I get asked about once a month if the Iowa (or Bismarck, Yamato, etc.) moves sideways when she fires a full broadside. To save myself some time, I've plagiarized Dick Landgraff's very good answer to this same question (what are friends for?). In addition, Greg Locock kindly pointed out an error in my math when I ineptly tried to calculate a real number for the motion.

What looks like a side-ways wake is just the water being broiled up by the muzzle blasts. The ship doesn't move an inch or even heel from a broadside.

The guns have a recoil slide of up to 48 inches and the shock is distributed evenly through the turret foundation and the hull structure. The mass of a 57,000 ton ship is just too great for the recoil of the guns to move it. Well, theoretically, a fraction of a millimeter.

But because of the expansive range of the overpressure (muzzle blast), a lot of the rapidly displaced air presses against the bulkheads and decks. Those structures that are not armored actually flex inwards just a bit, thus displacing air quickly inside the ship and causing loose items to fly around. Sort of like having your house sealed up with all windows and vents closed and when you slam the front door quickly the displaced air pops open the kitchen cabinets.

To calculate the velocity of the USS New Jersey moving sideways, what you need to consider is conservation of momentum. A 16" Mark 8 APC shell weighs 2,700 lbs. and the muzzle velocity when fired is 2,500 feet per second (new gun).

The USS New Jersey weighs about 58,000 tons fully loaded (for ships, a ton is 2,240 lbs.)

All weights must be divided by 32.17 to convert them to mass.

If the battleship were standing on ice, then:

Mass of broadside * Velocity of broadside = Mass of ship * Velocity of ship

9 * (2,700 / 32.17) * 2,500 = 58,000 * (2,240 / 32.17) * Velocity of ship

Solving for the ship's velocity:

Velocity of ship = [9 * (2,700 / 32.17) * 2,500] / [58,000 * (2,240 / 32.17)]

= 0.46 feet per second

So, ship's velocity would be less than 6 inches per second, on ice.

This analysis excludes effects such as (1) roll of the ship, (2) elevation of the guns (3) offset of the line of action of the shell from the centre of gravity of the ship and (4) forces imposed by the water on the ship. These are variously significant, and will all tend to reduce the velocity calculated above.

I need to point out that in Greg's masterful analysis he assumes that the guns are at zero degrees elevation, that is, the guns are pointed directly at the horizon. In actuality, they are almost never fired at this elevation as it would mean that the shells would only go a short distance before they struck the water. At a higher, more realistic elevation, the force of the broadside would also have to be multiplied by the cosine of the angle of elevation. This means that the horizontal velocity imparted to the ship would be even less than the numbers calculated above.

In the years since this essay was first published, various people have sent me notes complaining about the over-simplicity of this analysis, as it ignores the other factors involved, namely, the effects of the propellant gasses. I usually try pointing out that these are not significant compared to the projectile momentum and therefore do not have a significant impact on the solution above - by significant, I mean that including them is not suddenly going to change the ship's movement to six feet per second rather than the six inches per second that Greg calculated above. This sort of rational answer does not always satisfy my questioners. So, here is a little extra analysis for the purists out there.

First, a some background information from my friend, Leo Fischer, to help determine the effects of the propellant gasses and free recoil:

The total mechanical energy created when a 16"/50 is fired can be computed as follows:

Given:

• Projectile Weight: W_p = 2,700 lbs
• Charge Weight: W_c = 650 lbs
• Muzzle Velocity: V_o = 2,500 fps
• Weight of Recoiling Parts: W_r = 250,000 lbs
• g = 32.174 fps²

Projectile Kinetic Energy = 0.5 * ((W_p/g) * V_o²)

= 2.622*10⁸ ft-lb

Propellant Gas Kinetic Energy at shot ejection = 0.5 * ((W_c/g)) * (V_o² / 2)

= 3.174*10⁷ ft-lb

Empirical data has shown that only about half of the propellant gas has been accelerated to the muzzle velociy at shot ejection. The other half of the propellant gasses exits at a higher velocity once the projectile "corking" the barrel is removed.

To compute the kinetic energy of the propellant gases after shot ejection, we must know the average velocity of the gases as they escape the muzzle. Experiments have shown that this velocity varies between 1,200 and 1,400 mps, depending on the muzzle velocity of the weapon. For purposes of these calculations, we will use 1,200 mps or 3,937 fps.

