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    Re: Lunar Distances with Alex's SNO-T
    From: Fred Hebard
    Date: 2006 Oct 31, 16:48 -0500

    I was considering that I had messed up reading the contours on the
    topo map for Alex' apartment, but no.  However, I did not correctly
    identify Alex' apartment.  It may be about 680 or 690 feet above sea
    level rather than 670 (which would be an insignificant effect on
    refraction, by the way).  The same map shows the NE end of the small
    runway of the airport at 607 and a 600-foot bench about 3/4 of the
    way west on the large runway.  Google Earth is about 7 feet less that
    that at the airport.  Is it using a different datum from USGS maps?
    
    Topozone.com's maps are a great resource, as they cover the entire
    U.S.  They are bitmapped scans of USGS 7.5', etc, topo maps, the same
    map Bill is using.  The real map Bill is using is much more of a
    pleasure to use because one can see the whole thing rather than just
    a snippet, kind of like the difference between a chart and a GPS
    chart plotter.
    
    So I guess Indiana is not completely flat!  But the contours where I
    live (36 46.8N, 81 50.7W) are at 20' intervals, in the flatter spots!
    
    Fred
    
    On Oct 31, 2006, at 3:41 PM, Alexandre E Eremenko wrote:
    
    >
    >
    > Interesting:
    > Jean-Philippe says that the airport is 606.
    > Fred says that my appartment is 670.
    > Well, the road from the airport to my place goes up
    > indeed. By 54ft? A bit surprising.
    > And you say that your appartment is 707.
    > (Your appartment is 37 ft higher?
    > Could be, of course...)
    > What is your result for my appartment?
    >
    > Alex.
    >
    > On Tue, 31 Oct 2006, Bill wrote:
    >
    >>
    >>
    >> Alex
    >>
    >> I have a topo map for our area.  My apartment is just about 705
    >> feet above
    >> sea level.  Yours should be close, perhaps a bit lower.  The
    >> Wabash river in
    >> is approx. 520 ft in Lafayette/West Lafayette.  Will look up your
    >> place if
    >> you wish.
    >>
    >> According to USA Today:
    >>
    >> http://www.usatoday.com/weather/resources/askjack/wfaqpres.htm
    >>
    >> "The pressure reported for Denver, or any official observation
    >> station for
    >> that matter, is not the actual pressure on the surface, but rather
    >> is the
    >> pressure corrected to sea level. The reason this is done is so that
    >> meaningful maps of constant pressure lines, called isobars, can be
    >> drawn for
    >> stations across the USA. These maps are useful for picking out
    >> areas of
    >> relative high and low pressure. If pressure readings were not
    >> corrected,
    >> places like Denver would almost always have lower pressure than
    >> spots at
    >> lower elevations. Essentially, the map would reflect topography,
    >> rather than
    >> weather systems in the atmosphere."
    >>
    >> A while back Frank, if I recall, gave a method for converting the
    >> broadcast
    >> barometric pressure to local station pressure.
    >>
    >> On a calculator divide your altitude above see level by -32000 (maybe
    >> -34000?).  Then hit (on my TI-30XA) 2nd then LN.  (You are looking
    >> for e^x,
    >> the mathematically inclined can say it better, but that's the
    >> cookbook
    >> method).
    >>
    >> Multiply the broadcast pressure by the above result to get local
    >> (station)
    >> pressure.  I use it with inches Hg, so not sure if it also works in
    >> millibars, but don't see why it would not.  Perhaps Frank can
    >> clear up my
    >> fuzzy memory.
    >>
    >> There is some interesting information a well as a formula at:
    >>
    >> http://en.wikipedia.org/wiki/Atmospheric_pressure
    >>
    >> Bill
    >>
    >>
    >>>
    >
    >
    > >
    
    
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