Skyfield: HomeTable of ContentsChangelogAPI Reference


Many programmers work most efficiently when they can start with a working example and adapt it to their needs. Each of the following examples pulls together several Skyfield features to solve a general problem, that should provide readers with a basis for solving other similar problems of their own.

What time is solar noon, when the Sun transits the meridian?

The Earth travels fastest in its orbit when closest to the Sun in January, and slowest in July when they are farthest apart. This stretches out January days by around 10 seconds, shortens July days by the same amount, and means that your watch never reads exactly 12:00 when the Sun transits your meridian at “solar noon”.

You can compute solar noon by asking at what time the Sun transits the meridian at your location. To learn the time of solar noon today, for example, you might do this:

import datetime as dt
from pytz import timezone
from skyfield import almanac
from skyfield.api import N, W, wgs84, load

zone = timezone('US/Eastern')
now = zone.localize(
midnight = now.replace(hour=0, minute=0, second=0, microsecond=0)
next_midnight = midnight + dt.timedelta(days=1)

ts = load.timescale()
t0 = ts.from_datetime(midnight)
t1 = ts.from_datetime(next_midnight)
eph = load('de421.bsp')
bluffton = wgs84.latlon(40.8939 * N, 83.8917 * W)

f = almanac.meridian_transits(eph, eph['Sun'], bluffton)
times, events = almanac.find_discrete(t0, t1, f)

# Select transits instead of antitransits.
times = times[events == 1]

t = times[0]
tstr = str(t.astimezone(zone))[:19]
print('Solar noon:', tstr)
Solar noon: 2020-04-19 13:34:33

When will it get dark tonight?

Sunrise, sunset, and the several varieties of twilight are all available through the Almanac Computation module. Here’s the script I use when I want to know when it will be dark enough to see the stars — or how early I need to rise to see the morning sky:

import datetime as dt
from pytz import timezone
from skyfield import almanac
from skyfield.api import N, W, wgs84, load

# Figure out local midnight.
zone = timezone('US/Eastern')
now = zone.localize(
midnight = now.replace(hour=0, minute=0, second=0, microsecond=0)
next_midnight = midnight + dt.timedelta(days=1)

ts = load.timescale()
t0 = ts.from_datetime(midnight)
t1 = ts.from_datetime(next_midnight)
eph = load('de421.bsp')
bluffton = wgs84.latlon(40.8939 * N, 83.8917 * W)
f = almanac.dark_twilight_day(eph, bluffton)
times, events = almanac.find_discrete(t0, t1, f)

previous_e = f(t0).item()
for t, e in zip(times, events):
    tstr = str(t.astimezone(zone))[:16]
    if previous_e < e:
        print(tstr, ' ', almanac.TWILIGHTS[e], 'starts')
        print(tstr, ' ', almanac.TWILIGHTS[previous_e], 'ends')
    previous_e = e
2020-04-19 05:09   Astronomical twilight starts
2020-04-19 05:46   Nautical twilight starts
2020-04-19 06:20   Civil twilight starts
2020-04-19 06:49   Day starts
2020-04-19 20:20   Day ends
2020-04-19 20:48   Civil twilight ends
2020-04-19 21:23   Nautical twilight ends
2020-04-19 22:00   Astronomical twilight ends

As you can see from the above code, if the new light level is brighter then we say that the new level “starts”, but if the new level is darker then we say the previous level “ends” — so instead of saying “astronomical twilight starts at 21:23” we say “nautical twilight ends at 21:23.” That’s why the code keeps up with previous_e and compares it to the new level of twilight.

What phase is the Moon tonight?

The phase of the Moon is defined as the angle between the Moon and the Sun along the ecliptic. This angle is computed as the difference in the ecliptic longitude of the Moon and of the Sun. The result is an angle that is 0° for the New Moon, 90° at the First Quarter, 180° at the Full Moon, and 270° at the Last Quarter.

from skyfield.api import load
from skyfield.framelib import ecliptic_frame

ts = load.timescale()
t = ts.utc(2019, 12, 9, 15, 36)

eph = load('de421.bsp')
sun, moon, earth = eph['sun'], eph['moon'], eph['earth']

e =
_, slon, _ = e.observe(sun).apparent().frame_latlon(ecliptic_frame)
_, mlon, _ = e.observe(moon).apparent().frame_latlon(ecliptic_frame)
phase = (mlon.degrees - slon.degrees) % 360.0


What is the angular diameter of a planet, given its radius?

