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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?

Pondlife
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David Teahay
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    It depends on your climb or descent rate during the bank. – Ryan Mortensen May 20 '18 at 20:07
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    It can't be 0.5 g, when it's already 1 g in straight flight. – Jan Hudec May 20 '18 at 21:05
  • 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.

     – hemp May 21 '18 at 00:30
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    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. – Erin Anne May 21 '18 at 03:54
  • I read it the arc turned, so keeping all equal (bank, radious, velocity, attitude, climbing, etc) the experienced g can be the same (just for more time) – jean May 21 '18 at 16:39
  • @hemp Considering that gravity at 400 km above the surface of Earth (for example, as experienced by the International Space Station) is about 90% of that on the surface, and gravity falls as distance squared, I think it's safe to assume that the gravity at 25 km is, for our purposes, equal to that on the ground. It certainly is close enough to its value on the ground that the difference will be of mostly academic interest. Orbit isn't high up; orbit is fast. – user May 24 '18 at 09:36
  • @MichaelKjörling The question is mostly of academic interest, and the intent of my comment was to point out how pedantic many of the other comments were (apparently missed the joke). That said, I did check and gravity at 25km is approx 9.7239 m/s2. – hemp May 24 '18 at 21:15
  • @ErinAnne if you are inverted in level flight, your bank angle is zero, so I'm not sure how you think that's contained. Altitude does matter, it affects the force of gravity from the earth. A 45º banked turn on the moon would be a lot different, ask an astronaut. It is a rounding error in practice, for sure, but the question didn't indicate the level of precision required to answer it. Of course there may be a formula for "relative g-force" which would be unaffected by actual gravity. – hemp May 24 '18 at 21:18
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    @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. – Erin Anne May 24 '18 at 21:48
  • "it would make more sense for inverted-level-flight to be 180 degrees roll/bank" .. I'm not sure it makes sense in any context. The G-Force experienced by the pilot in level, inverted flight is 1G. It happens to be a negative G versus a positive 1G, but it's still 1G. The load factor experienced on the airplane depends on the airframe. In most non-aerobatic aircraft, it is higher when inverted than while upright due to bias in the elevator, but the question was about G force, not load factor. – hemp May 28 '18 at 21:35

2 Answers2

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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.

Peter Kämpf
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    This is true only for level, coordinated flight. – user May 20 '18 at 20:15
  • @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). – Erin Anne May 25 '18 at 03:12
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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.

user
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    Using three significant digits it is 2.00, using four it is 2.000 and so on... – DaG May 20 '18 at 22:11
  • @DaG Sure, but 60° has at most two significant digits. – user May 21 '18 at 07:12
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    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. – DaG May 21 '18 at 07:20