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我正在探索 Brandon Rhodes 的宏伟软件 Skyfield 的可能性。我已经制作了一个脚本来计算随机对象之间的右升序中的连接。我使用以下脚本：

from skyfield import almanac
from skyfield.searchlib import find_maxima, find_minima, find_discrete
from skyfield.api import Star, load
from datetime import datetime, date,timedelta
import pytz

planets = load('de430t.bsp')
earth = planets['earth']

x = [
['Aldebaran',[4, 35, 55.2],[16, 30, 33]],
['Regulus',[10, 8, 22.3],[11, 58, 2]],
['Pollux',[7, 45, 18.9],[28, 1, 34]],
['Antares',[16, 29, 24.4],[-26, 25, 55]],
]

ts = load.timescale(builtin=True)
t = ts.now()
tzn = 'Europe/Amsterdam'

tz = pytz.timezone(tzn)
now = datetime(2020, 1, 1, 0, 0, 0)
t0 = ts.utc(tz.localize(now))
t1 = ts.utc(tz.localize(now) + timedelta(days=+365))

def difference(t):
    e = earth.at(t)
    ra11, dec11, distance = e.observe(object).radec()
    ra12, dec12, distance2 = e.observe(planets[target1]).radec()
    diff = ra11.hours - ra12.hours
    return diff >= 0

difference.rough_period = 1.0

for count in range (len(x)):
    object = Star(ra_hours=(x[count][1][0], x[count][1][1], x[count][1][2]),dec_degrees=(x[count][2][0], x[count][2][1], x[count][2][2]))
    target1 = 'venus'

    t, b = find_discrete(t0, t1, difference)
    if len(t) > 0:
        print (f"{x[count][0]} and {target1}")

        for i, (a, b) in enumerate(zip(t, b)):
            e = earth.at(a)
            ra11, dec11, distance = e.observe(object).radec('date')
            ra12, dec12, distance2 = e.observe(planets[target1]).radec('date')
            print (f"Diff: {ra11.hours - ra12.hours}, ra_{x[count][0]}: {ra11.hours}, ra_{target1}: {ra12.hours}")
            print(f"{a.utc_iso()},{dec11._degrees - dec12._degrees}")
            print ("")

我相信这是在计算两个对象具有相同 RA 的时间实例。

不幸的是，我得到了这些结果：

Aldebaran and venus
Diff: 4.600705365706519, ra_Aldebaran: 4.6176105612629375, ra_venus: 0.016905195556418524
2020-02-07T21:20:06Z,16.962942031863825

Diff: -0.0014205156698450239, ra_Aldebaran: 4.617748605529588, ra_venus: 4.619169121199433
2020-04-17T20:25:49Z,-10.136618155596008

Diff: -0.000670093218860579, ra_Aldebaran: 4.617892691238783, ra_venus: 4.618562784457644
2020-06-08T07:56:08Z,-4.921187478324768

Diff: -0.0001286749609095139, ra_Aldebaran: 4.618000948409605, ra_venus: 4.618129623370515
2020-07-12T06:44:16Z,-0.962286810883981

Regulus and venus
Diff: 10.140247344361857, ra_Regulus: 10.157152539918275, ra_venus: 0.016905195556418524
2020-02-07T21:20:06Z,12.28436748615726

Diff: 5.852858068422506e-06, ra_Regulus: 10.157702949500333, ra_venus: 10.157697096642265
2020-10-02T23:42:45Z,0.0903429562716429

Pollux and venus
Diff: 7.758742204719277, ra_Pollux: 7.7756474002756955, ra_venus: 0.016905195556418524
2020-02-07T21:20:06Z,28.391501492522927

Diff: 0.001226225278400328, ra_Pollux: 7.776229220287632, ra_venus: 7.775002995009232
2020-09-01T16:39:22Z,8.682000412217121

Antares and venus
Diff: 16.493491164600684, ra_Antares: 16.510396360157102, ra_venus: 0.016905195556418524
2020-02-07T21:20:06Z,-26.059118330110437

