• You are using the old Black Responsive theme. We have installed a new dark theme for you, called UI.X. This will work better with the new upgrade of our software. You can select it at the bottom of any page.

Battleships move sideways on broadsides?

Status
Not open for further replies.

ZekeLuvs1911

Member
Joined
Dec 24, 2002
Messages
368
Location
Avenel, NJ
Do battleships move sideways on broadsides?

I've seen this question brought several times and I just had to post the answer again. :D
Moderators, it is gun related...........;)

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.). 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 about 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 cosign 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
 
Last edited:
Yeah, well, the cosine of 25 degrees is 0.90, so you gotta be shooting at a pretty good upward angle before it's significant. :) That's why uphill/downhill isn't much of a range-estimation factor in most hunting shots.

The big problem is persuading folks that "little motion" isn't the same as "no motion".

:), Art
 
This means that the horizontal velocity imparted to the ship would be even less than the numbers calculated above.
Which assumes a non-reactive medium...and salt water isn't.

Simple physics...there is indeed a lateral movement, something the fire-control computers would take into account.

The formula is a constant; it's the same for 16" naval rifles as it is for a three-pound 1911A1 being fired from a human platform. Take the variables into account and report accordingly.
 
Which assumes a non-reactive medium...and salt water isn't.

It's hardly a frictionless environment, either. As a matter of fact, I can't imagine a much better hydraulic brake than trying to push an 800'x30' steel wall through the water. ;)
 
A 16" Mark 8 APC shell weighs 2,700 lbs. and the muzzle velocity when fired is 2,500 feet per second (new gun).
Hmmm.... A 16 APC... Anybody sell a CCW holster for that pistola? :rolleyes:
 
It's hardly a frictionless environment, either. -- Tamara
Agreed...but doesn't change my point.

Let's assume a projectile weighing a ton fired at 2700 fps...

The formula for muzzle energy is:

(MVxMV ÷ 7000) ÷ 64.32 x BW

MV = muzzle velocity

BW = bullet weight in grains [7000 grains to the pound]

What does your calculator spit out?

Three sixteen-inchers fired simultaneously aren't going to change the rotation of the earth, but they are darn sure going to affect the attitude and course of a battleship.

Simple physics...
 
Zander,

No need to drag out my calculator, the equation's in the first post of the thread. In a frictionless environment, the ship would be accelerated to .46 fps. Figure the resistance of incompressible water into the equation, and that acceleration is reduced to "not worth measuring". ;)
 
Not worth measuring, and besides that, battleships have been decommisioned anyway:(
 
Actually, the New Jersey, Missouri and Wisconsin have been converted to floating museums. The Iowa is currently in inactive status with her repair materials for her #2 turret stored inside. She can come back if notice is given.

Here's a link for interesting info:
http://www.warships1.com/
 
There was a show in these big boats on TLC recently, I want to say they quoted an operational cost of $1 million per *day* for one Iowa class. Easier to mothball them and then bring them back later if they need...at a billion or so a year to keep all 4 operational.
 
As with the point brought up here and that I have seen elsewhere, it appears illogical that the guns could generate enough force to move such a big ship, right?

As such, it would make sense that I could not push a 55 ft ocean-going fishing boat loaded down with gear and people for an extended trip away from the dock with one arm, but I did. I could never do that on dry land, but water offers a unqiue environment for the drill. Similarly, I watched a 60 lb. little kid push a pickup truck (GMC) several feet with his daddy behind the wheel and the vehicle in neutral. It would appear that a 60 lb kid could not generate enough force to move such a heavy object (and his daddy wasn't exactly lean, either), but he did. Amazing.
 
The question appears to be moot anyway. I seem to remember that one of my old navy buddies said they have an interlock system where they can't fire all nine at once. I think one gun per turret at a time was what he said. :confused:
 
BigG,

They can fire full broadsides, but I think individual tubes next to eack other are staggered fractions of a second apart so that the shockwave from the muzzleblast and wake turbulence from the shell don't affact the trajectory of the nxt tube over...
 
I controlled a few fire missions from New Jersey in Vietnam. They were quite accurate at about fifteen miles range, adjusting fire as I called for it, firing for effect (!) at bunkers. I ran out of fuel before they ran out of ammo. :D

TC
TFL Survivor
 
griz,

Nice! We ought to start dropping full-color leaflets of that kind of thing (Army artillery, Air Force cruise missile strikes) all over Iraq. Maybe give a few of those "hardened veterans" something to think about...
:evil:
 

Heh...Neil, I was in the USMC. I wodnered if that use of the word would attract some attention.

Leatherneck: I was a fire controlman during Desert Storm. One thing I envied the Navy....they had the biggest guns. :mad:

Although in restrospect, I guess they don't make a truck big enough to tow a 16" piece.
 
Status
Not open for further replies.
Back
Top