations

Here the solid line shows the fractional change in your travel time you'd get by going 5 mph over the target speed at the bottom. Notice that it scales as $1/v$ just as the very top of the post. The dotted lines show fractional change in travel time induced by a 1 sigma variation in the behavior of the traffic lights, for different number of blocks. Notice that at around 40 mph, the time you would shave off by going 5 mph is comparable to the natural variations you would expect in travel times due to your luck with the traffic lights if you are travelling 10 blocks, and both of these are at about the 10% level. At this point it starts to become difficult to justify speeding as its effect will be hard to notice over the natural variation. But, notice that if you are travelling a longer distance, there is a clear gain given by speeding, as the variations in travel time start to be suppressed through averaging. On the flip side, for very short trips of a few blocks, the variations in your travel time given by your luck at the lights completely dominates any gain you'd get by speeding.

Speeding gains in urban environment going 50 in a 45

Here I've simulated traveling for 5, 15 or 50 blocks according to our model, both at 45 mph and going 50 mph. I ran the simulation 10,000 times and here I show the time gained from speeding in different trials. Here we can really see that there are no noticeable gains for going 5 over over 5 blocks, it's completely washed out by our luck with the lights, but a noticeable change in our expected time over 50 blocks.

The code used for this answer is available as an ipython notebook

Addendum: I've explore a more realistic model of power losses in cars in this more recent answer

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    $\begingroup$ @ChrisWhite from my reading of the graphs (I may be wrong) the efficiency improvement is a marginal increase where as the cost increase is total cost. i.e. the cost of accelerating from 60 to 65 mph is less than the cost of accelerating from 55 to 60 mph BUT the total cost of travelling at 65 is always greater than the total cost of travelling at 60. $\endgroup$
    – MD-Tech
    Commented Jul 9, 2014 at 10:23
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    $\begingroup$ @MD-Tech I see what's going on. Traveling 20 miles per hour is always better than traveling 10 miles per hour to get somewhere. Period. alemi, your second two plots show cost per hour, which is... not really useful in the real world. People (outside of LA) travel to get somewhere, not for the sake of being in their cars for a prescribed amount of time. Really, though, this is all tangential to what the OP asked - the question as phrased has nothing to do with fuel economy. $\endgroup$
    – user10851
    Commented Jul 9, 2014 at 10:29
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    $\begingroup$ So saving five minutes costs \$1.50 - do this 12 times and you've saved one hour and spent \$18. If you cost your company more than $18 an hour, you should really be speeding when on company time. $\endgroup$
    – corsiKa
    Commented Jul 9, 2014 at 16:44
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    $\begingroup$ Your question went into great detail about cost, but didn't address the speed question in relation to stop lights, traffic, and so on. If I speed between stop lights, but the slower drivers behind me always catch up, I gain no time. But if I slip through one light, then I'm one light ahead. $\endgroup$
    – ErikE
    Commented Jul 10, 2014 at 0:08
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    $\begingroup$ I'm with the people saying that this answer, while well-organized and pretty, doesn't address the question. The asker is specifically wondering how much time (not money) you can save by speeding, when taking into account the flow of traffic. This answer simply doesn't consider that context. $\endgroup$ Commented Jul 10, 2014 at 0:39

Aside from the effect on one driver you might consider the effect on traffic in general, that is what happens when everyone breaks the speed limit. As far as fluid dynamics is concerned, the side-effects of your speeding (if any) are felt by the people behind you.

Reaction distance increases linearly with speed, but stopping distance must include a term proportional to the square of speed, since with constant deceleration we know $v^2 = u^2 + 2as$

"Official" stopping distances are nominal anyway, since even under fixed driving conditions they don't account for variation among cars. However, the UK government defines "braking distance" as $v^2 * k$ where $k$ is 0.015 metres / mph2 [I know, we're not fully on the metric system either], then rounds it.

  • total stopping distance at 65mph: 83m
  • total stopping distance at 70mph: 96m

This means that if cars separate themselves by the stopping distance (which in point of fact they don't, but if), then this 7.7% speed increase increases separation by 15.7%, so in a given time 8% fewer cars pass any given point per lane of traffic. Faster traffic reduces the capacity of the road.

In fact cars don't separate themselves by the stopping distance (or even by a fixed fraction of it), so what actually happens is some trade-off between reduced capacity, and the whole system becoming more reliant on people's reaction times, which are not entirely reliable. This is somewhat more dangerous, but even assuming no collisions it still has an effect on overall progress of the traffic. Drivers are more likely to have to stamp on the anchors when the person in front of them does something surprising, because they're driving proportionally closer to their pure reaction distance.

This is one area where the fluid dynamics comes in. Sudden braking causes the person behind you to do the same, which transforms smooth (laminar) flow of traffic into turbulent flow. Much less traffic gets through the same point, and traffic backs up behind it. It's even said that just changing lanes in fast congested traffic has a certain chance of causing traffic to come to a halt some time later and of course many cars behind you. Once one of these spontaneous no-reason traffic jams has formed, they persist for quite a long time and slow down everyone's journey. I don't have figures on the magnitude of the effect, though.

