…does the coastline (or water level of the ocean) change noticeably?
I’m including any man-made water vessel to be a boat – ships, jet-skis, submarines, whatever. And this includes all the wrecked ships lurking on the bottom of the sea.
So. They’re all removed from the water. How much land is reclaimed from the sea, if any?
Interesting question about water displacement. To understand the question, first we need some details.
For starters, water displacement differs between floating boats and sunken boats. Sunken boats displace only their volume. This may include any sealed air pockets as empty volume, but any boat of any kind submerged for long will eventually leak, so at some point you end up with only displacing the physical volume of structure.
Whereas floating boats displace their weight - that is, the weight of the boats and all objects/people on the boats displaces an amount of water that weighs the same. This is called buoyancy. Submarines work exactly the same as surface boats. It does not matter that they are submerged, only that they are floating and not resting on the bottom of the ocean.
Okay, now that I’ve got that out of the way, let’s get to the answer. I don’t know. 
Seriously, using a gut feel intuitive guess, there’s a whole heck of a lot of ocean surface area out there, and most boats are pretty tiny in comparison. Sure, you’ve got aircraft carriers and oil tankers and even nuclear subs, but still…
I’m thinking that the sum total displacement will be negligible. But I don’t have any numbers to back that up.
Well, look at it from the opposite direction. We’ve only had lots and lots of really big heavy boats floating on the world’s oceans for the last, say, 100 years. If boats’ displacements did make a difference in the water level, then ocean levels today would be measurably higher than they were 100 years ago.
Which they are.
:eek:
Hey, maybe I’m onto something here–it’s not global warming that’s making sea levels rise, it’s the boats! It’s the boats!!
(Um, to address the OP: Nah. D’you have any idea just how big the oceans are? And how puny all the displacements of all the supertankers ever built, when added together, are in comparison?)
Okay, so I’m not just a know-it-all smart-ass, I’ll try to help a little. NO, I will not do the calculation, or even a back of the envelope estimate, but I will try to point you in the right direction and let you go dig up information if you really want to know.
Let us simplify to only floating ships, and ignore sunken hulks. I think this will make it easier, and the estimate will be so loose that I don’t think it will matter much. You can always tweak the values for a fudge factor.
Okay, so now you need some statistics and some reasonable guesses. These you will have to supply yourself.
Ws = weight of all ships, boats, submarines, watercraft, etc that are in the water all at one time, including their crews, payloads, etc. Make whatever assumption you want about how many there are, vs. how many are in dry dock. Any boat in dry dock is not in the water, so by definition it doesn’t count.
Ps = perimeter of shoreline on the ocean for the Earth. This should be some reasonable guess based upon all continents including Antarctica. You can pick low or high tide or an average or whatever you want.
Aw = area of the oceans. This is surface area, using the guidelines for Ps above. All ocean waters, ignoring lakes and rivers.
let Ww = the weight of all the displaced water.
I just said before that
Ws = Ws
so now you have some guess at the total amount of water that has been displaced, and thus is spread out to raise the shoreline.
Use the density of sea water (slightly less than 1) and then calculate the volume of water that is displaced. Vw
Now take that total volume of water and divide it by the area of ocean available to spread out. This will give you Delta height = H.
H = Vw/Aw
That will give you the total rise above a baseline that is caused by all the boats and ships, so you subtract H from the current sea level.
Now, given that shores are sloped and not equivalent at all, this can get funky for finding the difference, so you can for simplicity assume the shore is a cylinder of Perimeter = Ps and smooth perpendicular walls, and then H will show the water height difference.
Oh, in case you’re wondering, you could use Ps to help calculate Aw if you start with the amount of surface area of the entire Earth, and assume that cylinder or otherwise obtain a ratio for surface area land to surface area sea. Heck, you could probably estimate a 30/70 ratio, or 70% of Earth’s surface is sea, 30% land, and using the diameter of the Earth approximate your Ps and Aw.
Okay, you going to do the calculations and report back?
Well, yeah. My guess is that the difference is miniscule, if it’s able to be measured at all – that’s why I threw in everything man-made that’s floating on or sunk under the water, trying to make the human influence as great as possible.
It’d surprise me to find out that it’d make a measurable difference, but it’s an idle question, and I really had no idea of how to start figuring it out.
Thanks to Irishman, I’ve got an idea of how to start figuring it out. When I’ll finish figuring it out, I don’t know…
The answer is zero. The variation in the amount of water retained in the atmosphere and on land is much, and I mean, much, greater than all the ships combined. Heck a slight rise or fall in the Great Lakes is more than all the ships in the world. And the Oceans are quite a big bigger than the Great Lakes.
http://www.kiae.ru/rus/inf/new/new4.htm
So say there are the equivalent of 10,000 Russian underwater tankers out there on the world’s oceans. That’s 10,000 x 44,000. That’s 440,000,000 cubic meters. Yes? I did it on the calculator, it came out 44 followed by 7 zeros.
