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I am trying to find out the pressure at a given altitude but have been disappointed to find the following equation does not work at all: (h = height in meters)

enter image description here

Where

  • sea level standard atmospheric pressure p0 = 101325 Pa
  • sea level standard temperature T0 = 288.15 K
  • Earth-surface gravitational acceleration g = 9.80665 m/s2.
  • temperature lapse rate L = 0.0065 K/m
  • universal gas constant R = 8.31447 J/(mol·K)
  • molar mass of dry air M = 0.0289644 kg/mol

( from Wikipedia )

Upon completion of the equation, which I think spits out Pascal values, they don't match what a pressure would realistically look like at that altitude (comparing the output value to one from Wolfram). Any idea what the issue with this equation is?

EDIT: inaccuracy alone isn't the problem, it's not giving the right values for anything. I saw a value of 101800 or so for an altitude of 800ft, which is completely wrong, and then the output starts turning into negatives around 100,000ft. I do need something to work at that height, perhaps this isn't the equation for me?

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  • $\begingroup$ Lets see...it's only good as some kind of average, near ground level real air isn't particularly dry, the composition of air is a function of height, the real ground-level temperature is a function of space and time, as is the pressure...all of which put limits on it's applicability and make a mockery of values expressed to five or more significant figures. I mean, even the gravitation "constant" varies by several tenth of a percent from place to place, and at the one percent level over 30 km in altitude. All that said: how bad is it? And how high do you have to go before it diverges badly? $\endgroup$
    – dmckee --- ex-moderator kitten
    Sep 15, 2011 at 2:31
  • $\begingroup$ It's giving me values higher than 101325 at 800ft altitude, then it goes downhill from there, and it starts throwing negative numbers once the altitude gets to about 100000ft altitude. It's not like a small inaccuracy by any means, it just doesn't work. $\endgroup$
    – Kyle Hotchkiss
    Sep 15, 2011 at 2:40
  • $\begingroup$ Yeah, well it's not a first principles derivation (which would have messy integrals in there...) it is a local parameterization (that is, only good over some reasonable range of height (and 100,000 ft is not reasonable for this purpose: things get pretty thin up there)). That doesn't explain the bobble at the bottom....hmmm...the exponent isn't dimensionless (there is a spare meter in there...). That is a problem. $\endgroup$
    – dmckee --- ex-moderator kitten
    Sep 15, 2011 at 2:46
  • $\begingroup$ Well, I do need the equation to work up to 100000ft somehow. Perhaps this isn't the one? Perhaps international $\endgroup$
    – Kyle Hotchkiss
    Sep 15, 2011 at 2:54
  • $\begingroup$ FWIW, the exponent is basically 5.25. $\endgroup$
    – Kyle Hotchkiss
    Sep 15, 2011 at 2:54

2 Answers 2

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Use the Properties Of The U.S. Standard Atmosphere 1976, which is valid to an altitude of 86 km (282,000 feet).

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I see nothing wrong with the formula, if we take into account what dmckee fairly noticed, that it is valid in a certain range only.

A quick&dirty matlab program gives us the similar looking curve to the one shown on wikipedia. 

clear all; format long g; clc; clf;

p0 = 101325; %Pa  
L = 0.0065; %K/m
T0 = 288.15; %K
g = 9.80665; %m/s2
M = 0.0289644; %kg/mol
R = 8.31447; %J/(mol*K)

h = 0:100:10000; %m

A = (g*M)/(R*L);
B = L/T0;

p = p0.*(1-B.*h).^A

plot(p,h)

plot

My "Fundamentals of aerodynamics" lists the same one.

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  • $\begingroup$ So the issue I am dealing with makes sense then, it's a range. Any idea of a formula that would actually give pressure at different altitudes? Complexity is no issue, I just have no idea where one would exist. I would like to think that Wolfram uses a formula and not lookup tables to get the values in that. $\endgroup$
    – Kyle Hotchkiss
    Sep 15, 2011 at 3:21
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    $\begingroup$ @Kyle -Why not take an available table value? engineeringtoolbox.com/standard-atmosphere-d_604.html It would be much simpler than using that regression equation, especially if you have trouble even interpreting it. There is nothing wrong with table values; if you need some value in between you just interpolate. $\endgroup$
    – Rook
    Sep 15, 2011 at 3:44
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    $\begingroup$ Kylie - If you want proper values at high altitudes you have to work with gravitation intensity "g" as function of altitude. $\endgroup$
    – Crowley
    Sep 15, 2011 at 4:48
  • $\begingroup$ Thought accel of gravity only changed like .1m/s when at 100000ft? Or do you mean something different? $\endgroup$
    – Kyle Hotchkiss
    Sep 15, 2011 at 14:09
  • $\begingroup$ I do feel as if you're on something Crowley, could you post it as an answer? $\endgroup$
    – Kyle Hotchkiss
    Sep 15, 2011 at 14:14

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