I'm a little familiar with the physics and thermodynamics of the wind chill effect, but this question seems to come up from time to time:

Why, given two temperature sensors or thermometers in the same environment, do both report the same temperature if one is exposed to wind when the other is shielded from it?

People often ask because the temperature reported by, for example, their vehicle, doesn't seem to change as they drive at different speeds, etc. (Other than of course the changes from one actual temperature to another as they change geography.)

My understanding is that inert devices aren't endothermic like we are, so the effects of wind chill don't apply.

Can you explain this in layman's terms?

  • 1
    $\begingroup$ Why do you think the temperature of the wind is different from that of still air? $\endgroup$
    – Anixx
    Feb 11, 2013 at 2:37
  • 1
    $\begingroup$ @Anixx: Go stand outside in -10°C weather and tell me if you feel colder with a stiff breeze or not. $\endgroup$
    – JYelton
    Feb 11, 2013 at 3:11

3 Answers 3


It’s really pretty simple: The thermometer measures temperature, wind chill measures heat loss for a body warmer than the air. Wind makes more unheated air available to conduct heat away from a hot body, but with a body at air temperature no heat is being conducted away from the thermometer.

Say you asked a secondary question, “When I put a thermometer outside, how long must I wait until it has reached air temperature so that its reading is meaningful?” In that case, the windy conditions will decrease the time it takes to equilibrate.


The thermometer in wind will (at equilibrium) show a lower temperature than one in still air, if and only if it is a wet-bulb thermometer.

The wind's cooling effect comes from the wind evaporating moisture away. Your skin is moist, so a wind evaporates some of that moisture, taking away (on average) the higher-energy water, lowering the average energy of the remaining moisture - hence the cooling effect of sweating.

A dry-bulb thermometer has no moisture to evaporate away, and so records no wind-chill effect.

A wet-bulb thermometer (e.g. a normal mercury thermometer with its bulb wrapped in waterlogged muslin) will record a wind-chill effect.

  • $\begingroup$ +1 for being the first to mention the effect of evaporation on wind chill. $\endgroup$ Jun 17, 2011 at 12:47
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    $\begingroup$ +1 There's a chart relating dry-bulb, wet-bulb, and dew-point temperatures. $\endgroup$ Jan 28, 2012 at 14:56
  • $\begingroup$ what about decrease of temperature due to Venturi effect around the thermometer? $\endgroup$
    – ceillac
    Sep 27, 2015 at 8:38

I maintain that this isn't an absolute fact but is probably true under the outside conditions you are talking about. We should all agree that wind blowing does, in fact, increase the thermal coupling between the thermometer and the air. However, the device has no heat production. If there is no heat production, it doesn't matter how strong or weak the thermal connection between two isolated items is because they will eventually equilibrate to the same temperature.

A major caveat to that is that fact that the device and the air are not in an isolated system. So what else is there? Well, there are heat potential sources/sinks that exist at different temperatures. These include the night sky for instance, the sun, and the ground. There exists radiative heat transfer between all of these and they all constitute a different temperature than the air. To summarize in an equation, there is no heat production, $\dot{Q}$, but there is a heat flow, denoted by $\dot{q}_c(..)$ for conductive and $\dot{q}_r(...)$ for radiative, from everything that can thermally interact.

$$\dot{Q} = 0 = \dot{q}_c(T_\text{thermometer},T_\text{air})+\dot{q}_r(T_\text{thermometer},T_\text{sky})+\dot{q}_r(T_{\text{thermometer}},T_\text{ground})$$

And basically $q_c$ dominates, which could be written with Newton's law of cooling, proportional to $\Delta T=T_\text{thermometer},T_\text{air}$, which means that $T_{\text{thermometer}}=T_\text{air}$ ultimately.


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