In the video "Maneuvering during Slow Flight" the narrator states that while flying slow the airplane will be less responsive to aileron and other control inputs. He also says that the plane will turn quicker at a certain bank angle than it would if the plane was flying at normal speed. My question is: why does the plane turn quicker when flying at a lower speed?

So basically what I am asking is, as a plane gets slower, why would it start turning faster then it was when in fast flight?

The part of the video I am confused about is at 1:16


7 Answers 7


Rate of turn is dependent on the following two items:

  • The horizontal component of lift (centripetal force)
  • The tangential velocity of the aircraft (true airspeed)

The rate or turn is directly proportional to the horizontal component of lift and inversely proportional to the tangential velocity of the aircraft.

For a given angle of bank, the vertical and horizontal components of lift will be the same, regardless of airspeed in level flight.

Consequently the airplane will experience the same centripetal acceleration, regardless of airspeed.

Since the tangential velocity is slower, any kind of centripetal force will produce a greater rate of turn for a slower flying aircraft as opposed to a faster moving aircraft and this can be shown by the centripetal acceleration equation

$$a_c = \frac{v^2}{r}$$

so both slow flying airplane with a true airspeed $v_s = 100$ knots and fast flying airplane with a true airspeed $v_f = 200$ knots experience the same centripetal acceleration.

$$\dfrac{v_s^2}{r_s} = \dfrac{v_f^2}{r_f} = 4\ \dfrac{v_s^2}{r_f}$$

or, $$\dfrac{1}{r_s} = \dfrac{4}{r_f}$$

Consequently $r_s < r_f$; in this case $r_f = 4\ r_s$

Since the angular velocity is equal to the tangential speed divided by the radius.

$$\omega = v/r$$

the angular speed of the slower aircraft will be greater than the faster aircraft.

$$\omega_s = v_s/r_s$$


$$\omega_f = \dfrac{v_f}{r_f} = \dfrac{2 \ v_s}{4 \ r_s} = \frac{1}{2}w_s$$

So our twice as slow airplane turns twice as fast as the faster one does under these conditions.

  • $\begingroup$ @Krumia Only variables should be typeset in italics, therefore the indices need to be contained in \mathrm{...} $\endgroup$
    – bogl
    May 24, 2017 at 11:07
  • 8
    $\begingroup$ @bogl "should" according to? Remember that this is not a scientific publication with editorial guidelines. $\endgroup$
    – Federico
    May 24, 2017 at 13:28
  • 3
    $\begingroup$ According to improved readability. I would have used "shall" if it was mandatory. Note that my comment addressed @Krumia, because he put some effort into beautifying your answer, and I voted on his edit. Anyway, feel free to roll back. $\endgroup$
    – bogl
    May 24, 2017 at 14:59
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    $\begingroup$ You can see this same phenomenon in a car, no? Try making a turn at top speed vs. a slow one. Good luck making a 90º turn at 80 MPH! But we do it all the time at 5-10 MPH. $\endgroup$
    – Raydot
    May 24, 2017 at 16:44
  • $\begingroup$ It's there, but a car does not have a horizontal force component, only tire friction controls turn. But (infinite disclaimers), if you slam on brakes and wrench the wheel to skid/spin the car (without flipping it), you can turn faster. Skid and rollover limit car turns, not planes. $\endgroup$ Apr 2, 2019 at 22:52

Another way of explaining it in simpler terms would be:

Two vehicles, driving at 10m/s and 100 m/s respectively, both execute 180 degree turns to the left.

The catch is that each car must do the turn so that the driver only experiences 0.5G lateral acceleration.

For the car traveling 10m/s this will mean a turn radius of 20m.
This car will complete the turn in just over 6 seconds while covering 62.8m.

For the car going 100m/s, a turn radius of 2000m will produce the same sideways force. It will complete its U-turn in 63 seconds while covering a distance of 6283m.

In short, the slower moving car can make a U-turn much quicker.

The same thinking can be applied to flying.

  • 1
    $\begingroup$ Got to clue you all in on something, a car in a turn does not skid. This analogy, though well liked, is inaccurate in that it does not reflect the physics of turning in air in 2 ways. First the sum vector of forward motion and horizontal component of lift vector is the direction of flight. Secondly, the tail pivots the aircraft through the turn. Cars are attached to the ground, their turn rate is limited by the friction of the tires on the ground. The 100 m/s car would turn faster if it could pivot. $\endgroup$ Apr 2, 2019 at 18:02

The key word is "rate" of turn. It means that if you are travelling slower, it will take less time to complete a 360 degree turn than if you were going fast. It's the same as when driving a car.