Diff: 0.000832014094154232, ra_Antares: 16.51126040187071, ra_venus: 16.510428387776557
2020-12-23T00:34:39Z,-5.652225571050259

以“Diff”开头的行是监控输出有效性的行。Diff 代表计算得出的 RA 差异。它应该接近于零。其他两个值是两个对象的赤经。它们应该非常相似。第二行是我想要的结果，它是计算的时间和对象之间的距离（以度为单位）。正如您所看到的，出于某种原因，对于每组对象，我得到了时间实例的无效结果：2020-02-07T21:20:06Z，并且该实例的差异值肯定不会接近于零。如果我将对象金星更改为月球，情况会变得更糟，因为每秒钟的结果都是无效的。我对照 Skychart / Cartes du Ciel 软件和那些结帐检查了其他结果。

我无法弄清楚这里出了什么问题。有人能帮助我吗？

4

1 回答 1

1

好问题！我应该在https://rhodesmill.org/skyfield/searches.html中添加一个新部分，解释减去两个经度或赤经时出现的这种常见行为。解开这个谜团的关键是观察在你的输出中作为幻象连接出现的时刻的角度差会发生什么变化。我附上了一个脚本，它为你打印的第一个事件打印这个，在金星和毕宿五之间：

2020-02-07 Difference: -19.33880215224481 Venus RA: 23.93746881891146
2020-02-08 Difference: 4.5908654129343995 Venus RA: 0.007801253732248779

角度差在 -19.3 和 4.6 之间跳跃，这应该立即让我们感到怀疑，因为它们只是完全相同角度的两个不同名称！您可以通过将 24.0 添加到 -19.3 来确认这一点，您将得到一个非常接近 4.6 的角度（给予或接受金星在一天内完成的实际运动）。

为什么结果会在天空中完全相同的角度差异的两个别名之间跳跃？

答案就在上面印的第二个事实：金星的 RA。不连续性恰好发生在金星恰好穿过 0h 赤经的那一刻！尽管 23.93746881891146 和 0.007801253732248779 几乎是相同的角度，但它们相差 24.0，因为它们跨越了天空中我们重新命名直升的方式。

我下面的脚本还显示了一个说明情况的情节：

赤经图

您可以在上图中看到，正是在金星将其 RA 重置为零的确切时刻，RA 差异在相同的四个半小时内从一个别名跃迁到另一个别名 24.0 小时RA的差异。

解决方案？

角度差异需要限制在 [-12h, +12h) 之类的范围内，以强制为每个可能的角度值选择一个首选别名。在 Python 中，正如您在下面的脚本中所见，典型的操作是：

(ra1.hours - ra2.hours + 12.0) % 24.0 - 12.0

这在上面的第二个图中显示为“改进的差异”，它不仅正确隐藏了 2 月 7 日的不连续性，不再让它看起来像一个事件，而且它现在正确地识别出金星和毕宿五之间的对冲接近尾声2020 年的 2020 年（在图的右边缘）以前仅显示为超过 -12.0 的差异，但现在作为角度差异的关键时刻闪耀。

最后，此脚本会检查反对意见并将其从搜索结果中过滤掉。您还会发现一些可能的 Python 代码调整，当您继续使用 Python 编码时可能会考虑这些调整。开始：

from skyfield.searchlib import find_maxima, find_minima, find_discrete
from skyfield.api import Star, load
from datetime import datetime, date,timedelta
import pytz

planets = load('de430t.bsp')
earth = planets['earth']

stars = [
    ['Aldebaran', (4, 35, 55.2), (16, 30, 33)],
    ['Regulus', (10, 8, 22.3), (11, 58, 2)],
    ['Pollux', (7, 45, 18.9), (28, 1, 34)],
    ['Antares', (16, 29, 24.4), (-26, 25, 55)],
]

ts = load.timescale(builtin=True)
t0 = ts.utc(2020, 1, 1)
t1 = ts.utc(2021, 1, 1)

