Nothing. You said a .50 wouldn’t penetrate the wood and I showed that you are incorrect.
Ah, you quoted the wrong part of my post. You quoted the part where I was disagreeing that the Burke could ram the HMS Victory without damage.
But yes, you are correct the .50 could penetrate 'Pine Wood - 32"- but seasoned oak will give significantly different numbers. However, I will concede a .50 could indeed penetrate that wood, especially given full auto. Nice cites.
HMS Victory displaced 3,500 tons in 1805. Note: the linked cite is a British website, so these are presumably “long tons” (2,240 lb). A “short ton” (2,000 lb) is the commonly used measurement in the U.S. A “tonne” (or “metric ton”) is 1,000 kg, which is equivalent to 2,204.62 lb.
The displacement of an Arleigh Burke-class destroyer ranges from 8,184 to 9,600 long tons (9,166-10,800 short tons).
Short version: the displacement of an Arleigh Burke-class destroyer is approximately 3 times that of HMS Victory.
it would be doubtful if any of the ordnance on the Arleigh Burke would survive to fire back. One hit from a 24#- and there are about 50 such cannons would destroy any mount on the destroyer. And the HMS Victory could easily get off 4-5 such broadsides before the destroyer managed to react. Hell, even grapeshop would penetrate and there were nine of them per load.
The numbers are correct, but short tons are never (correctly) used for ship displacement, even in the US. US ship displacements have always been quoted in long tons, as they were in Britain. And outside the US now (basically) quoted in tonnes of 1000kg.
It’s the byproduct you see sometimes of ‘written by the masses’ encyclopedia (Wiki). It’s not an outright wrong thing to correct, converting long tons to short tons and showing the displacement of a DDG in short tons, but nobody involved in the navy or ship design would ever quote it that way and whatever person came up with that is still creating somewhat of a mistaken impression.
The OP would be better served by not trying to come up with ridiculous scenarios. Just pose the question as, “In a testing condition, imagine [some specific 1625 English ship] placed 100 yards away from an Arleigh Burke class destroyer with their broad sides facing each other. The 1625 ship fires all of its cannon simultaneously. Does the destroyer sink? Can the destroyer return to port under its own power? How much damage is done to the hull?”
Oh it would, even if it was 36 inches thick, same way it penetrates concrete and other non homogenous destructible materials.
Think of it as a very dangerous multi purpose saw.
That’s interesting. I was in the U.S. Navy (though not involved in ship design), and I was not aware that the “tons” used for ship displacement were actually long tons (or if I did, I’ve evidently forgotten).
I just looked at the public information guide with the relevant specs (of which I still have a copy) for my submarine from the early '90s as well as the current U.S. Navy data sheet here, and it simply refers to the displacement in “tons” (as well as the equivalent in metric tons). Doing the math, it’s apparent that in both cases that the tons are in fact “long tons.” Thanks for bringing this to my attention.
I concur. I where people brought in ships from much later periods which violates the OPs original scope. Please people, focus, ships from 1625 era not later (or earlier).
Something that I thought about is the alloy of the steel of the cannon balls versus the steel on the ship. This does matter as armor, assuming that is what the ship is using, has different properties than say hot or cold rolled steel. Namely, they try to prevent fragmentation when the armor is penetrated. It would probably be better to make an armor that would allow a round to pass through with significant reduction in energy without fragmentation versus an armor that stopped the round but then fragmented and sent steel shards everywhere. Somewhere in between is the ideal, exactly, I do not know.
Back to the alloy of the ball. It was designed to go against a wooden ship, at an ideal distance (to aim and allow for ball drop), said ship would have an estimated thickness as well. This would also affect the charge behind the ball. The result could be more disastrous than one would think. For that first salvo, probably go for the max charge. This could cause the ball to penetrate and then proceed to shatter.
