# Is this scene from the A Team explainable by Physics?

This scene (youtube link) from the movie The A-team, the four members are in the tank and its falling from the air, they fire the canon and it slows the tank from falling for a moment before falling again. Is this possible from a Physics point of view?

I am looking at this from a recoil point of view. Can the tank firing the rocket produce enough recoil for it to counter the force of gravity?

• Honestly, you ask if a american action movie is physically correct? +1 anyway – dani Dec 1 '14 at 11:50

Olin Lanthrop suggested a plausible approach but there was a lot of (inaccurate) guessing in his answer. I was going to write this as a comment to his answer but it got too long. Note - in the below I round to no more than 2 significant figures - the nature of the problem doesn't support more.

Let's take the famous Sherman tank as our example. A brief search tells us that it weighed 66,000 pounds (about 30,000 kg) - not clear if that includes full tank of fuel, ammunition and crew) and that its main cannon (the M3 L:40) could fire 6.7 kg rounds at a muzzle velocity over 600 m/s.

From conservation of momentum we conclude that firing 80 rounds per second would keep the tank from falling. The tank had 90 rounds, so that would work for about one second. It would also melt the barrel in a heartbeat.

So let's look at the other weapons on the tank. There was a .50 caliber Browning with 50 g bullets with muzzle velocity of 800 m/s and firing around 800 rounds per minute. That's a mean impulse of 40 Ns per round, or around 500 Ns firing at full tilt. That would be enough to keep a child airborne - not a tank. Point all three machine guns forward - it still doesn't even carry the gunner plus the guns. This should give you a sense of the enormous power of the main cannon: those M3 rounds are not something you want to stop with your Kevlar vest.

Movie physics - you don't have to get the math right. It's magic.

• Keeping things airborne by firing a gun at the ground has also been covered by xkcd's What if? – Josh Caswell Sep 11 '14 at 2:20
• Note that we're not talking about keeping it airborne but reducing it's acceleration. – slebetman Sep 11 '14 at 3:54
• @slebetman $10 m/s^2$ (one G, enough to keep an object airborne) for about one second is $10 m/s$. That's not much compared to the 200 miles per hour (90 m/s) Olin is talking about. Not insignificant either, true... – John Dvorak Sep 11 '14 at 8:04
• @slebetman the question as phrased in the last paragraph is about "countering the force of gravity" - in the limit that means staying airborne. – Floris Sep 11 '14 at 11:53
• Yes, but the context is the movie scene. So countering the force of gravity here means "countering the force of gravity like a parachute does" instead of "countering the force of gravity like a rocket engine does" – slebetman Sep 11 '14 at 12:55

I took a look at the clip and my take on it that the object of the exercise was not to slow down the tank, but to move it sideways and land in the lake a half a mile away. Hannibal says, "rotate the main gun to 82º" which I take to be sticking out sideways.

Background Information

The facts I could find were:

• An M1A2 Abrams weighs 69.54 tons or 63085.63 kg

• A M829A1 tank round (Wikipedia) has a total weight of 41.1 lb (18.6 kg) - 8.1 kg (18 lb) propellant = ~10 kg and travels at 1,670 metres per second (5,500 ft/s)

• The basic load of a M829A2 is 42 rounds (also from globalsecurity.org)

So my statement of the problem is: Can a 63,000kg object be moved a distance of 800m by firing the main gun before it hits the ground?

1. From the conservation of momemtum (p):

\begin{align*} p_{tank} &= p_{round}\\ m_{tank} v_{tank} &= m_{round} v_{round}\\ v_{tank} &= \frac{m_{round}v_{round}}{m_{tank}}\\ v_{tank} &= 0.2647\frac{\text{m}}{\text{s}} \end{align*}

1. Given the gun is not pointing straight sideways, horizontal velocity of the tank with the first round fired is reduced as per: \begin{align*} v_h &= 0.2647\frac{\text{m}}{\text{s}} - \frac{90°-82°}{90°} \cdot 0.2647\frac{\text{m}}{\text{s}} \\ v_h &= 0.2412\frac{\text{m}}{\text{s}}\end{align*}

2. Assuming the very best case where drag is negligible (I know, not a good bet since the tank is hanging from a drag chute), each round would increase the velocity by $0.2413\frac{\text{m}}{\text{s}}$.

