Let's say you want to perform a turn anywhere from 1 to 90 degrees. Which will cost less energy? A bank or a yaw?

Using the rudder does produce extra drag, because the control surface is deflected for a time.

Using ailerons to bank also produce extra drag, but more than that, at a roll angle the lift direction will not be perfectly vertical. You will either lose altitude (that needs to be regained), or increase speed.

That's why I didn't want to phrase the question as "less drag". Lost energy could be in the form of drag, worse lift angle, lost altitude, maybe other things I haven't thought of.

I'm interested in large commercial aircraft as well as "ideal" aircraft with no adverse-torques (all control surfaces are aligned with the center of mass).

  • 1
    $\begingroup$ I think the correct answer is to perform a coordinated turn using both rudder and ailerons, but I'm sure someone will answer with a more satisfying mathematic equation to explain why that is so. $\endgroup$ Aug 11 '19 at 2:02
  • $\begingroup$ Surely the least-energy turn is the one which doesn't make passengers spill their drinks, and ideally are unaware of the turn at all unless they're looking out the window. $\endgroup$
    – Criggie
    Aug 12 '19 at 0:01

When you turn by yawing you are skidding the airplane to point the nose to the side, to offset the trust vector to move the plane sideways as it's going forward. The rate of turn you can achieve this way is very low and there is a massive drag of being in a skid as you slither around like a car on ice with the fuselage side presented to the airstream. To the extent that the airplane has a roll/yaw couple (roll caused by yaw) you will have opposite aileron to keep the plane from banking into the skid; more wasted energy serving no purpose.

When you turn by banking you offset the lift vector to move the plane sideways as it moves forward. There is no sideslip so the drag is only increased by the small amount caused by the increased elevator and AOA, and the minor drag of the moderately displaced ailerons and rudder (to the extent they are displaced at all once the rolling action is complete).

Because you are moving through a fluid, a skidding turn is like turning a car by drifting it on a dirt track, a banked turn is like going around on a banked corner, where you don't even have to turn the steering wheel to keep aligned with the road as the car changes direction. Which way is more efficient?

To find out I tried it on my own plane, a homebuilt PL-2. I flew it at 2000 ft, holding 70kt, close to max L/D for this plane, and applied full rudder with a bit of opposite aileron to maintain wings level (not much aileron is required on this plane), adjusting power to hold altitude at 70kt in the skid.

My plane has a manifold pressure gauge even though the prop is fixed pitch, useful for knowing precise power settings. It took 19" MP @ 1900 RPM to hold a skidding turn with full rudder at 70 kt and the turn rate was about 90 deg in 30 sec, or about Rate 1. This is about 55 HP on the Lycoming O-290 D2.

I then took out the rudder and banked into the turn (about 15 degrees bank was required to maintain Rate 1) to maintain the same rate of turn as the skidding turn but in a coordinated bank. To keep from climbing I had to reduce power. I ended at at about 16" MP @ 1700 RPM, in a banked level Rate 1 turn at 70 kt, which is somewhere around 45 HP.

So it took somewhere around 20% less power to maintain a banked turn at Rate 1 than a skidding turn at Rate 1 when flying near max L/D. The extra drag of the skid was also quite obvious in the deceleration when applying the skid with rudder vs rolling into the coordinated turn.

With airliners, turning by skidding is out of the question because with swept wings you get a powerful roll rate as soon as you yaw a couple degrees and you will quickly run out of opposite aileron trying to keep the wings level.

  • 1
    $\begingroup$ A quantification would be good. $\endgroup$
    – Koyovis
    Aug 11 '19 at 3:57
  • 1
    $\begingroup$ Well that's a project in itself. "Way more drag" and "much crappier turn" is sufficient quantification for something like this, it being pretty obvious to anyone who flies. $\endgroup$
    – John K
    Aug 11 '19 at 4:13
  • 3
    $\begingroup$ I disagree. They sound like statements based on how it feels to the inner ear. Your banked car turn provides all the extra lift required at no extra power. $\endgroup$
    – Koyovis
    Aug 11 '19 at 4:18
  • $\begingroup$ Actually the car is under increased load from the G of its turn on a banked track, so tire compression and rolling friction is increased. Anyway, you would have to go out and do flight testing to determine turn rate X with side slip Y (and believe me even with full rudder it isn't much), measure the drag increase, and compare that with a banked turn of the same rate, in the same airplane, same conditions. I can tell you that the differences are obvious if you actually go do it (and I have played around with turns by skids in the past just for fun), so obvious that measurements are redundant. $\endgroup$
    – John K
    Aug 11 '19 at 14:52
  • $\begingroup$ +1 for the testing. $\endgroup$
    – Koyovis
    Aug 24 '19 at 15:40

A constant turn requires a constant centripetal force, i.e. perpendicular to the direction of motion. This force has to be generated aerodynamically (unless rocket engines are used).

