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Teak is pretty dense. In fact if you take a as-cut piece of teak it'll sink in water.

To sink in fresh, pure water at 4c, wood has to have a specific gravity of 1 or higher. True ironwoods meet that requirement. Teak is not an ironwood.

As salt water is ~3.5% denser than fresh water at the same temperature, something that doesn't sink in fresh water isn't going to sink in salt water under the same conditions.

That doesn't mean that it isn't possible to find a particular piece of teak that is heavier than 1 (though this is highly unlikely, teak typically tops out around .75) due to specific growing conditions, and of course water gets less dense as it gets warmer, but I will leave the math to you to figure out just how warm the water has to be for a freshly cut piece of teak to sink. Adding in unknown impurities to the water starts make it challenging (other than salt).

For more homework, examine what kinds of trees grow along coastlines subject to tsunamis (including trunks of trees that could find their way to the sea shore via rivers) and what their specific gravities are. Add in that a trunk of meaningful size can take years to fully saturate, and realize that flotsam comparable to that dock has been crossing oceans far longer than bipedal locomotion has existed.

The person who keeps spamming that trees will quickly absorb water and sink needs to be trolled out of existence (not for doing it once, for doing it repeatedly). Balsa, for example, has a specific gravity of under 200, and it takes longer for water to saturate it enough than it takes to cross the Pacific Ocean. Remember Kon-Tiki?
In fact, had Heyerdahl followed the advice of other sailors, both he and his crew would likely have perished. The sap in the balsa wood prevented the sea water from penetrating the logs and kept them buoyant. Also, as the waves lifted the boat, the ropes used to lash the logs together dug into the soft wood, forming grooves that protected the ropes. Steel cables or wire would have sawed through the logs and left the crew clutching small stumps to survive. Balsa is not the only wood (far from it) capable of such a feat.

Which has nothing to do with whether or not that floating dock should be sold to the highest bidder as scrap, and if the highest bidder wants to put it back into service it must be cleaned/scraped before doing so.

So how do you think normal driving for truck looks like ? In europe it is highway driving at constant speed, more or less constant load for hundreds miles.

To avoid each person having a different metric, we use standardized drivecycles. The drivecycle that the EU uses to model how people typically drive for vehicle mpg ratings is called the NEDC -- the New European Drive Cycle. It is a combination of urban and highway driving that approximates typical european driving patterns (which, by the way, are lower energy than typical US driving patterns -- hence the US uses FTP75 (city) and US06 (hwy), which are higher energy, and correspondingly leads to lower MPG figures for the same car in the US). You can see the NEDC here.

If you want to talk about pure highway driving, even that is not constant speed. Speed on the highway varies based on traffic density, random factors (passing, being passed, etc), current weather conditions, stops (gas, rest, etc), start and end accel/decel, exits (to surface streets or other highways), and driver randomness. Beyond speed, energy consumption varies based on weather and especially altitude changes. For an example, here are actual measurements taken from a vehicle in the US. Here's a test drive that starts with city and progresses to intra-urban freeway. Your mileage may vary.

(I have my own drive data recordings, but I am not at liberty to disclose them, so I'm linking to publicly available ones)

Highway driving runs an engine much more efficiently than city driving. You're closer to the peak efficiency (although not at it), you brake less, idling is basically eliminated, etc. Now, there's obviously a big downside -- your aero drag is *way* higher, and your rolling drag slightly higher (yes). In non-hybrid vehicles, the upsides outweigh the downsides (sometimes significantly). In hybrid vehicles, the downsides usually outweigh the upsides.

No this is not from wiki. It is from book called "Automobile fuels" (translated)

Right. Which is why I said, "If you had cited ... you would have..." instead of "You cited... you did." Understand? I'm pointing out that different sources give different numbers because there is no single correct number because they're not a single chemical mixture. You've picked one source to latch onto, when there *is no single answer*. Check other sources; you'll see what I mean. Mixtures vary from location to location and even day to day (for example, summer versus winter blends). They even change from year to year, as standards and refineries are always changing. Their energy densities vary, too. But overall, the *current global average* is about 15% denser for diesel than gasoline.

I don't know how many times I need to stress this, but let me do so once more: There Is No Single Fuel Called Gasoline Or A Single Fuel Called Diesel. How about this -- how about I cite a bunch of random sources?

Simetric: 820-950kg/m^3
Alan Harvey, National Institutes of Standards and Technology: 850kg/m^3 typical, but 825-890.
Engineering Toolbox: 810-960kg/m^3
MSDS: 810-880kg/m^3

Gasoline:
MSDS: 710-770 kg/m^3
Simetric: 737kg/m^3
Engineering Toolbox: 680-740kg/m^3

So how do you think normal driving for truck looks like ? In europe it is highway driving at constant speed, more or less constant load for hundreds miles.

