Taking off on a cold day (5°C), it can take 10 seconds to reach rotation speed but on a hot day (40°C) degrees it will take longer to reach the same speed.

Is this due to the dynamic pressure? How does temperature affect it?

  • $\begingroup$ Huh? Dynamic what? $\endgroup$ Jul 14 '19 at 19:30
  • $\begingroup$ Are you asking about how temperature affects ram air pressure and static air pressure (and as a result indicated airspeed) or are you asking about how engine performance changes due to temperature? $\endgroup$
    – Bianfable
    Jul 14 '19 at 20:09
  • $\begingroup$ how temperature affects ram air pressure and static air pressure. $\endgroup$ Jul 14 '19 at 20:33

Yes, temperature affects dynamic pressure by affecting air density: it reduces with increasing temperature.

Dynamic pressure q $ = ½ \cdot \rho \cdot V^2 $

The molar form of the ideal gas law: pressure p = $\rho \cdot R \cdot T$, with R = gas constant and T in degree K

$$\rho = \frac{p}{R \cdot T} \tag{1}$$

Eliminate $\rho$ from dynamic pressure:

$$q = \frac{p}{2R \cdot T} \cdot V^2 \tag{2}$$

From (2) we can see that if temperature increases, dynamic pressure decreases linearly. In a deleted answer @xxavier mentions correctly that Indicated Air Speed is a measure of dynamic pressure, so let’s compare the two situations. 5°C = 292K, 40°C = 327K, TO true airspeed on a cold day = $V_c$, on a hot day = $V_h$

$$ \frac{p}{2R \cdot 292} \cdot V_c^2 = \frac{p}{2R \cdot 327} \cdot V_h^2$$

$$ V_h^2 = \frac{327}{292} \cdot V_c^2 => V_h = \sqrt{\frac{327}{292}} \cdot V_c $$

$$ V_h = 1.058 V_c \tag{3}$$

So to reach the same IAS, TAS must be 6% higher. But there are more ways that dynamic pressure affects time to reach rotation IAS:

  • The propeller thrust is proportional to air density: simple momentum theory models thrust as $T = C_T \cdot ½ \rho A \cdot (\Omega R)^2$. Again density $\rho$ which decreases linearly with temperature according to (1). Propeller thrust reduces with factor (292/327), integrated over time until $V_h$ is reached. 11% less thrust to reach a 6% higher velocity.
  • The engine has less dense air to operate with. From this site:

    When your engine is not equipped with a turbo- or supercharger it will also suffer from the less dense air. Each intake stroke (which is by volume) will contain less air molecules and thus less power can be developed by the engine (due to the fixed fuel / air ratio).

So yes, dynamic pressure plays a role, but the real underlying cause is the less dense air everywhere. Also in the dynamic pressure.


  • All above comparisons at equal static pressure $p$.
  • $V_c$ and $V_h$ are actual true airspeeds, not indicated speeds.
  • $\begingroup$ I suggest a change, instead of inserting the values right away and then solving for vh just use the variable names and solve that equation. Then you get a formula v_hot = v_cold * sqrt( T_hot / T_cold ) and you can any values that you like, and it doesn't just the values given in the example $\endgroup$
    – Jan
    Jul 18 '19 at 12:25
  • $\begingroup$ A nitpick, "11% less thrust to reach a 6% higher velocity". 1.06 squared is around 1.12. This is the increase in drag going 6% faster. But it isn't doing that either (IAS). The Density Altitude (PV=nRT) is actual molecules per cubic foot that the prop bites and the plane moves through. True airspeed will be higher "high and hot", and with less thrust, it will take longer to get to IAS needed. Up to the point where not enough thrust is produced (too high and/or hot). $\endgroup$ Jul 18 '19 at 18:53
  • $\begingroup$ @Jan Yes I know, that is how we are taught at school. I chose not to do that here, for readability purpose. $\endgroup$
    – Koyovis
    Jul 19 '19 at 0:54
  • $\begingroup$ @RobertDiGiovanni 232/327 = 0.893 $\endgroup$
    – Koyovis
    Jul 19 '19 at 0:56
  • $\begingroup$ @RobertDiGiovanni Please read the answer again. It does not feature a state of the art pressure container. It does state that IAS needs to be reached in order to fly, and that TAS is 1.058 higher in the hot day case. And there is 11% less available thrust to reach the same IAS (higher TAS), with an equation to prove it. For the form of ideal gas law that contains density, please scroll down in this link, the fourth equation under Molar Form. $\endgroup$
    – Koyovis
    Jul 19 '19 at 9:16

Sure, when it's hot the indicated airspeed is lower than normal for a given true airspeed, due to reduced air density, and air density is indeed a factor in dynamic pressure. The relationship between air temperature and indicated airspeed is same as the relationship between air temperature and dynamic pressure, and both are due to the relationship between air temperature and air density.

Of course, power and thrust are also less on a hot day. That has to do with air density too, but not in a way that directly involves dynamic pressure.


Dynamic pressure q = $\frac{1}{2}\rho \cdot u^2$, with $\rho$ is density, u is flow speed. It is a measure of aerodynamic pressure at a given speed and is expressed in pascals or psi.

This value is the origin of the term "Max Q", which is the maximum aerodynamic stress a rocket experiences when launched and is accelerating through the atmosphere.

Since it has a density term, it is part of your answer, but the solution is not usually described with the term Dynamic Pressure, it is described as Density Altitude.

Density Altitude is Pressure altitude corrected for temperature. So you get your pressure altitude by setting your altimeter to 29.92 and reading the altitude.

Then Density Altitude is calculated as P + (120x(OAT-standard temperature))

Standard temperature is 15 Celsius at sea level, BUT it decreases 2 C with every 1000 feet of altitude.

So now pressure and temperature on a 40 C day make you "high and hot". What this does is give your prop less air to "bite" at max rpm, resulting in less thrust.

With less thrust, your plane will take longer to reach its rotation speed due to slower acceleration ( a = f/m ), and because it will have to accelerate to a higher speed to get enough IAS to lift off.

  • $\begingroup$ As with the conveyer belt question, ground speed at takeoff (wheel rotation) will be higher, so good tires and well greased bearings are in order. $\endgroup$ Jul 14 '19 at 23:02
  • $\begingroup$ Have added some MathJax, you can roll back the edit if you don’t like it. $\endgroup$
    – Koyovis
    Jul 15 '19 at 2:14

"where is this temperature"(ground, air, water, space), what is the medium(dirt vs asphalt vs rock vs mountain vs altitude vs clouds vs sun vs rain vs tides vs waves vs saltwater vs freshwater etc etc" and by "dynamic" what is the driving force?(propeller, impeller, turbo jet, ram jet, lifting body, wing loaded lofting body, rocket, hydrofoil, etc etc). Yes of course higher temperature equals greater pressure in a vacuum but there are "inversions" in physical reality where in fact pressure will drop as temperature increases and vice versa. Since a presume by "dynamic pressure" the question pertains to an object with a given forward velocity of some specified medium many right answers would suffice provided there is actual experience and pertinent physical data observed, collected, houses, studied etc none of which is presented here.

"Start with an instrument that can measure pressure" by way of example and go from there.

  • 1
    $\begingroup$ I don't think you know what dynamic pressure means... $\endgroup$
    – JZYL
    Jul 21 '19 at 6:49

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