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If the cone's constant atm pressure entry-rate is 100 square feet square at the top, (or 14,400 sq. ins.) and only 1 square foot square at the bottom, (or 144 sq. ins.), the direct and indirect atm top-hole surface pressures would be altered (ostensibly)by 100 to 1, because all 14,400 sq inches of 14.7 PSI each, would descend at the crater-wall angle upon the 1 foot square of Mercury surface at 14,400 times what is sitting on the water of the beach. That pressure would be 211,680 lbs., AKA 94.5 Metric tonnes.
The pressure at the bottom of the cone of mercury will be determined simply by the vertical distance between the bottom of the mercury and the top of the mercury.
A nice clear diagram might possibly explain what you are talking about.
Stay on a topic, and don't throw in red herrings. I'm sure there must be at least a few serious debaters out there that know something about the Einstein Constant, the Standard Model, and how to stick to a prime topic. Where are you please?With apologies fellows, I'm gone until they show up,or you get serious.Bye.Fleep
I want to start this topic once again to conclude a final result