To avoid each person having a different metric, we use standardized drivecycles. The drivecycle that the EU uses to model how people typically drive for vehicle mpg ratings is called the NEDC -- the New European Drive Cycle. It is a combination of urban and highway driving that approximates typical european driving patterns (which, by the way, are lower energy than typical US driving patterns -- hence the US uses FTP75 (city) and US06 (hwy), which are higher energy, and correspondingly leads to lower MPG figures for the same car in the US). You can see the NEDC here.

If you want to talk about pure highway driving, even that is not constant speed. Speed on the highway varies based on traffic density, random factors (passing, being passed, etc), current weather conditions, stops (gas, rest, etc), start and end accel/decel, exits (to surface streets or other highways), and driver randomness. Beyond speed, energy consumption varies based on weather and especially altitude changes. For an example, here are actual measurements taken from a vehicle in the US. Here's a test drive that starts with city and progresses to intra-urban freeway. Your mileage may vary.

(I have my own drive data recordings, but I am not at liberty to disclose them, so I'm linking to publicly available ones)

Highway driving runs an engine much more efficiently than city driving. You're closer to the peak efficiency (although not at it), you brake less, idling is basically eliminated, etc. Now, there's obviously a big downside -- your aero drag is *way* higher, and your rolling drag slightly higher (yes). In non-hybrid vehicles, the upsides outweigh the downsides (sometimes significantly). In hybrid vehicles, the downsides usually outweigh the upsides.

No this is not from wiki. It is from book called "Automobile fuels" (translated)

Right. Which is why I said, "If you had cited ... you would have..." instead of "You cited... you did." Understand? I'm pointing out that different sources give different numbers because there is no single correct number because they're not a single chemical mixture. You've picked one source to latch onto, when there *is no single answer*. Check other sources; you'll see what I mean. Mixtures vary from location to location and even day to day (for example, summer versus winter blends). They even change from year to year, as standards and refineries are always changing. Their energy densities vary, too. But overall, the *current global average* is about 15% denser for diesel than gasoline.

I don't know how many times I need to stress this, but let me do so once more: There Is No Single Fuel Called Gasoline Or A Single Fuel Called Diesel. How about this -- how about I cite a bunch of random sources?

Simetric: 820-950kg/m^3
Alan Harvey, National Institutes of Standards and Technology: 850kg/m^3 typical, but 825-890.
Engineering Toolbox: 810-960kg/m^3
MSDS: 810-880kg/m^3

Gasoline:
MSDS: 710-770 kg/m^3
Simetric: 737kg/m^3
Engineering Toolbox: 680-740kg/m^3

It wouldn't be that difficult to get the weight right, would it? I mean most coins are a copper core with a nickle covering. So if you could create a heavier core, it would compensate for the mass of the removed area. Actually, now that I look at it, lead would be the only non-expensive metal that's heavier than copper by volume, but it's not THAT much heavier (I'm not sure if it's enough to compensate for such a large void)... Sure, they could use something more exotic like Platinum or Tungston (or even Uranium or Plutonium, but if you use them in a coin, I think you have bigger problems than detecting a hidden microSD card), but how much would that thing cost then? http://www.simetric.co.uk/si_metals.htm

I think the losses in the CAES system are due to the fact that it is a non-adiabatic process (a diabatic process?, i.e. one where heat can be lost from the system). When you compress the air, the temperature rises, and some heat is lost to the surrounding ground. But if the cycles are fast enough, those losses may be reduced -- i.e., you allow the air to re-expand, which cools it, and it sucks heat back from the ground. Since heat moves slowly through the ground, you may be able to get a lot of it back before it goes anywhere. The innovation in the Alabama system was to use waste heat in the turbine's exhaust gases to replace this lost heat as well.

I think the solution you propose is isobaric (constant pressure) and isothermal (constant temperature), but still not adiabatic. Some of the energy used to compress the air is converted to heat, and that heat would be lost to the ocean instead of raising the temperature of the air.

A better solution might be to use pre-inflated air bags (or air boxes?) attached to pulleys on the bottom of the ocean. Use a motor to pull the other end of the rope, and you would draw the air bag downward, storing energy. Play out the rope and the rising air bag would turn the motor (now acting as a generator), generating electricity. You could also do this with stones or bags of silt/gravel, just raising and lowering them from the surface.

