<p>We can use the <a href="https://en.wikipedia.org/wiki/Sunrise_equation" rel="nofollow noreferrer">sunrise equation</a> to determine the Sun's hour angle at sunrise &amp; sunset and thence determine the daylight length. The sunrise equation is derived from the solution to the PZX navigation triangle:</p>
<p><span class="math-container">$\sin(alt)=\sin(\phi)sin(\delta)+\cos(\phi)\cos(\delta)\cos(h)$</span></p>
<p>where <span class="math-container">$alt$</span> is the Sun's altitude above the horizon, <span class="math-container">$\delta$</span> is the Sun's declination, <span class="math-container">$\phi$</span> is the latitude, and <span class="math-container">$h$</span> is the Sun's hour angle, where <span class="math-container">$h=0$</span> at solar noon.</p>
<p>Since we're ignoring refraction (and the Sun's angular radius), we set <span class="math-container">$alt=0$</span>. That is, we define sunrise &amp; sunset to be when the Sun's centre is on the horizon. This gives the simplified sunrise equation:</p>
<p><span class="math-container">$$cos(h) = -\tan(\phi)\tan(\delta)$$</span></p>
<p>That's the hour angle from solar noon to sunrise or sunset, so we need to double <span class="math-container">$h$</span> to get the daylight length. And to convert it to hours, we use <span class="math-container">$2\pi = 24$</span> hours.</p>
<p>Here's a plot for latitude 45°.</p>
<p><a href="https://i.sstatic.net/zq8Ib.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/zq8Ib.png" alt="Daylight hours vs declination" /></a></p>
<p>To find the arc length along the latitude circle we just multiply the daylight hour angle (in radians) by the radius of the latitude circle. However, the radius of that circle is <em>not</em> simply <span class="math-container">$R\cos(\phi)$</span>. Instead, we need to use the <a href="https://en.wikipedia.org/wiki/Latitude#Parametric_latitude_(or_reduced_latitude)" rel="nofollow noreferrer">parametric latitude</a> <span class="math-container">$\beta$</span> of the <a href="https://en.wikipedia.org/wiki/World_Geodetic_System#WGS_84" rel="nofollow noreferrer">Earth ellipsoid</a>.</p>
<p>The parametric and geodetic latitude are related by
<span class="math-container">$$\tan(\beta) = (1-f)\tan(\phi)$$</span>
where <span class="math-container">$f=1/298.257223563$</span> is the flattening of the WGS-84 ellipsoid.</p>
<p>Here's the plot for latitude 45°.
<a href="https://i.sstatic.net/MH6SF.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MH6SF.png" alt="Daylight arc length" /></a></p>
<p>Here's the <a href="https://sagecell.sagemath.org/?z=eJxtU1tv0zAUfvevOPIklExp1ybtuqEFAZpgQiAmhrQHhCo3cRozxw62s6Ug%5FjvHTloqIFIUxz7%2DLudCKYVXbWt0LxrmOJRsJ8W2dsBMAZKrravh0ULJCykUc0IrAvjcG%2DEcV7DZwe0HSD8JtYV0li6ms2w6XxJKKamMbgAxaxBNq42DViRgWCmYsglYoRIoNK7qXatdAgx%5FCDmB%2D7d3FwvgUorWalEiwjdtgPUCQ4WCh4YwyOE8W11M59nK33gjmdeCEkiFR3M4g%5FTyYpouV2maLc8zH4N08Mzzga7A1RxaZljDnRHoEm25ruSk5JUPxKgNdyxqaxE%5FD253CfSIHM1hAlUMpz4qHAcLYRXinjAo2In8jWHPcNcZBTtU9eRh8OMFfdxI8b0TbrcX5BPcOlEk8C6dzWZEcwRLs%2Dkiu0xXl4vhIeSlUI4bVrigdh253uU3rpGvdR%5FRqzp7cf1vAa%5FOcJ%2DiWHSaL5ajKVSNDGNBIjyKybHZvxIRjk7g%5FZgrKIQpJA%5FXOxsOEWGNv3iVYYb6AcwxtR6IcDUAhX0vHlttPQiMfHuNsgaeG91h0dUWKZgD2ykjLMfc4cpydwisPZsvwWRPdBqI9q4CbhwfAV%5FrboOgvhfUI8eudPooU4fAsWop4tX4jt4G7bdI2kos8h8DCUQTzRPQHLNc%2Dc7KP5sON7ZGlDg43OY0NDJNRgrWc7uWbMOlzb%5FQu04djxig8K3h3MY0AXqo6DiO0UMT06%5FJMcTaih88n%2DPo7fGxTJLnFf1PO0CFA7VveviJq1803Bprczu1tX6KYvIbMJ46YQ==&amp;lang=python" rel="nofollow noreferrer">plotting script</a>, running on the SageMathCell server. The code is plain Python, apart from the Sage graph plotting function (which is based on matplotlib). The script works on any latitude, but will give warning messages in polar regions for declinations where the Sun doesn't rise, or doesn't set.</p>

