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:
Solving for the ship's velocity:
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
Momentum including Propellant Gasses
Addendum added 12 July 2010, updated 04 August 2021
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 fps2
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
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
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:
Free Recoil Velocity:
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.
As can be seen above, the projectile kinetic energy is determined by the following equation: