Hello.

I can see the moon high in the northern sky. It is close to the meridian. The left hand side of it is brightly illuminated by the sun. The terminator is slightly clockwise of vertical. Yet the sun has already set, i.e. the sun is much, much lower than the moon as I look at it. Why isn't the terminator at right-angles to the line of the sun?

I've noticed this puzzling thing many times. It is most obvious when both the sun and moon are in the sky together, and the moon is visible in daylight. Often the terminator is totally skewed to the direction of the sun. The only time I've seen the terminator orthogonal to the sun was when I"ve been near the equator and the sun and moon are passing directly overhead.

What is the explanation please?

clop

I've always noticed that it is orthogonal. Next time you see the Moon and Sun together in the sky, hold a round ball in front of the Moon. You'll find that the phase of the ball matches that of the Moon. And the terminator on the ball will be orthogonal to the Sun line.

Do you live near the equator then?

It is certainly not orthogonal for me, round ball or no round ball. If the sun is below the horizon to my left, and the moon's terminator is slanted slightly to the right, which would imply that the sun were slightly higher than the moon in the sky, which it isn't.

clop

Which phase is it in? If its a waning gibbous phase, and the Moon is near the horizon after sunset you need to draw the line the other direction, from the Moon to its closest horizon, beneath the Earth, to the Sun, rather than across the sky to the western horizon. Just guessing, but I suspect you're not drawing the shortest line possible.

I live in San Francisco, at latitude 38, but even when I was in Antarctica I noticed the Moon's terminator was perpendicular to the Moon / Sun line.

See if you can take a picture to post here next time you notice the terminator pointing the wrong direction. Try to zoom out as much as possible, and include horizon in the shot. It should be pretty easy to figure it out from that.

Here's a picture I took in Antarctica. The Sun is not in the picture, but it is exactly where you'd expect to find it based on the Moon's terminator. It is about 90 degrees to my left, and slightly higher than the Moon.

Click here for the bigger version:

http://orbitsimulator.com/Antarctica...G_71891024.jpg
http://orbitsimulator.com/Antarctica...G_72191024.jpg

Can you provide a photo?

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I've noticed it for many years, too. It's caused by an
unconcious impression that the Sun and Moon are about the
same distance away from you, even though the Sun is actually
400 times farther. That huge error in judging the distance
makes the angle look wrong.

Here is a pair of diagrams I made to show the geometry:

http://www.freemars.org/jeff2/SunMoon.png

They are not to scale, of course. In both diagrams, the
observer on Earth is at the "top" of the globe, with the Sun
near the horizon on the left, as shortly after sunrise, and the
Moon high overhead.

The diagram at upper-right shows the Sun and Moon as the same
size, since that is how they appear. In reality, the Sun is
400 times larger than the Moon.

The thing that makes the angle of the terminator look wrong
is the direction that sunlight appears to be coming from,
compared to the direction it is actually coming from.

-- Jeff, in Minneapolis

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"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

When you say orthogonal are you projecting the line between the Moon and Sun as a straight line? If so that may be the source of youir confusion.

Remember, the Moon and the sun follow a path that appears from our point of view to be an arc. In the attached jpeg I've drawn a rough diagram that I believe represents the situation. The Moon's terminator is perpendicular to the path it traces. The arrow represents the direction you assume the sun to be in by projecting a line perpendicular to the terminator.

Attached Images

Sun and Moon.JPG (11.1 KB, 312 views)

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"The very powerful and the very stupid have one thing in common: They don't alter their views to fit the facts, they alter the facts to fit their views." The Doctor, Doctor Who: The Face of Evil.

In your diagram, shouldn't the side of the moon facing the sun be illuminated?

As Jeff Root says, others have noticed this illusion before.

shouldn't the side of the moon facing the sun be illuminated?

Nope. Go out and observe the progress of the Moon as it moves across the sky. You will find that the terminator is perpendicular to the path it traces.

Jeff's diagram explains phases nicely, but I can't see how it explains why the angle of the terminator is aparently off.

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"The very powerful and the very stupid have one thing in common: They don't alter their views to fit the facts, they alter the facts to fit their views." The Doctor, Doctor Who: The Face of Evil.

Yes I understand that the moon and sun follow an arc across the sky (roughly the ecliptic) and that the terminator on the moon is orthogonal to the tangent of the ecliptic at the point where the moon is located. But it makes for a really puzzling optical illusion when you can see the sun below the moon and yet the terminator indicates the sun is above the moon.

clop

I've always wanted to illustrate that so I can visualize the effect. Thanks, that's a great diagram.

