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How to calculate the satellite zenith angle of each pixel of a satellite image based on the known data in the VNIR band

Posted: Wed Apr 23, 2025 7:48 am America/New_York
by jinyangwang11
How to calculate the satellite zenith angle of each pixel of a satellite image based on the known data in the VNIR band of the AST_L1A file?

Re: How to calculate the satellite zenith angle of each pixel of a satellite image based on the known data in the VNIR b

Posted: Wed Apr 23, 2025 3:36 pm America/New_York
by LP DAAC - jwilson
@jinyangwang11 I've forwarded your inquiry to our science team for further assistance.

Re: How to calculate the satellite zenith angle of each pixel of a satellite image based on the known data in the VNIR b

Posted: Thu Apr 24, 2025 10:46 am America/New_York
by LP DAAC - jwilson
@jinyangwang11 While we are waiting for further guidance on your inquiry, there is ASTER documentation at this link that might be useful at: https://doi.org/10.5067/ASTER/AST_L1A.003
Thank you,
Janice

Re: How to calculate the satellite zenith angle of each pixel of a satellite image based on the known data in the VNIR b

Posted: Fri Apr 25, 2025 11:27 am America/New_York
by jinyangwang11
@LP DAAC - jwilson
Thank you, I've downloaded the ASTER documents, but they don't seem to work. The documents don't involve the satellite zenith angle.

Re: How to calculate the satellite zenith angle of each pixel of a satellite image based on the known data in the VNIR b

Posted: Mon Apr 28, 2025 8:54 am America/New_York
by LP DAAC - dgolon
Hello @jinyangwang11 We are still waiting to hear back from the ASTER Science team but as soon as we do, we'll let you know.

Re: How to calculate the satellite zenith angle of each pixel of a satellite image based on the known data in the VNIR b

Posted: Wed Apr 30, 2025 11:29 am America/New_York
by LP DAAC - jwilson
@jinyangwang11 Our science team replied with the following calculation for solar zenith angle.

cos(θz) = sin(φ)sin(δ) + cos(φ)cos(δ)cos(h)

θz = SZA

φ = latitude

δ = solar declination

h = hour angle

Best regards,
Janice