Shattering of the ice wall

I was unable to get a response to this on another thread so i thought i might post it again

volume of earth is 10^21, volume of water is 10^18

http://en.wikipedia.org/wiki/Orders_of_magnitude_%28volume%29

that is in meters, so in km it becomes Earth 10^12 water 10^9


one cc of water is 1 gram, so 1 cubic meter becomes 1000000 grams, 1000 kg, or a metric tonne

therefore there are 10^18 metric tonnes of water in the world.


now. lets consider the tensile strength of carbon nanotubes (the strongest material we have) which can theoretically go up to 300 GPa or 3x10^11 Pa, 63x10^9 Newtons per square meter. now the ice wall has circumference of 78225 miles, that is about 125000 km (using 1.6 km to the mile), lets assume that the water goes to the top of the wall, all 150 feet of it, 50 meters give or take (taking 3 feet per meter) so we have a surface area of 6.25 million meters squared, 6.25x10^6 meters squared. so lets divide the volume of the water by the surface area, rounding the SA up to 10^7 that gives us 10^11 meters of water affecting every unit area of the icewall. so using that F=ma, that water is going to give 10^12 Newtons per unit area (in otherwords 10^12 Pa. Therefore even a carbon nano tube ice wall would fail.


of course i have made certain assumptions, such that all of the gravitational force of the wate is being pushed towards the ice wall (and i'm meaning Gravitational force in the FE model too, water will be force outwards by the upthrust of the earth, and that tensile strength should be used for the icewall breaking apart (it should be compressive strength, however i have found not numbers for that - although compressive strength only affects when the crack forms, once the crack forms it is tensile strength that causes the crack to expand) however i also used a carbon nanotube, the strength of ice is mcuh less - steel is 4x10^8 Pa, water is 1.7x10^7 Pa

I have since found that the compressive strength for ice is around  10^8  Pa, so it turns out that tensile strength is a pretty good approximation.


It is important to not that the thickness which is important in these matters is the cross section, so for the icewall the thickness would be the surface area that it exposed to the water (so the depth of the water times the circumference) NOT the thickness of the ice sheet - the thicker it is the longer it would hold, but it would eventually crack and break down.

Sounds like you are a math buff. The Wiki site also lists the the volume of water in the Hoover Dam. Would you mind running those same mumbers on the Hoover Dam using concrete as your model rather than ice?

I am not saying your wrong, just interested.

so lets divide the volume of the water by the surface area, rounding the SA up to 10^7 that gives us 10^11 meters of water affecting every unit area of the icewall.

It sounds to me like you're assuming every ounce of water on earth is in contact with the ice wall; surely this is not the case.

Unimportant:
no but i'm assuming that they all put a force on each other, and a resultant force is put on the ice wall. it isn't an exact model and is just an approximation, but it will most likely be on 3 or 4 orders of magnitude out





Also to do hoover damn i would need the compressibility of its concrete, concrete can be made to varying compressibility (although once a crack forms i imagine it also depends on the tensile strength...), i might do it tomorrow.

Unimportant:
The claim you tried to make would be like saying, a man punching you only hits with the musles in his hand, not the whole body force behind it

Unimportant:
The claim you tried to make would be like saying, a man punching you only hits with the musles in his hand, not the whole body force behind it

If I hit you with my right fist, I doubt the muscles in my left arm are helping out. In fact, I think the only muscles contributing would be the ones in my right arm, those in my chest, abdomen, back, hips, and one of my legs. Certainly things like my neck aren't contributing.

But I wonder, by this logic would not all the continents of the world have been crushed long ago? Surely the tensile strength of an ocean-front cliff face cannot be greater than carbon nano-tubes, yet many of us have seen such cliffs.

My thinking is that somewhere along the line your force transfer logic is severly flawed. How, really, would the ice wall be different from any other shear face exposed to the oceans?

Im not too sure of my logic in this but hear are my thoughts anyways.

