Do Battleships move sideways when they fire? (2024)
Momentum from 16-inch guns
Original essay from 2 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 Loco*ck 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 calculated above.
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:
Given:
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 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
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:
Introduction: My name is Kimberely Baumbach CPA, I am a gorgeous, bright, charming, encouraging, zealous, lively, good person who loves writing and wants to share my knowledge and understanding with you.
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