rayleigh radiance computing problem

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lwk1542
Posts: 26
Joined: Tue Apr 03, 2018 9:52 am America/New_York

rayleigh radiance computing problem

by lwk1542 » Sat Apr 03, 2021 9:50 am America/New_York

Dr. :
I computing rayleigh radiance using a code that is similar to "rayrad.pro" (https://oceancolor.gsfc.nasa.gov/cgi/id ... r=atmocorr). However, the result is different with seadas product "Lr". the scatter in the following attachment file shows difference at eight bands. I find,after analysis in ENVI,the difference increase from center to edge of the scan line. the figure and their description and my code are as follows. Does anyone know the reason?

# -*- coding: utf-8 -*-
import numpy as np
from netCDF4 import Dataset
from scipy import interpolate
import general
def rayleigh(rayleigh_lut_path=None, sza=None, vza=None, saa=None, vaa=None, windspeed=None, pressure=None, F0=None):
"""
Args:
rayleigh_lut_path ():
sza (): solar zenith
vza (): sensor zenith
vaa (): sensor azimuth
saa (): solar azimuth
windspeed (): wind speed
pressure (): atmospheric pressure
Returns:
Lr_i
"""
reaa=vaa-180-saa
reaa[reaa<-180]=reaa[reaa<-180]+360
reaa[reaa > 180] = reaa[reaa > 180] - 360
rayleigh_lut = general.get_filelist(rayleigh_lut_path, 'rayleigh', 'iqu.hdf')
windspeed[windspeed > 30] = 30
sza[sza > 88] = 88
mu0 = np.cos(sza / 180 * np.pi)
mu = np.cos(vza / 180 * np.pi)
airmass = 1 / mu0 + 1 / mu
Taur = np.zeros(shape=(rayleigh_lut.__len__()))
Lr_i = np.zeros(shape=(sza.shape[0], sza.shape[1], rayleigh_lut.__len__()))

for i in range(rayleigh_lut.__len__()):
rayDtset = Dataset(rayleigh_lut[i])
taur = rayDtset.variables['taur'][:]
depol = rayDtset.variables['depol'][:]
senz = rayDtset.variables['senz'][:]
solz = rayDtset.variables['solz'][:]
wind = rayDtset.variables['wind'][:]
# sigma = rayDtset.variables['sigma'][:]
i_ray = rayDtset.variables['i_ray'][:]
# q_ray = rayDtset.variables['q_ray'][:]
# u_ray = rayDtset.variables['u_ray'][:]
Taur[i] = taur

Norder0 = np.zeros_like(sza)
Norder1 = np.ones_like(vza)
Norder2 = Norder0 + 2

ray_i0 = interpolate.interpn(
(wind.reshape(-1), solz.reshape(-1), np.arange(3).reshape(-1), senz.reshape(-1)), i_ray,
np.stack([windspeed, sza, Norder0, vza], axis=2))
ray_i1 = interpolate.interpn(
(wind.reshape(-1), solz.reshape(-1), np.arange(3).reshape(-1), senz.reshape(-1)), i_ray,
np.stack([windspeed, sza, Norder1, vza], axis=2))
ray_i2 = interpolate.interpn(
(wind.reshape(-1), solz.reshape(-1), np.arange(3).reshape(-1), senz.reshape(-1)), i_ray,
np.stack([windspeed, sza, Norder2, vza], axis=2))

ray_i = ray_i0 + ray_i1 * np.cos(reaa / 180 * np.pi) + ray_i2 * np.cos(2 * reaa / 180 * np.pi)

# from Wang menghua ;is from Seadas
p0 = 1013.25 # hpa
x = (-(0.6543 - 1.608 * taur) + (0.8192 - 1.2541 * taur) * np.log(airmass)) * taur * airmass
fac = ((1.0 - np.exp(-x * pressure / p0)) / (1.0 - np.exp(-x)))
Lr_i[:, :, i] = ray_i * fac * F0[i]

return Lr_i
Attachments
the scatter between my Lr and seadas Lr
the scatter between my Lr and seadas Lr
Lr_-Lr_ac2_300dpi.png (169.04 KiB) Not viewed yet
the difference of spectrum at left of scan line between my Lr and seadas Lr
the difference of spectrum at left of scan line between my Lr and seadas Lr
left.png (58.28 KiB) Not viewed yet
the difference of spectrum at center of scan line between my Lr and seadas Lr
the difference of spectrum at center of scan line between my Lr and seadas Lr
center.png (60.38 KiB) Not viewed yet

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amiribr
Posts: 4
Joined: Thu Nov 15, 2018 7:26 am America/New_York

Re: rayleigh radiance computing problem

by amiribr » Mon Apr 05, 2021 4:45 pm America/New_York

Hi, the Rayleigh table coefficients i_ray have the dimensions of wave mean square slope *sigma* and not *windspeed*. Before the interpolation convert windspeed to sigma using sigma = sqrt(0.00534*WS).

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