This question is taken from a similar one inside the mathematics stack exchange. I followed the suggestion given in the comment by Jyrki Lahtonen and posted said question here.


Recently, on a flight back from Charlottesville, NC to Logan Airport, MA, the Pilot stated this to his passengers over the intercom, I paraphrase:

As you can see out the window, we just saw some lightning strike within 3 miles of our plane. Because of this, they closed the ramp gates. Every time lightning strikes within three miles of our plane, we have to reset the wait timer to ten minutes.

This made me wonder if there was a mathematical connection between the distance of a lightning strike and the amount of time needed to wait for take-off to ensure another strike will not occur on the ramp.


Let the "area that causes a delay" when lightning strikes be represented by a circle of radius 3 miles. In addition, let us envision a scenario where lightning strikes on the circumference of the circle, i.e. 3 miles away from the plane, which is the farthest distance from the place that will trigger a 10 minute wait.

The intersection of these circles is a clean venn-diagram shape, and results in an area of (9π)/3 sq.mi. or 3π sq. mi., which is about 9.42 sq. mi, and the ceiling of which would be 10.


Given that the pilot stated that a wait time of 10 min would be triggered if lightning struck within 3 miles of the plane, and the area in between a strike on the circumference of 3 miles and the plane results in a ceiling of 10, is there any mathematical basis for a lightning strike wait time of 10 if given a "danger zone" of 3 miles?

  • 3
    $\begingroup$ I fail to see the relevance of mathematics to this question. 3 miles is almost certainly an arbitrary distance employed (and probably not that accurately tracked to be honest) to ensure the safety of ground operators. Are you really asking why they close for another arbitrary amount of time? $\endgroup$
    – Jamiec
    Commented Jun 30, 2020 at 15:04
  • 6
    $\begingroup$ The position of individual planes does not matter, and it is not the pilot deciding to wait. It is the airport that has rules regarding not having people work on the ramp and passengers walking around outside while a thunderstorm passes over the airport $\endgroup$ Commented Jun 30, 2020 at 15:10

2 Answers 2


No, because adding candles to candy does not make much sense.

The dimensions are different, so the relative numeric values depend on the choice of units. You could equally state the rule as you have to wait ⅙ of an hour after lighting strike within 5 km (if it's statue miles; 5.5 if nautical) and now the numbers won't match any more.

Besides, it is a 3 mile radius, so the area is about 28 square miles and the numbers never matched to begin with.

The real reason is two-fold:

  • When there is active storm right over the airfield, somebody could get struck by lightning. Inside aircraft and vehicles is safe, because the lightning will ground through the skin and not get inside, but out on the ramp it wouldn't be. 10 minutes is long enough to consider the storm no longer active.

  • The storm builds up static charge on the aircraft and vehicles. This is again not a problem for the occupants, because the conductive skin ensure the whole vehicle is at the same potential, but it could cause a static discharge strong enough to hurt someone or set something on fire when touching the vehicle from outside. 10 minutes is considered long enough for these charges to gradually discharge through the wheels to the ground and through the static wicks to the surrounding air.

    This would also apply to aircraft that flew near the storm just before landing, but that is normally avoided as storms generate dangerously strong wings.

  • The 3 mile radius is simply size of typical storm.


There most certainly is mathematics behind this 3mile/10minute rule you encountered. This rule likely comes from statistics, and it's purpose is to give a simple rule ro follow, and thus avoid increasing the risk of getting struck by a lightning and/or a microburst.

I bet you can twist this statistical model into a geometric form, but i think it will be a 2d curve of normal distribution, not a 3d form.

P.S. I beg to differ with a commenter above who stated that *3 miles is almost certainly an arbitrary distance...". There's hardly anything arbitrary in aviation rules, regulations and procedures. Most things, if not all, have been learned the hard way...

  • 1
    $\begingroup$ It is arbitrary in the sense like 40 mph speed limit is arbitrary (rather than a mathematically optimal, say, 43.2). If the ligtning strikes at 3.1 mi, there is suddenly no danger? How did they choose the risk threshold? It is almost certainly a standard practical rule with 'typical' parameters which are easy to estimate and remember. Aviation has lots of such simplifications. $\endgroup$
    – Zeus
    Commented Jul 1, 2020 at 0:43
  • $\begingroup$ How is 43.2mph mathematically optimal? Aviation parmeters are simplifications, yes, but not arbitrary ones. In his answer Jan explains this case further. $\endgroup$
    – Jpe61
    Commented Jul 1, 2020 at 8:22
  • 2
    $\begingroup$ If it's not arbitrary (as in "3 miles gives us roughly enough safety margin") then please demonstrate why it is 3 miles and not 2 or 4, or 10. Lots of things in aviation are arbitrary in that sense, in that they give enough safety margin without being able to calculate the exact figure required. eg, wind correction angles, descent profiles, course corrections etc etc etc. $\endgroup$
    – Jamiec
    Commented Jul 1, 2020 at 8:25
  • $\begingroup$ We seem to have profound disagreement on the meaning of the word arbitrary. I understand it as: dictionary.cambridge.org/dictionary/english/arbitrary $\endgroup$
    – Jpe61
    Commented Jul 1, 2020 at 15:02
  • $\begingroup$ @Jpe61, suppose that we somehow scientifically determined that the optimal speed limit in this place was 43.2 (it is unlikely to be a nice round number). But instead of posting this limit, we instead round it to an arbitrarily selected set of standard numbers (for various reasons, convenience basically). The set is arbitrary, because, in the first place, that the decimal system we use is arbitrary and is no more valid than an 11-base system, mathematically speaking. $\endgroup$
    – Zeus
    Commented Jul 2, 2020 at 1:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .