Generally flatearther gurus ask for left to right curvature because they know it's a globe and they know you can't fit the curvature in frame. But it is possible to pan across the curvature and capture it if a structure behind the horizon is projecting a great circle. You know, something like a line of transmission lines. Like this -> Source: https://www.youtube.com/watch?v=GypxnDT1oBQ

Wait till they ask you:"Ha! The buildings do not curve!" and knock yourself out... By the way amazing capture, your work and videos are just the best!

Big fan of your work, Soundly, and love what you and the guys are doing on youtube - but I'm wondering...is it actually the left to right curve? Or is the curve that we're seeing a result of distance, and therefore the 'head on curve'? The nearest pylons are 4 miles away, while the ones on the left are about 9.5 miles, and the ones on the right about 8 miles, if I've got that right. Therefore isn't their apparent height determined by distance from the camera?

Their apparent height would change with distance clearly. However, that's not the subject of the observation. Notice how the obstruction height of the bases of the pylons changes with distance. That's the bit that's predicted by the conventional model, not predicted for a flat plane geometry, and specifically denied to occur by flat-earthers. This observation in particular was made as a direct answer to such a claim. The claim was that the curvature witnessed in my other videos (head on shots) was due to an atmospheric effect that would not be visible from a 90 degree angle (if the earth were flat). From this altitude the horizon is practically flat and the transmission lines are projecting the surface curvature above that flat horizon. Because of this we see the curve revealed by the variable obstruction heights of the base of the pylons. (i.e. "clown frown" ™ Jesse Kozlowski ).

Right. Apparent height lower due to distance, and greater "hidden amount" also. Not sure what your camera height was, but at say 5 feet, the towers at 9.5 miles would have 24 feet less of the base showing than those at 4 miles. Also, the bulge height for the line of towers is 40.3 feet - which makes me think the curve of the towers is actually due to a combination of both the left-to-right and the head-on curves. I guess I need more of an explanation to get it...

It is definitely both but do not you think the curve is way to pronounce; caused by the distance from the camera? The left to right curve is very small. Head-on curve has to be the likely candidate as the towers at the edges are more far away.

How is that a left to right curve? I mean if the pylons are at equal distance there will be a "bulge", no hidden amount. Plus the towers at the edges are more far away than the middle ones so there should be a hidden amount, which again is only caused by their distance from the camera. You also said in post no.7:"Their apparent height would change with distance clearly. However, that's not the subject of the observation. Notice how the obstruction height of the bases of the pylons changes with distance." Obstruction height can only change due to the head-on curve which has nothing to do with left to right curve. If the subject was not the "apparent height" or "obstruction height" then you should have chosen the place where pylons were at equal distance. OR am I wrong? Explanation needed.

Thought experiment: let's say we're 10 feet above a lake looking at a 12.5-mile long line of 50-foot tall towers that run in an arc such that each tower is exactly 4 miles away. Hidden amount is zero, so we can see all the bases. If we take a video of the line and then compress the stills, what do we see? Probably they will all look the same: we've removed the hidden amount factor, but there's no left-to-right curve either, since we're basically standing in the centre of a semi-circle of towers. Next, take the outer towers and move them back, till they're 8 miles away. As they recede, the bases will become hidden, and the apparent height will grow lower. Moving all the towers until they're in a straight line will give us the image in the OP. Thinking about it like this helps me see that the 'head on curve' is playing a large factor, if not all of it - how to show the left-to-right curve while not having hidden amount or perspective interfere with the results seems to be the question. Maybe Walter Bislin's simulator could help? Or raising the altitude?

