On Dec 20, 7:37*am, Doug Miller
wrote:
nestork wrote in news:nestork.b187e08
@diybanter.com:



There's been a lot of talk about guns in here lately, and there's
something I've always wondered about...

I don't know how far a regular pistol or rifle will shoot,

A *lot* farther than you think it will. :-) Read on.

but this web
site about the battleship USS Missouri (which is the one that McArther
accepted the Japanese surrender on at the end of WWII)

http://en.wikipedia.org/wiki/USS_Missouri_(BB-63)

states as follows:

"Missouri's main battery consisted of nine 16 in (406 mm)/50 cal Mark 7
guns, which could fire 2,700 lb (1,200 kg) armor-piercing shells some 20
mi (32.2 km)."

20 miles! *Geez. *Obviously, that's a lot farther than any rifle or
pistol will shoot.

No, it isn't.

Is that because:

a) the gunpowder used in those cannons is more powerful than the stuff
used in regular bullets,

No. It is more powerful, but that's not the reason for the extremely long range.

b) or is it because the battleship's cannons are pointed upward to
maximize range,

Yes -- but if you do the same thing with a rifle, you'll get the same result: greatly increased
range, compared to firing it with the barrel horizontal.

c) or is it that the cannon's barrel is longer so the power in the
explosion acts on the projectile for a longer period of time, thereby
accelerating the shell more than a bullet

No. The muzzle velocity on those guns is in the neighborhood of 800 m/s, which is actually
15-20% *lower* than that of many common hunting rounds (e.g. typical muzzle velocity for a
30-.06 is around 880 m/s).

d) or something else entirely.

Yes -- because the battleship shells are so large, they are less affected by air resistance
than are rifle bullets. I have a .243 rifle which fires an 85-grain round at a muzzle velocity of
about 1000m/s. A quick back-of-the-envelope calculation shows that -- in the absence of air
resistance -- if the barrel is elevated at an angle of 45 degrees, the bullet will travel over 100
kilometers before striking the ground. Obviously that won't happen, because it *is* subject
to air resistance. While the battleship shell is subject to the same air resistance forces as
the bullet, the magnitude of those forces in proportion to a 1200-kilogram shell is obviously
much smaller than in proportion to a 6-gram bullet, and hence their effect on the trajectory of
the shell is correspondingly much smaller.



It just strikes me as odd. *I'm thinking that the ratio of lengths
between a bullet in a rifle barrel and a shell in a cannon barrel would
probably be pretty similar.

No, actually, it's not -- the ratio is a *lot* *higher* for a typical hunting rifle than it is for those
cannons.http://www.navweaps.com/Weapons/WNUS_16-50_mk7.htm

With an 800-inch barrel and a 64-inch projectile, the cannons have a ratio similar to that of
most *pistols*.

So, what accounts for the cannon having 20
times the range?

It *doesn't* have "20 times the range." What makes you think that a rifle can shoot only one
mile?

Is it just that the gunpowder used in the cannon is 20
times as powerful as than that used in bullets,

It's not.

or is it that the shell
stays in it's barrel 20 times longer than a bullet

It doesn't.

and therefore
receives 20 times as much of a push?

It doesn't.

It's all about angle of elevation, and the fact that battleship shells weigh two hundred
thousand times as much as rifle bullets and consequently aren't affected nearly as much by
air resistance.

If the cannonball is larger, it must be more affected by air
resistance so something is wrong with your conclusions. Now, if the
ratio of diameter to weight is different, then maybe what is
postulated could be true.