使用PyEphem计算阴影长度

Question

我正在使用PyEphem并且想要计算阴影的长度(假设一个单位长度的棍子被种植在地上).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).我不确定在太阳上使用哪个字段？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)

请检查我的解释如下:

1. 如果切线是无限的,则表示太阳直接在头顶,没有阴影.
2. 如果切线为零,则表示太阳位于地平线上且阴影无限长.
3. 我不知道如何解释cot(phi)的负面结果.有人能帮我吗？

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到下一次太阳将投射该长度的阴影.

我很感激这方面的帮助.

Answer 1

在太阳上使用什么领域？

这sun.alt是对的.alt是地平线以上的高度; 与北方以东的方位角一起,它们定义了相对于地平线的明显位置.

你的计算几乎是正确的.你忘了提供一个观察者:sun = ephem.Sun(o).

3. 我不知道如何解释cot(phi)的负面结果.有人能帮我吗？

在这种情况下,太阳低于地平线.

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到下一次太阳将投射该长度的阴影.

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()

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current time: 2011/4/19 23:22:52 next time: 2011/4/20 13:20:01