I have not found explicit details. It looks like the stills are fitted at the water horizon, which is closer than the nearest pylon, but are not normalised(scaled) for range

Ok, so I've been playing around with Walter Bislin's app and I think I've got the answer: Soundly's image does show a combination of the head-on curve and the left-to-right curve - as well as perspective - though it's mostly the head-on curve, and not so easy to differentiate between the left-to-right curve and perspective. If you click this you should get the settings I've used to simulate Soundly's experiment. There's a 15.5-mile straight line of 100-feet towers with the nearest one 4 miles from the viewer. Viewer height is set at 50 feet, to eradicate hidden amount being a factor. If you pan across the field of view (fourth down on the 'Views' tab) you can see what the towers would look like. Here's a snapshot of towers 1-6, plus 35, 37, and 39 (no need to do all 40): It looks like it does show a curve, and one that isn't being caused by hidden amount. But, to check, I had a look at what it would look like on a flat earth: Again, that's a bunch of the left-most towers, and then the ones closest to the centre. It looks basically the same as the round earth view - which seems to indicate the towers are shrinking because of perspective. I also had a look at raising the viewer height so it was level with the top of the towers. On the round earth we get this: While on the flat earth we get this: This illustrates that, in reality, more than perspective is occurring. In the flat earth view, the tower tops stay level, being as they're already on the same elevation as the camera and, therefore, the vanishing point, whereas in the round earth view the left-right curve is dropping the tower tops 'down': the apparent height of the towers is dictated by the left-right curve in addition to perspective. I think that's about right - no doubt others can clear up any errors I've made and fine tune it - but, in a nutshell, it seems that the image in the OP is mostly demonstrating the head-on curve, while perspective also reduces the towers' apparent height, and finally some left-to-right curve, as shown above. I do think it would be possible to demonstrate left-to-right curve using these towers, but that the camera would have to be level with the tops of the towers to do it best - and also that it would be difficult to use it as proof given the similarity of views between the two models (e.g., RE from the tops of the towers looks like FE from lower down; and FE from the tops of the towers looks like RE would from a little higher). Still, it is a cracking demonstration of the head-on curve, and maybe one of the best: it's basically like being able to play with all the variables without having to move, and having your target in all places at once.

so... if I'm looking out at the ocean and I have super vision, so I see this horizon line is that the 'head-on' curve or the 'left to right' curve?

Neither. That would be the curve of the horizon, like this guy's looking at here: Source: https://www.youtube.com/watch?v=XQ5Zp2fxTv8&t=1m27s The curve of the horizon is the one people were looking at detecting with levels and straight edges in this thread. The head-on curve would be the one from his feet to where the two lines intersect, and we can't really see that, only the effects of it, such as hiding distant landmarks, making boats disappear, etc (that's the one people usually talk about). The left-to-right curve, or the side-to-side curve, would be, for example, running across the top of the ball, perpendicular to the guy's line of sight.

Wait a minute, is not the tower 1 farther away than tower 6? Although you have taken the screenshot from a higher elevation so that there is no head-on curve but perspective will play with us as the farther tower will appear a little lower (perspective that is). The thing is, what EXACTLY is left to right curve? I thought it was what you said is called "curve of the horizon"

Yes. Spot on. Look at a ball. It's the curve that goes from left-to-right - or right-to-left, if preferred. Yes, I mentioned perspective, and the effect it has on the apparent heights. Mick's chess piece comparison is a really good practical demonstration of the difference between "left-to-right" and "head-on" curve: https://www.metabunk.org/posts/208430/ ADDENDUM: I need to do a little more work in the simulator. The apparent height of the towers are still being 'dropped down' due to head-on curve, even though they're not being hidden, so I haven't quite isolated left-right curve and how much of an effect it's having. Got an idea how to do this though, which I think should work. It may be that's it's not having any effect at all... The question is: to take Deirdre's diagram where, say, a is 4 miles away, and b and c are 8 miles away, is there any practical difference between b and c? After all, aren't they both being observed over the exact same curve?

Probably don't need to play with the simulator anymore: I think I was right with my thought experiment earlier. Basically, whether distant objects are lined up side-by-side or stretching into the distance seems immaterial - the curve between the viewer and each object is the same. In that sense, then, there is no left-to-right curve - or, to put it another way, the left-to-right curve and the head-on curve, relative to the viewer, are the same thing. Consider above: to see each tower the viewer will turn towards it and create a line of sight between himself and it; and the individual lines of sight aren't dependent in any way on the tower being in a line. Tower b at 8 miles away and tower c at 8 miles away will appear exactly the same. It still feels a bit weird - probably because I've been thinking for a long time that there was a left-to-right curve - but it's all simply dependent on distance, if I've got it right.

the point of what Soundly is showing is that the horizon line, when we look out at the ocean.. does in fact curve at the edges (like the spirit level experiments) even if our eyes alone cant see the subtleness of it when using the ocean. so it does matter if the towers are in a line because they are representing the 'horizon' flat earthers think they are looking at.. and they think is flat.

