# How can flying high minimize drag and increase fuel efficiency? [duplicate]

As a plane goes higher, the engines have to work harder to compensate for the air density, therefore it will require more fuel in order to provide the same power at lower altitudes.

But I always hear that flying high means less fuel burnt.

How can that be the case?

• Less air molecules means less drag... – Ron Beyer Oct 16 '18 at 0:25
• On what basis do the engines have to "work harder"? – Greg Hewgill Oct 16 '18 at 0:39
• Flying at high altitude produces the same amount of drag, because you must fly faster in order to get enough lift power. BUT, it allows you to cover more ground miles with the same power setting in given amount of time, compared to what would you cover flying at low levels. Why? Because the air density is lower at altitude. There's also a significant temperature benefit for the engines. They run more efficient in low temperatures. – Electric Pilot Oct 16 '18 at 0:59
• Related, maybe even a dupe? – Pondlife Oct 16 '18 at 1:03
• This question doesn't make sense unless you specify what kind of engine you are talking about -- reciprocating, turbofan or turboprop. – Juan Jimenez Oct 16 '18 at 9:57

Flying higher means less drag because the air is thinner; therefore, you can fly faster at altitude and hence travel farther on less fuel.

However, flying higher also means less oxygen available to burn your fuel, so available horsepower decreases with altitude.

There is a special altitude at which these two effects (drag reduction and available power reduction) achieve a crossover point. This is called the critical altitude and it is there where you achieve optimum cruise and best economy.

If you want to fly higher than critical altitude, you will fly slower because the power loss is greater than the drag reduction and you will spend more time in the air.

• You don't necessarily fly slower above that altitude, you just fly more inefficiently. The issue is that the stall speed increases at the same point you start approaching critical mach, so the "window" between these two speeds gets smaller, which is known as the coffin corner. – Ron Beyer Oct 16 '18 at 2:12
• Got it, thanks. Is it true that above critical altitude, to maintain altitude constant requires progressively larger amounts of pitch-up trim? if so, is the inefficiency issue related to increased induced drag? – niels nielsen Oct 16 '18 at 5:47
• In fact the aircraft is designed to fly at that sweet spot and ideally it would only fly at that altitude and speed. The design problem is to get the aircraft to fly for takeoff and landing speeds. Hence flaps. – mckenzm Oct 16 '18 at 6:14
• "so available horsepower decreases with altitude" I would stress this part - because if you want <some power> and can obtain that up to <some height> ... then you will save fuel all the way up to that height with limited tradeoffs to power. Above that - you start trading power for drag. – UKMonkey Oct 16 '18 at 15:20
• If you read other answers, it appears that the thrust reduction is because of the reduction of available mass rather than o2 – Antzi Oct 16 '18 at 15:55

Adding to the answer by Niels Nielsen, the special altitude changes (increases) during the flight as the power required reduces as the weight of the aircraft decreases due to fuel consumption. But these effects also depend on the type of propulsion (jet vs propeller). Basically, for propeller aircraft the plot should be power required and for jet aircraft it should be drag. Now, you draw a line tangent to the curve going through the origin. The point where the two lines meet is the optimal velocity for maximum range. Power required will shift down and to the left for decreasing weight and up and to the right for increasing altitude. The drag curve will approximately translate to the right for increasing altitude.

The approximate analytical equation for range is given by the Breguet equation:

$$R_{prop}=\frac{\eta_j}{c_P}\frac{C_L}{C_D}ln\frac{W_1}{W_2}$$

$$R_{jet}=\frac{V}{c_T}\frac{C_L}{C_D}ln\frac{W_1}{W_2}$$

$$\eta_j$$ is the propeller efficiency and $$c_P$$ and $$c_T$$ are the specific fuel consumptions, which remain approximately constant regardless of airspeed and altitude changes (relative to cruise altitude). But it does decrease with altitude. Propellers are also less effective at high altitude and also have a sweet spot similar to jet engines. Moreover, the required lift should be equal to the weight for a given time instant, which is given by: $$L = W = C_L\frac{1}{2}\rho V^2 S$$ where $$\rho$$ is the air density and S is the wing surface area.

Now to maintain the condition for maximum range and fly at a constant $$\frac{C_L}{C_D}$$ either the air density must decrease or the velocity must decrease. Decreasing the velocity is not preferred since the aircraft will fly slower and slower meaning longer flights. This flight profile is often referred to as cruise-climb flight.

A great book that treats a lot of these types of aircraft performance considerations is:

Elements of airplane performance by Ger J.J. Ruijgrok

• I see you cannot comment but if you improve your answer to show how maximize efficience for each type of engine I'm sure you wiull get enough upvotes to allow you to comment in the future – jean Oct 16 '18 at 10:56
• Adding on Ralph's comment: answers here should be self-contained. Adding and explaining one of these power required/available plots to your answer could achieve this. – Sanchises Oct 16 '18 at 14:51

This is a good question but first we must clarify something.

There is a major difference between propeller driven aircraft and pass through turbine propelled aircraft. Without getting to technical in this explanation, let’s leave propeller driven aircraft alone for the moment and keep our answer to the confines of high bypass ratio turbine driven engines.

The long definitive answer involves a great deal of physics, another subject we will not get to rhetorical about at this time. We will however consider Bernoulli’s principles concerning fluids and the airfoil or wing. This brings us to our first thought; consider the atmosphere, or air, around us as a fluid. Now that was simple enough.

When an airplane takes flight it must defeat its greatest foe, Newton’s contribution to this question, gravity. In its most simple terms, the ability to defeat gravity requires an opposing force that is greater than the weight of the mass being lifted. In aviation this comes from the lift created, of course, by the wings.

