I just read on reddit that you could use the recoil of the gun as the propelling force for the A-10 if pointed backwards. There is also a question here on Aviation.SE, saying that the gun has the same power as one single engine.

I'm now wondering how fast airspeed actually decreases when firing the gun.

  • Let's suppose the aircraft is in level flight with no acceleration in any direction. The gun is now being fired. What acceleration is being applied to the aircraft (In m/s or kt/s)?
  • Bonus question: If there was infinite ammo and nothing like gun overheat, how long would it take to slow the plane down to stall speed, if initial speed was a usual A-10 speed.
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    $\begingroup$ FYI, reddit came second: what-if.xkcd.com/21 $\endgroup$
    – Federico
    Commented Dec 4, 2017 at 14:48
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    $\begingroup$ Suggested additional bonus question: Could it take off while firing the gun? $\endgroup$
    – JollyJoker
    Commented Dec 5, 2017 at 17:24
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    $\begingroup$ So you're basically asking "What is the airspeed velocity of a gun-laden warthog?" $\endgroup$
    – Machavity
    Commented Dec 5, 2017 at 19:46
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    $\begingroup$ At 300 knot cruise, 23,000 kg (MTOW) = 3.5e6 kg*m/s, 13,000 kg (empty) = 2e6 kg*m/s. All 1350 projectiles, 0.4 kg each, at 1000 m/s = 0.5e6 kg*m/s. So in a vacuum, you'd lose a fair bit of speed, more than I expected. $\endgroup$ Commented Dec 6, 2017 at 8:59
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    $\begingroup$ @Sean It's just a theoretical question. $\endgroup$ Commented Aug 28, 2019 at 16:11

3 Answers 3


What acceleration is being applied to the aircraft (In m/s or kt/s)?

From XKCD we know that the gun

produces almost five tons of recoil force

that is about 50000 newtons (round numbers for simplicity, they won't overly affect the results)

The acceleration is force divided by mass, and the A-10 has an empty mass of 11000 kg, and a MTOW of 23000 kg.

This gives an acceleration between $4.545\ \mathrm{m/s^2}$ and $2.174\ \mathrm{m/s^2}$.

The Wikipedia page about the GAU-8 mentions that

While this recoil force is significant, in practice a cannon fire burst slows the aircraft only a few miles per hour in level flight.

and gives as a reference a book that I do not own, so I cannot verify.

If there was infinite ammo and nothing like gun overheat, how long would it take to slow the plane down to stall speed, if initial speed was a usual A-10 speed.

With the additional assumption of thrust balancing only the aerodynamic drag, we can say the following: cruise speed is $155.6\ \mathrm{m/s}$, and stall speed is $61.11\ \mathrm{m/s}$ (see Wikipedia link above), so this means that it would take between ~34 and ~71 seconds to completely stop the aircraft; or between ~20 and ~43 seconds to stall it.

If instead we want to be more realistic, we have to assume that we have infinite fuel, that ammo and fuel do not factor in the weight computation, and that the yawing/pitching moments do not overcome actuator authority.

With all that in mind, we should start from firing the gun at strafe speed, not at cruise speed, since we want to hit a specific target on the ground, so we want to move as slowly as possible, to have the highest chances of hitting it. This means that our starting velocity would be above stall speed but not significantly, leading to a smaller margin of continuous fire before stalling, unless the throttle is opened (if possible) to counteract the extra "drag".

If we ignore the discussion about strafing and hitting the target, and we fire from cruise speed (and this is what is meant by "usual speed"), it is possible that stall is not achieved, since at lower speeds you will have more excess thrust (but we are still assuming that the firing only produces "drag" or a negligible amount of yawing/pitching moments). This means that we could settle at a lower speed, but the determination of the exact speed requires the knowledge of the exact drag and thrust characteristics of the aircraft, in addition to it being a function of the initial speed, see Koyovis' answer for some basic numbers.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. Additional comments will be deleted without being moved to chat. $\endgroup$
    – Farhan
    Commented Dec 5, 2017 at 14:08

I asked this question of an A-10 pilot at the Stewart, FL airshow. His response was that the recoil from the guns wasn't noticeable. However, the exhaust generated by the spent shell casings (smoke, particulate) tended to get into the engines and cause issues. Hence the relatively short bursts followed by periods of non-firing to let the engines clear.

