The Moon Tilt & Terminator Illusions

[Mick West]

but in the video there was camera paning left and right.. you said that con/chem/trail looked straight.. I think if you paned non fisheye camera to the left and right of it it would still look straight.. just like when we see it.. is this effect due to fisheye lenses? or that effect when compositing panoramic images?

It IS straight. The curved effect here is because you can't render a 180 degree scene with straight lines.

You should try this:

If you imagine the moon being placed in the top left corner with the terminator perpendicular to that line then you can see that it would not point at the sun in the top right corner.

View attachment 22811

Sit on the floor about two feet from the middle of a wall (the longer the better), and look up at the corners of the room. See how the ceiling line goes UP from each corner when you look at that corner. Trace the line by pointing at it - you can see after a while how the line visually seems to go up and then curve down, just like these contrails do.

Go give it at go.

 

[Mick West]

Don't try to get the sun in the same frame. Not necessary or desirable. We're just trying to show that the sunlight here on earth is the same sunlight that shines on the moon. It's from a distant source with parallel rays.


The moon will be good from Saturday, December 3 to Saturday the 10th. (Up during the day and not too full or too thin.)

Like this:

http://www.skyandtelescope.com/astronomy-resources/what-are-the-phases-of-the-moon/

Dec 6, in the afternoon looks like a good time to duplicate that image in California:

 

[Mick West]

Actually you could do it today or tomorrow, in the mid-morning

 

[Mick West]

I've used a combination of the "room corners" and the flashlight with a ball to illustrate the tilt illusion from the OP:

Source: https://www.youtube.com/watch?v=AI4b_TAkcoM

 

[Mick West]

 

[Z.W. Wolf]

I'd like to get a larger image of the moon.

The YT video in the OP was uploaded on 4/26/16. There was a waning gibbous moon (84% illuminated) that day. The sun and moon positions match a local time of 6:30 a.m. DST in Ann Arbor, MI on that day. We just passed that phase of the moon 2-3 days ago - although the moon phase was about 12 hours offset from that in April.

The next day the moon will be in that phase will be December 17, which happily is a Saturday. The moon phase will also be less offset and a good match to April. I'll try to get some photos in similar conditions on that day -sun and moon altitude about the same. Here in Las Vegas 7:40 a.m. looks pretty good. Any one want to join in?


(The author of that YT video has edited his video, and from what he says it's seemingly in response to this thread. You might want to take a look at the new version.)

 

[Z.W. Wolf]

As Mick previously said, we've been mixing two issues here in this thread - and I was one of the prime culprits.

1. The Moon Tilt Illusion.

2. The Moon Phase Problem.

It's important to keep them separate and I apologize for not doing so. But photos with the sun-illuminated ball and the moon in the same frame will resolve both questions, especially if we catch the moon in different phases.

 

[Mick West]

There's an experiment we can all try that will demonstrate that sunlight and both the moon phase and terminator angle really do go together.

Put a ball on a stick, use a zoom lens, back off from the ball/stick, get the ball and the moon reasonably close together in the frame. I promise that the light and shadow on the ball will match the light and shadow on the moon.

Be sure to back away from the ball and zoom in. You want to get a large image of the moon, and keep both the ball and the moon in focus at the same time. You might have to get pretty low to the ground. A standard cell phone camera won't do.

Any ball will do, but some will probably be better than others because we are trying to catch a shadow in full daylight. I'm going to try a golf ball. I like the dimples and the white color.

Take multiple frames in different exposures. Trying to get the right one to get the moon well exposed and to catch the elusive shadow on the ball.

Don't try to get the sun in the same frame. Not necessary or desirable. We're just trying to show that the sunlight here on earth is the same sunlight that shines on the moon. It's from a distant source with parallel rays.


The moon will be good from Saturday, December 3 to Saturday the 10th. (Up during the day and not too full or too thin.)

