# Formula for comparing drag on planes flying at different speeds at different altitudes. Ex: Mach .80 at 40,000 ft, vs Mach 5 at 100,000 ft

I am interested in getting a sense of how much extra energy is needed to overcome drag at different speeds and altitudes. For example, for me a baseline would the drag on an airplane flying something like Mach .80 at around 40,000 ft. Then I would be interested to understand, that same exact airplane, how much energy would be required if it was more like a Concorde and flying at Mach 2 at 55,000 ft, or when we read about these sci fi hypersonic transports that Boeing and Airbus theoretically research, how much energy is required to overcome something like Mach 5 at 100,000 ft?

I found a table that gives various measures for densities and pressures at various altitudes (https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html)

I can express the speeds I am interested in as Mach, knots, km/h, etc.

Is there a simplified formula to be able to make these sorts of calculations?

No, especially not for the conditions you stated.

Even if we consider that "same exact ariplane" will not melt away at Mach 2 or 5, different phenomena start to appear at such speeds.

Let's restrict speeds to below about Mach 0.5 for now. At such speeds, compressibility of air does not affect characteristics significantly, and things become easier. Still, drag will depend on how exactly the airplane is built.

Most aerodynamic forces at such speeds, including drag, are proportional to dynamic pressure:

$$q = \frac{\rho V^2}{2}$$

That is, linearly proportional to air density and to square of speed.

Now, drag consists of two parts: one related to lift, called induced drag, and everything else, often called parasitic drag. This parasitic drag, at such low speeds, will indeed change proportional to $$q$$,1 and can be calculated easily using the standard atmosphere table you found.

But not so with induced drag. Your lift will also change with $$q$$, and if you want to fly level, you don't want that. So you aim to have the same lift. You do it by adjusting the angle of attack. But by doing that, you change the "aerodynamic efficiency" of the airplane: that is, you change induced drag disproportionally. Normally, there is one "best" angle of attack, and if you move away from it, you'll have more induced drag even for the same lift. How much more? This entirely depends on how the wing and the entire airplane is designed.

You can find conditions where $$q$$ will be the same: you can fly higher (=> lower density) and faster, and come up with the same dynamic pressure. Pilots will see it as the same indicated airspeed (IAS). In such conditions, you will fly at the same angle of attack, and drag will be the same.2

So, in general, some of the drag will change predictably in a more or less universal manner, and some will depend on the airplane. And this is for the trivial case of low speeds.

Now enter compressibility (and other) effects, and things get much more complicated. Even the parasitic drag will change non-linearly; it will acquire a component called wave drag, which depends not on $$q$$ but on the Mach number - and on the actual shape of the airplane. Some airplanes will not see much up to M0.8 or even more, some will suffer at 0.6. The effect is most complicated at transonic speeds; at high supersonic speeds, shape starts to matter less. Furthermore, the way lift is generated also changes, which may produce extra drag (see e.g. trim drag).

So, your best bet is to get approximate data for some "typical" airplane. Concorde is a good candidate: it has some published data regarding its efficiency at sub- and supersonic speeds.

1 To the first level of approximation: there are still some fundamental effects that change, but we are doing back-of-an-envelope calculations.

2 Note though that on a practical level it will still affect the "extra energy needed". First, this energy comes from an engine, and its efficiency changes not only with $$q$$. Normally, it becomes more efficient in colder air (i.e. higher). Second, you'll reach the destination faster when flying higher at the same $$q$$ (IAS) and thus at about the same power setting (yet, on the other hand, having spent some energy climbing).

• Thanks great answer. I am still left hungry to understand exactly how much more energy per seat mile is required for a typical subsonic airliner vs. Concorde/Boom Overture vs. Boeing/Airbus Mach 5+ hypersonic concepts. I know from experience with more common existing airplanes that there are circumstances where you are going to fly your CRJ900 at 450 kn at lower seat-mile fuel flow than your Q400 at 315 kn. So going higher is definitely helpful. Jun 14 at 19:49