Airplane taking off

When an airplane banks 60° during row, 2 g of force is induced. When an airplane pitches 60° during takeoff, is 2 g induced on the airplane?

Is the angle of rotation during takeoff related to the amount of g-force made during takeoff?

  • 1
    $\begingroup$ That is not 60° of pitch up. Not even close. In the US, more than 30° requires everyone to have parachutes except during some training maneuvers towards a certificate. I would bet other countries flying similar aircraft have similar regulations. US FAR 91.307 (c) Unless each occupant of the aircraft is wearing an approved parachute, no pilot of a civil aircraft carrying any person (other than a crewmember) may execute any intentional maneuver that exceeds— (1) A bank of 60 degrees relative to the horizon; or (2) A nose-up or nose-down attitude of 30 degrees relative to the horizon. $\endgroup$
    – CrossRoads
    May 25 '18 at 15:41
  • $\begingroup$ How many Gs do you experience when you are walking on a 60 degree slope? $\endgroup$ May 25 '18 at 15:48
  • $\begingroup$ 😂.....good point! $\endgroup$ May 25 '18 at 15:58
  • 2
    $\begingroup$ It's about 10° btw ;) $\endgroup$
    – ymb1
    May 25 '18 at 23:02
  • $\begingroup$ Just to give you an idea, here's a very crudely done superimposition of a 60°-up pitch on the image in your question: i.stack.imgur.com/JjTZG.jpg (If anyone wants to use this in an answer, go right ahead.) $\endgroup$
    – user
    May 26 '18 at 11:51

The increase in g-force in a banked turn is due to centrifugal force. That's a completely different situation than the g-forces during takeoff.

The forces during take off will be due to acceleration from thrust and, at rotation, an upward acceleration due to lift (and part of the thrust). The pitch angle doesn't have a direct relationship to g-force as the roll angle has in a banked turn. During climb the pitch angle can be as high as 15°, but once you've established a steady climb and a steady airspeed the g-forces are just the 1g of gravity.


No. The angle of attack is related to the force, to a degree.

The difference in this case is the angle between the direction the nose is pointing and the angle the aircraft is actually flying. Consider the second of the two diagrams; the aircraft's nose is pointed up at something like 15 degrees, but at that very instant it is still rolling down the runway, so its direction is flat. The angle between the two is the important one in this case. To further confuse things, the wings are often connected to the fuselage so they have some inherent angle, usually a few degrees positive, so in this case the angle of attack, or AoA, could be anywhere between 10 and 20.

Now that's half of it. The other half is the actual force that results, which is the g force you feel. That force, the lift, depends on the AoA, the speed, and the design of the wing. For any given single wing design, you can make a chart like this:


I cannot embed this for some reason. Anyway, the image shows the lift coefficient vs. AoA, which you can then multiply through the formula found on that page to calculate the lift. Add gravity, and there's your g.

  • $\begingroup$ "I cannot embed this for some reason" Wikipedia prevents it. you have to save the image on your pc and then upload it through the interface here $\endgroup$
    – Federico
    May 25 '18 at 16:42

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