Wind Shear

Overview

Engineering reference data for Wind Shear in miscellaneous.

Key Formulas

Unit Conversion

$y = x \cdot k$

Multiply by conversion factor.

Linear Interpolation

$y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}$

Estimate between two known points.

Percentage

$p = \frac{\text{part}}{\text{whole}} \times 100\%$

Part as fraction of whole.

Variables

Terrain Data Table

Example Calculation

This example demonstrates how to estimate the wind speed at a higher altitude given a measurement at a reference height, using the wind shear exponent for a specific terrain.

Given:

• Terrain: Several buildings
• Wind shear exponent, $\alpha = 0.25$
• Reference height, $h_0 = 10\ \text{m}$
• Wind speed at reference height, $v_0 = 5\ \text{m/s}$
• Height of interest, $h = 40\ \text{m}$

Find the wind speed $v$ at 40 m.

Solution:

1. Start with the wind shear formula rearranged to solve for $v$: $v = v_0 \left( \frac{h}{h_0} \right)^{\alpha}$
2. Substitute the given values: $v = 5\ \text{m/s} \times \left( \frac{40\ \text{m}}{10\ \text{m}} \right)^{0.25}$
3. Simplify the exponent term: $\left( \frac{40}{10} \right)^{0.25} = (4)^{0.25} = 4^{1/4} = \sqrt[4]{4} = \sqrt{2} \approx 1.414$
4. Calculate the final wind speed: $v = 5\ \text{m/s} \times 1.414 \approx 7.07\ \text{m/s}$

Result: The estimated wind speed at 40 m is approximately 7.1 m/s.

Physical Interpretation

Wind shear occurs because the Earth's surface and its features (vegetation, buildings, terrain roughness) create friction that slows down air molecules close to the ground. The effect of this friction diminishes with height above the surface. The wind shear exponent ($\alpha$) quantifies how rapidly the wind speed increases with altitude for a given terrain type; a higher $\alpha$ indicates a more rapid change in speed with height.

Interactive Charts

References

• Original Source