<p>Yes, see <a href="https://www.desmos.com/calculator/vq7djgkyir" rel="nofollow noreferrer">graph</a>.
At the March equinox, the Sun's ecliptic longitude $\lambda_\odot$ is zero, and the Sun's declination $\delta_\odot$ is zero and increasing.
At the June solstice, $\lambda_\odot$ is 90$^\circ$ and $\delta_\odot$ is at a maximum, 23.5$^\circ$ north of the equator.
At the September equinox, $\lambda_\odot$ is 180$^\circ$ and $\delta_\odot$ is zero and decreasing.
At the December solstice, $\lambda_\odot$ is 270$^\circ$ and $\delta_\odot$ is at a minimum of -23.5$^\circ$.</p>

CC BY-SA 3.0
2016-11-23T13:06:02.250Z
2016-11-23T18:16:28.453Z
2016-11-23T18:16:28.453Z
Mike G
<p>Yes, it is correct. For better precision:
Declination =23.5*sin(longitude)
This is true only if you don't take into account the eccentricity of Earth's orbit and other movement. It isn't easy to understand why, but I'll try ti explain it.
The different declination of the sun during the year is generated by the different angle of inclination of the Earth (the north pole point always in the same direction, so during the earth revolution the poles don't point always in the sun direction and the protection of the inclination on the sun-earth plane change during the year) 
So, what wee see of the sun is the projection of the circular movement of the earth on a line of sight, so an harmonic motion.</p>

CC BY-SA 3.0
2016-11-23T20:59:41.887Z
Why is this a community wiki answer?
Sir Cumference
CC BY-SA 3.0
2016-11-23T12:49:00.260Z
2016-11-23T12:49:00.260Z
pietro31700
<p>I encountered a book discussion that defines the Sun's declination as 23.5 degrees (inclination of Earth's equator to the ecliptic) times the sine of the Sun's longitude. Does this make sense? How do i visualize it?</p>

<positional-astronomy>
Declination of the Sun formula
Gamliel Beyderman
# Declination of the Sun formula

(Asked by: username_0 on 2016-11-23)

I encountered a book discussion that defines the Sun's declination as 23.5 degrees (inclination of Earth's equator to the ecliptic) times the sine of the Sun's longitude. Does this make sense? How do i visualize it?

**Tags:** positional-astronomy


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## 2 answers

### Answer by username_1 on 2016-11-23. Score: 1

Yes, see [graph](https://www.desmos.com/calculator/vq7djgkyir).
At the March equinox, the Sun's ecliptic longitude $\lambda\_\odot$ is zero, and the Sun's declination $\delta\_\odot$ is zero and increasing.
At the June solstice, $\lambda\_\odot$ is 90$^\circ$ and $\delta\_\odot$ is at a maximum, 23.5$^\circ$ north of the equator.
At the September equinox, $\lambda\_\odot$ is 180$^\circ$ and $\delta\_\odot$ is zero and decreasing.
At the December solstice, $\lambda\_\odot$ is 270$^\circ$ and $\delta\_\odot$ is at a minimum of -23.5$^\circ$.


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### Answer by username_2 on 2016-11-23. Score: 1

Yes, it is correct. For better precision:
Declination =23.5\*sin(longitude)
This is true only if you don't take into account the eccentricity of Earth's orbit and other movement. It isn't easy to understand why, but I'll try ti explain it.
The different declination of the sun during the year is generated by the different angle of inclination of the Earth (the north pole point always in the same direction, so during the earth revolution the poles don't point always in the sun direction and the protection of the inclination on the sun-earth plane change during the year)
So, what wee see of the sun is the projection of the circular movement of the earth on a line of sight, so an harmonic motion.

> **Comments:**
> **username_3**: Why is this a community wiki answer?  

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