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*Remote Sensing*

*Localization*

*GIS*

*Informatics*

geoinformatics:thermal-infrared

Using special sensors and the thermal radiation properties of physical bodies, we can measure surface temperature from a distance. In this project, we'll be using thermal infra-red (TIR) techniques to estimate surface temperatures of an iceberg using thermal camera ground measurements and a land parcel using satellite sensors.

Some basic notes on blackbody radiation:

- All bodies with temperature above 0 K emit radiation.
- 0 K =
**‒273,15 °C** - A blackbody is a theoretical object that completely
**absorbs all wavelengths**of electromagnetic radiation incident on it. - When a blackbody is heated to a temperature above 0 K, it emits radiation.
- Blackbody radiation at temperatures comparable to the temperature of the earth’s surface (~300 K) is the
**thermal infrared**(TIR) - “Good absorbers are good emitters”, meaning that a blackbody
**emits 100% of the radiation it absorbs**(so the absorbed radiation has no effect on the blackbody’s temperature)

Basic principles:

Principle | Description |
---|---|

Planck’s Law of blackbody radiation | Describes the electromagnetic radiation of a blackbody of a defined temperature |

Stefan-Boltzmann Law | Calculates the total electromagnetic radiation as function of a blackbody’s temperature |

Wien’s Displacement Law | Calculates the wavelength at which maximum spectral radiant exitance (emitted radiation) occurs |

Kirchhoff’s Law | • Emittance at a given wavelength = absorbance at the same wavelength • Blackbodies are theoretical; the behaviour of real objects can be described based on how close to being a blackbody they are. • Emissivity (ε): ε = 1 ⇒ blackbody ε < 1 ⇒ real-life material |

This law describes how blackbody radiation can be calculated by the object’s (absolute) temperature:

$$ M_\lambda=\frac{2\pi hc^2}{\lambda^5\left(e^{\displaystyle\frac{hc}{\lambda kT}}-1\right)} $$

Where: $M_λ$ = radiation, $λ$ = wavelength, $T$ = absolute temperature.

Therefore, at any given wavelength, we can have a spectral radiant exitance curve (“radiation”) according to the absolute temperature:

*Fig. 1; Source: http://ozonedepletiontheory.info/Images/plancks-law-basic-extended.jpg *

Describes the *total electromagnetic radiation* as a **function** of the *absolute temperature*:

$$ T_\mathrm{RadBB}=\sigma T{}_\mathrm{kin}^4 $$

where $T_\mathrm{RadBB}$ is the radiant flux, and $T_\mathrm{kin}$ the kinetic temperature.

*If kinetic temperature is measured in degrees Kelvin, then the temperature value is directly proportional to the average kinetic energy of the molecules in the substance.*

Describes the **wavelength** at which **maximum spectral radiant exitance** occurs.

$$ \lambda_\max=\frac AT $$

In Figure 1, the dotted line connects the peak radiances of the temperature curves.

Therefore, **hot lava is
red** because it radiates the red wavelengths more than other wavelengths.

For example:

- Perfect blackbodies are nowhere in this world!

- For a given wavelength, the emittance is equal to the absorbance

$$ \varepsilon_{\left(\lambda\right)}=\alpha_{\left\{\lambda\right\}} $$

- Conservation of energy: [absorption] + [reflection] + [transmission] = 1 (for a given wavelength)

$$ \varepsilon_{\left(\lambda\right)}+\rho_{\left(\lambda\right)}+\tau_{\left(\lambda\right)}=1 $$

- For opaque objects that do not transmit radiation: [absorption] + [reflection] = 1

$$ \varepsilon_{\left(\lambda\right)}+\rho_{\left(\lambda\right)}=1 $$

- It is therefore calculated that:

$$ T_{\mathrm{rad}}=\varepsilon^{\left(\frac14\right)}\cdot T_{\mathrm{\kappa\iota\nu}} $$

For real materials, ε < 1, so the radiance temperature **will always be lower** than the actual kinetic temperature on the surface of the material!

geoinformatics/thermal-infrared.txt · Last modified: 2020/08/04 14:34 by sotosoul