Local Time and Latitude

[roe-dl]

It is said that local time depends on longitude only. But if I calculate the hour angle of the sun for the same timestamp and longitude at different altitudes, I get different results. The following table shows the difference of the value at the given latitude to the value at latitude 0.0°.

latitude	hour angle difference
0.0	0.00000000
5.0	0.00000050
10.0	0.00000199
15.0	0.00000445
20.0	0.00000788
25.0	0.00001225
30.0	0.00001752
35.0	0.00002366
40.0	0.00003063
45.0	0.00003836
50.0	0.00004681
55.0	0.00005590
60.0	0.00006558
65.0	0.00007577
70.0	0.00008639
75.0	0.00009735
80.0	0.00010858
85.0	0.00011999
90.0	0.00013149

You see, the hour angle increases with latitude.

The script is:

#!/usr/bin/python3
from skyfield.api import N, S, E, W, Loader, wgs84
import numpy

load = Loader('.',verbose=False)
ts = load.timescale(builtin=False)
eph = load('de440.bsp')
sun = eph['sun']

lat = numpy.arange(0,91,5)
lon = 0
t = ts.utc(2026,2,11,9)

observer = eph['Earth'] + wgs84.latlon(lat[0],lon)
ha0, _, _ = observer.at(t).observe(sun).apparent().hadec()

print('latitude |  hour angle difference')
print('---------|------------')
for l in lat:
    observer = eph['Earth'] + wgs84.latlon(l,lon)
    ha, _, _ = observer.at(t).observe(sun).apparent().hadec()
    diff = ha.hours-ha0.hours
    print('%8.1f | %11.8f' % (l,diff))

Do you know why?

[brandon-rhodes]

It is said that local time depends on longitude only.

Alas, you must have gotten bad information. Can you track down the reference that made that claim? If Skyfield's documentation says that somewhere, I'll be sure to get it corrected!

Local time also depends on latitude, because as you travel up the curve of the earth, each location you visit is looking at the Sun from a slightly different position in space, and so sees the Sun against slightly different stars. We can estimate how big the effect would be on the Sun's position of moving from the equator to the poles by adding this to the bottom of your script:

# ... your script goes here ...

from skyfield.api import tau

distance_to_sun_m = observer.at(t).observe(sun).distance().m
print('Distance to Sun:', distance_to_sun_m, 'm')

o0 = wgs84.latlon(0, 0)
o90 = wgs84.latlon(90, 0)
distance_between_equator_and_pole_m = (o90 - o0).at(t).distance().m
print('Distance between observers:', distance_between_equator_and_pole_m, 'm')

print(
    'Approximate angle in hours of arc:',
    distance_between_equator_and_pole_m / distance_to_sun_m
    / tau * 12.0,
)

The output:

Distance to Sun: 147654394985.58572 m
Distance between observers: 9004939.28771482 m
Approximate angle in hours of arc: 0.0001164758231562671

So the Sun is shifted against the background of stars by around the same amount that you are seeing as a difference in HA. There are other tiny effects at work, like the fact that locations at the pole travel at a different velocity that locations at the equator, thus changing the aberration of light slightly. But the fact that the Sun is at a different location in the sky is the biggest factor.

You can also confirm the effect by moving your positions underground, down most of the way to the Earth's center, where the positions will be very close together, by providing a negative number for elevation_m that's nearly equal to the Earth's radius:

# edits to your script:

...
observer = eph['Earth'] + wgs84.latlon(lat[0],lon, elevation_m=-6356e3)
...
    observer = eph['Earth'] + wgs84.latlon(l,lon, elevation_m=-6356e3)
...

That will drive your "hour angle difference" most of the way to zero.

Let me know if there's anywhere Skyfield's documentation needs to be fixed to agree with this.

[roe-dl]

Let me know if there's anywhere Skyfield's documentation needs to be fixed to agree with this.

Don't be afraid. There is nothing wrong with the Skyfield documentation. It's from timekeeping, and I know, they use the mean sun, and Skyfield calculates the apparent position. I underestimated the difference, when I thought, if I look at one single moment of time only, I need not observe that.