[GRASSLIST:4452] Temperature from LANDSAT-TM5 ch.6

Hi,

I'm looking at some LANDSAT-TM5 data and I'd like to derive surface
temperature from channel 6 (thermal IR) as described in Neteler &
Mitasova (1st ed.) S9.4.3.

The results look all wrong though, I'm getting a range of -54 to +69 C.

Using the following formulae:

radiance = 1.238+(15.6-1.238)*tm6/255.0;
temp_kelvin = 1260.56 / log(607.76/radiance + 1);
temp_celsius = temp_kelvin - 273.15

we get:

tm6(0) = -69.7787 deg C
tm6(255) = 68.6640 deg C

Using d.histogram on the "tm24r_b6" 8bit GeoTIFF input raster, I see a
bimodal distribution with peaks at 75 (land) and 220 (ocean). That
translates to -6deg for the land and +57deg for the ocean...

is my satellite data wrong or is it the conversion formulae?

thanks,
Hamish

I'm looking at some LANDSAT-TM5 data and I'd like to derive surface
temperature from channel 6 (thermal IR) as described in Neteler &
Mitasova (1st ed.) S9.4.3.

The results look all wrong though, I'm getting a range of -54 to +69 C

Using the following formulae:

radiance = 1.238+(15.6-1.238)*tm6/255.0;
temp_kelvin = 1260.56 / log(607.76/radiance + 1);
temp_celsius = temp_kelvin - 273.15

we get:

tm6(0) = -69.7787 deg C
tm6(255) = 68.6640 deg C

Using d.histogram on the "tm24r_b6" 8bit GeoTIFF input raster, I see a
bimodal distribution with peaks at 75 (land) and 220 (ocean). That
translates to -6deg for the land and +57deg for the ocean...

is my satellite data wrong or is it the conversion formulae?

It appears my data is missing all calibration coefficients; I was
assuming these were standardized. Apparently not.

... back to the vendor

,
Hamish

On Sun, Oct 03, 2004 at 04:44:27PM +1300, Hamish wrote:

> I'm looking at some LANDSAT-TM5 data and I'd like to derive surface
> temperature from channel 6 (thermal IR) as described in Neteler &
> Mitasova (1st ed.) S9.4.3.
>
> The results look all wrong though, I'm getting a range of -54 to +69 C
>
> Using the following formulae:
>
> radiance = 1.238+(15.6-1.238)*tm6/255.0;
> temp_kelvin = 1260.56 / log(607.76/radiance + 1);
> temp_celsius = temp_kelvin - 273.15
>
> we get:
>
> tm6(0) = -69.7787 deg C
> tm6(255) = 68.6640 deg C
>
> Using d.histogram on the "tm24r_b6" 8bit GeoTIFF input raster, I see a
> bimodal distribution with peaks at 75 (land) and 220 (ocean). That
> translates to -6deg for the land and +57deg for the ocean...
>
> is my satellite data wrong or is it the conversion formulae?

It appears my data is missing all calibration coefficients; I was
assuming these were standardized. Apparently not.

... back to the vendor

Good to hear that :slight_smile:

In the new edition of the book (apparently available now,
http://mpa.itc.it/grassbook2/ - also cheaper)
I have improved the wording of the section but the formula
for TM5 and TM7 (ETM) are almost unchanged. I can re-check
for differences if needed.

Markus