There are 1.35 x10[sup]17[/sup] cubic meters of water in the world’s oceans.
http://witcombe.sbc.edu/water/physicsearth.html
Lake Superior has a volume of 12,100 cubic km, which is 12,100,000 cubic meters. Yes? No?
http://www.great-lakes.net/lakes/ref/supfact.html
So if you removed all 10,000 Russian underwater tankers, it would be the same as removing about 37 Lake Superiors from the oceans. I can’t visualize that as having much of an impact on a body of water that holds Ten to the Seventeenth Cubic Meters.
In this thread a little over a year ago, I took a stab at this very question. Pulling some numbers out of thin air, I came up (after finding an error the first time) with a depth change of 2 or 3 microns.
What if you included all man made objects including bridges, tunnels, landfill, beach restoration projects, dreaging, ect.
I hate when I do a search and still don’t find the earlier thread asking the same question.
(I blame my search terminology, btw, not the board – at least this time.)
I remember talking to a USN scientist/oceangrapher years ago who explained that during summer months (lotsa sun) that the Mediterranean Sea on any given day gave up approximately 1 meter of water (depth?) to the atmosphere due to evaporation.
That’s far more important than the 2-3 microns attributed to the tankers/boats/yachts/canoes. When will we ban the sun in the Med? This is horrible! Someone call Greenpeace!
No. You need to multiply by 1000^3, because it’s cubic, not just 1000. Thus 12,100 cubic km is 12,100,000,000,000 cubic meters. So you’d need 275,000,000 Russian tankers just to fill Lake Superior, which is absolutely miniscule compared to the ocean.
Whether or not the submarine is submerged does matter. When the sub is surfaced, it displaces a weight of water equal to its own weight, like all floating objects. If any part of the submarine is sticking out of the water as it floats (which is the case, of course), this weight of water will have a volume less than that of the sub itself.
When the sub submerges though, it displaces a volume of water equal to its volume. This is true whether it is neutrally buoyant, or is resting on the bottom.
If the weight of the sub is equal to the weight of the water it displaces while fully submerged, it is neutrally buoyant. If the weight of the sub is greater than the weight of the water is displaces, it is negatively buoyant.
So when submerged, a sub always displaces more water than when surfaced.
How does a sub surface and submerge, then? It changes its weight, by adding or removing water from its ballast tanks. The volume of the sub remains virtually constant (ignoring slight hull compression at depth).
Robby,
The water added and removed from the ballast tanks come from the ocean does in not? So I think it is still the weight of the submarine unless it is resting on the bottom in which case it can have more ocean water in it.
Don’t blame the boats. For every boat out there displacing water, there are thousands of fish doing the same. Those lazy little bastards just swim around taking up valuable space in our oceans and lakes. If those scaly good-for-nothings had any ambition they’d be up here on dry land, evolving like the rest of us. And don’t even get me started on those damn whales; harpooning’s too good for that crowd.
'Uigi: I remember talking to a USN scientist/oceangrapher years ago who explained that during summer months (lotsa sun) that the Mediterranean Sea on any given day gave up approximately 1 meter of water (depth?) to the atmosphere due to evaporation.
He was probably talking about a cubed meter… m^3.
The change in height of the ocean surface is independent of the length of the coastline. Which is fortunate, since the length of the coastline is pretty much whatever you want it to be:).
Not necessarily! More likely cm.
[back of the envelope calculation]
Amount of relative sun-hours/day: Lets say that the sunshine on the Med. corresponds to 5 hours==18 000s of perpendicular sunshine/day. (probably a bit high, but we’re pretty close to the tropic of Capricorn)
Intensity of Sunshine: 1.3kW/m[sup]2[/sup] (That’s what I’ve been tought.)
Energy required to evaporate 1g of water: 804.18+2200=**2.510[sup]3[/sup] J/g**
Amount of water evaporated per day/m[sup]2[/sup] = Sunshine * Suntime / Energy_required
== (with numbers above)9360g/m[sup]2[/sup]
Pretty close to 10 kg of water! (Which on a m[sup]2[/sup] gives 1cm)
No doubt I’ve made some silly error in there somewhere. If you find it, please post a correction.
One error found. The hunt continues:
That should of course be the tropic of Cancer. Not that it changes the result, but it was wrong.
There are always major currents in the Gibraltar Strait. I have seen semi-serious suggestions on building a hydroelectric farm across the strait which could generate enough power to run a large fraction of Europe and northern Africa. Cost prohibitive, but one of those “wow” engineering projects you hear about from time to time.