When we lick a finger and put it up in the air, the side that is cooled is the one facing the direction from which the wind is blowing. But consider this, if we take a cross section of the finger, ie look at it from above, we see the wind blowing at our finger and our finger can be considered to be a cylinder.

From examples like vortex street forming around a cylindrical object, we know that a low pressure exists behind the object. Lower pressure would normally help saliva evaporate faster and thus from this point of view one would expect that the side of the finger away from the wind would evaporate faster.

What is the explanation to this?


First, the reason why the finger becomes more wind-sensitive with some saliva isn't that the saliva evaporates but because the saliva, or water, is a good thermal conductor. The finger has to be warmer than the air so the heat flows from the finger to the air and a good thermal conductor such as saliva helps this flux to take place.

Second, because it's the temperature and not evaporation that matters, we must care about the temperature of the air because this is what is actually cooling the finger. Without any wind, the human body – which is warmer than the air – creates a thin layer of warmed air in the very vicinity of the skin which acts as a thermal insulator and slows down the cooling of the human body.

But the wind disrupts this insulating layer on the front side of the finger and the cooler air gets directly in contact with the finger (or the with saliva on the finger). The air on the opposite side of the finger is either stuck in vortices – that may keep on recycling some warmer air (previously heated by the finger), or if there are no vortices, the air on the opposite side of the finger is still warmer because it was heated up by the finger "recently" when the air was flowing around the finger.

  • $\begingroup$ Regarding your first paragraph, I don't believe that placing a good conductor between two objects will cause a greater heat flux than direct contact between the two objects. The intervening conductor will introduce some resistance to heat flow that wasn't present otherwise. $\endgroup$ – James Jul 8 '16 at 20:02
  • $\begingroup$ Not evaporation? Doesn't it work in warm air? What's your reference? $\endgroup$ – Mike Dunlavey Jul 8 '16 at 23:09
  • $\begingroup$ I think the thermal sensitivity disappears in the air of the temperature basically equal to the body temperature. Such a temperature of no sensitivity would exist even if the evaporation played a role. There would always be a temperature at which the warming of the finger by the surrounding air would cancel the cooling of the finger by the evaporation. The only question would be what is the temperature - equivalently, what's the ratio of the importance of the cooling by contact and cooling by evaporation. ... I don't have a reference. The very purpose of this page is to become one. $\endgroup$ – Luboš Motl Jul 9 '16 at 6:06
  • $\begingroup$ Above this temperature, the effect gets reversed. If you have a hot air flowing from an oven, the front side of the finger will surely feel warmer. $\endgroup$ – Luboš Motl Jul 9 '16 at 6:10
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    $\begingroup$ @LubošMotl: I have a lot of respect for your good answers, but I think you're mistaken on this one. People whose body temperature is 37C live (and generate body heat) in 40C places where fortunately the humidity is low. They shed the excess calories by evaporation (unless I'm badly mistaken :) $\endgroup$ – Mike Dunlavey Jul 12 '16 at 12:02

The percentage by which the pressure lowers on the leeward side of your finger is minuscule, so it barely affects the rate of evaporation.

  • $\begingroup$ But on the upwind side there is a thin boundary layer and air flow over that, carrying away evaporated H2O, much more than on the wind-shadow side, no? (I was in M.E. at MIT. Pp 420 in Rohsenow and Choi, Heat, Mass, and Momentum Transfer discusses wet-bulb thermometers. In particular there is a mass-transfer convection term $h_D$. Also pp 384 talks about boundary layers in convection. This doesn't discuss upwind/downwind issues, but I think basic aerodynamics does that.) $\endgroup$ – Mike Dunlavey Jul 12 '16 at 12:43
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    $\begingroup$ I assume there's higher wind flow, and thus more evaporation, on the upwind side. I don't know enough fluid dynamics to support this, though. The web bulb/dry bulb thermometer method of determining humidity shows that evaporation can lower temperatures by several degrees C. $\endgroup$ – Peter Shor Jul 12 '16 at 13:09
  • $\begingroup$ But by that logic even if the pressure drop is miniscule, there is increased pressure on the front side; this can cause increased evaporation on the backside. $\endgroup$ – Hazkaz Jul 13 '16 at 16:09
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    $\begingroup$ Right. There's the increased evaporation from lower pressure on the back side, and there's the increased evaporation from higher wind speed on the front side. One can easily find resources which confirm both of these effects. You might have to search the literature and find numerical values to figure out which effect is larger. (Or you could just determine this experimentally by holding up a wet finger on a windy day.) $\endgroup$ – Peter Shor Jul 13 '16 at 17:18

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