• Average outflow velocity of propellant gases following shot ejection: w = 3,937 fps

Gas Kinetic Energy after shot ejection = 0.5 * ((W_c/g) / 2) * w²

= 78.29*10⁶ ft-lb

To compute the Kinetic energy of the recoiling parts, we must determine the velocity that they would achieve if allowed to recoil with no retarding force. This is commonly referred to as the free recoil velocity. To account for the difference between the velocity of the projectile and that of the propellant gases, we will use the aftereffect coefficient B which is defined by the relationship:

w = B * V_o

therefore B = 1.5748

Free Recoil Velocity:

V_re = (((W_p/g) + B) * (W_c/g) / 2) + (W_c/g) / 2)) / (W_r/g)) * V_o

= 35.37 fps

Recoil Energy = ((W_r/g) / 2) * V_re²

= 4.86*10⁶ ft-lb

The rotational energy of the projectile is small by comparison and can be neglected.

The overall mechanical energy is only a part (40 to 50%) of the chemical energy of the propellant, since a considerable portion of the energy is carried off as heat by the propellant gases, or transmitted to the gun barrel.

Ref. Rheinmetall Handbook on Weaponry, 1982, chapter 9.

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As can be seen above, the projectile kinetic energy is determined by the following equation:

Projectile Kinetic Energy = 0.5 * ((W_p/g) * V_o²)

Projectile momentum is the derivative of this equation and is calculated as follows:

Projectile Momentum = (W_p/g) * V_o

Regarding the calculations for determining the propellant kinetic energy and momentum, I was recently contacted (June 2021) by Lt.Kol. Frederik Coghe IMM from the Dept. of Weapon Systems & Ballistics, Royal Military Academy, Belgium. He informs me that there is now a slightly different way of calculating the energy and momentum added by the propellant gasses. This method will be used below.

The energy created by igniting the propellant has two main components; the mechanical or kinetic energy that moves the projectile down the barrel and the heat energy released by the propellant as it burns. Only the kinetic energy is of interest here for calculating the recoil forces as the heat energy portion has no effect on the recoil and thus can be safely ignored.

The propellant gas kinetic energy has two components, the energy just as the projectile leaves the barrel (shot ejection) and the energy after the projectile leaves the barrel, which is usually known as the "gas after effect," which acts somewhat like the exhaust from a rocket. The total kinetic energy created by the propellant gas is thus the sum of these two factors and is called the "Total Propellant Gas Kinetic Energy after shot ejection."

Propellant Gas Kinetic Energy at shot ejection = 0.5 * (W_c/g) * (V_o² / 2)

Total Propellant Gas Kinetic Energy after shot ejection = 0.5 * ((W_c/g) / 2) * w²

However, these Propellant Kinetic Energy equations are really describing only the "apparent" global kinetic energy of the gases. The kinetic energy of the propellant gases is actually much higher due to internal vortices, gas flows, chemical conversions, etc. Since these energies are not necessarily all going in the same direction, they do not all contribute to the momentum balance. The end result of all of these items is that the above energy equations cannot be used to determine propellant gas momentum. Fortunately, empirical analysis of the recoil forces has shown that the following Propellant gas Momentum equation to be approximately correct:

Propellant gas Momentum = (W_c/g) * w

Now, with these equations in hand, here are the calculations to compute the broadside momentum using both the projectile momentum and the propellant gas momentum.

The momentum of a single projectile can be calculated as follows:

Projectile Momentum = (W_p/g) * V_o

= (2,700 / 32.174) * 2,500

= 209.800 x 10³

The momentum of the propellant gasses for a single shot can be calculated as follows:

Propellant gas Momentum = (W_c/g) * w

= (650 / 32.174) * 3,937

= 79.538 x 10³

Summing these:

Total Momentum = Projectile Momentum + Propellant gas Momentum

= 209.800 x 10³ + 79.538 x 10³

= 289.338 x 10³

The Broadside Momentum for 9 projectiles can now be calculated as follows:

Broadside Momentum = 9 * (momentum of projectile + momentum of propellant gasses)

= 9 * (209.80 x 10³ + 79.538 x 10³)

= 2.604 x 10⁶

Using Greg's formula, the velocity of an Iowa firing a 9-gun broadside can be recalculated as follows:

