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It's often mentioned that flying is the safest form of transportation, and no doubt this is true when considering only a long-haul trip. There's no doubt, for example, that driving (or taking a train or bus) 3,000km would be more dangerous than flying the same distance.

However, most flight crashes occur at take-off or landing and so I suspect there must be a break-even point after which flying becomes more dangerous than alternatives. For instance, if a flight is only 300km, would it be safer to drive?

For the purposes of this question, it can be assumed that I'm talking about OECD countries, which by and large have very high standards for both flight safety and the safety of other forms of transportation.

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    $\begingroup$ I suspect you'd have to hypothesize an airline flight on the order of a few miles before you could get the numbers to work out the way you're suggesting, and at that point economic reality would intrude, because such fights would be expensive beyond all reason. And no quicker than driving, given time taken to board & deplane & etc. Commercial aviation (in the developed world) is literally safer than living (in deaths per million man-hours), so even with the higher risks at takeoff & landing, it'll be hard to get the lines to cross. $\endgroup$
    – Ralph J ♦
    Commented Oct 18, 2019 at 1:11
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    $\begingroup$ Would regularly scheduled helicopter routes count? There are and historically have been airlines operating dedicated short (10-20min), regularly scheduled, and standard fare helicopter routes. Example JFK->NYC and SFO->OAK (Historical) $\endgroup$
    – crasic
    Commented Oct 18, 2019 at 16:28
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    $\begingroup$ The 737 MAX has some pretty short flights. $\endgroup$
    – Ed Plunkett
    Commented Oct 18, 2019 at 16:33
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    $\begingroup$ I'm assuming if your flight lasts less than 100 metres in a jetliner, it would be very unsafe indeed. $\endgroup$
    – Richard
    Commented Oct 18, 2019 at 20:07
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    $\begingroup$ This question seems to imply that touch and go's add danger to flying... $\endgroup$
    – Michael
    Commented Oct 18, 2019 at 21:55

1 Answer 1

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Note: This does not take travel to/from the airport into account.

The Wikipedia page on aviation safety has a nice table with deaths per journey, time and distance (based on data from the UK between 1990 and 2000):

  • Car: 40 deaths per billion journeys, 3.1 deaths per billion km
  • Aircraft: 117 deaths per billion journeys, 0.05 deaths per billion km

Assuming that car accidents are equally distributed along a journey, we only need to consider the 3.1 deaths per billion km. Further assuming that all aircraft accidents are related to takeoff and landing (i.e. once per journey), we only consider the 117 deaths per billion journeys. While the first assumption is probably reasonable (there are differences between types of road though), the second one is not quite true since aircraft can also develop problems during cruise.

With these basic assumptions, we can calculate when the expected number of deaths are equal:

$$ N_\text{death} = 117 \cdot 10^{-9} = 3.1 \times \frac{d}{10^9 \, \text{km}} \; \Leftrightarrow \; d = \frac{117}{3.1} \, \text{km} \approx 38 \, \text{km} $$

So any flight longer than 38km is safer than driving the same distance.

Let us now try to take the cruise part for aircraft into account. According to this page 20% of all fatalities are due to accidents during cruise. We can subtract 20% from the 117 deaths per billion journeys (so this only takes the parts of the flight into account that happen once per journey). Modifying the 0.05 deaths per billion km is not so straight forward. Since most of the distance is traveled in cruise, I am going to be conservative and just leave this number unchanged, therefore providing an upper limit on the deaths per km.

Now, we can make a more accurate comparison:

$$ N_\text{death} = 0.8 \times 117 \cdot 10^{-9} + 0.05 \times \frac{d}{10^9 \, \text{km}} = 3.1 \times \frac{d}{10^9 \, \text{km}} $$ $$ \Leftrightarrow \; d = \frac{0.8 \times 117}{3.1 - 0.05} \, \text{km} \approx 31 \, \text{km} $$

This is actually lower now, because the deaths during cruise are irrelevant for such a short flight. And this gives the final answer: Flights longer than 31km are safer by airplane than driving the same distance by car.

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    $\begingroup$ Nice answer based on statistic extrapolation. Of course, there is no evidence that the extrapolation is valid for such short flights. I would expect that landing 4 minutes after start is stressful and results in a higher accident rate. $\endgroup$
    – bogl
    Commented Oct 18, 2019 at 7:56
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    $\begingroup$ The fault in analises like this is always the same: it assumes an "average" driver. But the fact is that most car accidents, especially serious ones, are caused by gross negligence. Of course, this isn't a problem that this answer can solve but just once, I'd like to see driving statistics that exclude drunk drivers, morons who text and drive etc. and see what are the odds of an accident if I'm a person who keeps proper focus on driving. $\endgroup$
    – Davor
    Commented Oct 18, 2019 at 9:42
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    $\begingroup$ @Davor yes. Then again, even though most accidents are caused by negligence, even a careful driver still has a considerable risk of getting involved in such an accident. — Anyways, a more interesting comparison would be plane vs train, not car. $\endgroup$
    – leftaroundabout
    Commented Oct 18, 2019 at 9:47
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    $\begingroup$ @AleksG that's because most car journeys occur near (definition?) the driver's home $\endgroup$
    – Caius Jard
    Commented Oct 18, 2019 at 12:44
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    $\begingroup$ @user3067860 Indeed, the Earth has a nearly 100% fatality rate if you consider how long it's been travelling... $\endgroup$
    – Darrel Hoffman
    Commented Oct 18, 2019 at 18:37

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