# Influence of terrain elevation on air pressure

If I invert the pressure altitude equation so that I can get a nominal pressure for a given altitude, and compute the pressure altitude for Denver CO (elevation about $1600 \,m$ -- yes I know the wiki page uses feet), I get that the pressure at that elevation should be about $24.7\, \mathrm{in} \,Hg$ ($835 \,mb$); about $5\,\mathrm{in}\,Hg$, or $\sim1/6th$ less than the typical (or standard) pressure at sea level.

At the time of writing this, the pressure, according to the Weather Channel website, the air pressure in Denver is $29.57 \,\mathrm{in}\, Hg$, and the air pressure in New York City (pretty close to sea level) is $29.88\, \mathrm{in}\,Hg$. These are virtually the same. In my experience weather systems can only shift things $\pm 1$ or maybe $2 \,\mathrm{in}\,Hg$, and thus high/low pressure systems (probably) don't account for the similarity.

This suggests to me that there is some mechanism that tends to equalize the air pressure along the surface of the earth, if so what is this mechanism?

• For weather forecasts, they normalize the pressure so that the measurements are all comparable from region to region (i.e. a stranger will understand that a low pressure system is coming in). The actual pressure is indeed lower, or else my in-air 400kV accelerator (designed in the Netherlands) would actually be able to exceed 350kV at Denver-like altitudes... – Jon Custer Jun 10 '15 at 21:28

Read the Wikipedia article on air pressure, the section on Mean Sea Level Pressure: http://en.wikipedia.org/wiki/Atmospheric_pressure

"When barometers in the home are set to match the local weather reports, they measure pressure adjusted to sea level, not the actual local atmospheric pressure."

So weather reports do not report the actual local air pressure. Local air pressure does vary. I know this from flight training days -- we had to set the altimeter before every flight to the airport elevation to compensate for changing local air pressure.