If you hit something with your right fist then yes, the muslces in your left arm aren't contributing to the acceleration of your fist.  However if we modify that analogy a bit and say have you walking towards someone and running into them with your fist (football straightarm style) then the mass of your entire body would contribute.  Basic idea is conservation of momentum.  I think thats what DrQuak means...not sure though

also...while i think I understand what DrQuak's theory is about I feel that there are a couple gaps in the logic..too tired now to sit down and figure them out though

Unimportant:
The claim you tried to make would be like saying, a man punching you only hits with the musles in his hand, not the whole body force behind it

Perhaps we should test this theory!

yes i have only a very basic understanding of stress and strain, if anyone knows more about it than me please enlighten all of us (i imagine the engineer - if he is an engineer - should know something about this)


Also i couldn't find any information on the speciffice compressive strength of the concrete in Hoover dam.

To the destruction of land, in the RE there is more surface area to be affected by it (and there is no where to fall off)

And yes i admit this is a poor approximation - but i imagine it is only about 3 orders of magnitude out - which in real life is quite a lot, but with the differences between the two relative strenght is very little


i should also add that the amount of salination in the water affects the tensile and compressive strength - i don't know which way howevr (since it is an impurity it would lead me to think it might lower them, but i truely don't know)

Unimportant:

once again you stun me with your stupidity, FE theory clains the ice wall is 10miles thick (iirc).  This while the Ocean cliffs that you mentioned for RE theory have the whole width of the land mass in question, so if you are referencing the cliffs on the pacific ocean in california; you have thousands of miles.

and on my punching theory, if any of you have ever boxed, you learn that the muscles in fact  control the acceleration.  but the power behind it comes from the body mass that you can throw into a punch (by throwing the hip)

Unimportant:

once again you stun me with your stupidity, FE theory clains the ice wall is 10miles thick (iirc).  This while the Ocean cliffs that you mentioned for RE theory have the whole width of the land mass in question, so if you are referencing the cliffs on the pacific ocean in california; you have thousands of miles.

DrQuak made it very clear in his approximation that thickness does not make a difference, only the total area exposed to the force of the water.

It is important to not that the thickness which is important in these matters is the cross section, so for the icewall the thickness would be the surface area that it exposed to the water (so the depth of the water times the circumference) NOT the thickness of the ice sheet - the thicker it is the longer it would hold, but it would eventually crack and break down.

Please refrain from petty insults, especially when they are based on incorrect assumptions.

and on my punching theory, if any of you have ever boxed, you learn that the muscles in fact  control the acceleration.  but the power behind it comes from the body mass that you can throw into a punch (by throwing the hip)

That's what I said, thanks.

Great editing, and actually the ones in your neck would help, as they contribute to your total body mass, and are close enough to the torso to have an effect, but that quite itself (about thicnkess not being important) basically disproves the FE theory anyway, as the earth is quite old, and actually the thickness is quite important in the initial hold, because if it is not thick enough, then there would obviously be some major breaks, as dams have in the past

So your claim is that, since thickness only slows down the cracking, the ice wall would have broken at some point in the Earth's presumeably long history?

Again, I ask, how is this different from any other continental land mass? Surely after billions of years all the major continents would have been split and cracked into landmasses no larger than a small island by now.

first of all, erosion does happen, but in the RE theory, the landmass replenishes itself, by volcanic eruption and the sort, and RE theory is completely different in that respect, as the earth is round, it is like filling a basin, because gravity holds it to the earth,  whereas the FE theory, the ice wall is literally holding the water back, so its much different

sorry i should qualify my statement that thickness doesn't matter


it doesn't matter in the case that it starts to crack, however it does matter for how long the crach has to go before the whole thing shatters - but with the pressure being much greater than the compressive and tensile strength (as in many orders of magnitude) it will be rather explosive cracking.

you know what.... i have just found it how it would actually work without the icewall shattering.


one of the moons of jupiter, europa, has a crust of ice, and when that ice faces jupiter it cracks because of the same reason i'm siting here, however it rotates away and the cracks "heal" (and leave impressive scars across the surface)


in the FAQ it states that the earth wobbles to cause the tides (which no matter how ridiculous is there) this may theoretically cause the same effect if the icewall were thick enough, of allowing the cracked side time to "heal".


course it would need to be ridiculously thick...and i'm not sure if the off tide time would be enough to compensate for the cracking.

here is a link to the bbc site

http://news.bbc.co.uk/1/hi/sci/tech/449135.stm

You should be proud, this is one of the first instances I've seen of an RE'er actually taking the time to think through and debunk his own theory, rather than just posting some insignificant point and saying "I win".

