What is the relation between an airplane's altitude and the drag it is experiencing?

The Reynolds-Number $Re$ is defined as $Re = \frac{c \cdot L \cdot \rho}{\mu} = \frac{c \cdot L}{\nu}$, with

• the velocity $c~\left[ \frac{m}{s} \right]$,
• the reference length $L~\left[ m \right]$,
• the density $\rho~\left[ \frac{kg}{m^3} \right]$
• $\rho = \frac{p}{R \cdot T}$ for ideal gases
• pressure $p~\left[ Pa \right] = \left[ \frac{kg}{m \cdot s^2} \right]$
• temperature $T~\left[ K \right]$
• ideal gas constant $R~\left[ \frac{J}{kg \cdot K} \right] = \left[ \frac{m^2}{K \cdot s^2} \right]$
• the dynamic viscosity $\mu~\left[ \frac{kg}{m \cdot s} \right]$ and
• the kinematic viscosity $\nu~\left[ \frac{m^2}{s} \right]$, $\nu = \frac{\mu}{\rho}$.

From my understanding, drag increases with decreasing Reynolds-Numbers. Hence, drag increases with an increasing kinematic viscosity (see e.g. this book):

• Relation between altitude and kinematic viscosity according to the ISA (International Standard Atmosphere)

• With an increasing altitude, the density of the air decreases.

• The dynamic viscosity decreases with an increasing altitude of up to $11'000~m$, then stays constant to $25'000~m$ and increases from an altitude of more than $25'000~m$.
This is based on Sutherland's Formula for ideal gases, which in turn relies on the air temperature.
According to ISA, air temperature decreases with an increasing altitude of up to $11'000~m$, then stays constant to $25'000~m$ and increases from an altitude of more than $25'000~m$.

• Dividing $\mu$ by $\rho$, one can see that the kinematic viscosity increases with an increasing altitude.
See for example here or here for precise data.

• In summary, the Reynolds-Number decreases with an increasing altitude, which means that drag increases with an increasing altitude - assuming velocity and reference length are constant.

Does an airplane really experience more drag, the higher its flying altitude is, assuming otherwise constant parameters?

When researching for this question, one often stumbles across the statement that drag decreases with increasing altitude due to the decreasing density. However, no one seems to take the kinematic density into account.
Are there trustworthy charts available indicating drag over altitude?

• Note the difference between drag and the drag coefficiënt: the total drag is a function of $\rho\cdot c_d\cdot ...$, so the division of $\mu$ by $\rho$ gets cancelled when you calculate the actual drag. – Sanchises Jan 25 '16 at 21:15
• There are multiple types of drag experienced by an aircraft. You seem to be referring to skin friction drag. Sum total drag typically decreases with altitude, though individual types of drag, including skin friction and induced drag often increase, depending on the circumstances. I think that is where the confusion expressed in your last paragraph comes from. – J Walters Jan 26 '16 at 1:25
• @JonathanWalters: "Sum total drag typically decreases with altitude" This is only true if the lift coefficient goes up due to lower engine power with altitude. But changing the polar point makes all comparisons invalid. No, drag will go up with altitude if you compare apples with apples. – Peter Kämpf Aug 1 '16 at 20:46

Yes, it does - skin friction increases with an increasing altitude.

The mechanism is linked more to temperature than to density, but the reasoning in your question is correct. There might be a few cases where viscous drag increases with Reynolds number (like laminar airfoils which lose their laminar bucket when the Reynolds number goes up), but in general the observation is true.

First a plot from Sighard Hoerners book Fluid Dynamic Drag:

Flow properties over altitude, from page 1-11 in Fluid Dynamic Drag, 1965 edition. The ratio of the actual Reynolds number to the Reynolds number at sea level $\frac{R}{R_0}$ shows a clear downward trend over altitude (which is plotted on the x axis). At 60,000 ft the Reynolds number at the same flow speed is only 12% of that on the ground. If you look at the same dynamic pressure, you have to correct that by the square root of the density ratio $\frac{\rho}{\rho_0}$, which is 0.3 at 60,000 ft. In total, the aircraft will fly in 60.000 ft at 40% of the Reynolds number at sea level when the dynamic pressure stays constant. Lift is proportional to dynamic pressure, therefore it is best to keep the dynamic pressure constant for the comparison.

Now for the viscous drag over Reynolds number: From the same book I copied a plot of lots of experimental data which shows the trend nicely:

Skin friction over Reynolds number, from page 2-6 in Fluid Dynamic Drag. Note that both axes are logarithmic to produce a nearly linear trend line. k denotes a correction for a flow which starts laminar but then turns turbulent when the critical Reynolds number is reached (critical = kritisch in German, hence the k).

EDIT:

Your comments helped me to understand the source of your doubts. Drag and lift are proportional to dynamic pressure $q$, and this in turn is the product of speed $v$ squared and density $\rho$: $$q = \frac{v^2}{2}\cdot\rho$$ Reduce density, and you reduce drag. But this will also reduce lift. In the end, you want to support the same aircraft mass at lower density, so you need to do something to get lift back to its old level despite the drop in density. You do this by increasing speed. You restore the same dynamic pressure, in other words. Now the drag is also back to its old level, plus some more. Isentropic expansion means that temperature drops when density drops, and the thermodynamic law of isentropic expansion describes what happens to air in the atmosphere (isentropic = no change in entropy). Lower temperature means higher viscosity. Moving at the same dynamic pressure in lower temperature will generally cause more viscous drag.

• – user12485 Jan 25 '16 at 23:41
• @Thesis That article is probably simplifying things and using the term frictional drag to refer to all non-induced drag. – J Walters Jan 26 '16 at 2:28
• @Thesis: No, but the wrong reference. Flying at the same speed in thinner air does reduce drag, but creating the same lift in thinner (more precise: colder) air will cause more viscous drag. And in the end you want to keep the same aircraft airborne, thus create the same lift. – Peter Kämpf Jan 26 '16 at 7:21
• Peter's answer is of course very thorough, but just wanted to point out that for a 'fuel economy' comparison, you should look at the total energy to cover a distance and not just the momentary force as a comparison. Basically, the force goes up, yes, but you need it for less time cause you're going faster if you keep the same dynamic pressure. I did not do any math here, but while user12485's statement also takes this into account (constant velocity), Peter's does not. Hence Peter: it would be nice if you included that. – Arnout May 26 '17 at 23:40
• @Arnout: Oh, come on. The question is about drag, so I answered it accordingly. We have plenty of questions and answers about efficiency and transport performance around here. If I would expand every answer this way, they would be even longer. – Peter Kämpf May 28 '17 at 21:34