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Hi Dave and others,
Not to flog a dead horse too long here I hope -- most NNS-ers may
want to skip this post -- and of course I was being flippant and
chauvinistic in the earlier reply, just to defend home turf and get you
going a bit: apologies if you took it to heart, not intended. Both
methods we cooked up would work in principle, the question being
whether either would work in practice, and with what sensitivity and
what likely error if so, always important in designing experiments.
What I meant originally was that I actually hadn't tried to
understand your method because the result seemed so obviously out of
whack, while you don't derive your method for po = h + p (previous
e-mail), perhaps thinking it self-evident. Looking at it, aren't you
putting variables for water pressure and air pressure illegally in the
same units in the same equation and that's how you come up with such
amazing sensitivity?
Despite the decimal place correction you make here, you can't say that
you'll get anything like a 5 cm change for a 39 m rise, because (a) you
haven't said what the absolute volume of air is that you've trapped in
the manometer, and (b) something must be wrong with the derivation.
On (a), logic would dictate that if you have a column of air of 1000 ml
trapped in the fixed-diameter manometer tube, and the length of it
increased by 5 cm when you ascended to a certain elevation above sea
level, if you now reduced this to a volume of only 10 ml in the same
tube (same height ascended) the change now would be only 1/100 of this,
or 0.05 cm -- obviously, the actual change in cm must depend directly
on the volume trapped. On (b), as I said last time, I simply looked
this up on Wikipedia -> Atmospheric Pressure -> Barometric formulae,
where I picked equation (2) to calculate P, the atmospheric pressure at
different elevations (the constants are a bit different for the 6
altitude ranges given, and it would be appropriate to use the lowest
one here). I've just re-checked this to see if I made a mistake, and
so that you can check this for yourself if your interest is still
piqued, it is
P = Pb*exp-[g*M*(h-hb)/R*T],
where Pb is the standard atm pressure at sea level, hb is the reference
height for sea level (zero meters here), and h in meters is the actual
elevation you want to go to above hb. If you look at the bottom of the
Wiki page, as I guessed, this is indeed derived from the universal gas
law PV=RT, only they insert mass/density in place of volume V.
When you insert the standard values given for constants g, M, R, and
the particular absolute temperature T quoted (presumably some world
average), this evaluates as
P=Pb*exp-[0.00011857*(h-hb)]
When at sea level, (h-hb)=0, and exp-[0] = 1, so P=Pb as expected.
When at 200 m above sea level, you get exp-[0.00011857*200] = 0.977, so
P=0.977*Pb, a pressure drop of 2.3% in ascending 200 m, which in my
opinion would be hard to measure practically in a Tygon manometer as a
corresponding volume expansion of 2.3% -- my original point.
When at "about 39 m" up, the height you discuss here, it comes out as
P=0.995*Pb, a 0.5% change in pressure, or volume. So if you had a long
vertical column of trapped air 100 cm tall in your Tygon manometer, the
expected change in its length for a 39 m change in elevation should be
only 0.5 cm. To get a 5 cm change in volume for a 39 m ascent as you
describe, you'd need an air column 1000 cm tall, or 10 meters.
Atmospheric pressure ideally can support a column of clean water 10.3 m
high before the column collapses, so this might be just possible, but
tricky to support in a different sense if the wind gets up at all
during your ascent up the hill carrying a 10 m high manometer.
Maybe we should continue this off-line or risk an "Editor: this
correspondence is now closed" rebuke? One of us is missing something,
or as the schoolboy riddle goes, " Two scotsmen are shouting across at
each other from two tall buildings, but can't ever agree on anything.
Why not?"**
All the best,
Steve
**(A: they're arguing from different premises).
On 16-Feb-07, at 11:33 AM, David & Alison Webster wrote:
> Hi Jamie & All, Feb 16, 2007
> In my original post of Feb 8, I stated--
> "As a rule of thumb, a 5 cm difference from sea level would represent
> an elevation difference of about 3.9 metres."
>
> This should have read 'about 39 metres'. I apparently went from a
> spreadsheet value of 3934.9 cm, transposed this to the e-mail as 3.9
> metres and never looked back.
>
> This correction does extend the possible elevation range, e.g. an h
> of 40 cm from sealevel would represent an elevation of 320.3 M.
>
> Yt, DW
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