我正在使用PyEphem并且想要计算阴影的长度(假设在地面上种植了一根单位长度)。长度将由cot(phi)给出,其中phi是太阳高度角(如果我错了,请纠正我)。我不确定what field to use on the Sun?在下面的示例中,我使用角度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())
noon = o.next_transit(sun, start=sunrise)
shadow = 1 / math.tan(sun.alt)
请在下面查看我的解释:
最后,考虑到ephem.Observer(),我很困惑如何使用PyEphem从阴影长度向下工作,下一次太阳将投射该长度的阴影。
我很感激你的帮助。
答案 0 :(得分:9)
在太阳上使用哪个字段?
sun.alt
是正确的。 alt
是地平线以上的高度;与北方向的方位角一起,它们定义了相对于地平线的apparent position。
您的计算几乎是正确的。您忘记提供观察员:sun = ephem.Sun(o)
。
- 我不知道如何解释cot(phi)的负面结果。能够 有人帮我吗?
醇>
在这种情况下,太阳低于地平线。
最后,我对如何使用感到困惑 PyEphem从一个向后工作 阴影长度到下一次的时候 太阳会投下阴影 长度,给定ephem.Observer()。
这是一个给出函数的脚本:g(date) -> altitude
计算下一次太阳投射与现在相同长度的阴影时(方位角 - 不考虑阴影的方向):
#!/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