The real reason for the oceans of the world's rising. Not Global Warming. , page 5

Originally posted by Slipdig1
reply to post by Pigraphia

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In relation to the boats displacement, Did you see the Archimedes principle of floating in the OP?

I know the principals behind displacement.
You are saying that displacement of volume is the same as the force exerted on the object they are not.

I can design a hollow cube that will sink no matter what.
The volume of water displaced is very easy to calculate.
I can then take that same cube and fill it with more weight while maintaining the same displacement.
More water doesn't get displaced because more weight was added to the system.

The displacement of both are the same.

So your premise that the weight is affecting it wrong.

You even gave the example of a 1000kg item it will float as long as the upward force equals the downward force.
That is true, but I can take that 1000kg restraint and design several hulls each will be able to displace different volumes of water.

Like I said I am not attacking your theory, I am pointing out a glaring hole in the method you used.

reply to post by Slipdig1

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Sea waters density is 1027kg/m3 not 1000kg/m3 so thats one mistake.

The density of surface seawater ranges from about 1,020 to 1,029 kg•m−3, depending on the temperature and salinity. Deep in the ocean, under high pressure, seawater can reach a density of 1,050 kg•m−3 or higher.

en.wikipedia.org...

All in all, this difference will not effect my calculations more than +/- 3%

There are 10000 cargo ships world wide able to take an average 15000 teu each. A Teu is around 24000kg 0r 24 tonne. This would weigh one in at 360 000 tonnes and 10000 would be 3 600 000 000 tonnes

The calculations that I used were for DWT, Dead-weight Tons, which is approximately 1,000 kilograms.

I *OVERESTIMATED* the Dead-weight Tonnage of the shipping fleets of the world, by multiplying the amount of Freighters in the world by the *MAXIMUM FREIGHTER SIZE*.

So, in essence, the displacement I recorded in my calculations is *LARGER* than it would be with more accurate calculations.

How did you convert the square Kilometres to a measurement of volume by multiplying it by a measurement of weight?

Firstly, I used square meters.

Secondly, an area in square meters "A" can be converted to a "Sheet" of 1 meter in depth of the exact same number as the area measurements.

In other words... if you have an area that is 100 square meters, and you want the volume that this area would take up if it were 1 meter in depth, you have 100 CUBIC meters.

The conversion from area to volume was so simple and self explanatory, I didn't really think to include it in my calculations, So... there it is.

I estimated the bulk shipping fleets at 282,361,250 metric tons, while you estimate them at 3,600,000,000, now that is around one order of magnitude difference, but compared to the total volume of the oceans of the world down to one cubic meter, it is STILL very small... and thus my original point still stands.

Now, what we *COULD* do, is wax poetic about the minutiae of specific bulk shipping containers, bridge support displacement, Naval fleet sizes, etcetera...

OR, we could realize that:

The oceans surface area is 361 million km squared

And since there are one million square meters in a square kilometer (1000 * 1000) then the surface area of the ocean in METERS SQUARED is 361,000,000,000,000 (361 Trillion square meters)

Now, to a depth of one meter, that gives us a total volume of:

361,000,000,000,000 SQUARE METERS * 1 = 361,000,000,000,000 CUBIC meters.

Multiplied by 1,000 kilograms per cubic meter (Or, if you prefer 1,027 kilograms per cubic meter, which, by the way, makes you even MORE wrong.)

Gives us a mass of 361,000,000,000,000,000 kilograms to 370,747,000,000,000,000 kilograms

Now, let's take the most conservative estimate (which is what I did... the one that makes you less wrong) of 361,000,000,000,000,000 kilograms and divide that by YOUR estimate of the bulk shipping fleet displacement of:

3,600,000,000 tons.

Now, for this calculation we will assume the *TYPE* of tonnage that makes your point more correct, which would be the *TYPE* of tonnage that actually has more mass, that being a Metric Ton (which is a full 10% larger than a long ton)

So, we have

370,747,000,000,000 tons / 3,600,000,000 tons

Which leaves us with a quotient of: 102,985.278

OR, 0.00000971012577 meters or... about 10 microns.

OR, about 1/10th the width of a human hair.

/me bows

reply to post by Pigraphia

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You put on more weight you displace more water. Its pretty simple. Trust me if a hull weighs 1000kgs it has to be able to displace 1000kgs of water or one cubic metre of water to float. If it doesn't displace it it sinks.

Once it has sunk it is just a matter of working out volume. I'm fairly sure.

reply to post by WildWorld

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Thank you I have thought of these. I imagine they would displace alot.

It is going take me alot of figuring out, and since I'm not a scientist and not on it full time, it will take me a little while to get any findings.

It does seem like there are alot of things that could effect this though.

reply to post by Slipdig1

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Perhaps I am not being articulate enough, I will use external references to explain this better.

Buoyancy = weight of displaced fluid.

