If one makes a looping for example with a plane and the g-forces that the planes makes will keep the water in the glass, but if one is at the top of the loop is there still enough g-force to keep water in the glass?

  • $\begingroup$ OTOH you can empty the glass without inverting the aircraft at all, just by suddenly and forcefully pitching down. $\endgroup$ Commented Aug 7, 2017 at 0:20
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    $\begingroup$ Not only can you keep a glass of water full, you can pour one (or tea) while rolling the aircraft yourself if you're Bob Hoover! youtube.com/watch?v=V9pvG_ZSnCc $\endgroup$
    – andy-m
    Commented Aug 8, 2017 at 15:45
  • $\begingroup$ youtube.com/watch?v=g99ho_ExApU $\endgroup$ Commented Aug 8, 2017 at 23:46

3 Answers 3


Yes, as long as you continue the loop, and don't fly the plane in the upside-down position.

See, for example: Barrel roll with a glass of water on a glider

  • $\begingroup$ but the question was in 'upside down position' isn't it possible than? $\endgroup$
    – Marijn
    Commented Aug 6, 2017 at 18:48
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    $\begingroup$ @Marijn I think what Doppelfish is referring to when he says "don't fly the airplane in the upside-down position" is that you don't try maintain level inverted flight. In other words, don't ask the wing to generate lift opposing gravity at the top of the loop. $\endgroup$
    – Terry
    Commented Aug 6, 2017 at 20:33
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    $\begingroup$ It seems mandatory to post Bob Hoover Pouring Tea Roll: youtube.com/watch?v=V9pvG_ZSnCc $\endgroup$
    – Adam
    Commented Aug 7, 2017 at 14:36

Can one fly up side down while a glass of water keeps full?

Yes, but given the speed of the aircraft, it must fly a more or less tight loop. For an aircraft at 160 km/h the radius of the turn must be smaller than 200 m. On the other hand for a fighter at 500 km/h, the radius may be as large as 2 km.

In both cases the force due to the turn is 1G positive (assuming the turn is "nose up"), that is the body and objects within the aircraft are pulled towards the outside of the turn. When they are at the topmost point of the turn, this force is vertical towards the sky, this 1G balances exactly the gravity and everything is weightless, as this dog.

Aircraft flying a vertical loop with a glass of water:

enter image description here

The interesting instant is when the aircraft reaches the top of the loop. Forces on water:

  • Are gravity force and force due to normal (centrifugal) acceleration.
  • Are proportional to the mass of water.

We must balance gravity

To prevent water from flowing out of the glass, forces exerted on the water must have a resultant (sum of vectors) which is going away from the ground.

As both forces are proportional to mass and accelerations, we can remove the water mass from the equation, and just keep the accelerations. This means we need to have a normal acceleration equal to or greater than gravity acceleration.

Calculation of the normal acceleration

Gravity acceleration is indeed a fixed value: 9.81 m/s². Normal acceleration $a_c$ itself is proportional to the radius $r$ of the loop and the angular speed $\omega$ of the aircraft (the angular speed is the RPM, but expressed in radians per second): $a_c = \omega^2r$. So to prevent being watered, we need to have an angular speed such that $a_c \geq g$, that is $\omega^2r \geq g$

On the other hand we can express the angular speed as a function of the radius and tangential speed $v$: $\omega = v/r$ and now our constraint $\omega^2r \geq g$ becomes: $v^2/r \geq g$ or $v \geq \sqrt {r \cdot g}$.

Back to speeds and RPM

Given a radius, we must fly at some minimal speed to neutralize gravity. Or given a speed we must decrease the radius to a maximal value. Here is a table for some values of radius. At the corresponding speed normal acceleration balances gravity (the number of loop rounds per minute is also mentioned).

enter image description here

Example: For a radius of 50 m, we need to fly at 80 km/h or higher. Or if we fly at 80 km/h, the radius must be 50 m or smaller. The corresponding angular velocity to 80 km/h is 4.2 RPM.

Flying at a higher speed or with a smaller radius will just push water deeper in the glass.

Special request for @DeltaLima, when flying a loop the size of the Earth, (well in practical this means flying horizontally but inverted), then to prevent gravity to empty the glass, you need to fly in excess of 28,000 km/h. That's what they do in the ISS. I guess we closed the loop on the topic.

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    $\begingroup$ When you mention RPM, I immediately think of engine revs. $\endgroup$
    – Koyovis
    Commented Aug 7, 2017 at 21:17
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    $\begingroup$ It becomes interesting when the radius is equal to earth's radius. Then you can really fly upside down whilst keeping the water in the glass. You have to "fly" close to 8 km/s to achieve that. $\endgroup$
    – DeltaLima
    Commented Aug 7, 2017 at 22:13
  • $\begingroup$ ^28800km/h I'm wondering where you have found this "fact". 8km/s on a circumference of a supposed ball of 40075km. Very interesting, yet very theoretical and not proven, hence, not reality. $\endgroup$
    – NormLDude
    Commented Aug 12, 2017 at 17:55
  • $\begingroup$ @NormLDude: It's the result of the formula in the answer ($v = \sqrt {r \cdot g}$) which gives 7.9 m/s or 28,460 km/h at sea level). All LEO satellites are orbiting at this speed around our planet. For instance, the ISS is currently at MSL + 402 km, orbiting at 27,630 km/h, a bit slower because the gravity is a bit lower. $\endgroup$
    – mins
    Commented Aug 12, 2017 at 19:04

If the loop is continued, the G at the inverted position is a function of airspeed and AOA demand on the aircraft (and the resulting nose track). If the airspeed is allowed to decrease near stall speed, it is possible for the aircraft to experience less than 1 equivalent G (causing the water to pour out of the cup). If the pilot flys the loop to allow for 2Gs across the top of the loop, the pilot will feel 1G into the seat (2 total Gs minus the 1G of God's G), allowing water to stay in the cup. Really, 1.01 G inverted at the top of the loop will result in a 0.01 equivalent positive G, keeping water in the cup.


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