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Remote Sensing
- 1. REMOTE SENSING Thursday, 29 November 2017 J.Mwaura RS APPLICATIONS & SEMINAR
- 2. Linear Contrast Stretching 0-255 in an 8-bit system β’ π·ππ π‘ = 255 Γ (π·π ππβπ·π πππ) (π·π πππ₯βπ·π πππ) DN values range from 60-158 β’ 0-59 and 159-255 are not utilized
- 3. Histogram-equalized Stretching β’ Lets stretch values in the range [75-150] β π·ππ π‘ = 255 π=0 π ( π π π ) β’ π·ππ π‘ = ππβπππππ π·π π£πππ’π β’ ππ = ππ’ππππ ππ πππ₯πππ βππ£πππ π·π π£πππ’π ππ π‘βπ π π‘β πππππ, ππ π‘βπ ππππ’π‘ πππππ β’ π = πππ₯. ππ’ππππ ππ π·π πππππ ππ ππππ’π‘ πππππ β’ π = π»ππππ ππππππ ππ πππ₯πππ ππ ππππ’π‘ πππππ
- 4. Calculate Slope for Band and Radiance Metadata: β’ LMAX_BAND6 = 15.303 β’ LMIN_BAND6 = 1.238 β’ πππππ = (ππππ₯βππππ) πππ₯ β’ πππππ π΅6 = 15.303β1.238 255 β’ = 0.0551 β’ π πππππππ = π ππππ Γ π·π + ππππ β’ π π΅6 = 0.0551 Γ π·π π΅6 β 1.238 Β»e.g. DN = 69 β’ 0.0551 Γ 69 β 1.238 = 2.5639
- 5. Calculate Temperature (at-sensor/TOA) β’ Using Planck's Black Body Radiation Law βπ = π2 ln π1 π +1 , οΆπ = ππππππππ‘πππ ππππ π‘πππ‘, οΆπ = ππππππππ πππ ππππ = 2.5639 Β»π1 = 666.09 Β»π2 = 1282.71 β’ ππ΅6 = 1282.71 ln 666.09 2.5639 +1 = 195.54
- 6. β’ Calculating Surface Temperature ππ = ππ΅ 1 + Ξ» β ππ΅ π β log π β ππ΅ = π‘πππππππ‘π’ππ, π = ππππ π π’πππππ ππππ π ππ£ππ‘π¦ β Ξ» = π€ππ£ππππππ‘β, π = β ππππππβ² π ππππ π‘πππ‘ Γ π π ππππ‘π§ππππ ππππ π‘πππ‘ β’ LSE (Ξ΅) can be extracted by using NDVI considering three different cases β’ Bare soil β’ Vegetation β’ Mixture of bare soil and vegetation
- 7. Land Surface Emissity (LSE) β’ For Band 6 of Landsat TM β’ π = 0.004 Γ ππ£ + 0.986 β’ ππ£ is the proportion of vegetation which is given by β’ ππ£ = { ππ·ππΌβππ·ππΌ πππ (ππ·ππΌ πππ₯βππ·ππΌ πππ) }2
- 8. Land Surface Temperature (LST) β’ Parameters Β»ππ΅6 = 195.54 Β»π = 14380 Β»π = 1.2345 Β»π = 11.5 Β΅m β ππ = 195.54 1+ 11.5Γ 195.54 14380 Γlog 1.2345 = 192.780 π
- 10. Irradiance at-TOA?(1) β6.70 πΈ π = πΈ π Γ cos β π. πΆπππππ πΈπππ΄πππ π. πΆπππππππ π = 0.996 πππ‘πππππ‘π πΈ π = πΈπ‘ π2 β23.30 ?
- 11. Irradiance at-TOA?(2) Sky irradiance (πΌπ = 12π’πππ‘π ) Path radiance (π π = 2π’πππ‘π ) Pixel irradiance πΌ π = 6π’πππ‘π Sensor πΊπππ = 0.89, ππππ ππ’πππππ‘ = β19 reflectance (π = 0.62) Incident irradiance (πΈπ) π. π. π Atmosphere (t = 0.79) Sky radiance (π π = 14π’πππ‘π ) @π¬ππ§π¬π¨π« @π©π’π±ππ₯
- 12. Irradiance at-TOA?(3) 1. πΈπ‘ = 258+19 0.89 = 311.24 π’πππ‘π 2. πΈπ = 311.24 Γ cos(6.7) = 309.11 π’πππ‘π ο§ at Sensor Radiance 3. At pixel: β’ πΌπ = 0.79 Γ 12 = 9.48π’πππ‘π β’ πΌ π = 0.79 Γ 6 = 4.74 π’πππ‘π
- 13. Irradiance at-TOA?(4) 4. Reflected at pixel: β’ πΌπ = 0.62 Γ 9.84 = 5.88π’πππ‘π β’ πΌπ = 0.62 Γ 4.74 = 2.94π’πππ‘π 5. At TOA (@sensor): β’ πΌπ = 0.79 Γ 5.88 = 4.65π’πππ‘π β’ πΌ π = 0.79 Γ 2.94 = 2.32π’πππ‘π β’ π π = 0.79 Γ 2 = 1.58π’πππ‘π β’ π π = 0.79 Γ 14 = 11.06π’πππ‘π 6. Total = 19.61π’πππ‘π
- 14. Irradiance at-TOA?(5) 7. Reflected at pixel: β’ 0.79 πΈπ + 19.61 = 309.11π’πππ‘π β’ πΈπ = 309.11β19.61 0.79 = 366.46π’πππ‘π 8. At TOA (@sun): β’ πΈπ = 366.46 0.62 = 591.06π’πππ‘π β’ πΈπ@π‘ππ = 591.06 0.79 = 748.17π’πππ‘π β’ πΈπ@π‘ππ = 748.17 0.9962 = 754.20π’πππ‘π