Be careful to select the correct radius when predicting a planet’s angular diameter in the sky. Many web sites will quote some kind of “mean radius” that averages between a planet’s squat polar radius and its wide equatorial radius. But most astronomers instead want to know the maximum, not average, diameter across a planet’s visible face — so you will want to use the planet’s equatorial radius in your calculation.

For example, a good current estimate of Neptune’s equatorial radius is 24,764 km. We would therefore predicts its angular diameter as:

import numpy as np
from skyfield.api import Angle, load

ts = load.timescale()
time = ts.utc(2020, 12, 30)

eph = load('de421.bsp')
earth, neptune = eph['earth'], eph['neptune barycenter']
radius_km = 24764.0

astrometric =
ra, dec, distance = astrometric.apparent().radec()
apparent_diameter = Angle(radians=np.arcsin(radius_km / * 2.0)
print('{:.6f} arcseconds'.format(apparent_diameter.arcseconds()))
2.257190 arcseconds

This agrees exactly with the output of the NASA HORIZONS system.

When is Venus at its greatest east and west elongations from the Sun?

This example illustrates the several practical steps that are often required to both find events of interest and then to learn more details about them.

This example can serve as a template for many other kinds of custom search:

from skyfield.api import load
from skyfield.framelib import ecliptic_frame
from skyfield.searchlib import find_maxima

ts = load.timescale()
t0 = ts.utc(2019)
t1 = ts.utc(2022)

eph = load('de421.bsp')
sun, earth, venus = eph['sun'], eph['earth'], eph['venus']

def elongation_at(t):
    e =
    s = e.observe(sun).apparent()
    v = e.observe(venus).apparent()
    return s.separation_from(v).degrees

elongation_at.step_days = 15.0

times, elongations = find_maxima(t0, t1, elongation_at)

for t, elongation_degrees in zip(times, elongations):
    e =
    _, slon, _ = e.observe(sun).apparent().frame_latlon(ecliptic_frame)
    _, vlon, _ = e.observe(venus).apparent().frame_latlon(ecliptic_frame)
    is_east = (vlon.degrees - slon.degrees) % 360.0 < 180.0
    direction = 'east' if is_east else 'west'
    print('{}  {:4.1f}° {} elongation'.format(
        t.utc_strftime(), elongation_degrees, direction))
2019-01-06 04:53:35 UTC  47.0° west elongation
2020-03-24 22:13:32 UTC  46.1° east elongation
2020-08-13 00:14:12 UTC  45.8° west elongation
2021-10-29 20:51:56 UTC  47.0° east elongation

At what angle in the sky is the crescent Moon?

The angle of the crescent Moon changes with the seasons. In the spring, a crescent Moon will stand high above the Sun and appear to be lit from below. In the autumn, the Moon sets farther from the Sun along the horizon and is illuminated more from the side. What if we wanted to know the exact angle?

You can find the answer by asking for the Sun’s “position angle” relative to the Moon, an angle you can compute between any two Skyfield positions. The angle will be 90° if the Sun is left of the moon, 180° if the Sun is directly below, and 270° if the Sun is to the right of the Moon.

from skyfield.api import N, W, load, wgs84
from skyfield.trigonometry import position_angle_of

ts = load.timescale()
t = ts.utc(2019, 9, 30, 23)

eph = load('de421.bsp')
sun, moon, earth = eph['sun'], eph['moon'], eph['earth']
boston = earth + wgs84.latlon(42.3583 * N, 71.0636 * W)

b =
m = b.observe(moon).apparent()
s = b.observe(sun).apparent()
print(position_angle_of(m.altaz(), s.altaz()))
238deg 55' 55.3"

The position_angle_of() routine will not only accept the output of altaz(), but also of frame_latlon() if you want a position angle relative to the ecliptic’s north pole.

Beware, though, that radec() produces coordinates in the opposite order from what position_angle_of() expects: right ascension is like longitude, not latitude. Try reversing the coordinates, like:

print(position_angle_of(m.radec(), s.radec()))
282deg 28' 15.7"

Drat, but this angle is backwards, because right ascension increases toward the east whereas the other angles, like azimuth, increase the other way around the circle.

