Do Battleships move sideways when they fire?

By R. A. Landgraff and Greg Locock
Updated 02 August 2000

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.

Tony DiGiulian

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.

R. A. Landgraff

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.) - Emphasis added on 16 June 2006 in an effort to stop vision-impaired individuals from sending Emails about the "missing" 2,000 in the equations below.

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.

Greg Locock

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.

Tony DiGiulian

by Tony DiGiulian
Updated 12 July 2010

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:

          Projectile Weight: Wp = 2,700 lbs.
          Charge Weight: Wc = 650 lbs.
          Muzzle Velocity: Vo = 2,500 fps.
          Weight of Recoiling Parts: Wr = 250,000 lbs.
          g = 32.174 fps^2

Projectile Kinetic Energy = 0.5*((Wp/g)*Vo^2) = 2.622*10^8 ft-lb.

Propellant Gas Kinetic Energy at shot ejection = 0.5*(((Wc/g)/2)*Vo^2) = 3.174*10^7 ft-lb.

Emperical data has shown that only about half of the propellant gas has been accelerated to the muzzle velociy at shot ejection, hence the "extra" divide by 2 in the above equation.  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 = 0.5*((Wc/g)/2)*w^2 = 78.29*10^6 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*Vo, therefore B = 1.5748

Free Recoil Velocity: Vre = (((Wp/g)+B*(Wc/g)/2 + (Wc/g)/2))/(Wr/g))*Vo = 35.37 fps

Recoil Energy = ((Wr/g)/2)*Vre^2 = 4.86*10^6 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.

Now, with these factors in mind, here are some calculations to calculate broadside momentum.

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

    Projectile momentum = (Wp/g) * Vo
                                     = (2,700 / 32.174) * 2,500
                                     = 209.800 x 10^3

Momentum of the propellant gasses can be calculated using the same assumption that only half the propellant gasses have been accelerated to muzzle velocity at time of shot ejection with the other half exiting at velocity w following shot ejection:

    Propellant gas momentum = ((Wc/g) / 2) * Vo) + ((Wc/g) / 2 * w)
                                            = ((Wc/g) / 2) * (Vo + w)
                                            = ((650 / 32.174) / 2) * (2,500 + 3,937)
                                            = 65.022 x 10^3

Summing these:

    Total momentum = Projectile momentum + Propellant gas momentum
                              = 209.800 x 10^3 + 65.022 x 10^3
                              = 274.82 x 10^3

The Free Recoil momentum calculation can be used as a check, as it should be about equal to the sum of the momentums of the projectile and propellant gasses.

    Free Recoil momentum = (Wr/g) * Vre
                                        = (250,000 / 32.174) * 35.37
                                        = 274.83 x 10^3

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 10^3 + 25.253 x 10^3)
                                       = 2.115 x 10^6

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]
                                 = 2.115 x 10^6 / [58,000 * (2,240 / 32.174)]
                                 = 0.52 fps

So, the ship's velocity ON ICE with the guns firing at zero degrees elevation would be about 6.3 inches per second rather than the 6 inches per second calculated above.  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:


USS Iowa BB-61 firing a 15-gun broadside in 1984
U.S. Naval Historical Center Photograph DN-ST-85-05379


USS Missouri BB-63 firing a 15-gun broadside circa July - August 1987
U.S. Naval Historical Center Photograph NH 96814-KN

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.

And a further Addendum

Updated 28 November 2005

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 top 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.

And yet another Addendum

Updated 18 July 2010

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 ***

5-inch projectile:  55 lbs.
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 = (Weight of projectile/g) * Muzzle velocity of projectile
                                     = (55 / 32.174) * 2,600
                                     = 4.44 x 10^3

Momentum of the propellant gasses can be calculated as follows:

    Propellant gas momentum = ((Wc/g) / 2) * Vo) + ((Wc/g) / 2 * w) = ((Wc/g) / 2) * (Vo + w)
                                            = ((15.5 / 32.174) / 2 * (2,600 + 3,937)
                                            = 1.57 x 10^3

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 10^3 + 1.57 x 10^3)
                                       = 36.06 x 10^3

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

        Velocity of ship = Broadside Momentum / [Mass of ship]
                                 = 36.06 x 10^3 / [58,000 * (2,240 / 32.174)]
                                 = 0.009 fps

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

Page History

12 July 2006 - Benchmark
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


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