Lab 1: Surface temperature extraction from thermal remote sensing data

Goal and Background: 

The goal of this lab was to become familiarized with the concepts pertaining to thermal remote sensing data and working with thermal remote sensing data from various satellite data to find ground temperatures (assuming perfect blackbody emissivity of the surface). Part 1 required students to make various basic observations of the thermal data. Then, data covering a large area of central western Wisconsin was processed in Erdas Imagine using the Model Maker to first find at-satellite radiance from raw Digital Number (DN) data (Part 2, Section 1), and then to find the blackbody surface temperature from the at-satellite radiance data (Part 2, Section 2). While Part 2 worked with Landsat 7 ETM+ data, Part 3 worked with Landsat TM data and Part 4 worked with Landsat 8 data. These last two sections required the same processes as in Part 2 Sections 1 and 2, but in a fully integrated model form.

Landsat TM includes a thermal band as band 6 with a width of 10.40 to 12.50 micrometers, at a resolution of 120 meters resampled to 30 meters. Landsat 7 includes a thermal band of the same 10.40 to 12.50 micrometers on band 6, but has a resolution of 60 meters resampled to 30 meters. Landsat 8 has two thermal infrared bands, bands 10 and 11, at respectively 10.60 - 11.19 and 11.50 - 12.51 micrometers. These bands have a spatial resolution of 100 meters, but are also resampled to 30 meters.

To receive high enough amounts of energy that ensure that a reliable and high enough radiometric resolution reading is received, the radiometers bands in satellite remote sensors must have a coarser spatial resolution. If the spatial resolution was finer, measurements would be noisier and less reliable. Each thermal band pixel takes up a wider area on the sensor, leading to more energy being captured.

Methods:

In order to convert thermal data to blackbody surface temperature, two steps must be followed. First, the DN must be converted to at-satellite radiance. This requires the following equation:

Ll = Grescale * DN + Brescale

Lλ is the at-satellite radiance.
Grescale = (LMAX-LMIN)/(QCALMAX-QCALMIN)
Brescale = LMIN

LMIN, LMAX, QCALMAX, and QCALMIN are all values that can be found in metadata.

In a single model for this first step, the model built in Erdas Imagine Model Maker will look like this:

Figure 1

Figure 1 shows the model created for Part 2 Section 1.


Second, the at-satellite radiance is converted to blackbody surface temperature using:

TB = K2/LOG(K1/Ll + 1)

TB is the blackbody temperature in Kelvin.
K1 and K2 are calibration constants specific to each sensor. K1's unit is Wm^2 ster micrometer, and K2's is degrees Kelvin.

Landsat 7 ETM+ has a K1 of 666.09 and a K2 of 1282.71.
Landsat 5 TM has a K1 of 607.76 and a K2 of 1260.56.
Landsat 8 has a K1 of 774.89 and a K2 of 1321.08

What this step looks like in model builder is shown in Figure 2. This figure shows the model created for Part 2 Section 2.

Figure 2

These two models were combined for efficiency in creating rasters for Parts 3 and 4. The appropriate values were used, and only a temporary raster was saved for the intermediate stage of at-satellite radiance. Figure 3 shows the model for Part 3, while Figure 4 shows the model for Part 4.

Figure 3

Figure 4

In Part 4, an AOI file provided, but that is suspected of originating from merged county shapefiles then processed through Erdas Imagine, was used to create a new raster image file only covering Eau Claire and Chippewa counties. This smaller area of interest was what was then processed.

Results:

Results were best analyzed and shown in ArcMap. The ability to use the find feature and the ESRI geocoder made finding features to analyze easy along with the overlaying basemaps with altered transparency. Also helpful was the multitude of symbolization and classification schemes. The resulting data from Part 4 was used to create a more complete figure with important cartographic elements included.

Figure 5: Part 2 Final Result

Using the data from Part 2 shown above, it was found that the runway at the Chippewa Valley Regional Airport was 88.27 degrees Fahrenheit and Lake Wissota was 61.06 degrees Fahrenheit. This data was collected in the afternoon on June 9th, 2000.

Figure 6: Part 3 Final Result

Using the data from Part 3 above, it was found that Lake Wissota was 73.85 degrees Fahrenheit. This data was collected on August 3rd, 2011. The difference between the June 9th, 2000 value, and this new value was 12.79 degrees Fahrenheit. This difference was speculated to have been created by differences in snowfall the preceding winter, more or less spring precipitation, or differing temperatures in the months leading up to the measurements.

Figure 7: Part 4 Final Result

The above figure shows the blackbody temperature results from Part 4. After processing an area of homogeneous value 147.57 surrounded the area of interest. This area was removed from the figure by simple reclassification, however a more professional way to remove this area would be to use a shapefile of the area of interest and the Extract by Mask tool in ESRI softwares.

In all of these images it can be seen that different types of land cover result in different temperatures. For example, in these images which were all collected in warmer months, roadways and downtown areas are much warmer than rivers and lakes.

One concept that explains the different temperatures between objects is thermal inertia. While roadways and buildings are made out of materials with low thermal inertia, vegetated areas have greater thermal inertia. Bodies of water have the highest thermal inertia. A higher thermal inertia means that an object has a tendency to hold its temperature or to resist temperature change. Thermal inertia, combined with time of year and the time of day determines how different objects will stack up in temperature.

Objects with higher emissivity will absorb and emit more energy. Emissivity therefore plays as a factor in determining an objects thermal inertia. This is not the sole factor in determining thermal inertia however. Chemical and physical properties also play into determining an objects thermal inertia. For example water has a high emissivity, and also has a high thermal inertia. It is able to have this due to its fluid mixing qualities, its size as an object, and its ability to be vaporized instead of absorbing energy as an object.

Sources:

Landsat imagery is from the Earth Resources Observation and Science Center of the US Geological Survey. The area of interest file provided was derived from ESRI counties vector features.