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May 31st, 2014, 02:13 PM   #1
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Word problem

The city Villach lies 501 m over the sea level. They measure there an air pressure from p = 962 hPa. Near Villach rises the mountain Dobratsch at 2 167m with a air pressure from 790hPa. On the top of the mount Everest at 8 850m we have an air pressure at 326 hPa. I have to write a function for the air pressure and my ideas doesn´t work. I have thus problems to interpret the answer to the question.
h=0 -> p = 962
962=p(0) = a
790 = 962 e^lambda*1666
I have no idea in which way I can calculate the lambda value and why it is 1.18
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June 1st, 2014, 03:52 AM   #2
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A good dose of sleep and the problem is solved.

My conclusion:

I assumed that I should use only the scopes of high meters in which I have measure points. And the first measure point was at 501 meter with 962hPa this was my p(h) and 790hPa=a (However I have PROBLEMS to make such ASSUMPTIONS/rearrangements, PLEASE HELP).
Thus 962=790*e^(-lambda*1666)
790hPa next higher measure point. h=1666 meters is the difference between 2167-501.
Set lambda alone ->
962/790 = e^(-lambda*1666) <-> 1.2177 = e^(-lambda*1666)
The next hurdle was to switch e to the left side.
However the Natural log of e = ln was the answer.
ln(1.2177) = 1*-lambda*1666 PLEASE CORRECT me when my understanding or handling with the ln is not correct.
0.1969815 = -lambda*1666
0.0001182 = -lambda

Now I can use the given function: p(h)=a*e^(-lambda*h)

a=962hPa my start value, first measure point.
h=501m the height of Villach
h=2167m the height of Dobratsch
h=8850m the height of Everest
h=199m to test the function on h values under 501m and it works.

p(501)=962*e^(-0.0001182*501) = 906.78hPa
p(2167)=962*e^(-0.0001182*2167) = 744.94hPa
p(8850)=962*e^(-0.0001182*8850) = 338.56hPa
p(199)=962*e^(-0.0001182*199) = 939.67hPa
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June 1st, 2014, 06:09 AM   #3
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The function p(h) = a*e^(-lambda*h) is a reasonable model to try. In your first post, you omitted the "-".

However, you don't know what values of h and p to use to calculate the values of the parameters a and lambda.

As there are two unknowns, you can use two equations. However, you are effectively given three equations (three known pressures at three known altitudes). Some kind of compromise is required.

Putting h = 0 in the equation, one should get the atmospheric pressure at sea-level. That wasn't mentioned in the question, but it's generally taken to be 1013.25 (hPa). I'll further cheat (any compromise is a cheat) by using the figures given for Mt Everest to determine a value for lambda.

I get lambda = 0.000128138, so the model becomes p(h) = 1013.25e^(-0.000128138h).
That equation gives p(501) = 950 and p(2167) = 768 (both rounded to nearest integer).
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