If you want to complete the turn quickly at a high speed, you need a steeper bank angle compared to the angle you'd need at a low speed.

  • 1
    $\begingroup$ I don't see how this answers the question. The question was "given a constant bank angle, why do you turn faster when flying at a slower speed?" You've given a clear explanation of what "turn faster" means, but you don't seem to have said anything about why. $\endgroup$ May 24, 2017 at 11:04
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    $\begingroup$ @TannerSwett I took a guess that the OP had simply misunderstood what was meant in the video, and did not need a detailed explanation of the physics. I felt that the existing answer, whilst accurate, could have confused him if he did not have a strong understanding of physics. So I gave a simple answer which hopefully helps him understand the concept, even if it doesn't fully explain the 'why'. $\endgroup$
    – Ben
    May 24, 2017 at 12:25
  • $\begingroup$ Yeah, that makes sense. I think this is a good answer in that case. $\endgroup$ May 24, 2017 at 13:11
  • $\begingroup$ A steeper bank angle and higher g load ... $\endgroup$ May 24, 2017 at 15:55

Weight does not change for different speeds, so lift does not change, too, if you maintain the same bank angle. At lower speed, however, the kinetic energy, the direction of which needs to be changed in a turn, is smaller, so the same lift force has less work to do.

A banked wing creates a side force which is used as the centripetal force in a turn. This force is actually pulling the aircraft sideways into the new direction of movement. When banking into the turn, the centripetal force will accelerate the aircraft sideways and will decelerate its original speed component, such that the direction of the speed vector continually changes while its scalar value stays constant. If there is less speed to convert, turning can be made more quickly.

  • 2
    $\begingroup$ The most intuitively understandable answer, while still being correct. $\endgroup$
    – Waked
    May 24, 2017 at 11:23
  • $\begingroup$ I hope you mean "work" in the figurative sense... ;) Ideally, lift never performs any work. $\endgroup$
    – Sanchises
    May 24, 2017 at 15:41
  • $\begingroup$ @Sanchises "has less work to do" also means "overcoming fewer opposing forces" I think. $\endgroup$ May 24, 2017 at 15:54
  • $\begingroup$ @KorvinStarmast In physics, the term "work" is exclusively reserved for energy transfer in the form of force times distance. Plain English is not that strict; so you are of course right - I was just being pedantic in a semi-joking manner. $\endgroup$
    – Sanchises
    May 24, 2017 at 16:03
  • 1
    $\begingroup$ @Sanchises Yeah, this is the internet and Pedantry 'R Us. 8^D $\endgroup$ May 24, 2017 at 16:11

Please pardon my one-liner: because it is very hard to turn a speeding bullet.

enter image description here

Same bank angle => same turning force. Much less inertial energy to turn around when the plane flies slowly.


When you turn, acceleration is used to redirect your direction of travel. If your initial velocity is low (slow flight) then less acceleration (bank angle * time) is needed to redirect your travel.

If your initial velocity is high (SR-71 Mach 3.2 flight) then more acceleration (bank angle * time) is needed to redirect your travel.

Bank angle describes acceleration here because it effectively revectors some portion of "lift" into a horizontal direction, which causes the change in direction.

Of course flying straight and level, the lift is used to exactly counteract the force of gravity. In a level turn, some of the wing's lift is used to cause the direction of travel to change (horizontal acceleration), and also back elevator is added to increase the angle of attack, and cause a temporary increase in lift while turning. (That might slow you down a little in the turn by the way.)

So the simple answer is that there is less energy (acceleration * time) spent in turning a slow object than a fast object.


A slower plane at a constant bank angle will have to fly at a higher AOA to maintain the same amount of lift than the faster one. At a constant bank angle vertical and horizontal lift components are distributed in the same ratio fast or slow.

Why does the slower one turn quicker? Because of increased drag at higher AOA, there is more resistance to forward motion, while the side profile is identical to the faster one.

The slower one has equal lateral acceleration to the faster one (horizontal lift component) but less forward motion. The resultant vector (direction of flight) is more in the direction of the turn. Notice lateral G forces are the same. In order to do this the faster plane must turn slower!

Also skewing the result in favor of the slower, increased elevator deflection, turned sideways in the bank, helps turn the aircraft faster.

Finally, the prop blast over the tail surfaces (thanks John K) also creates greater turning forces.

The wing pulls the plane sideways, the tail turns it. Add power in a slow turn.


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