<p>The apparent curvature of the moon's transit across the sun is primarily due to the change in perspective of the viewer as s/he is rotated to the east by the earth's normal daily-cycle rotation - approximately 15 degrees absolute per hour.</p>

<p>At the beginning of the eclipse, the viewer is standing "upright," and sees the moon touch the sun at approximately 1 O'clock.  Over the next three hours, as the eclipse progresses, the viewer is "laid on her/his side" relative to his/her initial orientation due to the rotation of the earth; as if tilting his/her neck to the east. Since most of us were looking (essentially) towards the south for this eclipse, it was as if we had tilted our head to the left, so the exit of the moon - instead of appearing at approximately 7 O'clock as expected for a linear transit, the exit point appeared to be at 9 O'clock.</p>

<p>The magnitude of this effect is dependent upon several variables, including the viewer's latitude during observation.</p>

<p>The effect is also exaggerated by the fact that the sun transited through its peak azimuth between the beginning (rising towards its highest azimuth - approximately 63 degrees) and the end of the eclipse (setting away from its highest azimuth).  While the earth itself only rotated approximately 45 degrees absolute  during the 3 hour event, the apparent rotation of the moon's transit line across the sun was approximately 60 degrees (in the Kansas City simulation).</p>

CC BY-SA 3.0
2017-08-25T14:28:14.513Z
2017-08-25T14:28:14.513Z
Schnout
<p>You have misunderstood the website. When the eclipse begins, in the morning, the sun will be in south-east, at a bearing of 135 degrees. </p>

<p>At maximum eclipse (at about midday) the sun will be almost due south, at a bearing of 178 degrees.</p>

<p>When the eclipse ends, in the afternoon, the sun will be in the south-west, at about 217 degrees bearing.</p>

<p>The sun and the moon will move in the sky, due to the rotation of the Earth, this says nothing about the straightness of the path of the moon relative to the sun.</p>

CC BY-SA 3.0
2017-08-18T19:46:27.153Z
2017-08-18T19:46:27.153Z
James K
<p>Reviewing several simulations of the solar eclipse, it is apparent that the path of the moon relative to the sun curves. </p>

<p>Using <a href="https://www.timeanddate.com/eclipse/in/usa/kansas-city?iso=20170821" rel="nofollow noreferrer">https://www.timeanddate.com/eclipse/in/usa/kansas-city?iso=20170821</a> for example, the moon enters at 135 degrees and leaves at 217 degrees direction - that is almost a 90 degree right turn.</p>

<p>Why is the path curved not straight?</p>

CC BY-SA 3.0
2017-08-18T19:44:48.580Z
You may be misunderstanding what this “direction” value indicates. It tells you where in the sky the Sun (and in this case, also the Moon) is. At the beginning of the eclipse, it is SE; at the end, SSW. That’s just the normal apparent movement of the Sun as on every day (rises in E, sets in W) caused by Earth’s rotation.
chirlu
<solar-eclipse>
Why does the path of the moon in an eclipse curve (relative to the sun)
D Mac
# Why does the path of the moon in an eclipse curve (relative to the sun)

(Asked by: username_0 on 2017-08-18)

Reviewing several simulations of the solar eclipse, it is apparent that the path of the moon relative to the sun curves.

Using <https://www.timeanddate.com/eclipse/in/usa/kansas-city?iso=20170821> for example, the moon enters at 135 degrees and leaves at 217 degrees direction - that is almost a 90 degree right turn.

Why is the path curved not straight?

**Tags:** solar-eclipse

> **Comments:**
> **username_1**: You may be misunderstanding what this “direction” value indicates. It tells you where in the sky the Sun (and in this case, also the Moon) is. At the beginning of the eclipse, it is SE; at the end, SSW. That’s just the normal apparent movement of the Sun as on every day (rises in E, sets in W) caused by Earth’s rotation.  

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## 2 answers

### Answer by username_2 on 2017-08-25. Score: 4 | Accepted answer

The apparent curvature of the moon's transit across the sun is primarily due to the change in perspective of the viewer as s/he is rotated to the east by the earth's normal daily-cycle rotation - approximately 15 degrees absolute per hour.

At the beginning of the eclipse, the viewer is standing "upright," and sees the moon touch the sun at approximately 1 O'clock. Over the next three hours, as the eclipse progresses, the viewer is "laid on her/his side" relative to his/her initial orientation due to the rotation of the earth; as if tilting his/her neck to the east. Since most of us were looking (essentially) towards the south for this eclipse, it was as if we had tilted our head to the left, so the exit of the moon - instead of appearing at approximately 7 O'clock as expected for a linear transit, the exit point appeared to be at 9 O'clock.

The magnitude of this effect is dependent upon several variables, including the viewer's latitude during observation.

The effect is also exaggerated by the fact that the sun transited through its peak azimuth between the beginning (rising towards its highest azimuth - approximately 63 degrees) and the end of the eclipse (setting away from its highest azimuth). While the earth itself only rotated approximately 45 degrees absolute during the 3 hour event, the apparent rotation of the moon's transit line across the sun was approximately 60 degrees (in the Kansas City simulation).


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### Answer by username_3 on 2017-08-18. Score: 4

You have misunderstood the website. When the eclipse begins, in the morning, the sun will be in south-east, at a bearing of 135 degrees.

At maximum eclipse (at about midday) the sun will be almost due south, at a bearing of 178 degrees.

When the eclipse ends, in the afternoon, the sun will be in the south-west, at about 217 degrees bearing.

The sun and the moon will move in the sky, due to the rotation of the Earth, this says nothing about the straightness of the path of the moon relative to the sun.


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