I’m not familiar with the construction of an Arliegh-Burke, but, I’m pretty sure all ships now a days, war ships or not, have double-hulls. So, there is a gap of some sort between the “outer” and “inner” hulls. Assuming armor, the ball would more than likely penetrate the outer-hull, possibly penetrate the inner-hull at a significantly reduced amount of force, if it did. I imagine the ball would fragment if it did make it through the second hull and shred everything. Any ways, for sure, some serious damage to the struck side, I don’t think it would sink, sever list, yes, not not sink. Redundancy kicks in and the 1625’s ship (not before or later) is going to have a really bad rest of the day.
The 5" gun on the Arleigh-Burke uses a 70lb shell, 16-20 rounds per minute, with a muzzle velocity of 2500 ft/s. That gun, by itself, will fuck that other ship up all by itself. You better hope that 1600’s ships took that thing out. If not, the A-B has some torps on it, which will for sure finish the job.
I’ll admit, I’ve only performed some very basic research, and as someone mentioned the the HMS Victory of the 1620’s, it was a tiny thing, 109 feet long. It did have a lot of guns at 42, but that’s be 21 per side. But still, 109 feet long, say three rows, they can only be so big with a ship of that size. Depth of hold, I assume draft but I could be wrong, is 17 feet, that’s not much. Yeah, a formable ship in its day, but, after that first salvo it’s a goner.
I don’t think it’d be possible for the HMS Victory of 1620 to take out both the gun and the torps. As they did not have guns like that, they might think it’s something else. As the broadside of the ship is way easier to hit, and it’s simply larger, and you want the highest rate of success, you’d go there first.
We’re assuming that the olden ship is awfully close. Could the destroyer even use torpedoes against a target that close?
I don’t see it as a problem to bring in later ships or guns, if similar, if the purpose is to find hard data to answer the general line of question, especially as the exact question is entirely arbitrary. So as I mentioned twice there were actual trials of 32 pdr smoothbore cannon of the 19th century (not a lot different than 17th century ones, a period of very slow technical advance in weapons, mid 17th to mid 19th centuries) v. wrought iron armor. And we can then relate those results to the known differences in 19th century wrought iron armor and modern high strength steel, the known (quite modest) thickness of shell plating on a modern DDG, and easily conclude the largest smoothbore guns of the long period of naval technological statis from mid 19th back to at least mid 17th century would penetrate such structures with full charges and single shot at close range.
As to ‘alloy of steel’ solid shot of that era, aka ‘cannonballs’, were cast iron*, a form of iron/carbon composite with a relatively high % of carbon, no other metals mixed in specifically for their favorable properties, and relatively lesser control of impurities. High strength steel is an iron/carbon composite with low and closely controlled % of carbon, other alloying metals in small % specifically for their favorable properties, and tight control of impurities. And the hull plating of modern warships is pretty generic high strength steel, an alloy called HSLA-80 in the latest ones, though ‘rolled homogeneous armor’ and high strength structural steel are similar or the same thing in the limit. These plates are not specialized forms of armor with particular hardness properties to break up projectiles or specifically control spall (ie pieces of the plate breaking off the back if the armor stops the projectile, or from around the sides of the penetration if it goes through).
It’s just <1" thick high strength steel plate and not going to stop a 32 pound cast iron cannon ball moving at ~1500 ft/sec, given trials showing such projectiles denting 4" wrought iron plates to half or more their thickness, the difference in HTS and wrought iron properties being not that dramatic, see above. If considering longer range or smaller cannon, eventually it would get into having to calculate, or simulate, or do trials (but for what real reason I don’t know 
). 32 pdrs became common lower deck guns on ‘liners’ close enough to 1625 (if perhaps not quite that early) for government work as they say.
*at the close of the smoothbore period some round shot were wrought iron and had somewhat higher performance v/ metal targets.
Arleigh Burke destroyers are equipped at least one Phalanx CIWS canon and some have two, each of which can fire 75 rounds per second. The R2D2-shaped front and rear Phalanxes can be seen here on the USS Barry: USS Barry (DDG-52) - Wikipedia
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Phalanx can be used against surface targets and has several operating modes: standby, manual, automatic, etc. Obviously the OP scenario is highly dependent on the readiness state of each ship. If in auto mode, the Phalanx would track and fire within about 1 sec.