Again, good representation, but I had a hard time at first figuring it out. I'm not sure how I would make it clearer.

Maybe some combo of the two. I see the perception of distance and the location relative to the horizon as two seperate factors here.

I have a question about the ecliptic and didn't want to necessarily start a new thread:

I understand the ecliptic (the apparent paths of the sun, moon, and planets when viewed from a particular place on earth) will change seasonally as the earth progresses in its orbit and due to the tilt of the earth's rotational axis in relation to the so called plane of our solar system. Sometimes it will be higher in the sky with a steep angle to the horizons, sometimes lower with shallower angles.

Last month for me (40 degrees N lattitude) the Moon and at least a couple planets were very low in the sky. I tried figuring it out holding apples and oranges and thought I understood the seasonal change. However, less than a month later, the moon is extremely high in the sky for me, so much so that it looks like it rises north of east. I didn't think the change would happen so fast. Is there something about the orbit of the Moon I'm not understanding that makes its path across my sky vary so quickly? (Or am I nuts? I can accept that. )

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Thump

I've accepted mine a long time ago.
The moon's orbit has its own tilt in relation to the sun, or the equator, and none of them match up.

The seasonal change you're talking about happens once a year for the Sun as it moves around the ecliptic. When the Moon or planets are near the Sun, they'll be in the same seasonal position as the Sun: low in winter, high in summer. But when they're opposite the Sun, they'll be appearing around midnight, and their seasonal position will be reversed: high in winter, low in summer.
Since the Moon passes right around the ecliptic in a single month, you'll see it shift from low to high and back again in that time frame. In winter, it will be low in the sky during crescent phase, but high around the time of the full moon.

Grant Hutchison

When considering the Moon/Sun line, I always follow the shortest arc across the sky, rather than the straight line as depicted in Jason's diagram. That probably explains why I've never percieved the effect of the terminator pointing in the wrong direction.

btw. Today's a good day to see the Moon in the daytime sky if you have good weather. It's a waxing gibbous, just past 1st quarter.

I know that light travels in a more or less straight line through space from the sun to the moon, though. The only side that will be illuminated is the one facing the sun. That's irrespective of perspective, too.

When you talk about the path that the moon traces, are you talking about its daily path, or its monthly path? It would seem you meant the latter, right?

I just looked at Jeff Root's diagram too, and I don't like it either. The top half says it shows what your eyes see--but the text seems to indicate that the observer is under the moon, rather than away from it. I don't think you can explain this one with diagrams--it's just an illusion, it's not a real effect, just like the moon illusion.

Hmm. I totally disagree. There is no doubt that, if you draw the perpendicular to the terminator and extend on the bright side of the waxing gibbous moon shortly after sunset, it is pointing upward into the sunless sky, as diagrammed by Jason. Of course if you allow yourself to follow a curved line, as Jason and Tony do, you can convince yourself that there is no discrepancy, but it isn't sufficient in itself, particular as the curved line is almost defined by being at right angles to the terminator.

The issue may be a misinterpretation of relative distance to sun and moon as Jeff suggests but even that doesn't seem sufficient to me. I think it's more likely a misinterpretation of the absolute distance to both sun and moon. If they were fixed in a sphere about 5km (3 miles) away, ie. horizon distance, rotating about me personally, you might expect the moon's terminator to lie perpendicular to the line to the sun. Being off equator on a round planet that rarely obstructs the moon's illumination, the geometry is more than we can instinctively cope with. (Intellectually - that's different, of course).

I'm sure these shadow lines will not be angled in the same direction. Can anyone manage a photo of a rising moon 4-5 days before full (next couple of days), plus a tennis ball, just before sunset?

But they will! That's what's so cool about this trick. Hold the ball anywhere in the sky and it will show you what phase the Moon would have if it were hidden behind the ball.

I'll try to take the picture and post it here if these pesky clouds will ever leave.

It is an illusion, but I'm confident that a diagram can show
how the illusion is caused. My diagram is lacking in that it
is only two-dimensional, rather than three-dimensional. I'm
pretty sure that if you consider the geometry in 3-D, it will
fully explain the illusion.

I don't understand what you meant by my text indicating that
"the observer is under the moon, rather than away from it."
What does that mean?

The geometry I depicted is with the Sun and Moon 90 degrees
apart in the sky. I did that simply because a right angle
is familiar to everyone, and easy to visualize.

