I've found on another site evidence of some confusion in regards to what starships should do in water.
A modern oceangoing ship floats because, although her hull and superstructure materials are denser than water, the ship is largely hollow and thus filled with air. Gore a hole in the ship and allow it to fill with water, and the ship will sink.
This is, of course, what happened to the Titanic. Her hull was scraped along 220+ feet of her side, with an estimated opening of 12 square feet in "a series of bent plates, split seams and small holes" to quote another site. That's only 1.1 square meters of opening spread across 67 meters of the hull, which is why the ship took a good long while to fill with water and finally sink.
Notable is that despite becoming a water-heavy husk of steel, the ship still only cut through the water downward at 16km/h, or about 4.4 meters per second.
It's actually possible to work this out mathematically, as well. Much as I did recently with the Type-6 shuttlecraft terminal velocity, it is just as easy to figure out the terminal velocity in water.
So what we need is the the square root of ((2*m*g)/(ρ*A*C)). Our result will be the velocity. The "m" is the falling object's mass, "g" is gravitational acceleration, ρ is the density of the medium through which the object falls (air or water), "A" is the frontal area, and "C" is the drag coefficient.
Now, I don't have the first clue as to the drag coefficient of the Titanic or the mass of her halves, so for the moment I'm going to skip re-doing that math.
However, the reason this is so exciting is that we can do some estimates from "Gungan Attack"[TCW4].
In that episode, a small Republic cruiser of the type seen in TPM and "Jedi Crash"[TCW1] is exploded to bits just above the surface of the ocean of Mon Calamari. One of her three engines is propelled up and away from the rest of the ball of debris and, like the rest, falls into the water.
This engine becomes the ride down to the bottom of the sea for our heroes as they grab hold of it. We get to see it pass Ahsoka, who we will presume to be 1.5 meters tall. At a framerate of a hair under 30fps, it takes about 10 frames for the leading edge of the engine to pass her up as she remains relatively stationary. This implies a velocity of 4.5 meters per second . . . we'll round up to 5 meters per second for ease. (A smoking feature passes her by shortly thereafter and only takes about eight frames, but by that time she's moving upward, swimming at it to grab onto it.)
The engine had almost certainly reached its terminal velocity at that time, since it had been in the water for thirty seconds by that point. So, we have the result to the equation.
But, that means we can find out other details, which will be sufficient to learn something even if we're just rocking the proverbial backside of an envelope here.
To review what we covered thirty seconds ago (for those of you in Washington, D.C.), that result of five meters per second is the the square root of ((2*m*g)/(ρ*A*C)). The "m" is the falling object's mass, "g" is gravitational acceleration, "ρ" is the density of the medium through which the object falls (air or water), "A" is the frontal area, and "C" is the drag coefficient.
So, let's say we want to find the engine's mass. All we need, then, is the gravity on Mon Calamari, the density of their ocean water, the frontal area of the falling cylindrical engine, and a drag coeffcient for it.
"C", the drag coefficient, is relatively easy in this case. The engine is cylindrical, and per this site and very helpful graphs thereon, a cylinder that's less than twice as long as its diameter will have a drag coefficient above 0.81 . . . the graph itself suggests a figure of 0.88 or so. Unlike a cylinder where the face and side meets at a 90 degree angle, however, the engine has more rounded corners, which drives the figure downward. For instance, one simple shape, a slightly squashed sphere with the wide bottom taking the wind, has a drag coefficient of 0.59, and a rounded hemisphere (a la the front of the Y-Wing's engines) 0.42. Of course, it also has little widgets on it and, in this case, a few person-sized holes, so that drives the figure back up. I figure the drag coefficient ought to be somewhere between 0.75 to 0.88, so we'll go with 0.85 as a somewhat round and reasonable figure.
"A", the frontal area, is also relatively easy to figure. Based on schematics from Wookieepedia that look fairly accurate, and assuming the 115m figure for the vessel's length is about right, the starboard engine ought to be about 16.5 meters in diameter. That equates to a frontal area of about 214 square meters.
(For those curious, that diameter equates to an engine length of about 20.6 meters. As a result, we can confirm our velocity figure . . . the engine should take about four seconds to pass a single point when we see it fall by Ahsoka, and indeed it does.)
"ρ", the density of the medium through which the engine falls, could vary somewhat from normal water, but our safest assumption here is 1000 kg/m^3.
