I am amazed it agrees as well as it does; I CERTAINLY would not expect five digits of precision.
While there is nothing inherently wrong with naive models, (hell the standard atmosphere is a naive model), you have repeatedly replaced detail with averages and assumed it all averages out. Gravity varies around ±½% from pole to equator. The earth is an ellipsoid, the average radius you used is correct for volume, not necessarily for surface area. As the atmosphere is a gas, and can move, variations in the vertical temperature profiles are critical to the local pressure (and cause variation day to day). Finally the assumption that averaging works has a flaw that there is (mostly) sea below sea level, while mountains intrude into the air space and displace air all over the world; the assumption that a uniform mass of air is uniformly resting on a sphere has a BUNCH of problems. (I have NO idea how to solve them all.) We know air pressure varies day to day with weather, however, the total atmosphere is relatively constant. I believe a better way to proceed is to average a large number of simultaneously measurements at (or corrected to) sea level, all over the earth. The stations should be chosen to represent a good sampling plan. It is certainly true that the overly precise 101.325 kPa is nothing but 760 mm Hg expressed in SI. Wikipedia's article on the atmosphere has a nice map of 15 year average air pressure in December and June, http://en.wikipedia.org/wiki/Atmospheric_pressure I think it shows two things: *The average is NOT 101.325 kPa everywhere *The standard value of 101.325 kPa is much more in the middle of data than your value. ________________________________ From: Pat Naughtin <pat.naugh...@metricationmatters.com> To: U.S. Metric Association <usma@colostate.edu> Cc: U.S. Metric Association <usma@colostate.edu> Sent: Fri, February 12, 2010 12:17:05 AM Subject: [USMA:46597] Average air pressure Dear Bill and All, I have had another thought on finding the average pressure at (or near) the surface of the Earth. As you will see, I am trying to derive the average by calculation. Our previous correspondence is below. I would appreciate any thoughts you might have on this way of proceeding and particularly on my calculations. National Center for Atmospheric Research According to the National Center for Atmospheric Research, "The total mean mass of the atmosphere is 5.1480 x 1018 kg with an annual range due to water vapor of 1.2 or 1.5 x 10^15 kg depending on whether surface pressure or water vapor data are used; somewhat smaller than the previous estimate. The mean mass of water vapor is estimated as 1.27 x 10^16 kg and the dry air mass as 5.1352 ±0.0003 x 10^18 kg." Calculation Mass of Earth's atmosphere = 5.1480 x 10^18 kilograms Weight of atmosphere = mg = m x 9.80665 m/s2 = 50.48463 x 10^18 newtons Surface area of Earth = 4 π r2 = 4 x π x (6.378 137 x10^6)^2 m = 511.207 8 x 10^12 square metres Average pressure at sea level (Force ÷ area) = 98.755 58 x 10^3 pascals = 98.756 kPa Irrelevant information about the atmosphere and the Earth The average mass of the atmosphere is about 5 teratonnes (= 5.1480 x 10^18 kilograms). The average mass of the Earth is about 6 zettatonnes (= 5.9742 × 10^24 kilograms). The proportion of the whole Earth to its atmosphere is roughly 1 000 000 to 1, that is if the atmosphere was valued at 1 cent then the Earth would be valued at $10 000.00. Cheers, Pat Naughtin Author of the ebook, Metrication Leaders Guide, that you can obtain from http://metricationmatters.com/MetricationLeadersGuideInfo.html PO Box 305 Belmont 3216, Geelong, Australia Phone: 61 3 5241 2008 Metric system consultant, writer, and speaker, Pat Naughtin, has helped thousands of people and hundreds of companies upgrade to the modern metric system smoothly, quickly, and so economically that they now save thousands each year when buying, processing, or selling for their businesses. Pat provides services and resources for many different trades, crafts, and professions for commercial, industrial and government metrication leaders in Asia, Europe, and in the USA. Pat's clients include the Australian Government, Google, NASA, NIST, and the metric associations of Canada, the UK, and the USA. See http://www.metricationmatters.com/ to subscribe. On 2009/12/05, at 08:51 , Bill Hooper wrote: On Dec 4 , at 11:38 AM, Pierre Abbat wrote (in response to my earlier note): >(I wrote) Normal or >>> >>average atmospheric pressure is about 101.3 kPa. (There may be different >>> >>standards for identifying "normal" or "average" air pressure.) >>>(Pierre replied) >>101.325 kPa, to be exact, is the standard. > > >The value of 101.325 kPa is the exact value (the "standard") specified by CGPM >for the standard atmospheric pressure, as Pierre correctly points out. That >value is as close to a universal value as is possible. > > > > > > >In addition, however, there are other circumstances where different standards >are used. Here are two I've encountered (from Wikipedia): > > >... International Union of Pure and Applied Chemistry (IUPAC) recommended that >for the purposes of specifying the properties of substances, “the standard >pressure” should be defined as precisely 100 kPa ... rather than the >101.325 kPa value of “one standard atmosphere”. ... For natural gas, the >petroleum industry uses a standard temperature of 15.6 °C (60.1 °F), pressure >101.56 kPa (14.730 psi). (air pressure) > > >The CGPM standard is probably more precise than is reasonable for many uses, >such as reporting air pressure in weather reports. For such measurements, a >rounder value of 101.3 kPa or even 101 kPa might be suitable as a state >"normal" or "average". > > >Certainly, atmospheric pressure is affected by altitude and other factors. For >specific locations or specific special conditions, the CGPM standard >atmosphere may not be too close to the actual average of the ambient pressure. > > >This is not meant to be argumentative. I recognize the value of 101.325 kPa as >the best value to use for most technical purposes. I just felt it was >necessary to indicate in my note that there are different possible averages or >normal values that might be used. > >Regards, >Bill Hooper > > > > > > >============================== >If you have not already done so, >please note my new email address: > > billhoope...@gmail.com > >(Old address will still work OK temporarily.) >
