How long does a 16" battleship shell stay in the barrel, when fired?

A naval ballistics question—can anyone tell me how long a shell from a US Navy 16"/50 Mark 7 gun, as used on Iowa class battleships, stays in the barrel of the gun after being fired, but before leaving the muzzle?

Say—a 1900 lb “HC” shell, with a muzzle velocity of 2,615 fps.

That’s…it, I guess. Further technical details are available as needed. Can anyone help me out?

If you assume a constant acceleration from a standstill to muzzle velocity (which may or may not be true), you can calculate that the average speed through the barrel is half the muzzle velocity, of 1307 fps.

If you take the length of the barrel from your link, 66.7 ft (800 in), and divide by that velocity, you get 0.051 seconds.

A non-constant acceleration will change that answer, but I’m not an expert in internal ballistics, so I can’t help you refine that rough estimate.

The bore length is given as 800 inches (66.7 feet). Since the muzzle velocity is given as 2615 fps, the average velocity as it travels through the barrel is 2615/2 or 1307.5 fps (since it is accelerating at a more-or-less uniform rate from zero velocity). Simple algebra then tells us that the projectile took 66.7/1307.5=.051 seconds to exit the barrel.

Yow, killer!

Now, I’ve just been crunching some numbers—so, with my math skills, I’d figure my figures are dubious, at best—and it looks like to accelerate an 862 kg (1900lb) projectile to 797 meters per second, I’d need a force of 687,014 Newtons, right? And so…if I’m doing this right, it would take 20 times that force (or 13,740,280 newtons) to do that in 0.05 seconds, right?

Just to be rigorous about it, it takes 687kN to accelerate an 862kg mass to 797 m/s in one second. Your math is right, you just have to specify the acceleration involved since F=m*a (a=velocity/time). Your 13.7MN number is also correct when the time drops to 0.051 seconds.

According to this sitethe initial chamber pressure for larger naval rifles runs as high as 40000 psi.

Using that number for a 16" gun the total initial force 8.04 million pounds.That force will give a 1900 lb shell and acceleration of 1.36*10[sup]5[/sup] ft/sec[sup]2[/sup]. For that acceleration it takes 0.019 seconds to reach 2615 ft/sec.

However, the acceleration of far from constant. I suspect it falls off exponentially. Most things do. The pressure and temperatur of the gas in the barrel both decrease. The temperature goes down because the gas is expanding and it is doing work on the shell.

Actually solving for a theoretical barrel pressure curve and from that the velocity looks pretty difficult and I’m not sure my math skills are up to it.

I’ll fiddle around but I don’t expect much output.

On further thought, the pressure profile is even harder to work out that I first thought. The burning rate of the propellant charge counts. The burning rate could be somewhat matched to the geometric volume change so as to try to keep the pressure up as long as possible as the shell travels down the barrel. That would make the 0.019 seconds computed above close to correct because the site gives a max barrel pressure in the neighborhood of 40 K psi. If the pressure is maintained close to that value, that and the shell base area gives the force on the shell and it it’s approximately constant …

All right, that’s it. I do not want to be shot out of a battleship cannon.

I’m not sure you even want to be on deck when one goes off.

One of the survivors of the HMS Hood was at his battle station when the fatal shell hit. His battle station? Lying face down on the main deck. :eek:

Can he still hear?

OK. If the pressure is really constant the speed builds up linearly and the 0.051 sec. is correct.

However if the max pressure is 40 K psi and the pressure is constant then 0.019 sec. is correct.

Both can’t be true and the only conclusion is that the pressure can’t be constant.
It does appear that the answer is somewhere between the two answers. I don’t think we can determine what it is from the info we have.

The pressure is not maintained at that value. Guns, like piston engines, need to allow the gasses to expand, and the pressure to drop in order to convert a significant amount of the chemical energy into mechanical work. Powder is used most efficiently if it burns instantaneously, but this would rupture the barrel, so slower burning powder must be used as a compromise. This is tricky business, as modern powder burns progressively…higher pressure gives faster burn rates.

If the pressure were still 40Ksi when the projectile left the muzzle, the resulting concussion would most probably be debilitating to the crew…not that these guns are quiet as they are.

All of which is germane to my conclusion that the pressure can’t be constant.

I found this in a patent on a new type of rifle barrel. It’s way down past the claims section in the invention background part under “state of the art.”

Like I said. We can’t really answer the question.

Anybody know what the muzzle velocity of the 1917 'PARIS" gun was? This thing had a range of 75 miles! The late Dr. Gerald Bull was fascinated by this german gun-I wonder what could be made today-in terms of long range guns?

Not just modern powder. Some of the huge late black-powder-era shore guns used powder with grains the size and shape of hex nuts. This gave a softer initial combustion that sped up as it progressed. This was needed to overcome the inertia of those massive anti-ship shells.

I think I lurched into a way to get an approximate answer based on the information about barrel velocities in my post above.

You can get an exponential velocity profile by using the data that the velocity is 98% of the muzzle velocity at 90% of the barrel travel. I assume that the travel is 61 ft., the other 5.5 ft being taken up by the propellant charge.

That velocity as a function is distance is:

v(d) = 2615(1 - e[sup]-0.06387d[/sup])

Then the distance traveled can be broken up into 5 ft long segments. Make the assumption that the velocity varies linearly during the segment. The time to travel each segment is the disance travelled divided by the average velocity during the segment. Compute these times, add them all up and we have the approximate time in the barrel.

My answer is about 38 milliseconds.

We’d need someone of a higher caliber in any case.

Hardehar. Nevertheless, I will share this awesome picture with you, of the mighty USS Iowa firing a salvo:

http://en.wikipedia.org/wiki/Image:Uss_iowa_bb-61_pr.jpg

Hmmm…I know I’ve seen that picture before

As others have stated, if we assume a constant acceleration, we get a figure of 51 milliseconds. This is probably an upper bound on the time spent in the barrel, since the pressure probably monotonically decreases after it’s fired. Meanwhile, if we assume instantaneous acceleration at the beginning and a constant velocity, we get half that time. This is a lower bound, since the only way to get a shorter time would be for the projectile to slow down at some point before leaving the end of the barrel. And it’s probably too low by a fair bit, since if they could achieve muzzle velocity instantaneously, they wouldn’t have such a long barrel. With an upper bound that’s too high and a lower bound that’s too low, separated by only a factor of two, you can get a pretty good answer just by splitting the difference. I don’t know whether David Simmons’ assumption of an exponential velocity profile is correct, but any plausible assumption about the velocity profile will get you about the same answer, so it works.