Here is a thought experiment. Back away from the globe far enough so that you can place the transmission lines on the the top most edge. They are projecting a great circle. The curve is left to right / right to left if you prefer.

Hey Soundly, I currently think your pictures of curved bridges and power lines are the greatest contribution yet to this whole flat earth debate. Thanks. If you do any more, I'd love to see the full context of the location where they are taken with full 360deg shots. I'm not sure why, but it makes it feel more real to me when I can see the full 360. I don't know why they didn't have their flat earth convention at Lake Pontchartrain instead of North Carolina. I guess that would have ruined it. Apparently their next one is in Birmingham UK. They should hold it near the Bedford Level.

For there to be a left to right curve the towers must be at an equal distance from the camera, right? I did see the videos of the transmission lines where Soundly demonstrated the head-on curve and they were beautiful, but I do not seem to get this. The towers should be at an equal distance from the camera otherwise it would be either the curve of the horizon or the head-on curve. Maybe I am late to the game.

No, if the towers are in a straight line then they cannot all be on the horizon. The horizon is always the same distance from your eye, so it forms a circle, not a straight line. The towers will be closer in the middle and further at the edges of your field of view. They will be going "up and over" the dome of Earth in front of the horizon. Imagine a slice taken off a ball. The circular edge of the slice is the horizon. The thin section of spherical surface is the ground you are looking over to the horizon. Draw a straight line near one edge, from, say, the 12 o'clock point in the horizon to the 2 o'clock point. The line goes "up and down" over a thin part of the spherical surface. Or, using the diagram from earlier in the thread, the line of towers is a straight line (on a sphere so it looks curved on this diagram), which is closer to the viewer in the middle and further away at each end. It is not the same as the horizon circle (black line). So there will be two effects: the towers will look taller (from base to tip) in the middle, as they are closer to the camera, but also the bases will be higher in the middle, due to the curvature of the Earth (NOT the curve of the horizon, because the horizon doesn't curve!) No – if they were equidistant from the camera then they would be following the horizon (for a given camera height), and horizons don't have any left-to-right curve.

Or, to put it even more simply, the towers at each end are further over the curve of the Earth (i.e. beyond the horizon) than the towers in the middle. It's really just another manifestation of the "front to back curve" over the horizon - the further towers have more "front to back curve". The horizon itself never curves; it's simply that the relative distances of the towers and the (flat) horizon line are different depending on your direction of view. On a Flat Earth that wouldn't happen: the only effect would be that the towers would look smaller from top to bottom at each end.

Would not that be the head-on curve? I mean middle one is nearer than corner ones. How is that a left to right curve?

It's really the same thing as far as I can see. On a sphere, the curve is the same in all directions. A straight line perpendicular to you will be lower at each end of your field of view than in the middle. You can either picture the end towers as being "further over the horizon", or the middle towers as being "on top of the bulge of the Earth. In both cases they are higher because they are nearer to you. The horizon doesn't curve because it is always the same distance away from you, so it's not a straight line, it's a circle. If the Earth was flat then you would only have the perspective effect of the towers getting shorter with distance; on the globe you have the additional effect of their bases getting lower (and more hidden behind the horizon) at each end. It's kind of hard to illustrate in two dimensions though.

These two videos provide some context of the area (Start at 5:20) Source: https://www.youtube.com/watch?v=69_4E4WAZPg&index=8&list=PLwMxm_D26MuWhld7hRvNpYptpAjlqjaG6 (Start at 6:20) Source: https://www.youtube.com/watch?v=u61oOOjwm7Q&index=24&list=PLwMxm_D26MuWhld7hRvNpYptpAjlqjaG6