My attempt:

Ball is on an 8 foot PVC tube. Right now the moon is so high that you can't get far enough away to get them both in focus with my 500mm. Hence the above is a composite image. A single shot focussed in-between looks like:

Here's the layout, The terminator with the ball goes though the middle of the image, so it lines up with the sun.

Here's a better single image, 100mm at f/32

 

[Mick West]

A very slight difference between the two terminator lines is apparent. Probably due to a combinations of factors the sun light is not exactly parallel at the moon and the ball, and I'm closer to the ball, so there's some perspective distortion.

 

[scientific601]

A very slight difference between the two terminator lines is apparent. Probably due to a combinations of factors the sun light is not exactly parallel at the moon and the ball, and I'm closer to the ball, so there's some perspective distortion.
View attachment 22886

How would one go about calculating the angle of the terminator line? In the illustration above for example, I would say it is at roughly 64 degrees. It's driving me nuts - should I project sun's and moon's azimuth and elevation (from other sources) to derive 3D spacial coordinates and find an angle between them? There should be shortcut though. I can't seem to wrap my head around 3-D coordinate systems vs. appearance from the observer.

Please point me in the right direction... I don't mind the math/geometry formulas but I'm having trouble getting started. I have searched all over and no one seems to have solved this problem. This forum's topic has come closest to the problem so far. I've been trying to figure this out for years. If solved I would make a terminator angle webpage.

 

[Apollonius]

The moon tilt illusion seems to be the same illusion that people are subject to when they assume that because Chicago and Beijing are at approximately the same latitude, the shortest route between them will follow a parallel of latitude. On the celestial sphere, the shortest path between two points is an arc of a great circle, so if the Moon is in the east and the Sun is in the west, the shortest line between them, i.e., the projection of a straight line between Sun and Moon on the sphere, will appear to go up closer to the zenith and then back down. The creator of the video in the OP mistakenly assumed that the straight-line path between Moon and Sun, both relatively low and separated by a wide angle, would be roughly parallel to the horizon rather than rising and then falling as a great-circle path would, and as the ecliptic and celestial Equator do.

 

[Ross Marsden]

A very slight difference between the two terminator lines is apparent. Probably due to a combinations of factors the sun light is not exactly parallel at the moon and the ball, and I'm closer to the ball, so there's some perspective distortion.
View attachment 22886

I see that I "Liked" this comment in Nov 2016, but now I think that the red line drawn on the ball on the pole is not joining the points of the terminator exactly. One of the points (the lower one) is obscured by the pole, so we can't use that. The line should be tangential to the terminator line (curve) at it's mid point, not some place above that, as illustrated. Just saying.

 

[Trailblazer]

I took a simple hand-held picture using a ping pong ball and an iPhone when I was on holiday last month.

 

[Astro]

How would one go about calculating the angle of the terminator line? In the illustration above for example, I would say it is at roughly 64 degrees. It's driving me nuts - should I project sun's and moon's azimuth and elevation (from other sources) to derive 3D spacial coordinates and find an angle between them? There should be shortcut though. I can't seem to wrap my head around 3-D coordinate systems vs. appearance from the observer.

Please point me in the right direction... I don't mind the math/geometry formulas but I'm having trouble getting started. I have searched all over and no one seems to have solved this problem. This forum's topic has come closest to the problem so far. I've been trying to figure this out for years. If solved I would make a terminator angle webpage.

Here's the math. I incorporated it into this spreadsheet here:
http://dropcanvas.com/ecs9l
I presented it in a video here, but without going into too much detail about the math used:


Take the geocentric equatorial coordinates of the moon and add .25 degrees to declination (approximate radius of the moon). This will give you the equatorial coordinates of both the center point of the moon and the north point of the moon on the equatorial grid. Then calculate the position angle of the moon's bright limb relative to the north equatorial point. This is given by the following equation:

X = arctan((cos(declination of the sun)*Sin(right ascension of the sun - right ascension of the moon))/(cos(declination of the moon)*sin(declination of the sun)-sin(declination of the moon)*cos(declination of the sun)*cos(right ascension of the sun - right ascension of the moon)))

