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Calculate F_in based on 365 days #4

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Calculate F_in based on 365 days #4

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839f9b9
improve performance of OLR calculation; only predecimate when downsam…
Aug 15, 2024
eccd414
Merge branch 'dlr-pa:main' into main
jsmretschnig Sep 12, 2024
1d95205
Merge branch 'dlr-pa:main' into main
jsmretschnig Sep 8, 2025
349781c
Revert "improve performance of OLR calculation; only predecimate when…
Sep 8, 2025
fff4b1f
update gitignore
Sep 8, 2025
e25a4c0
fix day of the year
Sep 9, 2025
73e5ea1
to follow the definition that day of the year should be 1 for Jan 1
Sep 9, 2025
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1 change: 1 addition & 0 deletions .gitignore
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Expand Up @@ -48,3 +48,4 @@ Thumbs.db
*.mov
*.wmv

.eggs
20 changes: 11 additions & 9 deletions climaccf/accf.py
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Expand Up @@ -571,20 +571,22 @@ def get_Fin(self):
:rtype: numpy.ndarray
"""
Fin = np.zeros (self.ds.t.values.shape)
for t in range(self.nt):
date = str(self.ds['time'].values[t])
month = date[5:7]
day = date[8:10]
N = (int(month) - 1) * 30 + int(day) # Day of year
delta = -23.44 * np.cos(np.deg2rad(360 / 365 * (N + 10)))
S = 1360 # W/m2, Solar constant
theta = np.sin(np.deg2rad(self.lat)) * np.sin(np.deg2rad(delta)) + np.cos(np.deg2rad(self.lat)) * np.cos(np.deg2rad(delta))
S = 1360 # W/m2, Solar constant

days_of_year = self.ds.time.dt.dayofyear.values # 1 to 365 for regular years, and 1 to 366 for leap years
deltas = -23.44 * np.cos(np.deg2rad(360 / 365 * (days_of_year + 10)))
thetas = [
np.sin(np.deg2rad(self.lat)) * np.sin(np.deg2rad(delta)) + np.cos(np.deg2rad(self.lat)) * np.cos(np.deg2rad(delta))
for delta in deltas
]

for t, theta in zip(range(self.nt), thetas):
for l in range(self.nl):
if self.member_bool:
for m in range(self.nm):
Fin[t,m,l,:,:] = S * theta
else:
Fin[t,l,:,:] = S * theta
Fin[t,l,:,:] = S * theta
return Fin

def get_encoding_dict(list_name, encoding):
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