A 60° turn in an airplane doubles the g-force; how much g-force is experienced in a 15°, 30° and 45° turn?

Is it 0.5, 1 and 1.5g respectively?

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    $\begingroup$ It depends on your climb or descent rate during the bank. $\endgroup$ Commented May 20, 2018 at 20:07
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    $\begingroup$ It can't be 0.5 g, when it's already 1 g in straight flight. $\endgroup$
    – Jan Hudec
    Commented May 20, 2018 at 21:05
  • $\begingroup$ It also depends whether you're upright or inverted - and also the altitude (what's the G-force in a 45º bank in an SR-71 at 85,000ft?) Certainly assumptions can be made about the question or asserted in the answer, such as upright, level, coordinated flight, in an airplane, below 40,000ft. $\endgroup$
    – hemp
    Commented May 21, 2018 at 0:30
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    $\begingroup$ Upright/inverted is, generally, contained in the "bank angle." The altitude doesn't matter to the load factor until you're not actually flying anymore. Coordinated/level does matter. $\endgroup$
    – Erin Anne
    Commented May 21, 2018 at 3:54
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    $\begingroup$ @hemp in this context, and many others, it would make more sense for inverted-level-flight to be 180 degrees roll/bank, not zero. In a terrestrial context, load factor is already relative to the lift needed to loft the airplane, so constant gravity need not be assumed. And yes, a 45 degree banked turn on the moon would be different, since you wouldn't be generating lift using a wing and so none of this applies at all. Your humor is not only not coming through but it distracts from a worthwhile note that assumptions matter. $\endgroup$
    – Erin Anne
    Commented May 24, 2018 at 21:48

2 Answers 2


The formula for the load factor $n_z$ in a turn is

$$n_z = \frac{1}{cos\phi}$$

where $\phi$ is the bank angle. This makes the load factor in a 45° turn 1.414 or $\sqrt{2}$.

table and graph of load factor over bank angle

Since Michael correctly mentions the simplifications in his comment, let me expand the formula above. In case of a climbing turn, the load factor depends on the coordinate system used. If it is measured in airframe-fixed coordinates, the load factor will be split into a z-component ($n_z$, pointing down) and an x-component ($n_x$, pointing fore- or backward, depending on the aircraft's pitch attitude $\Theta$; nose-up is positive): $$n_x = -\frac{sin\Theta}{cos\phi}$$ $$n_z = \frac{cos\Theta}{cos\phi}$$

For an uncoordinated turn the value needs to be modified further with the relationship between bank angle change for a given sideslip angle, which is aircraft-specific.

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    $\begingroup$ This is true only for level, coordinated flight. $\endgroup$
    – user
    Commented May 20, 2018 at 20:15
  • $\begingroup$ @MichaelKjörling is it for level flight, or is it unaccelerated flight? I think the same applies to e.g. a steady-state climb, we just don't think of doing that sort of thing as often. (I typically also think coordinated/level for these but I just got to wondering). $\endgroup$
    – Erin Anne
    Commented May 25, 2018 at 3:12

The G forces (also known as the load factor) experienced by an aircraft and its occupants do not scale linearly with the bank angle of the aircraft.

Assuming constant vertical speed, coordinated flight (unchanging rate of climb or descent, ball centered), the load factor at a bank of 0° is 1.00. As the bank angle increases, this grows – slowly at first, then faster. At 15° the load factor is about 1.03; at 30°, about 1.15; at 45°, about 1.41. Increasing bank angle further, at 60° it's 2.00; at 75°, about 3.86.

In general, assuming level and coordinated flight, the load factor at a given bank angle $\theta$ is calculated as $$ n = \frac{1}{\cos{\theta}} $$

For example, for a bank angle $\theta = 45°$, we have $$ n = \frac{1}{\cos{45°}} \approx 1.41$$ and sure enough, the load factor for a level, 45° bank is just that.

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    $\begingroup$ Using three significant digits it is 2.00, using four it is 2.000 and so on... $\endgroup$
    – DaG
    Commented May 20, 2018 at 22:11
  • $\begingroup$ @DaG Sure, but 60° has at most two significant digits. $\endgroup$
    – user
    Commented May 21, 2018 at 7:12
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    $\begingroup$ It was just to say that, since that is an exact value (and a simple one: 2), it sounds a bit peculiar to mention how many significant digits are we considering. $\endgroup$
    – DaG
    Commented May 21, 2018 at 7:20

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