Quote from: "Unimportant"

You should be proud, this is one of the first instances I've seen of an RE'er actually taking the time to think through and debunk his own theory, rather than just posting some insignificant point and saying "I win".

STFU, dumb ass.

Couldn't agree more.

I am a dumbass for complimenting DrQuak's quest for knowledge? How juvenile.

I am a dumbass for complimenting DrQuak's quest for knowledge? How juvenile.

You're a dumbass for writing nonsense.
You're a dumbass for thinking you're right, using nothing but your fantasied and imagination as "science" and arguments for your belief.
You're a dumbass for acting superior when you really cannot argue in a sociable level.
You're a dumbass for making me write this waste-of-time post.

You're a dumbass for acting superior when you really cannot argue in a sociable level.

I find this humorous.

Also, please stop using the distasteful language. Iron Kong will surely be banned, but you seem somewhat interested in discussing FE theory and so you shouldn't sink to his level.

Quote from: "Xargo"

You're a dumbass for acting superior when you really cannot argue in a sociable level.

I find this humorous.

Also, please stop using the distasteful language. Iron Kong will surely be banned, but you seem somewhat interested in discussing FE theory and so you shouldn't sink to his level.

I beg of pardon.

(i imagine the engineer - if he is an engineer - should know something about this)

-I'm hurt by this acusation.        
Anyway, the main problem with your model is in the force of the water.  The force applied doesn't depend on the volume of water held back.  You can see this for yourself in a very simple experiment:  Take a plastic container (milk jug, or whatever) and fill it with water.  Take something sharp and punch a hole in the jug an inch or two below the water line.  Notice how far the water shoots from the hole.  Now patch it up and refill the jug to the original level.  This time poke a hole just above the bottom of the jug.  Notice that the water shoots out much further than it did from the top hole.  Now the volume of water did not differ at the start of each trial. Therefore, the volume of the water doesn't affect the force applied.

The pressure on a wall or dam is called hydrostatic pressure.  It is defined as P = pgd where P is the pressure, p is the mass density, g is gravity, and d is depth (or height of water column).  An important fact of fluid pressure is that at any point in a liquid, the pressure is the same in all directions.

Think about a diver in the ocean.  He will feel the same pressure on his nose as he will on both of his ears.
Thus, the prssure in any direction at a depth d in a fluid with density p is P=pgd

Take the pressure on the ice wall at a certain depth (say 100 ft).  I have made certain assumpions, mainly for simplicity, namely that pg (weight density) for water is 62.5lb/ft^3 (this is for fresh water and salt water is slightly heavier). So using the weight density:  P = (62.5lb/ft^3)(100ft).  P = 6 250 lb/ft^2 or 43.4 psi or 298.7kPa.

Now if you really want me to, I can do the calculus to find the force applied over the entire area of the wall.  However, as we can see, the pressure applied is rather small, so the force won't be anywhere near the force calculated by DrQuak.

Think about it.  Dams are built relatively thin at the top and very wide at the bottom.  If the force was constant, the walls would have to be uniform.  Since the force is greater near the bottom, the walls have to be thicker to resist the force.

Damn, this has to be my longest post...

The force applied doesn't depend on the volume of water held back.  You can see this for yourself in a very simple experiment:  Take a plastic container (milk jug, or whatever) and fill it with water.  Take something sharp and punch a hole in the jug an inch or two below the water line.  Notice how far the water shoots from the hole.  Now patch it up and refill the jug to the original level.  This time poke a hole just above the bottom of the jug.  Notice that the water shoots out much further than it did from the top hole.  Now the volume of water did not differ at the start of each trial. Therefore, the volume of the water doesn't affect the force applied.