CC BY-SA 4.0
2024-09-12T09:19:46.330Z
It seems that you did also ignore effect of the EoT for this simplified sunrise-sunset equation.
Jacob Miller
CC BY-SA 4.0
2024-09-12T10:34:28.673Z
@Jacob As Wikipedia says, "The sunrise equation or sunset equation can be used to derive the time of sunrise or sunset for any solar declination and latitude in terms of **local solar time** when sunrise and sunset actually occur". OTOH, I think the EoT is irrelevant here, because we're looking at the Sun's altitude at a *single instant in time* over a range of longitudes. That's a bit different to calculating the daylight length at a single location, where we have to take the change in EoT from sunrise to sunset into account.
PM 2Ring
CC BY-SA 4.0
2024-09-27T11:07:40.317Z
EoT effect is merely seconds (for the daylight duration) as you know, but I think it should still influence your calculation here even if the result is a minute change. I mean the daylight length is usually defined for a fixed longitude, especially since the equation you used is defined for a single point of observation. Declination changes throughout the year, with that EoT also. As the EoT results from the non-linearity of the Sun's apparent motion relative to a fixed observer on Earth, an observer based on any longitude will experience changes in daylight length due to the Equation of Time.
Jacob Miller
CC BY-SA 4.0
2024-03-15T11:09:44.480Z
How could I compute the Beta value, as the formula shown on Wikipedia is too difficult for me?
Geographos
CC BY-SA 4.0
2024-03-15T11:49:08.020Z
@Geographos I have the equation for beta in my answer, $$\tan(\beta) = (1-f)\tan(\phi)$$ And my code shows how to compute $\sin(\beta)$ and $\cos(\beta)$.
PM 2Ring
CC BY-SA 4.0
2024-03-15T05:11:04.390Z
2024-03-15T05:11:04.390Z
PM 2Ring
<p>I would like to calculate the distance between sunrise and sunset along a parallel of latitude, e.g., 45 degrees north.</p>
<p>What I've done is:
Earth's circumference at the equator is 40075 km.</p>
<p>Earth's circumference at 45 degrees north is:
<span class="math-container">$40075 \times \cos(45°) = 28337$</span> km</p>
<p>The distance between sunrise and sunset at the equinox (ignoring refraction) is simply:
<span class="math-container">$28337 / 2 = 14168$</span> km.</p>
<p>But, what is this distance when the solar declination is other than 0? I.e., the same latitude during summer or winter solstice?</p>
<p>Is there some calculator, which could do it for me, or at least some references?</p>