Then to calculate for field rotation, convert the geocentric equatorial coordinates of both the center point and the north point to horizon coordinates. This is given by the following formulae. For azimuth, the formula is:
tan(az)=(-sin(Hour angle in degrees +delta)*cos(declination))/(sin(declination)*cos(latitude)-cosnope(declination)*sin(latitude)*cos(hour angle in degrees+delta))
where delta is:
tan(delta)=(p*cos(theta')*sin(hour angle in degrees))/((distance of moon in km/6378.14)*cos(declination)-p*cos(theta')*cos(hour angle in degrees))
For altitude the formula is:
sin(altitude)=(sin(declination)*sin(latitude))/(cos(declination)*cos(latitude)*cos(hour angle in degrees))
Now once you have converted both the center point and north point to altitude and azimuth, take the difference of each (altitude and azimuth) between the north and center points to find delta alt and delta az. From there, calculate the angle of the line from the center point to north point relative to the horizon:
arctan(delta alt/delta az)
Then add the position angle of the moon (X). In order to express the resulting orientation angle of the moon relative to the horizon as I have in my spreadsheet (as an angle expressed as degrees from vertical from 0 to 90 degrees) follow these directions:
angle1 = arctan(delta alt/delta az)
If angle1 > 360, then take angle-360 = angle2, otherwise angle1 = angle2.
If angle2 >90 and less than or equal to 180, then take (90-angle2)+90 = angle3, otherwise angle2 = angle3.
If angle2 = angle3 and angle2 > 180 and less than or equal to 270, then take the absolute value of (180-angle2) = angle4, otherwise angle3 = angle4.
Finally, if angle4 = angle2 and angle 2 > 270, then take (90-(angle2-180)+90) = angle5, otherwise angle4 = angle5. Angle5 is the final answer and the apparent orientation of the moon relative to the horizon expressed as the angle from 0 to +90 degrees from vertical.

Hope this helps.

 

[Astro]

Here's the math. I incorporated it into this spreadsheet here:
http://dropcanvas.com/ecs9l
I presented it in a video here, but without going into too much detail about the math used:


Take the geocentric equatorial coordinates of the moon and add .25 degrees to declination (approximate radius of the moon). This will give you the equatorial coordinates of both the center point of the moon and the north point of the moon on the equatorial grid. Then calculate the position angle of the moon's bright limb relative to the north equatorial point. This is given by the following equation:

X = arctan((cos(declination of the sun)*Sin(right ascension of the sun - right ascension of the moon))/(cos(declination of the moon)*sin(declination of the sun)-sin(declination of the moon)*cos(declination of the sun)*cos(right ascension of the sun - right ascension of the moon)))

Then to calculate for field rotation, convert the geocentric equatorial coordinates of both the center point and the north point to horizon coordinates. This is given by the following formulae. For azimuth, the formula is:
tan(az)=(-sin(Hour angle in degrees +delta)*cos(declination))/(sin(declination)*cos(latitude)-cosnope(declination)*sin(latitude)*cos(hour angle in degrees+delta))
where delta is:
tan(delta)=(p*cos(theta')*sin(hour angle in degrees))/((distance of moon in km/6378.14)*cos(declination)-p*cos(theta')*cos(hour angle in degrees))
For altitude the formula is:
sin(altitude)=(sin(declination)*sin(latitude))/(cos(declination)*cos(latitude)*cos(hour angle in degrees))
Now once you have converted both the center point and north point to altitude and azimuth, take the difference of each (altitude and azimuth) between the north and center points to find delta alt and delta az. From there, calculate the angle of the line from the center point to north point relative to the horizon:
arctan(delta alt/delta az)
Then add the position angle of the moon (X). In order to express the resulting orientation angle of the moon relative to the horizon as I have in my spreadsheet (as an angle expressed as degrees from vertical from 0 to 90 degrees) follow these directions:
angle1 = arctan(delta alt/delta az)
If angle1 > 360, then take angle-360 = angle2, otherwise angle1 = angle2.
If angle2 >90 and less than or equal to 180, then take (90-angle2)+90 = angle3, otherwise angle2 = angle3.
If angle2 = angle3 and angle2 > 180 and less than or equal to 270, then take the absolute value of (180-angle2) = angle4, otherwise angle3 = angle4.
Finally, if angle4 = angle2 and angle 2 > 270, then take (90-(angle2-180)+90) = angle5, otherwise angle4 = angle5. Angle5 is the final answer and the apparent orientation of the moon relative to the horizon expressed as the angle from 0 to +90 degrees from vertical.

