2

I am getting very low reflectance values (mostly < 0.1) in Landsat 8 bands 2-7 on different scenes after rescaling from Digital Numbers (DN) to Reflectance. All scenes are in a tropical landscape with a lot of vegetation, secondary and old grown forests, agriculture (rice paddy) and urban areas & infrastructure. I see no reason that the majority of reflectance values should be that low.

I am using the formula for TOA planetary reflectance with correction for solar angle (provided on http://landsat.usgs.gov/Landsat8_Using_Product.php).

For example:

The mean DN value in Band 4 (red band) of my scene is 6937. Putting that into the formula for planetary reflectance

ρλ = ρλ'/cos(solar_zenith_angle)

where

ρλ' = Mρ * Qcal + Aρ

and

Mp = 2.0000E-05 (REFLECTANCE_MULT_BAND_4)
Qcal = 6937 (DN)
Ap = -0.100000 (REFLECTANCE_ADD_BAND_4)

gives me

(0.00002×6937-0.1)/cos(90-41.71087310) = 0.058

Is it possible, that the Ap value (-0.1) has to be "positive" (0.1) in the formula? If it was positive, my reflectance values would become more reasonable:

(0.00002×6937+0.1)/cos(90-41.71087310) = 0.36

For a comparison I also tried to use i.landsat.toar in GRASS 7 but I also get very low reflectance values (TOA, uncorrected) there. Here the mean reflectance values for each band after i.landsat.toar:

B2: 0.0898574
B3: 0.0747076
B4: 0.0582399
B5: 0.252395
B6: 0.160692
B7: 0.0808918

I am having this problem regularly also with other scenes from different locations.

1

The standard formula is correct. You should not turn +Ap into -Ap.

A value of 6500 in the red band is often seen over water, where a reflectance of 0.02 makes pretty good sense - over clear water, that may actually be too high. Furthermore, seeing DN values as high as 40000 are not that uncommon, and with your suggested formula, that would give a reflectance of 1.04, which is clearly wrong.

Furthermore, when looking at average reflectance values, make sure that you aren't considering the no-data areas, where the DN value is 0 as part of your statistics.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.