CC BY-SA 4.0
2024-09-27T06:17:59.147Z
It shall be noted that the sub-solar point never moves parallel to any latitude because the sun's declination is constantly changing. So for an earth-bound observer, if the sun rises directly over one latitude, at sunset of the same day, the sun would be directly above a slightly different latitude. So pedantically, the distance that the sun (sub-solar point) travels within a day is not parallel to any latitude.
Jacob Miller
CC BY-SA 4.0
2024-09-27T11:35:17.000Z
That's true. In fact, for the sake of generalization, it can be neglected, unlike the Moon, which changes its declination much more rapidly.
Geographos
CC BY-SA 4.0
2024-03-14T21:16:27.833Z
I'm not sure I agree with your calculation.  You seem to be measuring along the horizontal from east to west. But the distance is normally defined as the great circle distance, which is substantially less. The great circle distance would be via the pole and about 10000km
James K
CC BY-SA 4.0
2024-03-15T02:50:33.500Z
Why do you want the distance along the latitude circle, rather than the angle, or time? BTW, your latitude circumference calculation is for a sphere, the ellipsoid distance is slightly larger. Here's a calculator, in Python, giving the sphere & ellipsoid circumferences.
PM 2Ring
CC BY-SA 4.0
2024-03-15T02:50:46.290Z
https://sagecell.sagemath.org/?z=eJxtUl1LwzAUfc%2DvOESQTrpua%5Ffl2ER8UB8ciD7sUbIutdE2CWmGHeJ%5F97Yd4sRAPrj35Jxzb8I5x4Pwyu93EqlyadFt%2DzKTTupUwmj4XGJz99yfjyGLQtnKqB0DjY1T3kuN7QGPa8RPSr8iHsbjaJhEownjnLPMmRKl8DlUaY3zsCqEEzsldBWiUjpEauiUH6zxjJ1hLd6Mg6gVBZXGe8kEVpgms3k0SmYN4rYQjSqJsYxSIwwQX86jeDKL42QyTRoMEeO8YYbJWv9WOFFK71SK4lgv28msARJqK70IbK56i7auQ4iamIMR%2Dsh6uGhQbbo1255a3AeBWuNBc6OLOen3TuNArj4aGtoYu1baSydS32q%2DBL72q3tfFjemDvgyT65O3uCn%2DcsBpTipkuXVeHJ0R%5FKke%2DxhQKlOuKJgTF6tokXQPLFqHVkIMl7ZnLixwGe1iKbZFzWY99jvqv905H%5Fu%2DoS0%2DxXylPQb%5FoywAQ==&lang=python
PM 2Ring
CC BY-SA 4.0
2024-03-15T02:57:27.127Z
FWIW, here's a plot of the daylight length at 45°N, 0°E, both with & without refraction. The "no refrac" graph is for the Sun centre on the horizon. My calculated sunrise/set times are generally within a second or so of the values from JPL Horizons. https://i.sstatic.net/FSgJe.png
PM 2Ring
CC BY-SA 4.0
2024-03-15T10:57:08.303Z
@JamesK The distance between sunrise and sunset on the same latitude is only a great circle through the pole exactly on equinox dates. On other dates the great circle does not cross the poles.
planetmaker
CC BY-SA 4.0
2024-03-15T18:41:54.323Z
@planetmaker thanks for clarifying.  Yes, I was talking about the example in the question, which is on equinox day.
James K
CC BY-SA 4.0
2024-03-19T09:54:27.950Z
@JamesK I somewhat agree with you, although I am not calculating the shortest distance between point A and point B but just the section of given latitude which is sunlit for the time being.
Geographos
CC BY-SA 4.0
2024-03-20T21:45:48.813Z
Maybe this: https://f-droid.org/packages/com.forrestguice.suntimeswidget/ (Android, open source)
U. Windl
<earth><sunrise><sunset><solstice>
What is the distance between sunrise and sunset along a parallel of latitude for declination other than 0?
Geographos
# What is the distance between sunrise and sunset along a parallel of latitude for declination other than 0?

(Asked by: username_0 on 2024-03-14)

I would like to calculate the distance between sunrise and sunset along a parallel of latitude, e.g., 45 degrees north.

What I've done is:
Earth's circumference at the equator is 40075 km.

Earth's circumference at 45 degrees north is:
$40075 \times \cos(45°) = 28337$ km

The distance between sunrise and sunset at the equinox (ignoring refraction) is simply:
$28337 / 2 = 14168$ km.

But, what is this distance when the solar declination is other than 0? I.e., the same latitude during summer or winter solstice?

Is there some calculator, which could do it for me, or at least some references?

