Since normal air pressure is 14.5 pounds per square inch, that means that a column of atmosphere that is one inch square on the surface of the earth and extends to outer space weighs 14.5 pounds. The density of granite is
approximately 0.1 pounds per square cubic inch, so a column of rock that is one inch square would have to be 145 inches tall to weigh 14.5 pounds, or 12 feet tall. So to answer your question, in a pressure differential situation, where the weight of the atmosphere on one inch were halved, you'd need an additional six feet of rock to offset that change.
A very shallow quake might be on the order of a mile down. Most are much deeper. A column of rock that is one inch square and a mile tall would weigh 6336 pounds. In other words, in order to affect an effective vertical pressure on a fault at that depth, the air pressure would have to be 3 tons per square inch. I'll leave it to you to figure out how thick the atmosphere would have to be to weigh the same as 3 tons.
In structural geology calculations, where we derive the differential stress on a given plane, the weight of the atmosphere is so small it is thrown out of the calculations.
Off Topic » in reply to Re: What is your opinion, why is the increase happening?
Look at it this way:
The whole thread (31 posts)
- HEAR YE HEAR YE, The threatening THEORY explained again. (originally from Mysteries of the Multiverse)
- Re: HEAR YE HEAR YE, The threatening THEORY explained again.
- Re: HEAR YE HEAR YE, The threatening THEORY explained again.
- Re: HEAR YE HEAR YE, The threatening THEORY explained again.
- What is this again?