The observer is at the top of the Earth. That is where all
observers always are! The Moon is shown directly above the
Earth and the Sun is to your left. Observers in the northern
hemisphere can interpret that to mean that the Moon is on your
meridian directly south of you, shortly after sunrise. Those
in the southern hemisphere can interpret it to mean the Moon
is directly north of you, shortly before sunset. The Moon
could be low in the sky, or high in the sky.

None of that is important.

What is important is that there is a definite angle from the
Sun to you to the Moon (90 degrees in the diagram). Given that
the Sun and Moon appear to be roughly the same distance away,
your visual system interprets the angle of the path of light
from Sun to Moon as about 45 degrees (for the case diagrammed),
though it is actually well under one degree.

I think the diagram needs to be 3-D to be complete.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

I calculate that, when the angle from Sun to observer to Moon
is 90 degrees, the difference between the Moon's terminator
and a ball's terminator will be between 0.1 and 0.2 degree,
measured from the center of each.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

That's not much of a difference. The eye would never discern that difference. It would even be difficult in a photograph to measure a difference as small as 6-12 arcminutes.

As far as the eye can tell, the ball in front of the Moon has the same phase as the Moon.

What were you using as the ball / eye distance? Does this difference increase or decrease as the Moon approaches full?

When I attempt to do the photograph I'm going to back off a bit, much more than an arm length, so I can zoom in and attempt to get the ball and Moon to be comparable sizes.

I know. This is exactly what I'm saying. It's just so annoying and difficult to visualise. I can't take a photograph right at the moment but I've drawn this diagram to try and illustrate what I'm talking about.

This is what I expect to see



but this is what I actually see.



The way I see it, wherever you are in the solar system you ought to be able to draw a straight line (three-dimensional vector) to the moon and a straight line (another three-dimensional vector) to the sun, and there is only one plane that can be defined where both these vectors lie on the plane, and by definition the observer's position must also lie on this plane, and the moon's terminator ought to be perpendicular to the plane, which means the moon's terminator must appear perpendicular to a straight line drawn between the moon and the sun as seen by the observer, and this is irrespective of the relative sizes or distances of the objects. But it clearly doesn't.

This is starting to do my head in.

clop

Here's a walk-through in a toy model that might help with the visualization. We assume the Earth's orbit, the moon's orbit and the Earth's equator all lie in the same plane. This simplifies the visualization without doing any violence to the real situation.
Now let's deal with a waxing gibbous moon, halfway from first quarter to full.
We go to the north pole, and look at the moon and sun. Because of our simplifying assumptions, they're both lying on the horizon, with the moon 135 degrees away from the sun. Its fully illuminated side is facing the sun, terminator orthogonal to the horizon. We can draw a horizontal line on the sky, parallel to the horizon, connecting the sun and moon, and it, too, is orthogonal to the moon's terminator. This is the path light rays are travelling from sun to moon, and all seems consistent and normal.
Now go the equator. Find a point on the equator where the sun is sitting on the the western horizon. The gibbous moon is 45 degrees above the eastern horizon (ie, still 135 degrees from the sun). Its illuminated side is pointing towards the zenith: that is, it is facing upwards, despite the fact that the sun is on the horizon. The line we drew on the sky connecting the sun and moon now climbs vertically from the western horizon, crosses the zenith, and descends to the east until it hits the moon, still orthogonal to the moon's (now horizontal) terminator. After a bit of thought we can see that this is correct: the sun's rays are shining far over our heads to illuminate the moon, so we can see a little way "under" the illuminated face to the shadowed side.
Now, picture a line of observers strung out along the line of longitude that connects our equatorial observing point to the north pole. For each of them the sun is going to be on the horizon. For each of them, the moon's illuminated face is going to be pointing in some direction intermediate between straight up (the equatorial situation) and horizontally (the polar situation). So every single observer in those mid-latitudes will see the illuminated side of the moon tilted upwards to some degree, despite the fact that all of them see the sun on the western horizon. Wait a few moments for the sun to set, and all will have a view similar to the one described in the OP.
For these observers, the line we drew in the sky now appears to curve upwards from the sun's position, reach a high point due south, and then curve down to meet the moon orthogonal to its terminator.
But it's the same line. We thought it was straight when it was parallel to the horizon, and we convinced ourselves it was straight when it ran vertically. But our brain is unhappy with the convergence that is an inevitable part of seeing long parallel lines in a perspective that covers a big angular arc. A long line that converges on both horizons (as sun rays at sunset do), seems like it must have a curve in it somewhere. So in these intermediate latitudes, it seems like we ought to be able to draw a straighter line between moon and sun, undercutting the line in the sky we previously agreed to be straight!
But imagine (for the moment) you could do that. So draw a new line that undercuts our original line and looks straighter. Now head back to the pole, so that our original line is running along the horizon, connecting the sun and moon. Where is our new line? It must be leaving the sun, curving below the horizon and then up again to hit the moon! So it can't be straighter.