**Tags:** earth, sunrise, sunset, solstice
> **Comments:**
> **username_2**: I'm not sure I agree with your calculation.  You seem to be measuring along the horizontal from east to west. But the distance is normally defined as the great circle distance, which is substantially less. The great circle distance would be via the pole and about 10000km  
> **username_3**: Why do you want the distance along the latitude circle, rather than the angle, or time? BTW, your latitude circumference calculation is for a sphere, the ellipsoid distance is slightly larger. Here's a calculator, in Python, giving the sphere & ellipsoid circumferences.  
> **username_3**: https://sagecell.sagemath.org/?z=eJxtUl1LwzAUfc%2DvOESQTrpua%5Ffl2ER8UB8ciD7sUbIutdE2CWmGHeJ%5F97Yd4sRAPrj35Jxzb8I5x4Pwyu93EqlyadFt%2DzKTTupUwmj4XGJz99yfjyGLQtnKqB0DjY1T3kuN7QGPa8RPSr8iHsbjaJhEownjnLPMmRKl8DlUaY3zsCqEEzsldBWiUjpEauiUH6zxjJ1hLd6Mg6gVBZXGe8kEVpgms3k0SmYN4rYQjSqJsYxSIwwQX86jeDKL42QyTRoMEeO8YYbJWv9WOFFK71SK4lgv28msARJqK70IbK56i7auQ4iamIMR%2Dsh6uGhQbbo1255a3AeBWuNBc6OLOen3TuNArj4aGtoYu1baSydS32q%2DBL72q3tfFjemDvgyT65O3uCn%2DcsBpTipkuXVeHJ0R%5FKke%2DxhQKlOuKJgTF6tokXQPLFqHVkIMl7ZnLixwGe1iKbZFzWY99jvqv905H%5Fu%2DoS0%2DxXylPQb%5FoywAQ==&lang=python  
> **username_3**: FWIW, here's a plot of the daylight length at 45°N, 0°E, both with & without refraction. The "no refrac" graph is for the Sun centre on the horizon. My calculated sunrise/set times are generally within a second or so of the values from JPL Horizons. https://i.sstatic.net/FSgJe.png  
> **username_4**: @JamesK The distance between sunrise and sunset on the same latitude is only a great circle through the pole exactly on equinox dates. On other dates the great circle does not cross the poles.  
> **username_2**: @planetmaker thanks for clarifying.  Yes, I was talking about the example in the question, which is on equinox day.  
> **username_0**: @JamesK I somewhat agree with you, although I am not calculating the shortest distance between point A and point B but just the section of given latitude which is sunlit for the time being.  
> **username_5**: Maybe this: https://f-droid.org/packages/com.forrestguice.suntimeswidget/ (Android, open source)  
> **username_1**: It shall be noted that the sub-solar point never moves parallel to any latitude because the sun's declination is constantly changing. So for an earth-bound observer, if the sun rises directly over one latitude, at sunset of the same day, the sun would be directly above a slightly different latitude. So pedantically, the distance that the sun (sub-solar point) travels within a day is not parallel to any latitude.  
> **username_0**: That's true. In fact, for the sake of generalization, it can be neglected, unlike the Moon, which changes its declination much more rapidly.  

---

## 1 answers

### Answer by username_3 on 2024-03-15. Score: 4 | Accepted answer

We can use the [sunrise equation](https://en.wikipedia.org/wiki/Sunrise_equation) to determine the Sun's hour angle at sunrise & sunset and thence determine the daylight length. The sunrise equation is derived from the solution to the PZX navigation triangle:

$\sin(alt)=\sin(\phi)sin(\delta)+\cos(\phi)\cos(\delta)\cos(h)$

where $alt$ is the Sun's altitude above the horizon, $\delta$ is the Sun's declination, $\phi$ is the latitude, and $h$ is the Sun's hour angle, where $h=0$ at solar noon.

Since we're ignoring refraction (and the Sun's angular radius), we set $alt=0$. That is, we define sunrise & sunset to be when the Sun's centre is on the horizon. This gives the simplified sunrise equation:

$$cos(h) = -\tan(\phi)\tan(\delta)$$

That's the hour angle from solar noon to sunrise or sunset, so we need to double $h$ to get the daylight length. And to convert it to hours, we use $2\pi = 24$ hours.

Here's a plot for latitude 45°.

[![Daylight hours vs declination](https://i.sstatic.net/zq8Ib.png)](https://i.sstatic.net/zq8Ib.png)

To find the arc length along the latitude circle we just multiply the daylight hour angle (in radians) by the radius of the latitude circle. However, the radius of that circle is *not* simply $R\cos(\phi)$. Instead, we need to use the [parametric latitude](https://en.wikipedia.org/wiki/Latitude#Parametric_latitude_(or_reduced_latitude)) $\beta$ of the [Earth ellipsoid](https://en.wikipedia.org/wiki/World_Geodetic_System#WGS_84).