The problem is, you would need a lot of air bags or stones to store any significant amount of energy. If the stones or gravel have a density of 2000 kg/m^3 (similar to "Gravel, wet" according to http://www.simetric.co.uk/si_materials.htm, higher than "Clay, wet excavated" (1600) but lower than concrete (2400)), then they will have a net weight in water of about 1000 kg/m^3 (i.e., a downward force of about 10000 Newtons per m^3). Air bags would exert a similar force upward. If you can find a near-shore location with a depth of 1 km, you could store 10000 N * 1000 m = 10 MJ of energy per cubic meter of material, which is about 3 kWh/m^3. A 100 MW wind farm (presumably closer to shore) would generate 100,000 kWh of electricity per hour when the wind is blowing, so if you wanted to store 6 hours of energy from this wind farm, you would need to raise and lower about 6 * 100,000 / 3 = 200,000 cubic meters of stone or air (e.g., 200 large chunks, each 10 meters across). I suppose it could be done...

What kind of brick?

• A common red brick has a mass of 1922 Kg/m^3 (ref)
• A standard brick has dimensions of 3 5/8" x 2 ¼" x 8" (ref)
• The smallest side (and therefore ideal side for max terminal velocity calculations) is 3 5/8 x 2 1/4, and therefore 8.15625 square inches
• The mass of the brick would be 2.05510989 Kg (Google calculator ref
• The coefficient of drag would be 2.1, according to this calculator site (which, incidentally, gives a nonsensical result of 3mph - which is why I did the math myself below)
• The density of air at 35 celsius is 1.146 kg/m^3(ref), and 95 F seems pretty optimistic for the salt flats this time of year.

So, Let's do the math with Google so I don't have to convert units, using the formula here. Since Google complains about square roots with more than a number inside, we'll find the answer in mph^2 first, then take the square root of that. Here's the first calculation, which gives 15 932.3342 mph^2. Take the square root, and you get 126.2 mph. So, you're not going to do better with a common red brick. Maybe something with a smoother surface, higher density, or a smaller surface area would be better... :)

I'll bet that's closer than you expected.. ;)

makes no difference to sea levels.

EVERY time this comes up I have to debunk this stuff.

Do you understand why things float in water? Because the mass of water they displace is equal to the mass of the thing floating.

So now you have very dense saltwater, and much less dense freshwater ice (do you understand why ice is freshwater? It forces the salt out as the surface freezes, so the saltwater below it is even saltier and denser) If you have 1kg of ice, it displaces 1kg of saltwater. Simple enough right? Now let's hit it with the math.

Density of fresh water at 0C: 999.9 kg/m^3
Density of ice at 0C: 915.0 kg/m^3
Density of Ocean: We'll take 1020kg/m^3, the minimum on the site, even though at the pole due to the salt concentration noted in the first link the density of the saltwater will be way higher, but any density over 999.9kg/m^3 means that the water level shall rise as I show below:

1 cubic meter of ice (915.0 kg) displaces 915.0kg of saltwater. 915.0kg of saltwater is 0.897m^3 (915kg/(1020kg/m^3)), which means that our 1m^3 of ice has .103m^3 above the surface of the water (so says the old sailor's adage of icebergs being 9/10ths below water).

Now, let's say the ice were to suddenly vanish. There would be a "hole" in the ocean with 0.897m^3 of air in it. Water would of course rush into the "hole" and the water level would drop by 0.897m^3 spread out over the entire surface of the ocean.

But let's say the ice were to melt. Our 915kg of ice would become 915kg of fresh water, which would occupy about 0.915m^3 (915kg/(999.9kg/m^3)). The hole the ice occupied previously was only 0.897m^3 large, which leaves us with .018m^3 more water than we began with. This .018m^3 would spread out over the surface of the ocean, raising the water level ever so slightly. (sorry, your "no difference" myth has just been busted.)

Don't forget that this tiny amount will be joined by water running off of Greenland, Antartica and other polar landmasses with ice on them, 100% of which will raise the water level.

I answered most of your issues in another post.

The highlights-

Powers of 10 ? Sure, but we are dealing with real materials. Real water has a density of 915kg/m^3 in solid form (aka ice) at 0C.

As I said in the original post, we don't have to deice the entire wing, just the leading edges and a foot or two back. For a 747 the total is ~160m^2. So yeah- it is a huge overestimate in our workload.

I did forget heat of fusion, so call it a car battery.

Its reasonable to consider raising the temp only 1 degree for two reasons. Heat of fusion is >> heat to change temp. More seriously, if there were reason to believe they had to change the temp that much they'd deice more often, that's why I came up with the ludicrous notion of deicing 1000 times a flight. That once every 18s on a 5 hour flight.