# Exploring the first bad result for Venus and Aldebaran.

star = Star(ra_hours=stars[0][1], dec_degrees=stars[0][2])

for utc in (2020, 2, 7), (2020, 2, 8):
    t = ts.utc(*utc)
    ra1, _, _ = planets['earth'].at(t).observe(star).radec()
    ra2, _, _ = planets['earth'].at(t).observe(planets['venus']).radec()
    print(t.utc_strftime('%Y-%m-%d'),
          'Difference:', ra1.hours - ra2.hours,
          'Venus RA:', ra2.hours)

# Plot showing how to protect an angular difference against discontinuity.

import matplotlib.pyplot as plt

t = ts.utc(2020, 1, range(365))
e = planets['earth'].at(t)
star = Star(ra_hours=stars[0][1], dec_degrees=stars[0][2])
ra1, _, _, = e.observe(star).radec()
ra2, _, _, = e.observe(planets['venus']).radec()

fig, (ax, ax2) = plt.subplots(2, 1)
ax.plot(t.J, ra2.hours, label='Venus RA')
ax.plot(t.J, ra1.hours - ra2.hours, label='RA difference')
ax.set(xlabel='Year', ylabel='Hours')
ax.grid()
ax.legend()
ax2.plot(t.J, ra2.hours, label='Venus RA')
ax2.plot(t.J, (ra1.hours - ra2.hours + 12.0) % 24.0 - 12.0,
         label='Improved difference')
ax2.set(xlabel='Year', ylabel='Hours')
ax2.grid()
ax2.legend()
fig.savefig('tmp.png')

# So we need to force the difference into the range [-12 hours .. +12 hours]

def difference(t):
    e = earth.at(t)
    ra11, dec11, distance = e.observe(object).radec()
    ra12, dec12, distance2 = e.observe(planets[target1]).radec()
    diff = (ra11.hours - ra12.hours + 12.0) % 24.0 - 12.0
    return diff >= 0

difference.rough_period = 1.0

for name, ra_hms, dec_dms in stars:
    object = Star(ra_hours=ra_hms, dec_degrees=dec_dms)
    target1 = 'venus'

    t, b = find_discrete(t0, t1, difference)
    if len(t) == 0:
        break

    print (f"{name} and {target1}")

    for a, b in zip(t, b):
        e = earth.at(a)
        ra1, dec1, _ = e.observe(object).radec('date')
        ra2, dec2, _ = e.observe(planets[target1]).radec('date')
        if abs(ra1.hours - ra2.hours) > 6.0:
            continue  # ignore oppositions
        print(f"Diff: {ra1.hours - ra2.hours:.4f}, ra_{name}: {ra1.hours}, ra_{target1}: {ra2.hours}")
        print(f"{a.utc_iso()},{dec1._degrees - dec2._degrees}")
        print()

现在打印的事件是：

Aldebaran and venus
Diff: -0.0014, ra_Aldebaran: 4.617748605529591, ra_venus: 4.619169121320681
2020-04-17T20:25:49Z,-10.136618155920797

Diff: -0.0007, ra_Aldebaran: 4.617892691238804, ra_venus: 4.618562784294269
2020-06-08T07:56:08Z,-4.921187477025711

Diff: -0.0001, ra_Aldebaran: 4.618000948409604, ra_venus: 4.618129623302464
2020-07-12T06:44:16Z,-0.9622868107603999

Regulus and venus
Diff: 0.0000, ra_Regulus: 10.157702949500333, ra_venus: 10.157697096607553
2020-10-02T23:42:45Z,0.09034295610918264

Pollux and venus
Diff: 0.0012, ra_Pollux: 7.776229220287636, ra_venus: 7.775002995218732
2020-09-01T16:39:22Z,8.68200041253418

Antares and venus
Diff: 0.0008, ra_Antares: 16.511260401870718, ra_venus: 16.51042838802181
2020-12-23T00:34:39Z,-5.652225570418828

我相信它可以解决并纠正您的问题！

于 2020-07-10T11:53:10.563 回答