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Wed 7 Sep, 2005 10:11 pm
when the olymic games were held outside mexico city in 1968, there was much discussion about the effect the high altitude (7340 ft) would have on the athletes. assuming air pressure decays exponentially by 0.4% every 100 ft, by what percentage is air pressure reduced by moving from sea level to mexico city?
I put down P (7340)= 100 (0.004)^ (7340/100), but I got 0 at the end...
If it decays by 0.4% every 100 ft, then it will drop to 99.6% of its previous value for each 100 ft you climb.
Given the equation
dP/dh = -0.4/100.0, (P: percentage of pressure; h:height in foot)
the percentage of the pressure P is given as
P = -(0.4/100.0)h + initial value
where initial value is 100(percent).
And if h = 7340, then
P = -(0.4/100.0)7340 + 100
= 70.64 (percent).
I believe pc6817 had it right except that .004 should be .996. That gives you the final pressure. You still have to compute the percentage of reduction.
satt_fs's equation will go negative at high altitudes.
P=100%*exp[(-.004/100)h]
Where h is elevation in feet
This assumes that the expansion is isothermal; it will be greater if adiabatic.
Rap
Well, the expression
P = -(0.4/100.0)h + 100.0
is the linear approximation of
P = 100*exp{-(0.4/100.0)},
and the latter is more exact for "exponential equation."
(Linear approximations apply only to local range of variables, and if they were applied globally results often yield absurd values. )
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