When is a body or fixed coordinate above the horizon?

The following code will determine when the Galactic Center is above the horizon. The Galactic Center is an example of a fixed object, like a star or nebula or galaxy, whose right ascension and declination can be plugged in to a Star() object. The code will also work with a body from an ephemeris, like the Sun, Moon, or one of the planets.

from skyfield.api import N, Star, W, wgs84, load
from skyfield.almanac import find_discrete, risings_and_settings
from pytz import timezone

ts = load.timescale()
t0 = ts.utc(2019, 1, 19)
t1 = ts.utc(2019, 1, 21)

moab = wgs84.latlon(38.5725 * N, 109.54972238 * W)
eph = load('de421.bsp')
gc = Star(ra_hours=(17, 45, 40.04), dec_degrees=(-29, 0, 28.1))

f = risings_and_settings(eph, gc, moab)
tz = timezone('US/Mountain')

for t, updown in zip(*find_discrete(t0, t1, f)):
    print(t.astimezone(tz).strftime('%a %d %H:%M'), 'MST',
          'rises' if updown else 'sets')
Sat 19 05:51 MST rises
Sat 19 14:27 MST sets
Sun 20 05:47 MST rises
Sun 20 14:23 MST sets

What is the right ascension and declination of a point in the sky?

An observer is often interested in the astronomical coordinates of a particular position in the sky above them. If the observer can specify the position using altitude and azimuth coordinates, then Skyfield can return its right ascension and declination.

from skyfield import api

ts = api.load.timescale()
t = ts.utc(2019, 9, 13, 20)
geographic = api.wgs84.latlon(latitude_degrees=42, longitude_degrees=-87)
observer =
pos = observer.from_altaz(alt_degrees=90, az_degrees=0)

ra, dec, distance = pos.radec()
13h 41m 14.65s
+42deg 05' 50.0"

What latitude and longitude is beneath this right ascension and declination?

Most Skyfield calculations, like an observation of a planet or an Earth satellite, directly produce a vector position centered on the Earth. You can pass such a vector to the subpoint() method of a standard geoid to compute latitude and longitude.

But sometimes the right ascension and declination of the position are known already. Instead of creating a Star with those coordinates and asking it to compute its position, there is a simpler approach: creating the position directly.

from skyfield.api import load, wgs84
from skyfield.positionlib import position_of_radec

ts = load.timescale()
t = ts.utc(2020, 1, 3, 12, 45)

earth = 399  # NAIF code for the Earth center of mass
ra_hours = 3.79
dec_degrees = 24.1167
pleiades = position_of_radec(ra_hours, dec_degrees, t=t, center=earth)
subpoint = wgs84.subpoint(pleiades)

print('Latitude:', subpoint.latitude)
print('Longitude:', subpoint.longitude)
Latitude: 24deg 10' 33.5"
Longitude: 123deg 16' 53.9"

Which geographic location is farther from Earth’s center?

After I hiked Mount Bierstadt in Colorado, a friend suggested that its 14,000 feet of elevation might have carried me farther from the Earth’s center than I had ever traveled before. It was a romantic thought: that under my own power I had hiked farther from my home planet’s core than ever before.

But there was a problem. I knew that I had once visited a city only a few degrees away from the Earth’s equator, and that the Earth’s equatorial bulge might push even modest elevations at that latitude out farther from the Earth’s center than a mountaintop in Colorado.

So I wrote a quick Skyfield script to compare the distance from the Earth’s center to both Accra, Ghana, and the top of Mount Bierstadt in Colorado.

from skyfield.api import N, W, wgs84, load
from skyfield.functions import length_of

ts = load.timescale()
t = ts.utc(2019, 1, 1)

bierstadt = wgs84.latlon(39.5828 * N, 105.6686 * W, elevation_m=4287.012)
m1 = length_of(

accra = wgs84.latlon(5.6037 * N, 0.1870 * W, elevation_m=61)
m2 = length_of(

assert m2 > m1
print("I was", int(m2 - m1), "meters farther from the Earth's center\n"
      "when I visited Accra, at nearly sea level, than atop\n"
      "Mt. Bierstadt in Colorado.")
I was 4211 meters farther from the Earth's center
when I visited Accra, at nearly sea level, than atop
Mt. Bierstadt in Colorado.