Sunrise Sunset Calculator (2024)

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Sunrise and Sunset times

The term sunrise time refers to the time when the sun first appears or when daylight has arrived. Similarly, we define sunset time as the moment when the sun disappears below the horizon. For a given location having latitude $\phi$ϕ and n^th day of the year, we can estimate the sunrise time using the hour angle for sunrise/sunset, $\omega$ω.

$$\footnotesize\omega \!=\!\arccos\left(\frac{\cos(z)\!-\!\sin(\delta)\!\cdot\!\sin(\phi)}{\cos(\delta)\!\cdot\!\cos(\phi)}\right)$$ω=arccos(cos(δ)⋅cos(ϕ)cos(z)−sin(δ)⋅sin(ϕ)​)

$$\footnotesize\delta = \arcsin(0.39872\!\cdot\!\sin(L))$$δ=arcsin(0.39872⋅sin(L))

where $L$L is the true latitude of the Sun (its projected position on the Earth). Note that there are multiple ways to calculate the Sun's declination.

In our case, we calculated it from $L$L, with $L$L being given by:

$$\footnotesize\begin{split}L &= M+1.916\sin(M)\\&\quad+0.02\sin(2\!\cdot\!M)+282.634\end{split}$$L​=M+1.916sin(M)+0.02sin(2⋅M)+282.634​

$M$M is a measure of the Sun's mean anomaly, a quantity that expresses the discrepancy between the true elliptic orbit of Earth and the ideal circular orbit with an identical period for a given moment of the year:

$$\footnotesize M = 0.9856\cdot t - 3.289$$M=0.9856⋅t−3.289

The value of $t$t depends on the day number (the number of days elapsed since the first of January) and the time of the day:

$$\footnotesize t =\begin{cases} n + \frac{6-\lambda_{\mathrm{hour}}}{24}\ \mathrm{(sunrise)}\\[1em] n + \frac{18-\lambda_{\mathrm{hour}}}{24}\ \mathrm{(sunset)}\end{cases}$$t={n+246−λhour​​(sunrise)n+2418−λhour​​(sunset)​

The second part of both equations is an approximation of the local sunrise and sunset times. In the formula, $\lambda_{\mathrm{hour}}$λhour​ is nothing but an expression of the latitude in hours. To find it, divide the value of the latitude in degrees by $15$15.

Once we have the hour (angle for sunrise), $\omega$ω, we can calculate the local time of the sunrise and sunset by applying the following formula:

$$\footnotesize T = \frac{\omega}{15} + \mathrm{RA}- (0.06571 \cdot t) - 6.622,$$T=15ω​+RA−(0.06571⋅t)−6.622,

where $\mathrm{RA}$RA is the Sun's right ascension, that we found with:

$$\footnotesize\begin{split} &\mathrm{RA} = \frac{1}{15}\!\cdot\!\Big(\arctan(0.91764 \cdot \tan(L))\\&+( L\!\!\!\!\mod 90\degree)\\&-(\arctan(0.91764 \cdot \tan(L))\!\!\!\!\mod 90\degree)\Big)\end{split}$$​RA=151​⋅(arctan(0.91764⋅tan(L))+(Lmod90°)−(arctan(0.91764⋅tan(L))mod90°))​

Back to business: we need to adjust the times we calculated, taking into account the local latitude and the time zone. To do so, we subtract the former and add the latter:

$$\footnotesize T_\mathrm{local} = T-\lambda_\mathrm{h}+\mathrm{tz}$$Tlocal​=T−λh​+tz

Remember to consider daylight savings if the desired location observes such time. Simply add $1$1 hour to the calculated time in the summer if this is the case!

To calculate the daylight hours, we compute the difference between the time of the sunset and the time of the sunrise:

$$\small t_\mathrm{daylight} = T_\mathrm{local,\ sunset} - T_\mathrm{local,\ sunrise}$$tdaylight​=Tlocal,sunset​−Tlocal,sunrise​

While the above methodology is correct, it does not consider atmospheric refraction. This effect occurs due to the sunlight refracting through the atmosphere and making the sun appear higher above the horizon than its actual position. A small angle is included in the sunrise and sunset formula to take this phenomenon into account. Therefore, the corrected sunset and sunrise equation is:

$$\!\footnotesize\omega \!=\!\arccos\!\left(\!\frac{\cos(90+a)\!-\!\sin(\delta)\!\cdot\!\sin(\phi)}{\cos(\delta)\!\cdot\!\cos(\phi)}\!\right)$$ω=arccos(cos(δ)⋅cos(ϕ)cos(90+a)−sin(δ)⋅sin(ϕ)​)

where $a$a is the altitude angle having the value of 0.83°.