Hope this helps.

Meh, actually scratch that above, mostly. It's really quite a specific "use case" for the moon relatively near the horizon. Here's a much easier general approach you can use in a variety of calculators and spreadsheets (just be mindful of working in degrees, not radians). For the moon, just plug in topocentric coordinates as needed (take the center of the moon and add the apparent radius of about .25 degrees to get the north equatorial point, and subtract to get the south equatorial point). For the moon and other solar system bodies you'll still need to add this position angle formula to your result:
X = arctan((cos(declination of the sun)*Sin(right ascension of the sun - right ascension of the moon))/(cos(declination of the moon)*sin(declination of the sun)-sin(declination of the moon)*cos(declination of the sun)*cos(right ascension of the sun - right ascension of the moon)))

Here's a spreadsheet that does everything described below:
http://dropcanvas.com/x35q1
Now to calculate the field rotation for any imaginary line in the sky, the first step is to convert the right ascension and declination coordinates of both ends of the line to altitude and azimuth:
Altitude = asin(sin(Dec)*sin(latitude)+cos(Dec)*cos(latitude)*cos(Hour angle in degrees))
Where Dec = declination
latitude = observer's latitude
Hour angle in degrees = (Local Sidereal time in hours - (Right ascension in degrees /15))*15 degrees/hr

In my spreadsheet also included some correction factors for atmospheric refraction, but it's not really important unless you're dealing with areas of the sky very close to the horizon. See cells E25 and F25 of my spreadsheet for those equations.

A' = acos((sin(dec)-sin(latitude)*sin(altitude))/(cos(latitude)*cos(altitude)))
If sin(hour angle in degrees)<0, then altitude = A', otherwise altitude = 360-A'.

Once you have converted both sets of coordinates to local altitude and azimuth, use the following to calculate the angle of the line:
delta(a) = altitude 1 - altitude 2
delta(az) = (azimuth 1 - azimuth 2)*cos((altitude 1 + altitude 2)/2)

Then finally, the angle of the line relative to the horizon = atan2(x = delta(a), y = delta(az))

 

[siqjohnny]

A very slight difference between the two terminator lines is apparent. Probably due to a combinations of factors the sun light is not exactly parallel at the moon and the ball, and I'm closer to the ball, so there's some perspective distortion.
View attachment 22886

A very slight difference between the two terminator lines is apparent. Probably due to a combinations of factors the sun light is not exactly parallel at the moon and the ball, and I'm closer to the ball, so there's some perspective distortion.
View attachment 22886

Nice capture, but it could be much better if the yellow ball were to be either in front or behind the Moon, not at either below or above, so that both would be equally perpendicular to the Sun. Also, take the picture during Sunset or Sunrise when the Sun is almost touching the horizon, as the atmosphere's refraction effects are stronger on/at the Earth's surface. The red line on the ball is not aligned perfectly with its t-line, I can see by slightly diverging my eyes to align the Moon's t-line with the ball's t-line, they appear to be perfectly parallel, except the ball's red line.

 

[Z.W. Wolf]

I took these in December of 2016 and never got around to posting...

I wanted to get some distance and use a telephoto to minimize perspective distortion so I hung a tether ball from a 20 foot high pole.




