Earth Pressure

I have only one thing to say...seek help

<...6355.94PSI approximately...>

That figure is just not approximate enough. After all, the <...mile...> is only 1 significant digit.

<...turbulent flow...>

So this is in seawater, not rock? That distinction would vary the approximation above by a factor of 5. Fortunately for every living thing on land, rock is very viscous at everyday temperatures.

<...mile and a half...>

So, what's the pressure at that depth, approximately?

<...constant pressure energy...>

WTF is that? It didn't appear at all in A-level Physics!

-o-o-o-

Try switching to decaf.

The bertha rogers drill hole in oklahoma, at 5.9 to 6. miles depth, according to wikipedia was almost 25000PSI, so 6355.94PSI is relitively accurate.

I would say that in solid granite the turbulence would be strictly on the magnecular structure of the grain boundaries and would likely not be on the kolmogorov scale, but rather a smaller amount of turbulocity, more on the macular flux level.

The effect this would have on the grain structure of the metal in question, unless the metal is pure Cargonite, would be minuscule and not magnescule as one might expect.

Once the metal has begun to solidify and the grain structure becomes more or less morphologically entwined within its own boundary layers, the pressure exerted on it by the overlying structure will become negated by the interstitial space between each magnecule, thus leaving one with a metallic chunk of crap.

I hope this helps. If not, call me at BR-549, and we can discuss in excruciating detail.

Well... 6500PSI isn't all that hard to achieve up here on the surface of the earth. My Hydraulic test bench can do it no problem. So why not take your metal you want to test, place it into a pressure vessel, and run it up to 6355.94PSI and see what happens?

That sounds a heck of a lot easier than carving out a granite mold a mile and a half down and filling it with molten metal.

Surely the "pressure" is the same as at the surface otherwise miners would need special suits to work.

Now the "weight" of the rock at that depth is a different thing.

Or have I got things completely wrong

John