While I agree with your conclusion, this particular experiment is not a valid refutation.

Your argument is structured: "~P and Q therefore P =/=> Q"; in your particular case, P = "The water's volume changed" and Q = "The force applied changed".  This is not a valid argument, since "P ==> Q" is consistent with "~P and Q".  For example, let P = "I take the boss to dinner" and Q = "I get a promotion," and suppose P ==> Q.  However, even if I don't take the boss to dinner (~P), I might still get a promotion (Q).  Thus from ~P and Q you cannot conclude P =/=>Q.

In other words, you're saying, "If you change the volume of the water, the force on the wall will not necessarily change."  However, your experiment didn't involve changing the volume of the water, so you can't draw any conclusions about what would happen if you did.

Instead, your experiment should go as follows: take two similarly-shaped containers, the only visible difference being that one is wider than the other.  Place them next to each other, fill them with water to the same height, punch a hole in the front of each at the same height, and measure the distance the water shoots out.  Observing that the water shoots the same distance in each case, we can conclude that changing the width of the reservoir does not alter the force on the walls of the container.

Another good experiment: despite the ocean being millions of times more voluminous than a swimming pool, the ocean (at a given depth) seems not to crush my compressible air spaces (lungs, middle ears, sinuses) any more than a swimming pool (at the given depth).

Another good experiment: despite the ocean being millions of times more voluminous than a swimming pool, the ocean (at a given depth) seems not to crush my compressible air spaces (lungs, middle ears, sinuses) any more than a swimming pool (at the given depth).

This is a great example.

Instead, your experiment should go as follows: take two similarly-shaped containers, the only visible difference being that one is wider than the other. Place them next to each other, fill them with water to the same height, punch a hole in the front of each at the same height, and measure the distance the water shoots out. Observing that the water shoots the same distance in each case, we can conclude that changing the width of the reservoir does not alter the force on the walls of the container.

Ok, I see what is meant here.  Now if you combine the above experiment with mine, one would be able to asertain what it is that affects pressure.

However, like I said, the ocean/swimming pool was an even better example.

Ok, I see what is meant here.  Now if you combine the above experiment with mine, one would be able to asertain what it is that affects pressure.

Yeah, definitely; you need both experiments for a complete "theory of pressure".

I would say that the reason the ocean doesn't crush you anymore than a swimming pool is the same reason our windows don't shatter in the atmosphere - there is equal force on both sides.


I was trying to find numbers on the base thickness of the hoover and the three gorges dam (hoover dam is about 200 m tall and three gorges is about 190 m tall, however the 3 gorges holds back a lot more water) - however i could only find numbers on the hoover dam - 660 m at the base

oh and thank you for responding Engineer - as i said i know little about materials science - not my area - so i was just winging it

I would say that the reason the ocean doesn't crush you anymore than a swimming pool is the same reason our windows don't shatter in the atmosphere - there is equal force on both sides.

You could say that, but you'd be wrong.  You see, both the ocean and the pool do crush you -- it's just a question of which crushes you more.  Unless you equalize the pressure in your body's compressible spaces by adding more material (air) into them, the outside pressure of the ocean would cause those spaces to collapse until the pressures were equal.  The greater the external pressure, the smaller the spaces will collapse before the pressures are equal.

I can verify from personal experience that even without equalizing pressures, my compressible spaces are not crushed with millions of times more force in, say, the Pacific Ocean than in an olympic-size swimming pool.

I was trying to find numbers on the base thickness of the hoover and the three gorges dam (hoover dam is about 200 m tall and three gorges is about 190 m tall, however the 3 gorges holds back a lot more water) - however i could only find numbers on the hoover dam - 660 m at the base

Water pressure in rivers is not hydrostatic, since the water is moving.  More data needs to be taken into account before the numbers you've found can be relevant.