The setup for this is complex, so I'll begin by stating the core question:
In a heavy ship engagement, using energy-dissipation armor, about how much warmer will the sea get, and will that have any gross effects?
In the late 1700s, as the Age of Sail is at its height (and getting close to its collapse), things are pretty much as you’d expect…except, of course, for the use of magic, principally defensive. It comes in many forms, and can have a number of effects, but for purposes of the present question, I want to focus on the following system of armor:
- A coarse network of steel cable is wrapped around the upper hull of a ship, running from the gunwales to the waterline, and is attached to the copper plating.
- Given some warning, an expert magician hidden away in the orlop can activate this network to absorb large, sudden bursts of energy—including, most particularly, bursts caused by the impact of cannon-fire.
- The energy is dissipated, not especially efficiently, as heat, radiating through the copper plating, where it heats the water under the ship. Thus this cable “armor” relies on the ocean to act as a huge heat-sink.
So my question was:
- In a heavy engagement, about how much warmer will the sea get, and will that have any gross effects?
By dint of a fair bit of calculation and some judicious estimates, I have found that, taken in large aggregate, naval warships of roughly 1800 tended to produce an average muzzle energy of about 16Kj/rated pound. That is to say, a 24-pounder ball departed the gun with nearly 400Kj of energy.
[Note: This average is not especially accurate cannon by cannon, though it does work passably for 24-pounders, but when you spread this across the total metal of a variety of different military vessels in ordinary distribution, it works pretty decently.]
Now in a heavy engagement, you’re firing between 2 and 3 broadsides per 5 minutes. The length of an engagement can vary tremendously. The amount of firing that happens in an engagement varies a great deal as well: sometimes there’s a lot of yardarm-to-yardarm smashing, and sometimes it’s hours and hours of long-bowling with 9-pounders. Usually there’s something of a mix.
In addition, of course, whatever energy is dumped into the sea by this magical system will dissipate pretty quickly, because the ships are moving (not especially fast if they’re really going at it, but still) and the sea is so darn big.
So I calculated like this:
For a heavy engagement I took the Battle of Trafalgar, measured by broadside weight and total number of ships and so on, figuring this was such a large engagement that it would average out pretty decently. At Trafalgar, there were 47 tons’ broadside weight firing, spread across 60 ships. This means that during the hot parts of the battle, when they were really going all guns blazing, they were putting out some 2Gj muzzle energy per 5 minutes.
For the water, I spread this incredible sum across the displacement tonnage of the combined fleets (again, a combination of research and judicious estimation). I estimated the total displacement tonnage at 96,000 tons.
For the temperature change, I simply dumped the energy in joules into the water in grams, since 1J will cause 1g water to increase its temperature by .24K.
My result is that, if armor systems like I’m proposing had been in use at Trafalgar—and in fact generally for significant ship battles—you’d find that during the hot parts of the battles the sea temperature would rise about 5 degrees C.
Now I had to make a number of much more problematic guesses to get this result—more problematic in that, while they sound kind of reasonable to me, I don’t actually know what I’m talking about, and I don’t know how to figure out whether I’m full of it.
I guess that, given that the ships are always moving to some degree, there is always current, and the sea is a colossal heat sink, the effect of this water heating at the copper plating will tend to dissipate over the course of about 5 minutes, give or take. In other words, if you keep on firing for half an hour, the sea temperature still only rises 5C total, and that only within a very short range of each ship.
I guess that the fact that we’re talking about seawater won’t make a really significant difference.
I guess that the ambient ocean temperature won’t make much difference either.
As to gross effects, my guess off the cuff is that you’d get a lot of fog. I went looking at things like ocean temperature, dew point, air temperature, and fog. I found that you get fog when the air temperature and the dew point get within a couple of degrees C of one another. This happens a lot in the morning because the air temperature bottoms out before sunrise while the dew point keeps rising until the sun gets going to dry the air out (that’s my rough-and-ready interpretation, anyway).
It looks to me as though ocean temperatures have a lot to do with dew points at sea, but I haven’t found any good indication of how that relationship works. Looking at the Caribbean, where the ocean is pretty much always 72-74F, the dew point 55-70F, air temp 65-80F, it looks to me as though the ocean temperature is sort of the crossing-point. That is, you get fog when the air temp is about the same as the water temp.
If that’s correct, then raising the water temperature during a battle by 2-5C (=4-9F) is going to have a huge effect. If I’m at all in the ballpark about the relationship of water temperature to dew point, a heavy engagement at noon in the Caribbean is likely to be enveloped in a very thick fog, and remain so for quite some time.
My Preliminary Answer
Therefore, my preliminary answer to my own question is:
In a heavy engagement, the ocean temperature will rise by 3-5C in the immediate vicinity of the firing ships, and no more. The gross effect of this will be more or less heavy fog.
So here’s my question to you:
Am I doing this right? Reality check, please?
[Please note: If you wish to debate my math about the ships and their metal, i.e., whether I have my facts straight about broadside weights and muzzle velocities, I’d prefer that this go in the comments. I’m pretty sure I’ve got that straight, and if I don’t, it’s largely incidental to the question I’m asking.]