Mass of broadside * Velocity of broadside = Mass of ship * Velocity of ship

As the Mass of broadside * Velocity of broadside term is equivalent to Broadside Momentum, this formula can be restated as follows:

Broadside Momentum = Mass of ship * Velocity of ship

Solving for the velocity of the ship and using the above calculated momentum figures:

Velocity of ship = Broadside Momentum / [Mass of ship]

= 2.604 x 10⁶ / [58,000 * (2,240 / 32.174)]

= 0.645 fps => 7.74 inches per second

So, when calculating for both the projectiles and the propellant gasses, the ship's velocity on ice with the guns firing at zero degrees elevation would be about 7.74 inches per second rather than the 6 inches per second calculated above for just the projectiles. When one considers that any sideways motion of the ship through water is actually resisted by the wall created by the hull of the ship, whose wetted surface is about 860 feet long and 38 feet deep, then it can be easily understood that Dick Landgraff's comment above, "theoretically, a fraction of a millimeter," is closer to the truth.

Hopefully, this further analysis will satisfy the purists and other doubters out there. But, if not, then please take a look at these two photographs:

Take a close look at the wakes of these two ships. See how straight they are? If these battleships had really been pushed sideways by their broadsides, you would see a "kink" in the wakes indicating the movement.

A couple of days ago, I ran across a forum that linked to this essay. To my surprise, I found that people were asking:

"Why do the pictures say '15-gun broadside' when Iowa and Missouri only have nine guns?"

Let me quote from "Gun Data and Definitions - Part 3" data page:

Broadside

Firing in a single salvo all guns that can bear on an abeam target. This may involve more than one caliber of weapons.

"Broadsides" comes from the days of sail, when ships often carried multiple calibers of guns. An order to fire a broadside meant that the gunners fired every weapon which could bear as the enemy ship came abeam.

Now, look very closely at Iowa in the above photograph. Note the small "puff balls" of smoke in the center of the photograph. These are from the six starboard 5"/38 secondary guns. In other words, Iowa and Missouri are firing all guns that can bear on the broadside - nine 16-inch guns and six 5-inch guns or 15-guns total.

Update of 04 August 2021 uses new Propellant gas Momentum formula used above.

It was bound to happen, I guess. I received an Email today asking why there were no calculations for the extra momentum generated by firing the six 5-inch guns.

Sigh.

• Weight of 5-inch projectile: 55 lbs.
• Weight of propellant charge: 15.5 lbs.
• New gun muzzle velocity: 2,600 fps

The momentum of a single projectile can be calculated as follows:

Projectile Momentum = (W_p/g) * V_o

= (55 / 32.174) * 2,600

= 4.44 x 10³

The momentum of the propellant gasses for a single shot can be calculated as follows:

Propellant gas Momentum = (W_c/g) * w

= (15.5 / 32.174) * 3,937

= 1.897 x 10³

The Broadside Momentum for 6 projectiles can be calculated as follows:

Broadside Momentum = 6 * (momentum of projectile + momentum of propellant gasses)

= 6 * (4.44 x 10³ + 1.897 x 10³)

= 38.022 x 10³

Solving for the velocity of the ship and using the above calculated momentum figures:

Velocity of ship = Broadside Momentum / [Mass of ship]

= 38.022 x 10³ / [58,000 * (2,240 / 32.174)]

= 0.009 fps => 0.113 inches per second

So, the six 5-inch projectiles add about a tenth of an inch (0.1") per second to the velocity calculated above.

In the ten years since I last updated this Tech Essay, I have still occasionally received Emails claiming that the battleships really do move 30 feet (10 m) when the guns fire. Usually, I gently mention that if this did happen then there must be video of it. I then suggest that the questioner should look at the following video, find the places where the camera is pointing down at the bow guns when they fire as part of a broadside and then give me the time stamp of where they see this movement occur. I rarely hear back from them.

USN 16-Inch Gun Training Film

I am going to do a little "critical thinking" in this section to try to further show that large movements do not occur when a battleship fires a broadside. Do not worry, there will be little math in this section, just some observations and deductions.