3. From the fact that the basic load (the standard quantity carried) is 42, you could, at best, get the tank moving sideways at:

$$v_{h,max} = 0.2412\frac{\text{m}}{\text{s}} \cdot 42 = 10.13\frac{\text{m}}{\text{s}}$$

1. The time to move 800m at the above rate is right at 79 seconds.

2. Would there be enough time to move the required distance? How high (height denoted by $y$ in the following) would the tank have to start its descent to have 79 seconds?

From the basic time/distance/gravity equation:

\begin{align*}t &= \sqrt{\frac{2y}{g}}\\ y &= \frac{t^2 g}{2}\\ y &= 61120 \text{m}\end{align*}

So the airplane from which the tank falls would have to have an altitude of over 200,000ft. Since:

1. The only aircraft the U.S. Armed Forces has that can carry an M1 tank is the C5 Galaxy and

2. The C5's service ceiling is 35,700 ft at 279,000 kg gross weight (that is, 5.6 times too low).

The flying tank scene in the A-Team movie could not be real - but it's still great movie-making.

Do the math. I'm no military expert, so I'll guess at some parameters, but I think it will show the effect is so small that it doesn't matter even if the numbers were considerably more favorable.

Let's say the tank weighs 50 tons, which I think is rather light for a tank. That's 100,000 pounds, which puts its mass at 45,000 kg. I don't know what the mass of a tank shell is, but let's say 10 kg. That sounds high to me. That puts the ratio of tank mass to shell mass at 4,500:1. Let's say the shell is fired at the speed of sound, 330 m/s. Dividing by the mass ratio, that means the velocity change of the tank would be 70 mm/s, or .16 miles/hour. I doubt the forensic team trying to reconstruct the remnants of your mangled skeleton will notice the difference between you having hit the ground at 200 miles/h or 199.8 miles/h.

I think my numbers were conservative, so the likely velocity decrease is even smaller. Still, even if I was off by a factor of 10 the other way, I don't see it making any practical difference.

An object under the force of only gravity will have a center of mass that accelerates at a constant rate (downwards). If you manage to shoot (a small) part of the mass down very quickly, one may be able to let the rest of the object fall a little slower. However, I strongly doubt that firing a rocket will be 1) efficient 2) safe (think of rotational effects) 3) significant in any way in this particular situation.

• firing a rocket would have little effect as the rocket's "equal and opposite reaction" is it's own expelled gasses. Firing the cannon would have recoil in the opposite direction (on the tank). Others have sketched some calculations about how inadequate that recoil would be. Your point about rotational effects is yet another problem - the cannon is unlikely to be in line with the tank's center of gravity, so it would spin far more than move. – Zeph Sep 11 '14 at 7:19
• The parachute should be able to dampen the rotation induced by shooting the canon, however there does not seem to be any oscillation in the movie, so it does not seem to be entirely physical correctly implemented. – fibonatic Sep 11 '14 at 14:08
• Without slipping the load will just act as a pendulum when a sudden force is exerted on it. The wind is always a greater factor than even a massive, sudden shove on the load (a gradual, steady force, on the other hand, can overcome the pendulum effect). I edited my answer to elaborate a bit more on how round chutes are controlled during descent. – zxq9 Sep 11 '14 at 15:45

tl;dr: No. But it was pretty awesome in the movie anyway.

The chute will resist horizontal motion on the load, as a load under a parachute has a tendency to convert lateral force into a pendulum action and results in very little gross movement.

To move under a round parachute you have to "slip" by lowering the skirt of the canopy on the side of the direction you want to go. This is usually done by pulling down on the risers/suspension lines on that side, either by hand or with a steering robot. The alternative is to open a vent on the opposite side or close a vent on the same side, but most cargo canopies lack that type of vent. This forces air to escape on the opposite side of the canopy and pushes you in the desired direction.

It would have been more believable for the men to climb out of the tank and up the suspension lines on one side to force the skirt down in the direction they wanted to go. That would have worked, but on a roughly 60t load it would have been very difficult to influence it without getting all the way up to the canopy and manually furling it a bit on the edges. Parachutes under load are remarkably stiff.

I'm not sure how the calculations of firing the main gun under canopy work out -- they are actually rather complicated in this case with a non-trivial number of variables -- but this is what you learn dealing with round parachutes in practice (which I spent several years in the Army doing...). In any case, the wind (which tends to change direction as you descend) would have had a vastly overwhelming effect without slipping the chute.

## protected by Qmechanic♦Sep 11 '14 at 13:26

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