To help make this possible on airplanes the wings are shaped with a leading edge, that is the part that first hits the air, then a highly rounded upper surface, known as camber, and much flatter lower surface which come together at the trailing edge. (Just a moment more, I’m getting to your answer.) Now we apply Bernoulli’s principal.

Looking at air as a fluid, in principle Bernoulli’s theory is this. The air traveling over the top of the wing is moving at a faster speed than the air traveling under the wing. In theory, this would cause greater pressure under the wing (lift) and less pressure over the wing as the air is rushing from the leading edge to the trailing edge. The reasoning here is that when the air molecules are separated by the leading edge of the wing, the molecules passing over the top, which has a longer surface due to the added camber, must move faster in order to reach the trailing edge at the same moment as the air passing below the wing. Now we get to the part concerning thicker (heavier) air mass and thinner (lighter) air mass and the effects on “lift”.

In order to for the wings to produce the lift needed to defeat gravity you must have thrust and lots of it! Now we enter into the power (engines) production that is required to create flight. As we all know, an airplane must be moving into the air mass in order to obtain the effects required to produce flight also known as the lighter than air effect. So in principle, the shortened version, the air passing over and under the wings must produce a lift equivalency greater than the weight of the aircraft. There is a formula but we’ll skip that here and just state this, the airmass flowing around the wings must be equal to or greater than the mass of the airplane. There, we have flight... well sort of.

You see the weight of air at sea level is much greater than the air at say 30,000 feet. Here again, in principle, the amount of air flowing across the wing on takeoff, by mass, is more at mean sea level than it would be at 5,000 feet. Since this is a principle of mass verses mass (in this case in order to defeat the effects of gravity), the fluid principles of Bernoulli’s would apply well here. So what are the effects on engine performance?

Modern power plants today are engineered to direct less airflow through the engine or turbine section of the power plant. The ratios have changed dramatically in the last decade with most engines diverting 80% or more of the air flow around the core and exiting as additional thrust at the tail. These engines, as one could imagine, consume extreme amounts of air as it flows through and around the core, or power unit, and exits the tail as super heated energy and thrust. To maintain this thrust the amount of air mass must remain fairly constant. So, how is this accomplished at 38,000 feet? And what about the thinner air passing over the airfoils?

Well, here is where physics and the dynamics of air mass come into play but in a much simpler way than you might expect. Remember the role that the wing plays in all of this? Air pressures above and below the wings can still be easily maintained at higher altitudes. Why? Because of the much greater amount of air passing over the wing surfaces at speeds often over 500 mph. You see, it’s the air molecules that hold the secret to air mass and density. At speeds like this the air density is super compressed to allow far more lift than needed to maintain flight. Then there is the engine question.

Since air is composed of gasses (mainly nitrogen with about 20% oxygen and other gasses mixed in) at the molecular level, they too can be “squeezed” under tremendous pressure and therefore compressed to create an even mixture through a jet engine. This ability to compress huge amounts of air while flying at high altitudes again makes up for loss of mass and density. Since the engine is moving between 500 and 600 mph, the intake or ingesting of plenty of air through the core and surround is no problem.

The added bonus is this; thin air creates less friction on the aircraft and much cooler air temperatures (normally around -60 degrees) as the air flows through the engine. There is also a slight reduction in fuel cost flying at these higher altitudes as we normally reduce our throttles 10 to 20 percent while at cruising altitude. This of course is dependent on the aircraft type and any head or tail wind!

• the molecules passing over the top, which has a longer surface due to the added camber, must move faster in order to reach the trailing edge at the same moment as the air passing below the wing. This is the classic "equal transit" fallacy. Please see NASA website for further explanation. – TomMcW Oct 16 '18 at 18:03
• Furthermore, "at speeds like this the air density is super compressed" - not true. In fact, in most flight regimes compressibility is almost negligible. Overall, most of these explanations (including about the engine, throttles, etc.) are not only wrong, but irrelevant. The actual answer is simpler: at altitude, to have the same required lift (exactly, and not "greater", as the weight) you must fly faster due to lower air density, while having about the same drag and thus thrust (depending on wing optimisation). Finally, you don't need engines to "create flight": just look at gliders! – Zeus Oct 17 '18 at 2:50
• First, “equal transit” and Bernoulli’s principle. Really? Let’s just change all the dynamics of fluid flow. Second, I just fly them. Big ones. I wouldn’t understand anything about flight yet my airline puts me at the controls of something I don’t understand. There is a factor of air compression while at high altitude flight. Without it we would starve for oxygen. Our throttles, though automated while on AT, are adjusted based on head and tail wind. Drag, weight, lift, thrust, cord, deflection, angle of attack, lift coefficient, etc. Are we in flight school? What gets the glider up there? – Joe Oct 17 '18 at 20:25
• Joe, please don't get offended. Pilots don't need to understand (and rarely do) all the details of aerodynamics. But "equal transit" is such a wrong explanation of Bernoulli principle that it's beyond the lie to children stage. The length of the top and bottom has nothing to do with lift. How would a flat plane create lift? As for compression, engines compress air for themselves (hence the need for turbine) and "we" get it from them. But aerodynamic compression is not a significant factor until about M>0.6, and depends only on M, not altitude. – Zeus Oct 17 '18 at 23:51
• Zeus, I started my career at Pan Am as a flight engineer on a B-727. That was over thirty years ago. We were required to have a very good understanding of flight. Not so today. Since that time NASA has “re-engineered” the theory of powered flight. “A”, “B” and “C” no longer apply. The four constants still remain; drag, weight, thrust and lift. As I extend my leading edges and flaps on the 777, I add 30% more surface to my wings to compensate for a greater attack angle at lower speeds. At cruising, the wings are engineered to reduce drag but maintain lift. Aerodynamic center is relatively new. – Joe Oct 19 '18 at 3:48