An aerodynamically generated force is always accompanied by induced drag - even if the force is generated by the fuselage (due to sideslip) instead of a traditional airfoil. For minimal energy expenditure, you want to minimize the induced drag for a given aerodynamic force. In other words, you want to have a high lift/drag ratio (L/D).

For virtually all heavier-than-air aircraft, a large part is dedicated to fighting the force of gravity (either fixed wings or a rotor). The surfaces used for fighting gravity are thus invariably the surfaces with the highest L/D.

By rolling the aircraft, the lift generated by these surfaces is tilted inwards, and it gains a force component perpendicular to the direction of motion. Rolling thus uses the most efficient means of generating an aerodynamic force - the wings or a rotor - to generate the centripetal force.

It gets better, still. Even if you had a fuselage that somehow had the same L/D in yaw as your fixed wings, a roll is still more efficient. This is because a single tilted force always has a lower magnitude than the sum of the horizontal and vertical components per the triangle inequality.

Use yaw to coordinate your turns, and use roll to do the actual turning. In a coordinated turn, the most efficient speed may be somewhat higher because the balance between induced drag (reduces with higher speed) and parasitic drag (increases with higher speed) is shifted. Compare this with a glider, which has a higher speed for optimal L/D (optimal range) with increased ballast.

  • $\begingroup$ Note this is not a violation of "pitch controls speed, power controls altitude", it's more like extra AOA is more efficiently added to turn, power is added to compensate for additional drag while turning. Now we can examine "should we maintain optimal wing AOA and speed up in a roll and turn, or increase wing AOA to a lower L/D, but not speed up as much (maintain airspeed)"? Due to increase in parasitic drag, it may be most efficient to add only enough thrust to maintain airspeed, while using the wing and rudder in a coordinated turn. $\endgroup$ Aug 11 '19 at 14:03
  • $\begingroup$ The roll method is the most efficient. It requires only a small amount of extra lift from the wings to sustain level flight in the turn. Yawing puts the fuselage at an angle to the wind in order to provide "sideways lift", and produces much more drag in the process because the fuselage is not shaped to provide lift efficiently. Indeed "side slipping" is sometimes used as a way to slow down in an emergency. $\endgroup$
    – Chromatix
    Aug 12 '19 at 4:35
  • $\begingroup$ @Sanchises Never mind mate, I edited my answer, should have done that in the first place. $\endgroup$
    – Koyovis
    Aug 12 '19 at 13:31
  • $\begingroup$ @Koyovis No problem! $\endgroup$
    – Sanchises
    Aug 12 '19 at 14:33
  • 1
    $\begingroup$ @RobertDiGiovanni "Pitch controls speed, power controls altitude" is a teaching aid, not a rule in physics (basically, it is to make sure the student pilot adds power proactively instead of retroactively). I addressed the rest of the comment, regarding optimum speed, in my answer. $\endgroup$
    – Sanchises
    Aug 12 '19 at 14:53

Depends on the speed: it costs very little extra energy to yaw a hovering helicopter.

From an old uni book on stability & control

  1. Yawed turn. A yawed turn increases sideslip angle $\beta$, creating a sideslip velocity which results in an aerodynamic force $Y = C_Y \cdot ½ \rho V^2 S $. Picture above is from measurements of an F-27 model with no flap deflection $\delta_f$, and shows an almost linear relationship between $C_Y$ and sideslip angle. The lift vector stays pointing straight up in a purely yawed turn, no additional lift required.

  2. Banked turn. The lift vector points into the turn, meaning some of the vertical force is lost. Angle of attack needs to be increased for the extra lift, creating more induced drag.

That is about the force vectors, the question is about energy = force * distance or force * velocity * time, and that is where an exact workout becomes laborious. This data is available from simulator aerodynamic packages, but there are limitations on data reproduction.

Qualitatively speaking, the difference between the two is:

  • Drag due to sideslip in yawed turns is a parasitic drag, and increases quadratically with airspeed;
  • Induced drag reduces with airspeed.

So at low airspeeds, a yawed turn requires less additional power. At higher airspeeds, a banked turn requires less additional power. The crossover speed may be quit low, intuitively we would tend to believe that above stall speed a bank wins.

Of course, the above does not mention the comfort zone, our brains and bodies are way more comfortable experiencing normal forces than they are experiencing side forces.