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    $\begingroup$ A guy I work with flew earlier A-10s and said the recoil was overplayed. He never offered numbers, but clearly poo pooed the effect on flight. That supports the comments of your contact. $\endgroup$
    – mongo
    Commented Dec 5, 2017 at 2:52
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    $\begingroup$ Specifically, the gases generated by the gun firing are oxygen poor, so if they get into the engine it can cause a flame out due to lack of oxygen. In order to counteract this, A-10 pilots are recommended to fire in 3-4 second burts and the pilot light for the engine is triggered to engage when the gun is firing. $\endgroup$
    – SGR
    Commented Dec 5, 2017 at 9:32
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    $\begingroup$ A pilot isn't really a good authority on this. I can drive up a hill in my car and while I may know I'm pressing a little harder on the accelerator, I have no frame of reference to estimate how much more fuel is being consumed to counteract the upward incline. $\endgroup$ Commented Dec 5, 2017 at 18:06
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    $\begingroup$ @DeanMacGregor I would disagree. A typical car driver isn't really a good comparison to a pilot. Pretty much every pilot has a much-better-than-average knowledge of their vehicle. An Indy car driver would be more applicable, and I'd bet an Indy driver could give you a pretty close estimate of gas consumption on that hill. $\endgroup$
    – Shawn
    Commented Dec 5, 2017 at 22:05
  • $\begingroup$ Note that the guns are fired in bursts only (per designer specs, pilot training, and pilot anecdotes). The thrust from a short burst may not act long enough to be noticeable by the pilot, but sustained firing for the durations discussed in other answers (30-120 seconds) would be. Still upvoting because this demonstrates the typical real-life usage of the gun; other answers already cover the theoretical effects. $\endgroup$
    – brichins
    Commented Dec 6, 2017 at 18:12

from http://www.fiddlersgreen.net/models/aircraft/Fairchild-Republic-A10.html

Initially, the aircraft flies with constant speed and the engine thrust equals the aerodynamic drag. Then the gun starts to produce negative thrust. The answer would be really easy if all forces stayed the same until the aircraft comes to stall speed, but of course this is not the case. We can't just compute the deceleration at time = 0 and project that into the future: the aircraft slows down because of the recoil thrust, therefore the aerodynamic drag reduces because it is a quadratic function of airspeed. But the propelling thrust is still the same. Deceleration changes constantly as a function of time as well.

It's a complicated situation that can be worked out not with high school physics but with differential equations, in which perhaps only a mathematician will find delight. But it can also be solved with a numerical solution: a real-time computation of the aircraft state. This is what flight simulators do, at every clock tick they update the aircraft state with the latest data.

The method to do this for this question, is:

  • Find the aircraft data. In simulators used for pilot training, this data is supplied by the manufacturer, or from an actual aircraft rigged out with transducers. In our case, we can deduce the aerodynamic constants that we need from Wikipedia.
  • Take an initial speed, work out the thrust required for trimmed flight, then at t=0 subtract 50,000 N recoil thrust from the gun. This results in the initial deceleration, say -2.78 m/s$^2$
  • The next second, t=1, the initial speed has decreased with 2.78 m/s. Compute the new drag at this speed. Thrust is still identical to that at t=0. At the lower drag, we find a lower deceleration value, say -2.66 m/s$^2$

1. Aircraft data from Wikipedia:

  • Powerplant: 2 × General Electric TF34-GE-100A turbofans, 9,065 lbf (40.32 kN) each. 80,000 N total.
  • Maximum speed: 381 knots (439 mph, 706 km/h) at sea level, clean. Equates to 196 m/s.
  • Weight: Loaded weight: 30,384 lb (13,782 kg) Anti-armor mission weight: 42,071 lb (19,083 kg).
  • Wing area: 506 ft² (47.0 m²)
  • Wingspan: 57 ft 6 in (17.53 m). So aspect ratio A = $b^2/S$ = 6.54

