使用PyEphem来计算阴影长度
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我正在使用PyEphem，并想计算阴影的长度（假设在地面上植有一根长度为单位的棍子）。长度将由cot（phi）给出，其中phi是太阳仰角（如果我错了，请纠正我）。我不确定要在太阳上使用哪个字段？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())
noon = o.next_transit(sun, start=sunrise)
shadow = 1 / math.tan(sun.alt)
没有找到相关结果
已邀请:
1 个回复
臂哦
是正确的。是地平线以上的高度；连同北方以北的方位角，它们确定了相对于地平线的视在位置。 您的计算几乎是正确的。您忘记提供观察员了：。 我不知道如何解释cot（phi）的负面结果。能够 谁来帮帮我？ 在这种情况下，太阳在地平线以下。 最后，我对如何使用感到困惑 PyEphem从 到下一次的阴影长度 太阳将蒙上阴影 给定一个ephem.Observer（）的长度。 这是一个提供功能的脚本：计算下一次太阳将投射与当前长度相同的阴影的时间（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()