Destroyers are essentially blue-water vessels and their tactics, radar and armaments are oriented around that engagement style. The OP scenario is a “brown water” or “littoral combat” scenario. Normally a destroyer would not go blindly creeping past a rocky hill next a coastline. If they were ordered to do that, the possible risk would be obvious, and they might have Phalanx on full automatic.
More likely the destroyer would have a small aerial drone scouting ahead, such as ScanEagle. This would identify the threat, the destroyer would pull back and take necessary action: http://www.naval-technology.com/projects/scaneagle-uav/
Oh, sure, the destroyer has plenty of weapons that would be quite effective against a wooden sailing ship, as well as means to prevent being surprised. I was just questioning whether torpedoes specifically would be effective.
And if we want a non-test situation for the wooden ship to be able to get close enough, I think the simplest is a war-starting sneak attack. The wooden ship and the destroyer are scheduled to be maneuvering in the same waters as part of some sort of photo op showing the progress of technology, or the long history of peace between two countries, or some nonsense like that, so the crew of the destroyer aren’t surprised or alarmed to see the wooden ship right next to them. But they are surprised when, for some reason unknown to them, the wooden ship suddenly fires a full-out broadside at them.
What are the detonators of those torps? magnetic?
Getting that close would mean raking the wooden ship with a .50 machine gun would kill much of the crew and cause serious top-side damage.
Guns:
1 × 5-inch (127 mm)/54 Mk-45 Mod 1/2 (lightweight gun) (DDG-51 to -80); or
1 × 5-inch (127 mm)/62 Mk-45 mod 4 (lightweight gun) (DDG-81 onwards)
2 × (DDG-51 to -84); or
1 × (DDG-85 onwards) 20 mm Phalanx CIWS
2 × 25 mm M242 Bushmaster cannons
And those two Bushmasters?
“The standard rate of fire is 200 rounds per minute. The weapon has an effective range of 3,000 metres (9,800 ft), depending on the type of ammunition used.”
Short burst at the waterline over a mile away and you have a rapidly sinking ship.
Ships in built in that era tended to serve decades, the 1620 Victory was eventually rebuilt to a 80 gun ship of the line.
None of which would survive 5 broadsides.
You think those cannons are that accurate? Do you think no one would be shooting back before then?
Remember it is postulated as a surprise attack.
As Quercus pointed out: "Stipulating a very surprised peacetime (but fully crewed and supplied) destroyer with a triple-decker (say HMS Victory) at point-plank range. It doesn’t seem unreasonable that it would take the destroyer 4-5 minutes to get a response going, and then another 5 minutes to get moving. That’s like 5 to 6 broadsides for a crack crew,…"
Even grape shot would destroy all but the 5" gun.
The torpedoes carried by USN DDG’s are intended to attack submarines rather than surface ships. And at one time such ‘light weight torpedoes’, launchable from the ship’s own tubes or by its helicopter(s), literally could not attack surface targets, even surfaced submarines. An RN helicopter dropped an early model Mk.46 against the surfaced Argentine ex-US WWII submarine Santa Fe in 1982 and it just circled at minimum depth below the submarine.
Later models of the Mk.46 and the current Mk.54 based on it have guidance modes and settings to attack surface targets, though again this would generally be surfaced or periscope depth subs in practice. But at least in the Mk.46’s case there was still an automatic minimum of 50’ when launched from surface tubes (so it wouldn’t turn and home on the launching ship), plus the torpedo gained anti-surface capability when dropped by a/c by going into passive acoustic mode near the surface, to avoid surface reflection effects when in active acoustic mode. And a sailing ship would not make much noise. The exploders on these torpedoes are dual influence/contact so lack of magnetic signature doesn’t rule out hitting a target and exploding on contact. The other issue would be the torpedo hitting and sticking in the bottom if launched in too shallow water, as alongside a dock.
The Mk.54 is newer so there less declassified discussion of its capabilities in detail, how the seeker operates near the surface for example, though Raytheon literature claims it has capability against targets from the surface on down.
Anyway again these weapons are not generally intended for use against surface ships.