# 1.10: Hypsometric Equation

- Page ID
- 9737

When the ideal gas law and the hydrostatic equation are combined, the result is an equation called the hypsometric equation. This allows you to calculate how pressure varies with height in an atmosphere of arbitrary temperature profile:

\(\ \begin{align}z_{2}-z_{1} \approx a \overline{T_{v}} \cdot \ln \left(\frac{P_{1}}{P_{2}}\right)\tag{1.26a}\end{align}\)

or

\(\ \begin{align}P_{2}=P_{1} \cdot \exp \left(\frac{z_{1}-z_{2}}{a \cdot \overline{T_{v}}}\right)\tag{1.26b}\end{align}\)

where \(\ \overline{T_v}\) is the average virtual temperature between heights z_{1} and z_{2}. The constant a = ℜ_{d} /|g| = 29.3 m K^{–1}. The height difference of a layer bounded below and above by two pressure levels P_{1} (at z_{1}) and P_{2} (at z_{2}) is called the thickness of that layer.

To use this equation across large height differences, it is best to break the total distance into a number of thinner intervals, Δz. In each thin layer, if the virtual temperature varies little, then you can approximate by T_{v}. By this method you can sum all of the thicknesses of the thin layers to get the total thickness of the whole layer.

For the special case of a dry atmosphere of uniform temperature with height, eq. (1.26b) simplifies to eq. (1.9a). Thus, eq. (1.26b) also describes an exponential decrease of pressure with height.

**Sample Application (§) **

What is the thickness of the 100 to 90 kPa layer, given [P(kPa), T(K)] = [90, 275] and [100, 285].

**Find the Answer **

Given: observations at top and bottom of the layer

Find: Δz = z_{2} – z_{1} Assume: T varies linearly with z. Dry air: T = T_{v}.

Solve eq. (1.26) on a computer spreadsheet (§) for many thin layers 0.5 kPa thick. Results for the first few thin layers, starting from the bottom, are:

P(kPa) | T_{v} (K) |
\(\ \overline{T_v}\)(K) | Δz(m) |

100 99.5 99.0 |
285 284.5 284 |
284.75 284.25 etc. |
41.82 41.96 etc. |

Sum of all Δz = __864.11 m__

**Check:** Units OK. Physics reasonable.

**Exposition: **In an aircraft you must climb 864.11 m to experience a pressure decrease from 100 to 90 kPa, for this particular temperature sounding. If you compute the whole thickness at once from ∆z = (29.3m K^{–1})·(280K)·ln(100/90) = 864.38 m, this answer is less accurate than by summing over smaller thicknesses.