I got frustrated with the point and shoot snap shot camera I had at the time that doesn't allow you to set f-stop and shutter speed. I wasn't happy with any of the shots I got, so I just stopped.

 

[Z.W. Wolf]

I also used a six inch nerf ball, with a rough surface that would better simulate the Moon's surface and minimize specular reflections.

The ball in better focus

The Moon in better focus

Same frame, but cropped

I've always been puzzled as to why the ball and the Moon exhibit slightly different illumination. Not terminator slant. I'm talking about the area illuminated. The ball is in a slightly different phase with a larger proportion of the surface illuminated.

I toyed with the idea that it's because the Moon is significantly farther away and perhaps the illumination is different because of the geometry of the situation.

Looking at this again, I think it's because of atmospheric extinction. The luminance of the Moon's surface drops off near the terminator. I think atmospheric extinction has reduced the luminance of that dimmer surface to invisibility in the photo. Thus we don't see the true terminator in the photo.

 

[Z.W. Wolf]

I'd forgotten how much I dislike the quality of these photos. Everything is off. I developed a real dislike for that camera.

Here are some extras...
I was under the path for flights departing McCarran. The plane is partially hiding the Moon here.

This is a big balloon holding up a banner advertising something or other. The ad banner was below and not visible here.

The interesting part is the specular reflection of the Sun on the shiny, or glossy, surface of the balloon versus the diffuse reflection of the Sun on the rough, or matte, surface of the Moon.

At the time (2016), Flat Earthers were going on about the missing "hot spot" on the Moon. Their argument was that a sphere always shows a "hot spot" due to the inherent properties of that shape. Because there is no "hot spot" visible on the Moon's surface at night, the Moon cannot be a sphere illuminated by a single light source.

This puzzled me for a long time, before I learned how to think in the mental language of a Flat Earther. I finally figured out that they expected to see a specular reflection of the Sun on the Moon if it were truly a sphere. Because all spheres display specular reflections, you see.

They were confusing the properties of the shape with the properties of the surface. The Moon has a rough surface and we see a diffuse reflection of the sunlight.

 

[Trailblazer]

This puzzled me for a long time, before I learned how to think in the mental language of a Flat Earther. I finally figured out that they expected to see a specular reflection of the Sun on the Moon if it were truly a sphere. Because all spheres display specular reflections, you see.

They were confusing the properties of the shape with the properties of the surface. The Moon has a rough surface and we see a diffuse reflection of the sunlight.

Funnily enough there are also Flat Earth memes saying the exact opposite - that the moon can't be rock because "rocks don't reflect light". I'll leave it to the reader to spot the flaw in that argument.

 

[Z.W. Wolf]

I speak Flat Earthish; I'll take a shot at translating.

In Flat Earthish...

Reflect:
1. to produce a bright and recognizable image of the light source
2. to shine

In English that's a specular reflection. A diffuse reflection doesn't count as a reflection. So what is a diffuse reflection called in Flat Earthish?

In Flat Earthish there's no word for diffuse reflection. Which sounds like it could be the title for a Malcolm Gladwell type bestseller.

The idea of diffuse reflection doesn't exist. "Things can be 'lit up' ... I guess.... now that you mention it. But that's not a reflection. A reflection is like when your going down the street and you see a reflection of the Sun in a window."

System One thinking in action. Fast, automatic, frequent, emotional, stereotypic, unconscious.



A guess as to the point of this particular meme...

The rock in this picture is not reflecting the light source. The rock is "lit up" - but it's not shining. The Moon at night is silvery bright and shiny, which equals reflection, which is impossible for a rock to produce. If the Moon were a rock - and just lit up like a common rock - it wouldn't look so silvery and bright. It would look dull and drab.

This is reflecting light

This is not

Duh!!



Which one does this look like? The mirror or the rock? Duh.

I can't be sure, but this particular Flat Earther was probably trying to push the idea that the Moon is self-luminous; emitting light from within itself.