"g" is gravity, and again our safest assumption is one g, or 9.8m/s^2.
"m", for mass, is the figure we actually want to find . . . we already know the velocity is five meters per second.
So, shall we begin? I'll try to show my work a little better this time:
5m/s = sqrt ((2*m kg*9.8m/s^2)/(1000kg/m^3*214m^2*0.85))
5m/s = sqrt ((m kg*19.6m/s^2)/(181900kg/m))
Truth be told, I hate the whole dividing fractions thing. But this is the part where you can basically flip the bottom fraction and multiply, and your units start cancelling, to put it in layman's terms, to the following:
5m/s = sqrt (m*0.00010775151181968114m^2/s^2)
Now we'll square both sides for ease:
25m^2/s^2 = m*0.00010775151181968114m^2/s^2
So, the mass of that part of the ship ought to be about 232000 kilograms, or 232 metric tonnes, give or take.
But, there's a corrective measure we need to consider here. In air there's not a whole lot of buoyancy to consider, but in water, one should. If we had the mass and wanted to determine a terminal velocity, we would simply "massage" the mass by subtracting the mass of the water that the item would displace. However, since we're doing things bass-ackwards, we have to take the 232,000 kilograms as a pre-corrected value.
That said, however, we're playing around a few unknowns here. As noted, the engine had some man-sized holes in it and these were seen to be full of water as the engine sank. How much of the internal volume of the engine was hollow and filled with water? We don't know. If half the engine was empty space, for instance, our displacement figure would be way off.
Now the question is about the volume, though, so we can at least start making guesses about the water displacement. It'll take a bit longer than I have at the moment to go into Sketchup and cut off that part of the ship and really do it right, but we can guesstimate for now.
As noted, the engine diameter was 16.5 meters, and the length was 20.6 meters. Assuming a proper cylinder of those dimensions gives us a volume of pi*(8.25^2)*20.6, or about 4400m^3. But, that's a bit too large given the rounded-off edges of the cylinder. I'd ballpark the correct figure at closer to a cylinder of 13.9 meters diameter, which gives us a volume of pi*(6.95^2)*20.6, or 3125m^3. We'll just call it 3500m^3 for now.
3500 cubic meters of water masses 3500000 kilograms. That means we need to add that or some fraction of that to our 232000 kg to get the complete picture. Assuming a perfectly un-holed watertight engine, for instance, we'd be looking at 3732000 corrected kilograms. If half the engine was waterlogged, we'd be looking at 1982000 corrected kilograms.
So, basically, we're looking at engine density of 1982000/3500 kg/m^3 or 3732000/3500 kg/m^3, which work out to engine densities of 566 and 1066 kg/m^3, respectively.
That's pretty amazing, since I had previously estimated that "Star
Destroyer density probably falls somewhere in the 500-1000kg/m³ range". And here we have figures that are almost exactly that after a chain of math and reasoning.
But, the fact is, the displacement correction in which I used half-water and no-water meant it couldn't have had any other result, because the 232 tonnes was such a small figure by itself compared to the 3,500 tonnes of 3500 cubic meters of water.
Let's ponder some alternatives and any weaknesses in the above, just in case. If nothing else, this will help anticipate inflationist objections. I say that because they argue that Star Destroyers ought to be super-dense, as in "25 times denser than water" super-dense. That's basically crazy-pants dense, and doesn't make any sense given hydrofoamed permacrete and whatnot.
Let's take the third line of calculation:
5m/s = sqrt (m*0.00010775151181968114m^2/s^2)
Now, the 0.00yadda bit is pretty solid, being based on 19.6/181900. The 19.6 bit can't be messed with too much unless an inflationist objector could claim Mon Calamari had way less gravity than Earth. But let's assume 0.1g . . . that gives us 1.96/181900, or 2,320,000 kilograms . . . corrected, that'd be 5,820,000 kilograms (which is a familiar number, being the volume of a Galaxy Class ship in cubic meters, but that's neither here nor there right now). That mass would result in a density of just 1660 kg/m^3, or over half again the density of water. That's still not a huge increase. And, put simply, it doesn't fly given the way we see clones and Ahsoka drop into the water . . . if anything, one could argue for a higher gravity, not less, which would drive down the mass, and thus the density.