Wiki page for induced drag

  • $\begingroup$ Angle of attack needs to be increased for the extra lift, creating more induced drag. I thought airspeed must increase in order to maintain enough lift, or else you sacrifice altitude. Even if AoA alone is enough, it would seem to require more engine power at the higher AoA. Either way seems to be adding energy from engines. Btw I always figured that pulling back on the stick during the bank is only to make the turn go faster. If you dont pull back, the bank will still accomplish the turn. $\endgroup$
    – DrZ214
    Aug 11 '19 at 7:41
  • $\begingroup$ If you fly trimmed, then bank without pulling back, the vertical force is lift * cos $\beta$, so diminishes with bank angle. Vertical force is less than weight and the aeroplane descends. Pulling back on the stick increases the lift vector, and no more altitude is lost when L*cos$\beta$ = W. Indeed, the horizontal component is now larger as well and the turn is tighter. $\endgroup$
    – Koyovis
    Aug 11 '19 at 9:16
  • $\begingroup$ Are you really suggesting that when flying near, say, the speed for minimum sink rate or max climb rate, you can turn most efficiently by skidding rather than by banking? If this were true, it would be the preferred strategy of glider pilots circling in a thermal updraft. This is not the case at all in reality-- in fact there is an argument that a slight amount of slip is helpful while circling in a thermal updraft. $\endgroup$ Aug 11 '19 at 12:49
  • $\begingroup$ @quietflyer no I'm not setting the transition point near minimum sink rate or max climb rate, merely using the graph as an illustration of the different nature of parasitic and induced drag. $\endgroup$
    – Koyovis
    Aug 11 '19 at 12:53
  • 2
    $\begingroup$ @Koyovis You have to associate a given yaw angle with a turn rate, and compare that total drag in that condition to total drag at the same rate of turn in a banked turn and you'd have to go out and test it. The yaw angles required to get anything close to a decent turn rate are substantial. Next time I'm out in my plane I'll try to do that. I'll see what turn rate I can get with a flat skidding turn at a constant altitude and a constant 80 kt with full rudder, note the RPM/MP, and compare the power setting required to meet the same conditions in a banked turn at the same turn rate. $\endgroup$
    – John K
    Aug 12 '19 at 0:38

The best way to look at this IS to look at drag, that is what "costs you energy".

We have arms and legs. Why don't we walk around on our arms? It's theoretically possible! I could even write an equation for It! Let's try some common sense first.

Turning requires motion, as does heavier than air flight. There for the hovering case cannot be considered, or any variable airspeed. Let's look at flight at a given airspeed. In ALL cases lowest drag is most desirable.

Assuming no change in altitude or airspeed executing the maneuver, how do we do it with the lowest drag? Certainly not by turning the fuselage sideways into the airstream and using the engine to pull us in a new direction. Banking the aircraft uses the far more efficient horizontal component of the wing lift vector to create side ways motion. This is the "turn".

Coordinating the turn with rudder keeps the thrust vector and empennage "following" the new direction of flight, with the best use of thrust vector, and least amount of drag, as possible.

Breaking the turn down to many instantaneous steps can help clarify the thought. If you are worried about that huge wing creating drag by moving side ways, have a good laugh. At any given instant it is accelerating horizontally from V zero (very little drag) while maintaining the vertical lift component. This is why turning is considered acceleration, even though airspeed is constant.

Comparing the graceful slow wing to a furiously churning prop, few could argue for the latter.

  • 2
    $\begingroup$ -1 for dubious assumption. Assuming no change in altitude or airspeed executing the maneuver... This is impossible in a bank. If your lift vector is tilted sideways, you will sink or increase speed to increase the vertical lift component to match your weight. This is precisely why I phrased it "least energy" and not "least drag". In a bank, you could just sacrifice altitude. Your speed is the same so there's no extra drag from that, but you lose energy due to decreased altitude. $\endgroup$
    – DrZ214
    Aug 11 '19 at 7:46
  • $\begingroup$ Whoops, no I see what you're saying now. Increase AoA but at the same airspeed. This increases lift so altitude can stay the same too. And ofc higher AoA is higher drag too. +1 instead...except my vote is locked :-( New SE rules maybe? $\endgroup$
    – DrZ214
    Aug 11 '19 at 8:28
  • $\begingroup$ @DrZ214 No, no new rules. Once voted you have 5 minutes (?) to change your mind but if the post is edited you can again change the vote. Maybe fix some minor typo (like "there for" vs. "therefore") yourself and then re-vote ;-) $\endgroup$
    – PerlDuck
    Aug 11 '19 at 8:51
  • $\begingroup$ My fault for not being clearer. Wing or prop can change direction, but only prop can add energy to the system. To maintain speed in an extended turn you add power, to maintain altitude and turn you add AOA. Rudder coordinates, savings is from wing being more efficient to accelerate side ways unless ..... the wing and prop were the same size airfoil like .... (pause for drama) .... birds! $\endgroup$ Aug 11 '19 at 10:11
  • 3
    $\begingroup$ @MichaelHall Different people come with different interests. Some are interested in how to fly a plane, some are interested in how a plane flies. Sometimes questioning what seems obvious may lead to new insights, which seems like an excellent use of a Q&A website. $\endgroup$
    – Sanchises
    Aug 12 '19 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.