I've had experience trying to explain - at star parties and such - why the Moon looks so silvery at night, even though it's surface is Cocoa Puffs colored. Or maybe better, old asphalt.

Brightness and contrast.

 

[Eburacum]

I've seen the Moon Tilt illusion several times. The first time I had no idea how it worked, but it seemed obvious to me that (just about sunset) the gibbous Moon was quite clearly visible in the sky, and the terminator was pointing upwards, even though the Sun was much lower than the Moon and very near the horizon.

It wasn't until I became familiar with the concept of Great Circles that I realised what was going on. If you follow the light rays from the Sun along the great circle that goes through the Moon, you can see that the light from the Sun is striking the Moon at right angles.

A diagrammatic representation of the Great Circle effect - although the circle is tilted, the observer (you, or in this case, me) is at the centre of the circle. Of course, in this image I've exaggerated the size of the Moon and the Sun by about x 20.

 

[FatPhil]

I've seen the Moon Tilt illusion several times. The first time I had no idea how it worked, but it seemed obvious to me that (just about sunset) the gibbous Moon was quite clearly visible in the sky, and the terminator was pointing upwards, even though the Sun was much lower than the Moon and very near the horizon.

It wasn't until I became familiar with the concept of Great Circles that I realised what was going on. If you follow the light rays from the Sun along the great circle that goes through the Moon, you can see that the light from the Sun is striking the Moon at right angles.

A diagrammatic representation of the Great Circle effect - although the circle is tilted, the observer (you, or in this case, me) is at the centre of the circle. Of course, in this image I've exaggerated the size of the Moon and the Sun by about x 20.
View attachment 81621

I don't see any projection that turns curves into straight lines (east-to-south-to-west), and straight lines into curves (the inexplicable "great circle" connecting the moon and the sun) to be explanatory at all.

The explanation is mostly based around the fact that the sun is behind you. The vector pointing from the centre of the moon to the sun has a positive component towards you, as more than half of it is illuminated, and that, because of the huge distance to the sun, even a smallish component of the vector being towards you ends up with the sun being behind you. To see why behind you's important, put on a hat with a nice wide circular brim. Align that brim with the illumination of the moon - so in your example you'd be looking up with your head tilted left. Where's the back of your hat brim pointing? Much of it is pointing towards the ground, including the part out to the right side behind your right ear that corresponds to the vector from the moon to the sun. Which explains why the sun is below ground level - i.e. set. The explanation works best with a pointlike sun at infinity, but we're only fractions of a degree away from that ideal.

 

[Mendel]

It wasn't until I became familiar with the concept of Great Circles that I realised what was going on. If you follow the light rays from the Sun along the great circle that goes through the Moon, you can see that the light from the Sun is striking the Moon at right angles.

I used to watch an amateur youtuber (can't find his channel right now), an old fart, exposed to FE claims, work his way through them. He ended up putting a length of wood on a stepladder, aligned that perpendicular to the moon terminator, and then tracked it across the sky to end up at the sun.

The moon tilt paradox is only paradoxical when your reference is the horizon, but the horizon isn't really a straight line, it's a circle wrapped around you, and the direct connection between two points in the sky typically doesn't follow the horizon.

If you don't try to understand this phenomenon on a 2D computer screen, but go outside and track it in 3D across the sky, it's easily verifiable.

I think you could demonstrate it in the lab by suspending a "moon" and a "sun" from the ceiling above, and then vary the camera position to make the moon look lit "from above", but I expect it'd be difficult for most people to imagine in 3D without trying it out.

 

[Trailblazer]

I think you could demonstrate it in the lab by suspending a "moon" and a "sun" from the ceiling above, and then vary the camera position to make the moon look lit "from above", but I expect it'd be difficult for most people to imagine in 3D without trying it out.

I've noticed a similar illusion in my bedroom, but I haven't tried taking photos of it.

Above my bed there is a ceiling light hanging from a length of cable. In the corner of the room, to the right of my head if I am lying looking up at the ceiling, is a bedside light.