All that said, I don't think 65/70/75mph is the critical region in congested traffic. The UK uses realtime variable speed limits more like 50mph to improve total throughput on congested roads. So the difference between 65 for everyone and 70 for everyone is likely negligible, since 65 already causes lots of turbulence when the road is "full". You'll likely find it almost impossible in a temporary speed limit to do any speed other than the same as everyone else, which is kind of the idea.

[Edit: for economists, this is an inefficient market. Each driver benefits from driving slightly faster, but loses if everyone drives slightly faster. They "should" all contract to stick to some speed below the fixed limit, but don't have the means to do so. Only the owner of the road (the government) has the information to choose a "better" limit (closer to a Pareto-optimal point) or the means to communicate it to the drivers. Hypothetical private roads of course could likewise impose whatever T&Cs they chose on drivers, including banning or surcharging them for failing to follow anti-congestion measures, and arguably might be more responsive to customer feedback than the government is :-)]

  • $\begingroup$ A city traffic center (I don't remember which one) stated that a congestion slowdown during rush hour has lasted 3 hours after the initial event. $\endgroup$
    – LDC3
    Commented Jul 11, 2014 at 13:16

If the speed limit is 60 mph, it would take 60 min to go 60 miles. To go the same 60 miles at 65 mph, it takes 55.4 minutes. A time savings of 4.6 minutes. Is it really worthwhile to speed if all you are going to save is a few minutes (even less for shorter distances).

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    $\begingroup$ To avoid hard calculations: Distance = time * speed. So 65 minutes at 60mph and 60 minutes at 65mph cover the exact same distance, no calculator needed. $\endgroup$
    – gnasher729
    Commented Jul 9, 2014 at 11:41
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    $\begingroup$ @gnasher729 Do you take a different route to get somewhere just so you can drive faster? $\endgroup$
    – LDC3
    Commented Jul 10, 2014 at 1:05
  • $\begingroup$ @LDC3: no, he's taking a different (5 mile longer) route in order to make the calculation easier when comparing the two speeds, not in order to drive faster :-) $\endgroup$ Commented Jul 10, 2014 at 9:59

Fluid dynamics models might have a practical value in heavily congested areas, but then you can't speed, save for some random short bits after traffic signals, rendering it irrelevant. Guess they're more of a traffic distribution models, smaller roads will attract more traffic if the counterpressure (congestion) gets higher on the main pipe (road).

It really depends on if you can keep the average speed higher, which is unintuitively hard. If you're driving far away, driving faster can make a big difference. On 400 mile trips on 60 mph roads even constant +5 mph would save about half an hour, assuming you can keep the average up. I don't know if your trucks are limited to 50 mph (80 kph), but overtaking one at the beginning could likewise save 1 h 20 min.

In partially congested areas where you can occasionally drive at +5, but then spend the same proportion of time behind someone at the limit, the average speed is less than +2.5. Say, your trip is 25 miles, and you spend in total 15 minutes at 50 mph (covering half the distance), and in between those sections drive at 55 mph, you get to your destination only 1 min 20 seconds earlier, and the average is under 52.4 mph (in reality even a bit lower, because your speed doesn't change instantly 50->55.

If the limit is still lower, and there are signal controlled intersections, the local conditions start to be more important. If you know the extra speed on some section will save you one red light, or that happens by chance (either by being there just before the light changes, or by having a better position in the queue so you don't have to stop twice), you might save an extra 30-90 seconds for each such intersection. If the same 25 mile trip is in a 40 mph zone, and you stop in 10 red lights for 40 seconds each time, your average speed is 33.9 (assuming instant speed change). Had you gone 45 and avoided the last three red lights, the average would depend on the distance from the 7th signals to the destination; with evenly spaced signals the average could be 41.6 mph - assuming you would have had to wait more (65 seconds, so it's plausible such a road could exist) in the first seven signals, i.e. you'd have passed on the same signal cycle. That's 44 minutes vs. 36 minutes. In an extreme case, if the signals are far apart enough and the +5 helps you avoid a stop in all of them, you'd get to your destination in just over 33 minutes. Usually, though, they try to time the signals close to each other so that any "reasonable" speeding won't get you through consecutive signals any faster than those who catch up with you on the next green cycle from the previous signals. Likewise, if going +5 would have gotten you past some other three signals on the way, depending on how or if their timing is interconnected and how long the cycles are, whether the guy going 40 would be there even within the same signal cycle of the last traffic signals + time difference from those signals to the endpoint.


[Preamble not really answering the question: Sounds almost like a classical philosophy question. To turn it around, you might ask whether its worth travelling at the speed limit of 5mph slower. And 5 mph slower than that. Or should you just walk? "]

The accepted answer computes the time saving and the fuel cost.

In the airline industry, this is formalised by defining the "airline cost index". This the ratio 'price of time' divided by 'price of fuel' and is an input to the flight management computer.

So, ignoring the legal issues around the speed limit, the drag equation means that the higher the price of time, the faster we should go.

It seems to me the same applies when driving a car.

Saving 5 minutes to get to a party running all evening is different to saving 5 minutes so as to get to a job interview on time, or getting someone to a hospital.

On top of that are the safety issues. Breaking distance is roughly quadratic in speed, so a 10% speed increase means a 20% increase in braking distance. This is turn relates to your question about "where you speed". Speeding just to get through lights might not be the safest place to do so.


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