Alternately, an inflationist objector could go after the 181900 bit, but again, that's pretty solid. It's based on water's density, which doesn't change that much, the frontal area of the ship's engine, and the drag coefficient, which we've estimated pretty well, I think. But, for fun, let's assume that I got the area all wrong it should be higher, and that the drag coefficient should be like 2.00 or something. That gives us 500000 instead of 181900, and so 19.6/500000 yields 0.0000392. The mass figure would then be 637755 kg, and the corrected mass about 4150000 kg, and the density would even then be only 1200 kg/m^3.
So, the only way to go is to try to attack the speed.
One objection would be that when we see an engine hitting bottom first (our heroes seem to ride the second, more banged up engine), it appears to be going faster than five meters per second. (The engines were pretty much the first objects to arrive at the bottom, and they really ought to have been the densest, least-dragged sections of the ship all around.)
This is actually true, but only for that final scene. I used the speed reference against Ahsoka because I had not yet, at the time, done the figuring of the engine size. The second engine, for instance, seems to take about 23 frames to pass a particular line during the hitting-bottom scene, suggesting (assuming 20m) a velocity of 26 meters per second.
That's hellaciously fast, and frankly suggests something is wrong with the hitting-bottom scene. After all, as noted, when the engine passes Ahsoka, it had been in the water for 30 seconds. That is, as far as I am aware, plenty of time for it to have reached terminal velocity. (In air, it takes about 8 seconds for a skydiver to reach 90% of terminal velocity, and 3 seconds to reach 50% . . . why would an object require ten times that long to reach a fifth of the speed?)
There's also a limit at some point simply based on the fact that the people were able to hold on to the engine on the way down. The forces could not have been that great if Amidala could hang on.
The drag force on a human being at 5 meters per second would be .5*ρ*v^2*A*C . . . assuming half a square meter and a drag coefficient of 0.5, then we get
F = .5 * 1000kg/m^3 * (5 m/s)^2 * .5 m^2 * 0.5
F = 3125 kg m / s^2
That's 3125 newtons, or over 3kN. A 70kg human being's weight is a force of about 700N and some people would be hard-pressed to even hold themselves up hanging on a bar for more than a minute. This would require Amidala to be holding the equivalent of herself and four other people, which is bad enough.
At 26 meters per second, the force gets larger. 84,500N, to be precise. That's like holding 120 people by your fingertips. Good luck with that.
Put simply, it's not possible that it was going that fast while folks were on it. Even if we assume some sort of protective effect due to well-chosen spots to hang onto (not unlike hanging on to a car's trunk rather than the hood if you're trying not to mess up your hair), it's much more realistic to believe they might be able to hang on at 5m/s as opposed to 26.
Also, that's not to mention that it takes some time for the engine to pass them after they let go. Skywalker lets go and swims nearby, Ackbar lets go and is able to swim over to another spot, and the folks who let go then go belly-down like a skydiver wanting more hang-time don't recede at a high rate . . . a few meters per second at best. Had the engine been doing 26m/s, and the terminal velocity of a sinking human belly-down being rather slower (near zero, I would think), the deceleration would've been tremendous . . . unless they weren't going that fast to begin with.
So, I'm thinking that the hitting-bottom scene was at a higher speed than the others . . . the opposite of slow-motion, in other words. It is demonstrably inconsistent with every other scene that's part of the sinking event.
There's really no other way around it.
I mean, one could try to argue that when the first bottom-hitting engine cylinder hits bottom, it does so while turned almost sideways, meaning the drag coefficient is significantly reduced. That's true, but doesn't help much. If at a perfect sideways angle, the drag coefficient would be somewhere in the 0.45 range, and the area would be approximately 20*16.5. So, let's run the numbers again:
V m/s = sqrt ((2*232000 kg*9.8m/s^2)/(1000kg/m^3*330m^2*0.45))
V = sqrt (4547200/148500)
V = sqrt (30.621)
V = 5.5 m/s
Oops. The engine's extra length when sideways added to the frontal area and basically nullified the drag coefficient decrease.
Put simply, there's really no way for an inflationist to take this example and wind up significantly north of water's density without ending up in crazy-town. While that hasn't stopped them before, I'm comfortable going with the old figure that I had before . . . Star Wars vessel densities ought to fall in the 500 to 1000 kg/m^3 range.