Is the downward pressure exerted by raindrops in even a strong storm on a lightweight aircraft negligible?

Someone better informed may likely improve the following reasoning, cobbled together from haphazard sources.

Heavy rain is 3 inches per hour, or 0.021 mm/s of accumulation. (Maybe a lot more while flying through a hurricane, but I couldn't find hard numbers.)
Per m$^2$ of impacted horizontal surface, that's 21000 mm$^3$ / s / m$^2$, or 21 cm$^3$..., or 21 g/s/m$^2$.
All else being equal, faster falling raindrops should exert more pressure.

Now use a few facts from the paper "Experimental determination of forces applied by liquid water drops..." (Yu & Hopkins, 2018), which models and measures raindrops falling onto smooth glass, similar to an airplane's skin.
From fig. 3, a fast terminal velocity of a raindrop is 9 m/s, for a (largish) 4.5 mm raindrop. (Some argue that larger drops break up; others, that they are squashed rather than spherical; but this recent paper seems trustworthy.)
For such a drop at such a speed, the first row of table 3 gives values of C$_1$=2.7186, $α_1$=3.3367, $β_1$=1.5027 to plug into eq. 33, the raindrop's force $F$ (N) as a function of elapsed time $t$ (s) since impact:
$F(t) = C_1$ exp(−(ln $1000t$ +$α_1$)$^2$ / $β_1^2$) N
$F(t) = 2.7186$ exp(−(ln $1000t$ + 3.3367)$^2$ / 2.2581) N.

Integrating that from $t=0$ to $0.004$ s, the duration over which $F(t)$ is non-neglible, yields an impulse of 0.0004526 N$\cdot$s/drop.

Each 4.5 mm spherical raindrop is 0.0477 cm$^3$, thus 0.0477 g.
So at 21 g/s, that's 21/0.0477 = 440 raindrops per second.
So the overall force is 440 drop/s $\cdot$ 0.0004526 N$\cdot$s/drop = 0.2 N.
Per m$^2$, that's a pressure of 0.2 N/m$^2$.

The airplanes with the lightest wing loading are paragliders, at about 3 kg/m$^2$. In terrestrial gravity, that's about 30 N/m$^2$.

So flying one of those through a heavy downpour would require some extra lift, but only 1/150th as much as the wing is already making. A Boeing 747, at 650 kg/m$^2$, would need only 1/30000th extra.

John Hunter's answer suggests that the weight of water buildup matters more than the pressure of impinging raindrops. I found confirmation in a 1982 NASA report:

For 500 mm/h rain (6 times more than my guessed maximum),
an airliner gets 2600 N of extra downforce (and 70x more drag force) (p. 23), while the water film is 0.6 to 0.8 mm thick (p. 29), an extra weight of 2340 kg (p. 30).

So at least for this case, downward pressure is about 5 N/m$^2$, while extra weight is 4.3 kg/m$^2$ or 43 N/m$^2$, much more.

This report describes many other effects of rain on airplanes. More recent (paywalled) papers may have measurements, not just models.

  • 2
    $\begingroup$ You could probably simplify your estimates by noting that Newton's second law actually says "force = rate of change of momentum," rather than trying to calculate the deceleration force on individual raindrops. $\endgroup$
    – alephzero
    Commented Oct 2, 2021 at 19:02

1 Answer 1


The impact of new drops even in heavy rain has only a small effect.

From force x time = change in momentum

$$F\times 1 = 0.021 \times 10^{-3}\times 1000\times 9$$

where the 1000 is for the density of water, the $0.2N/m^2$ is confirmed.

The weight of a thin layer of water lying on the wings may be more significant.

Even for 0.1mm, the weight is about $0.1 \times 10^{-3} \times 1000 \times 9.8$, about $1N/m^2$. So you could look into the thickness of the water on the wing.

As the water would be swept off for most aircraft, this factor would be more significant for vertical take off and landing aircraft, but still low.

To summarise, the 'downforce of rain on aircraft' seems to be negligible.

  • 2
    $\begingroup$ Sorry, what are the units for all those numbers? Between the m and mm and the $^2$, I'm not sure that a factor of a thousand or more isn't being dropped somewhere. Not likely, I'm just lost in the units. $\endgroup$ Commented Oct 3, 2021 at 1:43
  • 1
    $\begingroup$ @ Camille Goudeseune Hi, all units are SI units. The $0.021 \times 10^{-3}$ is the 0.021mm/s from line 5 of the question, (similar for the 0.1mm), the 1000 is for 1000 kg per cubic meter for the density of water and the 9 is for 9m/s from line 12 (the calculation is done for each square meter) $\endgroup$ Commented Oct 3, 2021 at 7:57
  • $\begingroup$ Thanks, that helps! Then the equation means Ns = m/s * kg/m^3 * m/s, or kg m/s = kg/(ms^2). So I'm still missing something here. $\endgroup$ Commented Oct 3, 2021 at 16:00
  • 1
    $\begingroup$ @ Camille Goudeseune the force is per square meter. It's N/m^2 = m/s * kg/m^3*m/s $\endgroup$ Commented Oct 3, 2021 at 16:07
  • $\begingroup$ @ Camille Goudeseune or if you prefer Ns/m^2 = m * kg/m^3*m/s $\endgroup$ Commented Oct 3, 2021 at 16:14

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