Inertia

Let us think for a moment about what would happen if a battleship actually moved 30 feet (~10 m) when she fires a broadside. For the 16"/50 guns on the Iowa class, when the guns are fired with full charges at a +15 degree elevation, recoil lasts 0.43 seconds and counter recoil (runout) lasts 0.90 seconds. This means that if the ship actually moved 30 feet, then it would have do this in the half a second that the guns are recoiling as all force pushing the ship ends when the guns reach their recoil limits. Now, think of what would happen to the gun crew should this occur. The ship moves 30 feet in half a second, but, due to inertia, the gunners do not move from where they were in regards to the earth's surface prior to the instant before the guns fired. In other words, to an observer securely fastened to the gunhouse deck, it would appear that the gun crew suddenly moved in the opposite direction from where the guns were pointing and were thrown violently into the front of the gunhouse. Similarly, any crewmen on the weather deck would suddenly move 30 feet towards the firing side of the ship or, worse, be tossed overboard. Likewise, any loose equipment on the ship would also move 30 feet. However, none of this actually happens. Again, look at the above training film, this time at the crewmembers standing around when the guns fire. Do they suddenly get thrown 30 feet from where they are standing? Why not? They should if the ship suddenly moved 30 feet. The answer is it is because the ship simply does not move 30 feet, it actually moves only "a fraction of a millimeter."

Displacement

Next, let us consider the water surrounding the ship. As mentioned above, an Iowa displaces about 58,000 tons of water and the wetted hull of the ship is about 860 feet long and 38 feet deep. Now, a cubic foot of seawater weighs about 64 lbs. (29 kg). If the ship moves 30 feet sideways through the water, then that means that it would displace 30 x 860 x 38 cubic feet of water, which works out to about 63 million pounds of seawater that needs to be pushed out of the way. This is roughly about half of the weight of an Iowa. Do not forget, you still need to move the 58,000 tons that the ship weighs along with the weight of the displaced water in order to move the ship 30 feet. Does this seem likely to you?

You should also note that moving the ship 30 feet broadside through the water in about half a second means that we should see a huge "hull wave" alongside the ship, with much of this water flooding up and over the side of the ship and onto the deck. Looking at the photograph at right, you can see that this simply does not happen and, in fact, the water on the opposite side of the ship barely moves. Why not? The answer again is that it is because the ship simply does not move 30 feet, it actually moves only "a fraction of a millimeter." ¹

Bow Flex

Finally, let us talk about the construction of an Iowa class battleship. When an Iowa is at speed, it is well-known that their bow flexes around due to the structural discontinuity where the armored citadel ends and her long, narrow bow begins. Think about what would happen should she move 30 feet broadside while moving at 30 knots. Would not the bow bend or break at the discontinuity? For that matter, would not the entire hull be subject to tremendous stress from all that water being pushed aside? Should we not expect to see dished-in or warped plates along the sides where the pressure of moving 63 million pounds of seawater overwhelmed the hull plates which were 0.75-inch (1.9 cm) thick in many places? Again, why not?

Summary

This section is meant to give some practical examples of what it would really mean should a battleship suddenly move 30 feet when her guns fire. Inertia does not go away just because the ship is in water nor does displaced water suddenly become as thin as air when something as large as a battleship moves through it. The math in the previous sections explain why this movement does not happen, this section was meant to show why the math is correct.

NOTE

• ^If you click on this photograph of USS New Jersey and use the magnifying tool, you can look along the waterline and see several places where the bilge pumps are working. Looking at the one in between Turrets I and II (the bow turrets) and the one just behind Turret II, you can see several feet of white water where the bilge water hit the ocean as the ship traveled along the coast. Note how these bilge discharges form straight lines. If the ship had moved any distance sideways at all when she fired her guns, you would see a "kink" in these streams. The fact that there is no noticeable kinks here, just like there is no kink in the wakes of USS Iowa and USS Missouri in the photographs above, is the clearest evidence possible that battleships move sideways only a minuscule amount when their guns fire.

12 July 2006 - Benchmark
16 June 2006 - Emphasis added to the ton to pounds conversion factor in Greg Locock's comment in an effort to stop vision-impaired individuals from sending Emails about the "missing" 2,000 in the equations that follow
18 July 2010 - Added new information and calculations provided by Leo Fisher in Addendum and Yet Another Addendum regarding Kinetic Energy and Momentum of propellant gasses
27 June 2016 - Converted to HTML 5 format
05 May 2020 - Added Inertia and Displacement section
04 August 2021 - Updated propellant gas momentum formulas and added photograph of USS New Jersey (BB-62)