I find this all fascinating, and I'm glad to see I'm not the only one pondering this concept. I guess the possibility of seeing this 'left-to-curve' is something that has always been assumed, on both sides, but just that it was too difficult to see. After all, isn't it the basis for the famous B.O.B. question? Mostly when I'd seen people address this they'd used equations based on a sagitta or bulge, and pointed out that it's too small to see (as well as the obvious fact of the terrain, etc) - but now we're questioning the notion of it. On the one hand, it's clear that the line of the towers in the image in the OP is curved because of distance and hidden amount. But what if we remove hidden amount? Ie, raise the viewer elevation high enough so that the towers are all nearer than the horizon, and are therefore only considering those within the blue scrawls: In one sense, the viewer is 'disembodied' - the eyes are floating in 3D space; the line of sight doesn't depend on the ground between him and the towers, and we could 'cut away' the earth so that all we had was a 3959-mile radius flat, vertical circle with a line of towers running across its rim. I believe this is what Soundly meant by "the towers project a great circle", and in that sense, what we'd be looking at would be the left-to-right curve. But, at the same time, the outer towers also appear 'smaller' because of perspective, while the eyes are 'raised' because of the head-on curve. I think I go back to my earlier idea: the reason the towers appear lower in the OP image is a combination of perspective, head-on curve, hidden amount, and left-to-right curve. If we raise the viewer's eye level to 50 feet, we eliminate hidden amount. If we raise the viewer's eye level to the height of the towers, and consider only the tops of the towers, we eliminate perspective, as the flat earth view shows all the tower tops would be at eye level. Now, to eliminate the head-on curve, I'm trying to imagine the earth as a 3959-mile radius cylinder. On the one hand, this feels like it works, being as it 'drops' the viewer height to the surface of the cylinder, so that when he's looking at the closest tower, at 4 miles, he's now 10.7 feet lower in 3D space than he was on the sphere. Or, in degrees, for a 100-foot high camera, looking at the base of a 100-foot high tower, he's looking down at 0.271°, whereas the sphere earth viewer is looking down at 0.3°. Similarly, when looking at the tower 8 miles from the closest tower, the angle on the cylinder is 0.173°, while on the sphere it's 0.186°. I'm not quite sure yet what the difference between the two angles signifies. Apologies: I know I'm fumbling here, and extrapolating way more from the OP than was ever intended. It's like an itch what needs scratching. A riddle and a puzzle. I feel I may be close, or I may be barking up completely the wrong tree. No doubt superior minds will clarify. PS Anybody know how tall the towers are? And how far above the water the camera was?

The line of towers is along a great circle. The horizon is along a small circle. When you look at these things, what you are seeing is a bit of that 3D circle, from a particular 3D position. When a line of towers goes behind the horizon you are seeing the intersection of the two circles.

I doubt it's measurably different. In Google Earth when you use the measuring tool (or the path tool) it's making great circle segments.

Been playing with some cylindrical geometry, since that makes sense to me as a way to isolate the left-right curve. It was a bit roughshod, but seems to work. Where v is the viewer; t1 and t2 are towers; d is the distance from the viewer to the first tower; h is the height of the viewer, and s is the position the viewer would be on a sphere. I calculated for v to t1 at 4 miles, and t1 to t2 at 8 miles, and came up with dip angles (from viewer at 100 feet to tower base) of 0.271° and 0.173° for the cylinder, and 0.3° and 0.186° for the sphere. I can't quite get my head around it. What does the 7-10% difference represent? That the curve illustrated by the towers (once perspective and hidden amount have been eliminated) is 90% due to head-on curve, or to left-right curve? Or something different? Completely mad but (I think) working spreadsheet attached.

I think it will be better if someone could draw some visuals showing how high and in what arrangement towers are (great circle segments as Mick mentioned) and show the circle of the horizon like all the processes (if it is easy to show) in one diagram, and try to explain that how is this a classic example of left to right circle rather than head-on curve (what I think). I think I am feeling like Rory; not thinking critically here. P.S. Google Earth is stuck at its loading; hardware acceleration maybe.

Wouldn't the circle of the horizon just confuse things? Because then you also have to work out the hidden amount, whereas calculating for towers that aren't hidden means just figuring out the two curves.

On a cylinder, the "circle" under your feet and the "circle" under the towers would be parallel to each other. On a sphere that is not the case. But I don't know if the difference at this scale would be enough to account for the variation.