At maximum speed, level flight, sea level, clean config, 18,000 kg, the aerodynamic lift coefficient $C_L$ and drag coefficient $C_D$ are:

$$C_L = \frac{2\cdot W}{\rho \cdot V^2 \cdot S} = \frac{2\cdot 18,000 \cdot 9.81}{1.225 \cdot 196^2 \cdot 47} = 0.16 \tag{1}$$

$$C_D = \frac{2\cdot T}{\rho \cdot V^2 \cdot S} = \frac{2\cdot 80,000}{1.225 \cdot 196^2 \cdot 47} = 0.072 \tag{2}$$

$$C_D = C_{D_0} + \frac{{C_L}^2}{\pi \cdot A \cdot e} \Rightarrow C_{D_0} = 0.072 - \frac{{0.16}^2}{\pi \cdot 6.54 \cdot 0.8} = 0.07\tag{3}$$

2. Trimmed horizontal flight

We use the aerodynamic constants found under 1. for computing lift, aerodynamic drag and thrust (which equals drag).

  • Take a weight: 18,000 kg, close to an anti-armour mission.
  • Take a speed: maximum speed = 196 m/s.
  • Procedure: compute $C_L$ from equation (1), then compute $C_D$ from equation (3), then compute initial drag from D = $C_D \cdot \frac{1}{2} \rho V^2 \cdot S$. This is also the engine thrust because the aircraft has constant velocity.

3. Open fire: the cannon produces 50,000 N reverse thrust

enter image description here

Image above shows the first 8 seconds, acceleration has decreased from 2.78 to 2.0 m/s$^2$. If we plot velocity over time, we see that speed decelerates to 110 m/s in 165 seconds, stall speed is never reached.

enter image description here

If we repeat the exercise for an initial speed of 160 m/s = 576 km/h = 311 kts, and at the same weight of 18,000 kg, we reach stall speed of 61.11 m/s after 66 seconds, if the pilot keeps the thrust at the same setting. They have plenty time to speed up. The plot also shows what we find if we use the initial acceleration only, a time of 36 seconds.

enter image description here

After reaching stall speed it will take a very short time before the aircraft comes to a stop: it falls out of the sky. A shallow dive cures that: gravity now produces a thrust component as well, with $T_G = m \cdot g \cdot sin(\Theta)$. At 18,000 kg, a dive angle of 16° is enough to maintain airspeed without adding thrust while firing the gun: the gravity component now exactly compensates for the thrust of the gun.

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    $\begingroup$ Sorry, I am a little confused about the table. I expect D is drag, which equals thrust (since the first value is about ~80kN), but what then is T? From the end point, it would appear T is thrust, since you imply the aircraft won't decelerate any further once D=T, but then T should be 80kN at the start. Also a little lost in the jump from eqn. 1,2,3 as to how to calculate a, and D (which is a function velocity)? Thanks in advance! $\endgroup$
    – Penguin
    Commented Dec 5, 2017 at 10:47
  • $\begingroup$ @Penguin have added an explanation on the table, plz let me know if it helps. A = aspect ratio, computer from $b^2/S$. Drag is computed analogous to (2): $D = C_D \frac{1}{2} \cdot \rho \cdot V^2 \cdot S$, it is a function of velocity squared. $\endgroup$
    – Koyovis
    Commented Dec 5, 2017 at 11:42
  • $\begingroup$ Yes! That makes it crystal clear. Thanks very much. $\endgroup$
    – Penguin
    Commented Dec 5, 2017 at 16:32
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    $\begingroup$ "Starting at 120 m/s, a dive angle of 16° is enough to maintain airspeed" but only for a time shorter than your initial altitude in meters divided by 33, afterwards you will have a really close encounter with the ground. $\endgroup$
    – Federico
    Commented Dec 6, 2017 at 8:24
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    $\begingroup$ As an A10A avionics tech, I would only add that the Stability Augmentation System adds 1.9° up deflection to the elevators while the trigger is depressed to counter the downward vector of the gun being fired. Most pilots are not aware of this, but it would tend to reduce the 'feel' of the gun recoil. $\endgroup$ Commented Aug 26, 2019 at 9:24

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