If the bedside light is on and the ceiling light is off, it casts a shadow of the ceiling light and its cable. But if I follow the line of that shadow back across the ceiling, it doesn't appear to line up at all with where the bedside light is - it seems to pass way beyond/above the light. That's because the light is not in the same plane as the shadow.

I would take some photos but my ceiling currently has a large patch in it thanks to a slightly misplaced foot in the loft...

 

[Eburacum]

...put on a hat with a nice wide circular brim. Align that brim with the illumination of the moon - so in your example you'd be looking up with your head tilted left. Where's the back of your hat brim pointing? Much of it is pointing towards the ground, including the part out to the right side behind your right ear that corresponds to the vector from the moon to the sun. Which explains why the sun is below ground level - i.e. set. The explanation works best with a pointlike sun at infinity, but we're only fractions of a degree away from that ideal.

That's right. In this case, the 'wide circular brim' is effectively a 'Great Circle'.

A Great Circle is basically a circle which cuts the sphere of the heavens in half, and becomes an imaginary diameter - but to be honest, your explanation is easier to understand, and doesn't need visualisation of the solid geometry involved.

 

[Z.W. Wolf]

To tell the truth, I had a hard time even understanding what the point of this Moon Tilt & Terminator Illusion thing was and I concentrated on the phase problem instead, because that was what I thought the original post was about. Because I don't see this terminator tilt illusion at all. Not in the sky, not in photos, or with objects/light sources inside the room.

I've tried to see this so called terminator tilt illusion... but I just can't. The Moon looks just fine to me in whatever part of the sky it is. Ditto, stuff inside the room. I just don't see what you guys are banging on about.

 

[Trailblazer]

To tell the truth, I had a hard time even understanding what the point of this Moon Tilt & Terminator Illusion thing was and I concentrated on the phase problem instead, because that was what I thought the original post was about. Because I don't see this terminator tilt illusion at all. Not in the sky or with objects/light sources inside the room.

I've tried to see this so called terminator tilt illusion... but I just can't. The Moon looks just fine to me in whatever part of the sky it is. Ditto, stuff inside the room. I just don't see what you guys are banging on about.

It's easy enough to see the "illusion". If you see a crescent moon and mentally join the two points with a straight line to give you the angle of the terminator, then project a line perpendicular to that, it doesn't appear to point at the sun.

Post 18 in this thread and the subsequent replies make it pretty clear.

 

[Z.W. Wolf]

Do you all see a mispatch between the terminator angle and the Sun in this photo? I don't. It looks just fine.

It's easy enough to see the "illusion". If you see a crescent moon and mentally join the two points with a straight line to give you the angle of the terminator, then project a line perpendicular to that, it doesn't appear to point at the sun.

Post 18 in this thread and the subsequent replies make it pretty clear.

Not for me. I've tried. Not everyone sees the "Moon Illusion" - the moon appears significantly larger when near the horizon. Maybe ditto here. Maybe everyone doesn't see this so called terminator tilt illusion.

I've been emphasizing for years that the brain creates everything we see (and hear and think and feel). It's a personal creation. Maybe there's variance with how people see this situation.

 

[FatPhil]

It's easy enough to see the "illusion". If you see a crescent moon and mentally join the two points with a straight line to give you the angle of the terminator, then project a line perpendicular to that, it doesn't appear to point at the sun.

Post 18 in this thread and the subsequent replies make it pretty clear.

However, whilst the terminator may define a line in 3D space, "perpendicular to it" only defines a plane that points in all directions away from it. It's selecting which one of those vectors is the appropriate one to be paying attention to that seems to be the problem.

In apparent contradiciton to what Mendel posted above, I think we're *less* likely to fall for it being above the surface of a globe with a conical horizon than if we were on a flat earth (the region of space where the sun set is larger in the flat earth case, and is a strict superset of the globe earth cases, so there are more times when the sun can be down but look as if it is up). Do people living in Quito or La Paz fall for this illusion less, and people from Baku see it more? (Better: do people who do fall for this illusion, such as me, see it less when they're in La Paz and more when in Baku. If anyone's willing to fund the science I'll do the experiment!)

