Python 使用PyEphem计算阴影长度
我使用的是PyEphem,希望计算阴影的长度(假设在地面上种植了一根单位长度的木棍)。长度将由cot(φ)给出,其中φ是太阳仰角(如果我错了,请纠正我)。我不确定?在下面的示例中,我使用的是角度alt:Python 使用PyEphem计算阴影长度,python,shadow,astronomy,pyephem,Python,Shadow,Astronomy,Pyephem,我使用的是PyEphem,希望计算阴影的长度(假设在地面上种植了一根单位长度的木棍)。长度将由cot(φ)给出,其中φ是太阳仰角(如果我错了,请纠正我)。我不确定?在下面的示例中,我使用的是角度alt: import ephem, math o = ephem.Observer() o.lat, o.long = '37.0625', '-95.677068' sun = ephem.Sun() sunrise = o.previous_rising(sun, start=ephem.now()
import ephem, math
o = ephem.Observer()
o.lat, o.long = '37.0625', '-95.677068'
sun = ephem.Sun()
sunrise = o.previous_rising(sun, start=ephem.now())
noon = o.next_transit(sun, start=sunrise)
shadow = 1 / math.tan(sun.alt)
请在下面检查我的解释:
sun.alt
是正确的alt
是地平线以上的高度;它们与北偏东的方位角一起定义了相对于地平线的方位角
你的计算几乎是正确的。您忘记提供观察者:sun=ephem.sun(o)
g(date)->aighty
计算太阳下次投射阴影的时间,该阴影的长度与现在相同(不考虑方位角——阴影的方向):
输出
上面的例子暗示地球上有一个观测者,是吗?如果观察者是环绕地球飞行的卫星呢?地球的阴影会被解释吗?@AjanO:是的,观测者在给定的纬度、经度和时间在地球上。
#!/usr/bin/env python
import math
import ephem
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as opt
def main():
# find a shadow length for a unit-length stick
o = ephem.Observer()
o.lat, o.long = '37.0625', '-95.677068'
now = o.date
sun = ephem.Sun(o) #NOTE: use observer; it provides coordinates and time
A = sun.alt
shadow_len = 1 / math.tan(A)
# find the next time when the sun will cast a shadow of the same length
t = ephem.Date(find_next_time(shadow_len, o, sun))
print "current time:", now, "next time:", t # UTC time
####print ephem.localtime(t) # print "next time" in a local timezone
def update(time, sun, observer):
"""Update Sun and observer using given `time`."""
observer.date = time
sun.compute(observer) # computes `sun.alt` implicitly.
# return nothing to remember that it modifies objects inplace
def find_next_time(shadow_len, observer, sun, dt=1e-3):
"""Solve `sun_altitude(time) = known_altitude` equation w.r.t. time."""
def f(t):
"""Convert the equation to `f(t) = 0` form for the Brent's method.
where f(t) = sun_altitude(t) - known_altitude
"""
A = math.atan(1./shadow_len) # len -> altitude
update(t, sun, observer)
return sun.alt - A
# find a, b such as f(a), f(b) have opposite signs
now = observer.date # time in days
x = np.arange(now, now + 1, dt) # consider 1 day
plt.plot(x, map(f, x))
plt.grid(True)
####plt.show()
# use a, b from the plot (uncomment previous line to see it)
a, b = now+0.2, now+0.8
return opt.brentq(f, a, b) # solve f(t) = 0 equation using Brent's method
if __name__=="__main__":
main()
current time: 2011/4/19 23:22:52 next time: 2011/4/20 13:20:01