I used surface areas according to where I looked up the info, but even if I used planforms it would be fine. This is /. not Boeing. Plane wings are pretty thin. Once you do the math right there are fudge factors aplenty elsewhere. Its close enough.

Despite my 'hideously' wrong math my point remains cogent- this is feasible.

Lastly 'innacurate' is supposed to be inaccurate. Get it ? Humor, in this case self effacing. I thought the silly concept of a floral bonnet (Firefly reference to boot) just might have clued you in. I use OS X and Safari. Safari underlines incorrectly spelled words- its pretty hard to make a mistake like that.

Now if only I had mathcheck in the browser.

Well, thanks and you were right up to a point. Of course the point is rather moot, as I said before they have already flight tested the system. (Read Petrenko's page for what little there is.)

Let's forget converting units and start again.

541.2m^2*0.000003=0.0016236m^3

The density of "solid water" at 0C is 915kg/m^3. Reference

That means we have 1.485kg of solid water to turn to ice. That's 1.485kcal or 6.217kJ.

As someone else pointed out I forgot heat of fusion. That works out to ~497kJ. So, our total energy for a de-icing cycle is 503.3kJ.

If we actually feel the need to de-ice the plane 1000 times in a NYC to LA flight of about five hours... well that's once every 18 seconds. I picked it to be pie in the sky high. I don't think planes of any sort run that many de-icing cycles. In any case 503MJ isn't unreasonable for a 747 at all. A 747 uses 2.5TJ in a 5 hour flight. Yeah- terajoules. What's a few megajoules one way or another ?

We could just burn fuel from the aircraft to charge a capacitor, since aviation fuels range from 33-37MJ/liter. We could consider a system that combines that with batteries too. A truck battery holds ~5MJ, so you could use those for a few de-icing cycles. D-Cell's was wrong but still, even a Cessna can do the job of deicing a 747 wing.

Energy and power are not what will keep this off new planes. Cost of refits may keep it off old planes.

I picked a 747 on purpose. It has a huge wing area. Remember that a plane really only needs to deice the leading edges during flight. The wing area of the 747 dwarfs the area of the leading edges of the wing. The total leading edge area, going back about two feet for a 747-400 is about 160 m^2.

For comparison The total wing area (both wings planform top and bottom) for a Cessna 140 is ~32m^2. Thats less than an sixteenth of the area we were calculating.

32m^2*0.000003m*915kg/m^3=0.087kg of ice, that requires 367J+29.1kJ to overcome heat of fusion... 29.47 kJ.

That can easily be handled by a car battery, and in a pinch we can use a few D-Cells. Oh, and we are again deicing the whole wing, not just the leading edges.

The real point was and remains that this is entirely feasible. The GP was wrong regarding its central point, which was that the concept was entirely infeasible due to energy restrictions.

Pure water is 1000kg/cu.m like others have said already, but apparently sea/salt water is slightly heavier (cuz of heavier minerals?).

Sugar is 800kg/cu.m and Salt is 1200kg/cu.m... (I'm assuming powder compacted by gravity and regular shaking)

Stone is around 2500kg/cu.m.

A cubic meter of steel, bronze or iron might be around 8000kg/cu.m.

Uranium and gold are close to 19000kg/cu.m.

Dalekenium (fictional metal alloy used in Daleks) might be around 600-700kg/cu.m (about 1/4 of aluminium) and this happens to be the density of wood, which the first Dalek models were made of.

Found this link some time ago. Weights of stuff
Works well in conjunction with Volume calculator page

TDP cheats, the energy used to remove the water is actually the same energy used to break down the organic molecules. TDP requires water. The water is heated for pressure to break down the organics, it is then quickly depressurised causing it to boil off rapidly. Instant dry! :-)

It is laughable to say that the body has pulled out most of the caloric value, the body cant process any fiber, and is no where near 100% efficient in using the energy inside sugars, fats, and protiens. Also the sewage wouldn't produce "net" oil. But it does.

Also on the trash front, yes it is possible to put a bound on the ammount of oil produceable by TDP. It is directly proportional to the world's output of garbage, sewage, and other waste products. I'll give a rough estimate below.