 

[FatPhil]

Do you all see a mispatch between the terminator angle and the Sun in this photo? I don't. It looks just fine.

It's not gibbous, so it's not the case we're talking about.

 

[Ann K]

Quite simply, there is never a time when an observer on earth can see both the sun and the moon as if seeing them both at the same distance (you'd have to be standing way out in space to do that). Two-dimensional diagrams can't explain it — and three-dimensional geometry baffles a lot of people.

(Note the em dash. I plan to use it a lot more, metaphorically shaking my fist and saying "damn you, AI, you can have my punctuation when you pry it from my cold, dead hands!")

 

[Trailblazer]

Not for me. I've tried. Not everyone sees the "Moon Illusion" - the moon appears significantly larger when near the horizon. Maybe ditto here. Maybe everyone doesn't see this so called terminator tilt illusion.

So you don't see a mismatch here? Source

 

[Eburacum]

It doesn't happen all the time, and is strongest when the Moon and Sun are on opposite sides of the sky (but you can't see it at full moon, for obvious reasons). So it is strongest when the moon is half to gibbous.

 

[FatPhil]

So you don't see a mismatch here? Source

View attachment 81655

What FoV and projection is that?
I can take some weird-arse photos with my fish-eye lens to demonstrate the effect, and that would prove *absolutely nothing*.

 

[Eburacum]

Here's the 'great Circle' path of the sunlight, superimposed onto Trailblazer's image. This corresponds to the brim of FatPhil's hat.

 

[Eburacum]

What FoV and projection is that?

The field of view looks to be somewhere near 180 degrees, perhaps a bit less. Of course, the Great Circle crosses the horizon at points exactly 180 degrees distant from one another.

The great circle followed by the sunlight is roughly equivalent to the Ecliptic, or to the Zodiac, except that the Moon's orbit is not exactly aligned with the ecliptic, so there is not a perfect alignment (nevertheless, they are all great circles centred upon the observer).

 

[FatPhil]

The field of view looks to be somewhere near 180 degrees, perhaps a bit less.

If so, then @Trailblazer should have told @Z.W.Wolf to bend his monitor around his head before viewing, in order to match the projection.

 

[FatPhil]

Here's the 'great Circle' path of the sunlight, superimposed onto Trailblazer's image. This corresponds to the brim of FatPhil's hat.
View attachment 81656

There is no "great circle" in that image. You've just drawn a curve that fits what you want it to fit, which proves absolutely nothing. Define your axis, your plane, and your radius, and then we can talk orthodromes - until then, you're adding more noise than you are signal.

 

[Eburacum]

There is no "great circle" in that image. You've just drawn a curve that fits what you want it to fit, which proves absolutely nothing. Define your axis, your plane, and your radius, and then we can talk orthodromes - until then, you're adding more noise than you are signal.

I realise this. I hoped my diagram might help to show how great circles are involved in this phenomenon, but in order to do so, I've had to manually add a curve that approximates to a great circle. So it is not accurate (although I'd guess it is within a few tens of degrees of showing the correct path).

However, I expect you fully understand how the light follows a great circle in this situation; your 'hat' explanation demonstrates this quite nicely. The brim of your hat is effectively a great circle, which you can tilt at any angle.

Astronomy is absolutely full of great circles; the ecliptic, the celestial equator, the plane of the galaxy, and the line that light follows across the sky; any orbit that is centred upon the location of the observer is also a great circle, but this isn't strictly true for the Moon, since its orbit is centred upon the Earth/Moon barycentre, so it is generally a bit south of observers in the Northern Hemisphere. But this deviation is relatively minor. Orbiting commercial satellites deviate quite strongly from a great circle, because they are so close by.