How to calculate the times of sunrise and sunset for tomorrow?

Example: Using the sunrise sunset calculator

$$\footnotesize \enspace \enspace t =\begin{cases}75 + \frac{6-1}{24}=75.2917\ \mathrm{(sunrise)}\\[1em]75 + \frac{18-1}{24}=75.7917\ \mathrm{(sunset)}\end{cases}$$t={75+246−1​=75.2917(sunrise)75+2418−1​=75.7917(sunset)​

$$\footnotesize \begin{split}\enspace \enspace M_{\mathrm{rise}}&= (0.9856 \cdot 75.2083) - 3.289\\&=69.9329\\M_{\mathrm{set}} &=(0.9856 \cdot t) - 3.289\\&=70.4257\end{split}$$Mrise​Mset​​=(0.9856⋅75.2083)−3.289=69.9329=(0.9856⋅t)−3.289=70.4257​

$$\footnotesize \begin{split}\enspace \enspace L_{\mathrm{rise}}&= 354.3794\degree\\L_{\mathrm{set}} &=354.8776\degree\end{split}$$Lrise​Lset​​=354.3794°=354.8776°​

$$\footnotesize \begin{split}\enspace \enspace \mathrm{RA}_{\mathrm{rise}}&= 23.6560\degree\\\mathrm{RA}_{\mathrm{set}} &=23.6865\degree\end{split}$$RArise​RAset​​=23.6560°=23.6865°​

$$\footnotesize \begin{split}\enspace \enspace \delta_{\mathrm{rise}}&=\arcsin\left(0.39782\sin(L_\mathrm{rise})\right)\\&=-2.2335\degree\\\delta_{\mathrm{set}}&=\arcsin\left(0.39782\sin(L_\mathrm{set})\right)\\&=-2.0359\degree\end{split}$$δrise​δset​​=arcsin(0.39782sin(Lrise​))=−2.2335°=arcsin(0.39782sin(Lset​))=−2.0359°​

$$\!\footnotesize \omega \!=\! \arccos\!\left(\!\frac{\cos(90.833)\!-\!\sin(\delta)\!\cdot\!\sin(\phi)}{\cos(\delta)\!\cdot\!\cos(\phi)}\!\right)$$ω=arccos(cos(δ)⋅cos(ϕ)cos(90.833)−sin(δ)⋅sin(ϕ)​)

The result is:

$$\begin{split}\omega_{\mathrm{rise}} &=293.5714 \degree\\\omega_{\mathrm{set}} &=89.1423 \degree\\\end{split}$$ωrise​ωset​​=293.5714°=89.1423°​

Use the following formula to calculate the sunrise and sunset times:

$$\footnotesize\begin{split}T = &\frac{\omega}{15} + \mathrm{RA}- (0.06571 \cdot t)\\[.5em] &- 6.622 -\lambda_\mathrm{h}\end{split}$$T=​15ω​+RA−(0.06571⋅t)−6.622−λh​​

We can ignore both daylight savings and timezone: they both are equal to 0. The results are:

$$\begin{split}t_{\mathrm{rise}} &=7.223\ \mathrm{h} =7\ \mathrm{h}\ 13\ \mathrm{min} \\t_{\mathrm{set}} &=19.093\ \mathrm{h}=19\ \mathrm{h}\ 5\ \mathrm{min}\end{split}$$trise​tset​​=7.223h=7h13min=19.093h=19h5min​

Subtract these values to find the number of daylight hours:

$$\begin{split}t_\mathrm{daylight}& = t_{\mathrm{set}}-t_\mathrm{rise}\\&=19.093-7.223\\ &= 11.87\ \mathrm{h} \\&=11\ \mathrm{h}\ 52\ \mathrm{min}\end{split}$$tdaylight​​=tset​−trise​=19.093−7.223=11.87h=11h52min​

🔎 You can also quickly estimate the day duration with Omni's hours calculator.

FAQ

What is atmospheric refraction?

The phenomenon that celestial objects appear higher above the horizon than they actually are is known as atmospheric refraction. This occurs because of the refraction of light reflected from the objects by the atmosphere.

What is the angular velocity of earth?

How do I calculate sunrise hour angle?

How do I calculate daylight hours?