The parametric and geodetic latitude are related by
$$\tan(\beta) = (1-f)\tan(\phi)$$
where $f=1/298.257223563$ is the flattening of the WGS-84 ellipsoid.

Here's the plot for latitude 45°.
[![Daylight arc length](https://i.sstatic.net/MH6SF.png)](https://i.sstatic.net/MH6SF.png)

Here's the [plotting script](https://sagecell.sagemath.org/?z=eJxtU1tv0zAUfvevOPIklExp1ybtuqEFAZpgQiAmhrQHhCo3cRozxw62s6Ug%5FjvHTloqIFIUxz7%2DLudCKYVXbWt0LxrmOJRsJ8W2dsBMAZKrravh0ULJCykUc0IrAvjcG%2DEcV7DZwe0HSD8JtYV0li6ms2w6XxJKKamMbgAxaxBNq42DViRgWCmYsglYoRIoNK7qXatdAgx%5FCDmB%2D7d3FwvgUorWalEiwjdtgPUCQ4WCh4YwyOE8W11M59nK33gjmdeCEkiFR3M4g%5FTyYpouV2maLc8zH4N08Mzzga7A1RxaZljDnRHoEm25ruSk5JUPxKgNdyxqaxE%5FD253CfSIHM1hAlUMpz4qHAcLYRXinjAo2In8jWHPcNcZBTtU9eRh8OMFfdxI8b0TbrcX5BPcOlEk8C6dzWZEcwRLs%2Dkiu0xXl4vhIeSlUI4bVrigdh253uU3rpGvdR%5FRqzp7cf1vAa%5FOcJ%2DiWHSaL5ajKVSNDGNBIjyKybHZvxIRjk7g%5FZgrKIQpJA%5FXOxsOEWGNv3iVYYb6AcwxtR6IcDUAhX0vHlttPQiMfHuNsgaeG91h0dUWKZgD2ykjLMfc4cpydwisPZsvwWRPdBqI9q4CbhwfAV%5FrboOgvhfUI8eudPooU4fAsWop4tX4jt4G7bdI2kos8h8DCUQTzRPQHLNc%2Dc7KP5sON7ZGlDg43OY0NDJNRgrWc7uWbMOlzb%5FQu04djxig8K3h3MY0AXqo6DiO0UMT06%5FJMcTaih88n%2DPo7fGxTJLnFf1PO0CFA7VveviJq1803Bprczu1tX6KYvIbMJ46YQ==&lang=python), running on the SageMathCell server. The code is plain Python, apart from the Sage graph plotting function (which is based on matplotlib). The script works on any latitude, but will give warning messages in polar regions for declinations where the Sun doesn't rise, or doesn't set.

> **Comments:**
> **username_0**: How could I compute the Beta value, as the formula shown on Wikipedia is too difficult for me?  
> **username_3**: @Geographos I have the equation for beta in my answer, $$\tan(\beta) = (1-f)\tan(\phi)$$ And my code shows how to compute $\sin(\beta)$ and $\cos(\beta)$.  
> **username_1**: It seems that you did also ignore effect of the EoT for this simplified sunrise-sunset equation.  
> **username_3**: @Jacob As Wikipedia says, "The sunrise equation or sunset equation can be used to derive the time of sunrise or sunset for any solar declination and latitude in terms of **local solar time** when sunrise and sunset actually occur". OTOH, I think the EoT is irrelevant here, because we're looking at the Sun's altitude at a *single instant in time* over a range of longitudes. That's a bit different to calculating the daylight length at a single location, where we have to take the change in EoT from sunrise to sunset into account.  
> **username_1**: EoT effect is merely seconds (for the daylight duration) as you know, but I think it should still influence your calculation here even if the result is a minute change. I mean the daylight length is usually defined for a fixed longitude, especially since the equation you used is defined for a single point of observation. Declination changes throughout the year, with that EoT also. As the EoT results from the non-linearity of the Sun's apparent motion relative to a fixed observer on Earth, an observer based on any longitude will experience changes in daylight length due to the Equation of Time.  

---

