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H2Otown_WX

Correcting Station Pressure to Sea-Level Pressure

10 posts in this topic

So I was trying to correct the measured pressure from my weather station at my house to sea-level pressure using the equation P = Po*e^-("roh"g/Po)y where Po is atmospheric pressure (1.013 x 10^5 N/m^2), "roh" is the density of air (1.29 kg/m^3), and y is the elevation of the station (in my case, about 680 feet or 207 meters). Anyway, in going through the calculation, I came up with that I should have to add 26 mb to my station pressure to obtain a corrected value. However, based on SPC mesoanalysis, it would seem that I only need to add exactly half that amount, 13 mb. The only thing I could come up with is that my weather station pressure reading is too high. I'm pretty sure the math was done correctly because I was not off by magnitudes of 10. If anyone has any ideas, let me know.

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you have to do a temperature correction on the pressure as well, depending on whether the barometer is inside or outside. if the barometer is inside where the temp doesn't vary too much, it's not that much an issue as long as it's within the specs (STP is i think 15C/288k, but please correct me if I am wrong). but if it's located outside, it may be a big issue. because the temperature will vary the density of the air as much as the elevation, depending on what medium is used for the barometer to measure the pressure.

because as P=n R T / V
n being the amount of particles
V being the volume
R is a constant
T = Temp

and n/V would = rho (density)

so P = rho R T

or rho = P / R T

so you have the correct volume, but you may not be correcting for the right density (amount of particles per volume, or in this case mass).

this is where when you set up the barometer, you try to get the closest barometer at a local airport and get the initial from there. but most observation sites will have temperature and standard height corrections to take in account the total density, not just height corrections.

hope that helps you out a bit.

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[quote name='H2Otown_WX' timestamp='1347038890' post='1734154']
So I was trying to correct the measured pressure from my weather station at my house to sea-level pressure using the equation P = Po*e^-("roh"g/Po)y where Po is atmospheric pressure (1.013 x 10^5 N/m^2), "roh" is the density of air (1.29 kg/m^3), and y is the elevation of the station (in my case, about 680 feet or 207 meters). Anyway, in going through the calculation, I came up with that I should have to add 26 mb to my station pressure to obtain a corrected value. However, based on SPC mesoanalysis, it would seem that I only need to add exactly half that amount, 13 mb. The only thing I could come up with is that my weather station pressure reading is too high. I'm pretty sure the math was done correctly because I was not off by magnitudes of 10. If anyone has any ideas, let me know.
[/quote]
This website should help.

[url="http://san.hufs.ac.kr/~gwlee/session3/reduction.html"]http://san.hufs.ac.kr/~gwlee/session3/reduction.html[/url]

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For sea level pressure reduction you'll need the temperature from the previous 12 hours plus you have to make a moisture correction since moisture also affects density.

Steve

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[quote name='H2Otown_WX' timestamp='1347038890' post='1734154']
So I was trying to correct the measured pressure from my weather station at my house to sea-level pressure using the equation P = Po*e^-("roh"g/Po)y where Po is atmospheric pressure (1.013 x 10^5 N/m^2), "roh" is the density of air (1.29 kg/m^3), and y is the elevation of the station (in my case, about 680 feet or 207 meters). Anyway, in going through the calculation, I came up with that I should have to add 26 mb to my station pressure to obtain a corrected value. However, based on SPC mesoanalysis, it would seem that I only need to add exactly half that amount, 13 mb. The only thing I could come up with is that my weather station pressure reading is too high. I'm pretty sure the math was done correctly because I was not off by magnitudes of 10. If anyone has any ideas, let me know.
[/quote]

What kind of station are you using? I know Davis "attempts" to compensate for elevation. Even though I'm at 5,075 feet, my display reads 29.71 inches, which is impossible, but obviously that's an attempt by the console to convert real pressure to sea level pressure.

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[quote name='famartin' timestamp='1347304069' post='1739049']

What kind of station are you using? I know Davis "attempts" to compensate for elevation. Even though I'm at 5,075 feet, my display reads 29.71 inches, which is impossible, but obviously that's an attempt by the console to convert real pressure to sea level pressure.
[/quote]


It uses a very simple formula the same way wristwatches that estimate elevation do, and just subtracts .01" of mercury for every 10 feet.

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The Davis stations approimate the altimeter setting whereas SLP is a more complicated affair especially at higher elevations where a plateau correction involving the mean annual temperature in degrees Rankin (459+ T(F)) by using that and the T12 plus current T using the circular slide rule one determines an "R" factor to use to get the SLP. I have a program that does this but it's an old one for use on DOS based systems.

Steve

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When I was at Clark AB we found that we had to adjust our altimeter setting by about 0.01in for every 100 ft elevation when forecasting the altimeter for Baguio which was at 5000' elevation vs our 470.

Steve

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Thanks for all the responses guys. I understand that making corrections to account for moisture and temperature could be causing a margin of error but I still feel that it should not be as significant as what I am encountering. I'll fiddle around with P = rhoRT a little bit and see if that helps.

Famartin - My station is a crappy Oregon Scientific so it doesn't have that feature lol.

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