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

The only justification was that using the exponential equation gives number for velocity at half and 90% of the barrel travel that’s close to that cited in the post on the barrel patent. Data from post: at half the barrel travel v = 0.8 muzzle velocity, equation = 0.85. At 90% of barrel post = 98% and equation 97%.

There is another effect that we have ignored and that’s friction. There are copper rotating bands on the shell that are slightly oversized for the lands on the rifling. The heat buildup as that band is forced into the rifling melts the copper and might even turn it to a gas. This would result in a static friction term until the copper begins to melt and a fluid friction term proportional to velocity thereafter.

My guess is that gun designers don’t simulate what goes on inside the barrel too closely and rely on instrumented barrels and test firings in the design. I’ve been at the naval gun center at Dahlgren, VA several times. Had I known the question would arize 50 years later I would have gone over and talked to their engineers so as to be prepared. :wink:

Ooops! I hadn’t actually clicked on the OP’s link. Too cool, though. Love the depressions in the water.