And yes TDP will if nothing else be a better way to take care of waste reclamation problems. however I think you grossly misunderestimate the amount of waste the world produces.

a quick google produces

http://www.newton.dep.anl.gov/newton/askasci/1993/ environ/ENV005.HTM/

"Your first question on the average number of pounds of trash produced by Illinois residents has a rather shocking answer. My sources say that we produce about five pounds of trash per person per day. While this may not seem too extreme, consider that the people of India produce only 1/2 pound of trash per person per day, 10 times less"

and here

http://www.metro-region.org/article.cfm?articleid= 5579/
"Each individual generates about 1.5 tons of solid waste per year - about 4.5 pounds per person, per day. If we continue this pattern, we will have each created 90,000 pounds of trash in our lifetimes.
Environmental Protection Agency, "Resource Conservation Challenge: Reducing Waste and Recovering Energy," EPA 530-F-02-033, 2002"

So lets assume the world average is 1 lb per person per day. assuming America is the highest and india is on the lowend. (and we will not consider the industrial waste products which are actually many many times more massive than the indvidual waste so this is a VERY lowball estimate)

6 billion people * 1 lb * 365 day / year = 2,190,000,000,000 ~ 2.1 billion lbs of waste per year (i dont believe this even includes sewage)

Assuming we used TDP to convert it all to oil at the sewage rate of 26% we get 569,400,000,000 ~ 500 billion lbs of oil per year.

http://www.simetric.co.uk/si_materials.htm/

Oil, petroleum (881 kg) / (cubic meters)

google calculator...

(881 kg) / (cubic meters) = 7.35230135 pound / US gallon

500 billion / (7.35230135 pound / US gallon) = 68 Billion gallons per year ... / 42 gallons per barrel = ~ 1.6 billion barrels of oil per year

1.6 billion / 365 = ~ 4,383,561 barrels of oil per day ...

The world pumps ~ 30 - 40 million barrels of oil per day. (OPEC is at 25ish so I estimated)

TDP can provide 10% of the World oil production. And as we get onto the other side of the Hubbert peak, that percentage will grow higher and higher.

-----------------

Does this solve the complete energy problem? No.

Is it a big step? Yes.

Will purposefully created fuels(biodiesel) be part of the equation also? Yes, but rememebr with TDP we can reclaim some of the waste at each cycle.

Will TDP provide better, more sustainable ways to clean sewage & provide drinking water, dispose of trash, and keep more energy in the loop? YES

Man that took way too long....

There seem to be about four different kinds of coal (Anthracite Solid and Broken, Bituminous Solid and Broken), and the previous link doesn't specify a type of coal, I'll go with the average of the cleaner burning two Anthracite coals (+-1300kg/m3).

The GPP doesn't state what kind of swimming pools we're talking about, but a single olympic swimming pool is (50*25*3*1300) 4875000 kg of coal, which is (4875000*8) 39000000 KWh. Assuming the coal doesn't burn completely, but only say 90% to it's potential, that's (39000000/365) 106849 years worth of continuous computing pleasure.

Assuming the GPP was talking about a backyard swimming pool, it seems they are about 70m3 on average, or 249 years worth of computing pleasure.

The inevitable smart-ass question of "Oh, but that electricity has to come from somewhere!!".

Consider this:
Energy content of gasoline: ~45 MJ/kg
Density of gasoline: 737 kg/m3
1 cubic meter = 264.172051 gallons, equals 2.79 MJ/gallon.

Now 1 kWh is exactly 3.6 MJ. Electricity costs (let's exaggerate) 30 cents per kWh.

What do you pay for gas?

Now add to that the facts that:
1) It is easier to clean up a handfull of power-plants than a millions cars distributed over the whole country.

2) Electricity doesn't have to come from fossil fuel sources

3) Even if it does, power plants still produce energy more efficiently than an automobile engine.

you could probably get a better deal if you were buying bulk, but i believe the profit margins are actually fairly tight in the gasoline market.

this site claims, that the average mass of a fish is 18,4 pounds. Or 8.34609961 kilograms or 8346,08861g.
Avogadros constant is 6.02214199 × 10**23, so the molar mass of fish is 5,02613307 × 10**27 or 5026,13307 septillion
Not sure what use it is though.

To save the other metric readers the trouble googling, 1 mils = 1/1000 inch = 2.54e-5 metre. conversion

Actually, water does change density based on temperature. It is most dense at 4 degrees C. As it gets warmer, the density decreases. This site has some numbers. Based on the density variation from 20-40 degrees, a 1 degree global change would increase the volume of water in the worlds oceans roughly 300,000 cubic kilometers (I used 1 billion for current volume, for simplicity). Or (again, pretty rough calculations) about enough to raise sea level around 10 cm. Probably 1/2 to 1/4 of that is more realistic, though since the density change over that range is more significant than the lower range, where most ocean water probably fits.