# 9.0: Sea-level Pressure Reduction

Near the bottom of the troposphere, pressure gradients are large in the vertical (order of 10 kPa km–1) but small in the horizontal (order of 0.001 kPa km–1). As a result, pressure differences between neighboring surface weather stations are dominated by their relative station elevations zstn (m) above sea level.

However, horizontal pressure variations are important for weather forecasting, because they drive horizontal winds. To remove the dominating influence of station elevation via the vertical pressure gradient, the reported station pressure Pstn is extrapolated to a constant altitude such as mean sea level (MSL). Weather maps of mean-sea-level pressure (PMSL) are frequently used to locate high- and lowpressure centers at the bottom of the atmosphere.

The extrapolation procedure is called sea-level pressure reduction, and is made using the hypsometric equation:

\ \begin{align} P_{M S L}=P_{s t n} \cdot \exp \left(\frac{z_{s t n}}{a \cdot T_{v}^{*}}\right)\tag{9.1}\end{align}

where a = ℜd/|g| = 29.3 m K–1, and the average air virtual temperature $$\overline{T_{v}}$$ is in Kelvin.

A difficulty is that $$\overline{T_{v}}$$ is undefined below ground. Instead, a fictitious average virtual temperature is invented:

\ \begin{align} \overline ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Meteorology/Book:_Practical_Meteorology_(Stull)/09:_Weather_Reports_and_Map_Analysis/9.00:_Sea-level_Pressure_Reduction), /content/body/p[7]/span, line 1, column 2  =0.5 \cdot\left[T_{v}\left(t_{o}\right)+T_{v}\left(t_{o}-12 \mathrm{h}\right)+\gamma_{s a} \cdot z_{s t n}\right]\tag{9.2}\end{align}

where γsa = 0.0065 K m–1 is the standard-atmosphere lapse rate for the troposphere, and to is the time of the observations at the weather station. Eq. (9.2) attempts to average out the diurnal cycle, and it also extrapolates from the station to halfway toward sea level to try to get a reasonable temperature.

Sample Application

Phoenix Arizona (elevation 346 m MSL) reports dry air with T = 36°C now and 20°C half-a-day ago. Pstn = 96.4 kPa now. Find PMSL (kPa) at Phoenix now.

Given: T(now) = 36°C, T(12 h ago) = 20°C, zstn = 346 m, P(now) = 96.4 kPa. Dry air.

Find: PMSL = ? kPa

Tv ≈ T, because air is dry. Use eq. (9.2) : $$\overline{T_{v}^{*}}=$$

= 0.5·[ (36°C) + (20°C) + (0.0065 K m–1)·(346 m)]

= 29.16·°C (+ 273.15) = 302.3 K

Use eq. (9.1):

PMSL = (96.4 kPa)·exp[(346 m)/((29.3 m K–1)·(302.3 K))]

= (96.4 kPa)·(1.03984) = 100.24 kPa

Check: Units OK. Physics OK. Magnitude OK.

Discus.: PMSL can be significantly different from Pstn