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I live at roughly $(52.4^\circ,-2.1^\circ)$. On sunny evenings I've often looked at the Moon and the Sun and noticed that the light part of the Moon does not appear to line up with the Sun. For example, at about 17:00 GMT on 13 Mar 2011, I noticed the half Moon was facing toward a point roughly $10^\circ-20^\circ$ above where the Sun appeared to be. Why?

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    $\begingroup$ Just because this effect can be explained using lots of words and 2D diagrams, doesn't mean it should. Unless you enjoy headaches, see: curvilinear perspective. $\endgroup$ – Nick Nov 21 '12 at 0:50
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I think it is a parallax effect/optical illusion, and I'm not confident of explaining this clearly but here goes!

The normal vector to the illuminated portion of the moon is pointing generally away from the Earth/moon system towards a point over our horizon. At low altitudes (evenings) the sun will be close to the horizon and this can lead to the brain interpreting it as closer than it is and messing up the geometry. This is similar to the enlarged moon illusion when close to the horizon. Basically the normal vector appears to overshoot the sun as we interpret the sun as closer than it is.

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This is what you expect, in terms of the moon pointing towards the sun:
half moon illuminated by sun
That is, the line across the moon appears perpendicular to the line towards the sun.

Now the above is a flat drawing. The sky appears curved (i.e. the dome of the starry sky). So that curve may introduce some apparent distortion.

To make a drawing that avoids the curvature issue, consider a drawing that only includes the sun, the moon, and a small amount of sky around the line connecting the two. (By the way, even for the curved bowl of the sky, the shortest line connecting the two is well defined except in the case of a new or full moon.) That drawing is approximately flat and will show the above relation.

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  • $\begingroup$ I think the parallax arises because depending on the phase of the moon what one sees from the earth is not the total lit up semicircle of the moon. Part of it is hidden from the human observer so the apparent bisector is not the real bisector. $\endgroup$ – anna v Mar 16 '11 at 4:34
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    $\begingroup$ @anna v; The sun and moon are approximately spherical so the system has bilateral symmetry. Thus the perpendicular is the only possibility (except for distortion due to optics or the human mind). Personally, I've never had any trouble figuring out where the sun is based on a partial moon so I'm not sure where the human effect comes into play. And I think the optical effect is so small that it will be smaller than the human ability to estimate angles. $\endgroup$ – Carl Brannen Mar 16 '11 at 4:41
  • $\begingroup$ @Carl Brannen . Yes, your plot is correct. But the observer on the plane you draw, does not see the full 180degree illuminated side. Depending on geometrical location he/she sees a projection and part of the illuminated region is hidden. The bisector of this projection is not the same as the true bisector which should point at the sun. $\endgroup$ – anna v Mar 16 '11 at 7:15
  • $\begingroup$ @anna v; Don't see how you violate the bilateral symmetry. $\endgroup$ – Carl Brannen Mar 17 '11 at 0:30
  • $\begingroup$ @Carl Brannen Go to your figure. Have the observer sit on the right of the line that makes the two hemispheres. What does he/she see? The further to the right, the projection gets smaller, because the dark side of the moon hides the top bright part, and the part seen bisected will not point to the sun. $\endgroup$ – anna v Mar 17 '11 at 14:59
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I think the impression of not lining up correctly is due to our intuition, which is based on everyday scales of distances. Intuitively we expect the visible objects to be of similar sizes and at similar distances. But the Sun is much farther from both the Earth and the Moon, than the Moon is from the Earth: compare Sun-Earth distance of $1\,\text{AU}$ and Earth-Moon distance of $0.0026\,\text{AU}$ — it's $390$-fold difference! Thus although they form a triangle, the shading of the Moon doesn't seem to follow this, instead looking as if the Sun is infinitely far away.

Here's what you would normally expect based on our everyday intuition:

intuitive expectation

But if we take into account that the Sun is really large and really far away, we'll get something like this:

more accurate picture

See how the shading of the Moon changed.

As an example, consider the following situation: the Sun is setting, and the Moon is in the opposite azimuth about $35^\circ$ above the horizon. You look at the Moon and see it as if it were lit from slightly above. This weirdness is simply because of the angle between your direction of sight and the light rays which light the Moon. Here's how it looks:

simple example of the Sun and the Moon at the opposite azimuths

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I am puzzled by you question. When one has two points, the sun and the moon, one can always find a line connecting them, by definition of line.

If you mean why the earth is not part of that straight line, it is because the moon has an orbit around the earth, and the angle of the line earth-moon changes. It is the reason the moon has phases. Earth, moon and sun are on the same line during full moon, and the moon rises while the sun sets.

Edit: If as Ted Bun says you mean the bisecting line from the center of the moon, then the drawing given in wikipedia gives an angle because of the motion of the moon around the earth and the motion of the earth around the sun, except at full moon and new moon (if the rays shown are a correct depiction of the sun's direction).

Edit2: If one looks at the drawing In Carl Brannen's answer in combination with the wiki drawing, I think the parallax arises depending on the phase of the moon, because what one sees from the earth is not the total lit up semicircle of the moon. Part of it is hidden from the human observer so the apparent bisector is not the real bisector.

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    $\begingroup$ I think the point is that you would expect a line going directly away from the lit-up side of the Moon (in other words, a line that bisects the lit-up side symmetrically) to point right towards the Sun. I have to admit that I haven't noticed this effect. Georg's explanation sounds plausible. Like Omega Centauri I would've guessed that the effect would be smaller than what the questioner is describing, but I don't know for sure. $\endgroup$ – Ted Bunn Mar 15 '11 at 17:44

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