Wiki
Density=Mass/Volume so Volume=Mass/Density
Wiki2
From the first wiki
D(hull)/D(water) = Weight(hull)/Weight(hull) - Weight(apparent immersed)
Immersed weight=Weight of Water displaced
Dw= 1000kg/cubic meter
____________________________________________
Hull A=C=1000kg
Volume A= 10Cubic meters
Volume C= 1/3A
From this we get the densities.
Da= 100kg/cubic meters
Dc= 300kg/cubic meters

____________________________________________

Now calculate the apparent weight displaced by Hull A which equals the weight of the water displaced
Da/Dw = Wa/(Wa - Wapp(a))
100/1000 = 1000/ (1000 - Wapp(a))
1.0*10^5 -[ 100 * Wapp(a)] = 1.0*10^6
-100 * Wapp(a) = 9.0*10^5
Wapp(a) = 9.0*10^3
So the apparent weight displaced by Hull A = 9.0*10^3
_______________________________________________
Now calculate the apparent weight displaced by Hull C which equals the weight of the water displaced
Dc/Dw = Wc/(Wc - Wapp(c)
300/1000 = 1000/ (1000 - Wapp(c)
3.0*10^5 -[ 300 * Wapp(c)] = 1.0*10^6
-300 * Wapp(c) = 7.0*10^5
Wapp(c) = 2.3*10^3

So the apparent weight displace by Hull C = 2.3*10^3

Two Hulls, same mass, alter the volume of the hull and the water displaced changes.

I hope that the calculations are clear, I did them with paper and pencil and had to type them up.
I hope nothing is lost in my representation.

reply to post by Pigraphia

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I will try to figure out what you just posted. I guess you know boats?

Sorry I guess thats a little more I got to learn.

So do you think this would increase or decrease the displacement caused by the ships?

There must be a whole lot of mathematics i will have to rehash.

If only boats were the problem, that would be easy to fix. Unfortunately, the reality is far more complex, scary, and unstoppable.

I'm not going to try and give an in depth explanation of what is actually going on because their is no way to simplify it. In fact, the truth is that there is still a lot of conjecture within the science community as to how and why certain alarming changes have been occurring on our planet recently. The thing that is beyond doubt and now widely accepted is that whatever is happening is serious and accelerating at a rapid rate.

For those that are genuinely interested in the subject, I have included some links below that will get you started with some of the latest studies and scientific papers. For those that are lazy, here is a very brief outline of some facts -

- Postglacial Rebound still continues since the days of the last ice age. Earth as had many ice ages which occure like clockwork throughout history.

- rotational instability as ice shifts, also internal an external gravitational and magnetic forces . Earth is actually changing shape

- 70% of the earths active volcanoes are submarine and these volcanoes first observed erupting in 2004 - 2006 in the pacific. Most active near New Zealand and Japan

Cox and Chao [2002] reported the detection of a large anomaly in the time series of Earth's dynamic oblateness J 2, the lowest-degree gravity spatial harmonic, in the form of a positive jump since 1998 overshadowing the decreasing secular trend in J 2 caused primarily by the postglacial rebound (PGR). Here we report that recent data show that J 2 has been rapidly returning toward “normal” (with PGR considered) since early 2001. In search of the geophysical and climatic causes for this “1998–2002 J 2 anomaly,” we report an oceanographic event that took place in the extratropic north and south Pacific basins that was found to match remarkably well with the time evolution of the anomaly. We examine the leading (nonseasonal, extratropic Pacific) Empirical Orthogonal Function/Principal Component modes in the sea-surface height (SSH) data from TOPEX/Poseidon, sea surface temperature (SST) data from the National Center for Environmental Predictions, and output fields of the Estimating the Circulation and the Climate of the Ocean (ECCO) ocean general circulation model (OGCM), including ocean bottom pressure (OBP) and temperature and salinity profiles. The phenomenon appears to be part of the Pacific Decadal Oscillation, and temporal correlations are made. However, quantitatively, the OBP field of the ECCO model predicts a J 2 anomaly that is smaller in magnitude than the observed by a factor of about 3. We discuss various possibilities for reconciling this discrepancy in terms of inadequacies of present OGCMs and considering other geophysical contributions; a complete resolution of the J 2 enigma awaits further studies.

large scale mass redistributions of the oceans water

ring of fire

www.agu.org...

Enjoy, these link lead to plenty more

reply to post by Slipdig1

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I know a little about boats.
I am an applied physics, and mechanical engineering major.

Your theory is interesting and if someone could gather all the data on it it would be interesting to look into.

I was just pointing out that the weight of the cargo alone isn't accurate enough because different boat hulls displace different amounts of water.

You just need a lot more precise data to work through your theory and I was trying to give you the basics of what extra data you need.

Oh, I also can't remember if you accounted for the average displacement on the oceans.
Cargo goes on the ocean but it eventually comes off so the levels rise and fall.
That extra data would be more accurate and you cold see the mean level increase per year.

reply to post by Pigraphia

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Agree and have thought aboutthe rise and fall, as well as a mean or average. I am training myself evernight with this and will eventually have an answer I guess.

reply to post by milkyway12

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So you come into a thread and call "it could've been 3 inches or 3 metres" there is a big difference here. Which one was it?

Slipdig1 actually started this around 2 years ago, so that scientist ripped him off. Which magazine was it in?

I'm sure Slipdig would like to know so he can claim his revenue.

That is if you really read a magazine like that?

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