# How much thrust is needed by an aircraft to have vertical takeoff?

In some airshows, I have seen some big aircraft (like 737) takeoff almost vertical.

Generally, fully-loaded passenger/cargo planes don't do that, I guess due to safety issues but I think there needs a lot of extra thrust to have an almost vertical takeoff.

So is a vertical takeoff possible in every kind of airplane (without excessive weight and giving full thrust) or only some specific ones can do it?

• To have vertical lift you need as much thrust as the weight of aircraft. Most figter airraft can do this but not commercial plane. – vasin1987 Jan 13 '15 at 11:50
• @vasin1987 figter airraft? Is that a kind of raft that floats in the air upon which fig trees grows? Or did you mean Fighter Aircraft? – haneefmubarak Jan 14 '15 at 5:33
• The important thing to note is that the aircraft isn't gaining altitude any faster than usual: it just stays on the runway longer before pulling up sharply because it looks more impressive. If anything, this usually results in a lower overall climb rate (measured as time-to-altitude). – Jon Story Jan 14 '15 at 11:46

First, let's agree on terminology: What you saw in airshows is a vertical flight path. Flying horizontally first, the airplane pitched up until the nose was pointing straight into the sky.

Surprisingly, no thrust is needed to perform this maneuver. Even gliders can do it. What happens is that kinetic energy is converted to potential energy, the rate of potential energy increase being proportional to flight speed and aircraft mass. If you start fast enough, this vertical flying can be maintained for several seconds, until the aircraft runs out of speed and stops in midair, followed by an uncontrolled drop. Skilled pilots orient the aircraft in the right direction by starting a rotation around the vertical axis at the top of the climb, so the following drop lets them pick up speed again with the correct nose-down attitude. Now potential energy is converted back into kinetic energy until speed is sufficient for a pullout. In aerobatics, this maneuver is called a stall turn or a hammerhead stall.

A few conditions apply, however. The airplane must be able to fly fast enough to have the needed potential energy to sustain the maneuver through the pitch-up phase. This is helped if its engines add energy, so the kinetic energy bleeds off more slowly. Also, at the top of the maneuver it is flying at zero g, and this requires at least that all items on board are securely fastened. Lastly, the pitch-up needs a load factor bigger than 1 g, and the higher the maximum load factor is, the tighter this pitch-up can be flown.

Now the question has been changed: The vertical flight path is flown right after take-off. This limits the entry speed for the maneuver, and gliders will not be able to do this. If we take the 737 from the question and fly it with no payload and little fuel, the flight mass $m$ of a 737-700 is 40 tons, and the installed thrust is about 200 kN (sea level static). Let's assume that the pilot accelerates after takeoff to a horizontal speed $v$ = 100 m/s (194 KTAS) while retracting the flaps, the kinetic energy ($0.5\cdot m\cdot v^2$) is equivalent to a potential energy ($m \cdot g \cdot h$) of an altitude gain h of $$h = \frac{v^2}{2\cdot g} = 510 m$$ The engines deliver less thrust with increasing speed; maybe 40% of the weight, so the airplane will still accelerate for the first 18° - 20° of the 90° flight path change. This will delay the point when speed has been bled off and add maybe 150 m to h. At 100 m/s a pull-up with a radius of 500 m will add a load factor of 2 g. The pilot needs to pull less first and harder at the end of the maneuver in order to stay within the maximum load factor of 2.5. When speed bleeds off, so will wing lift, and in the second half of the pull-up the wing will not create enough lift to change the flight path enough in order to reach the desired vertical attitude. Also, the aircraft will be very low for a safe recovery.

This makes it rather doubtful that an airliner can be pulled up to a vertical climb after takeoff. If the maneuver is started at a higher speed and with a little more distance from the ground, I see no reason why it should not safely be possible.

• From a physics standpoint, you're right, and this is how it works for sufficiently-maneuverable aircraft, but are airliners actually sufficiently maneuverable for a truly vertical flight path without stalling before they get to 90 degrees? Personally, I'd tend to guess that what the OP really saw was something more like 30-45 degrees nose-up attitude, not really 90. If anyone could shed some light on whether any airliners are really capable of safely accomplishing a (briefly) vertical flight path, though, it would be appreciated. – reirab Jan 13 '15 at 16:34
• Nitpick: the kinetic energy will not bleed off more slowly. It will bleed off (i.e. be converted into PE) at the same rate; the engines will just "add" (i.e. convert from fuel) more kinetic energy into the system, so converting all of it takes longer. – imallett Jan 13 '15 at 17:49
• @imallett: This is very nitpicky! I was thinking about "speed will bleed off more slowly", but that would have required to translate speed to kinetic energy in the mind of the reader. So I settled for the easier to understand (and still not wrong) term. – Peter Kämpf Jan 13 '15 at 20:27

For a craft weighing x kg you need g*x Newtons of thrust minimum for sustained vertical flight (ref.: high-school physics). In other words for each metric ton of weight you need around 9.81 kN of thrust. The A380 has a operational empty weight of 276 t so it would need 2707 kN of thrust to sustain a vertical climb. Its 4 engines each producing 320kN don't even come halfway.

However pitching up to vertical and continuing on while shedding speed is possible without thrust.

• I couldn't easily find any references so don't want to make the edit willy-nilly, but I believe it would be more accurate to say that the engines produce 320 kN each? – a CVn Jan 13 '15 at 13:32
• @MichaelKjörling Yes, you're right. A380 has 288,000 lb of max thrust and an empty operating weight of 610,000 lb. Its MTOW is 1,268,000 lb. So, at absolute best (really better than the best, since you can't take off with no fuel,) it would have a thrust/weight ratio of 0.47 and, at MTOW, it would be 0.23. So, no sustained vertical flight for the A380. :) What would be fun would be to get a 737 (with re-enforced wings and greatly lengthened landing gear) outfitted with GE90-115b's. Each one of them would be capable of accelerating the aircraft in a vertical climb by itself. :) – reirab Jan 13 '15 at 16:47
• An empty concord would be at 0.85, it should be able to fly vertical for quite some time :D – Antzi Jan 14 '15 at 2:41

Despite the hype that some news articles like to add to their stories, with talk of "near-vertical" takeoffs and "terrifying" turns, those maneuvers are less extreme than they look. While certainly higher than normal, 30 degree pitch after takeoff or 60 degrees of bank are not vertical. Yet the air show still said that was too much, so typical displays of non-aerobatic planes are going to be even less than that.

To answer the question in the title, take a look at the AV-8B Harrier II attack jet. It has a max vertical takeoff weight of 20755 lb, and its engine produces 23500 lbf of thrust. This corresponds to a thrust-to-weight ratio of about 1.13:1, since the thrust must exceed the force of gravity in order to accelerate the plane upwards. In contrast, the 787-9 from the above example has 142,000 lbf of thrust, and weighs between 304,000 lb and 557,000 lb. If the plane weighed 400,000 lb on that takeoff, the thrust-to-weight ratio was about 0.36:1. Although converting kinetic energy to potential energy will allow a plane to climb faster, the amount of kinetic energy available to an airliner at takeoff is more limited.