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> where Pb is the standard atm pressure at sea In case anyone is wondering if 'others' are interested in the 'elevation' posts. I have read every single one from beginning to end and enjoyed them immensely. I did not understand a single thing except that I would like to see someone running up a hill with a 10 m high manometer. Will you please post when and where this is going to be attempted. Virginia Steve Shaw wrote: > 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 > > -- The most courageous act is still to think for yourself. Aloud. Coco Chanel
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