So it's all just a trick of perspective. If, at some intermediate latitude, you could do an experiment in which you held your head dead steady, and stretched a piece of string from your right thumb (covering the sun on the horizon) to your left thumb (covering the gibbous moon high in the sky), you'd find the string did seem to pass higher than your head and then descend towards your view of the moon, meeting it at right angles to its terminator.

Grant Hutchison

Thanks, NEOWatcher and grant hutchison.

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Thump

I haven't checked those figures, but regardless, they could not possibly account for the illusion--no one would ever be able to reliably discern that small difference.

And the diagrams mostly deal with the sun below the horizon, or at the horizon, which would make that exercise a little more difficult.

If the earth were transparent though, and you imagined a straight line from the sun (even below the horizon) to the moon, that line would be perpendicular to the terminator to the degree necessary for discussion of this problem. That's my objection to the diagram--the line between the sun and moon is not perpendicular in the diagram, but we know that it is.

I agree that it would be difficult to mentally draw that line between a nearly full moon and the sun--because we as observers would be between them--but it's not like we look at the full moon and infer that the sun is shining behind us no matter which way we turn.
Who are you gonna believe, Euclid or the diagrams?

Thank you Grant, I respect your knowledge and explanation but I'm still confused. I'm sure this apparent stupidity on my part is simply a lack of intelligence or understanding but so be it.

From your explanation I have concluded the following

1) that from the earth's north pole, the moon's terminator appears vertical and perpendicular to the sun
2) that from the earth's equator, the moon's terminator appears horizontal and perpendicular to the sun

but I'm having trouble with the part that says

3) that between the earth's equator and the earth's north pole the moon's terminator, rather than being perpendicular to the sun, should be perpendicular to a point higher than the sun.

Imagine I'm at the earth's north pole and the moon's terminator is vertical, the eastern (left) side of the moon is dark and the western (right) side of the moon is lit. If we freeze time and I walk southwards towards the moon, the sun will stay on the horizon and the moon will rise in the sky. When I reach the equator the moon will be high overhead in the eastern sky and the eastern side will be dark and the western side will be lit, and the terminator will appear horizontal. But if you analyse the transition of the terminator from vertical at the pole (dark side east, or left) to horizontal at the equator (dark side east) the terminator has simply rotated clockwise by 90 degrees. At no point does it rotate anti-clockwise. And what I am seeing out of my window implies an anti-clockwise rotation of the terminator.

clop

Darkside left (from the pole) to darkside bottom (at the equator) implies a 90 degree anticlockwise rotation, doesn't it?

However, my toy model was only intended to demonstrate the general principle that the tilted terminator is to be expected when viewing from midlatitudes.
To the toy model you have to add the fact that in the real world the moon may have a more northerly or southerly declination than the sun, depending on the season and the moon's orbital tilt. (Whereas in my model the declination of moon and sun are always equal and zero.)
So in the real world, even when observed from the pole, there may be a pre-existing tilt to the terminator, which would then interact with the effect I've described as you moved south.

Does that help?

Grant Hutchison

I think what is complicating this discussion is what has been mentioned-- the need to draw a 2D picture of a 3D situation. So both Jason Thompson's figure, and hhEb09'1 's objection to it, are correct! The figure is right, because it is the best you can do in 2D. hh's objection is right if the figure is interpreted as what your eye could see without turning your head, which would always have to look like a straight line that is perpendicular to the terminator (if you don't turn your head, the curvature of the Earth is irrelevant). The problem is, in clop's illusion, you do have to turn your head, so what Grant is explaining is how your perceptions get skewed when you turn your head and reorient yourself with respect to the horizon (i.e., your mind tells you that parallel to the horizon is a straight line, and it's not.)

Weather was pretty, sun about to set and moon rising high on the sky, so i went out to take some pictures. Sun was quite close to west and moon in south east, so couldnt quite fit them on same frame so i took 5 photos with tripod and stitched them together.
Its too big to attach here (185k jpg )so clicky here

Nice picture.. if I had a landscape monitor i'd make it my desktop background,
might be worth submitting it to APOD/EPOD

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