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Index of Subjects Hi Virginia and All An old story that I read tells of surveyors going across India with chains to measure the distance and some sort of manometer to determine elevation. Results were pretty accurate to but it did take time! Have a nice winter Paul --- Virginia and Terry <vredden@imlay.ca> wrote: > 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 > ____________________________________________________________________________________ Finding fabulous fares is fun. Let Yahoo! 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