The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Theory => Topic started by: SiDawg on June 16, 2018, 08:34:58 AM
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The main EA debate is turning in to a bit of a doozey, hopefully this is OK to split in to a separate debate in the interests of debating one point at a time.
So, to my mind, given that it's accepted that "light rays travel away from the sun in all directions", then the sun would disappear top first in the EA model. If light rays are affected by an upward pull, they can be thought of as trajectories. They can be graphed and modelled as trajectories. For example, here's a plot of multiple rays with multiple trajectories, inline with how EA affects light rays. I've only chosen rays emanating from the centre of the sun in this example
(https://i.imgur.com/pi3iKYE.png)
Obviously, the rays of light will terminate when they hit the ground... but that's not the point of this image: i'm just showing that I'm using trajectory formulas, i.e. i'm not fabricating curves to suit my argument, i'm using the same exact formula just with different angles of light.
Now consider below: different angles of light, and different starting points of light, namely the top and the bottom of the sun.
(https://i.imgur.com/CruoTjA.png)
I put to EA proponents, that the point at which light rays from the top of the sun can no longer reach an observer, there will still be rays of light from the bottom of the sun that can reach the observer. For example: blue rays are going over the observers head, green rays are still reaching the observer.
I realise of course that this is diagramatic only: there would be other "blue" rays that would reach the observer... however the point is that as the observer gets further from the sun, the light rays from the top of the sun cease to reach the observer before the light rays from the bottom of the sun. Ergo, the top disappears first.
It's like two archers: one on the ground, one at the top of a building. They can shoot arrows at any angle they like, but they both fire arrows at exactly the same speed. Everything else being equal, then the archer at the top of the building will reach further than the archer on the ground. How could the same not be true of the sun? The difference is EA is an upwards force (the archers have gravity which is a downwards force), so the situation is reversed: the trajectories are upside down, and the "archer" at the bottom of the sun can always reach further than the top of the sun...
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Despite past clarifications, you continue to misunderstand the cause of sunset on this model. You stop seeing the sun because the earth obscures it (green rays), not because the blue rays bend away from the observer - the latter are almost entirely irrelevant. In your diagram, the sun will continue moving back until no rays of light can reach the person. As you've illustrated, this will first affect the far end of the sun, which will appear to be the bottom.
In fact, the very fact that you focused on the sun's centre is the problem here - you're treating it as a point, which doesn't make much sense when applied to non-point objects. Instead, try to illustrate the rays which reach an observer from multiple points of the sun, without any redundant rays obscuring them.
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I don't agree you have clarified why "sunset happens because light is obscured by the earth". Imagine you've duct taped yourself to a ceiling, and an archer is trying to hit you. If you're far enough away, he will no longer be able to hit you. He WILL be able to hit the ceiling in front of you, at any number of points on the ceiling. But to say "the archer is missing you because he is hitting the ceiling" is not as true as "the archer is missing you because he can not reach you". Regardless, we can accept your description if you like: the point at which the light rays from the top of the sun are obscured by the earth occurs earlier than the light rays from the bottom of the earth.
I haven't focused on the suns centre, I've focused on two points: the top and bottom of the sun. The first image is purely from the centre to describe the trajectories used for the second image.
I understand that light rays will emanate from multiple points in the sun and multiple angles, but for the sake of argument, we can choose to focus on any two points. Here i'm focusing on the top and bottom point. When trying to answer the question "which edge of the sun will disappear first" when dealing with a vertically orientated force, does it not make sense to focus on the the top and bottom points? From those top and bottom points, i've drawn a few angles of light. In retrospect it would be more helpful if i drew a shape which represented the total area that light was able to reach from any point (i.e. as described by drawing a line from EVERY possible angle from a point). I'll try to do that later.
You may have a point (excuse the pun!) that the two points i've used are bad: the bottom point is also clearly further away, so that will ensure the light rays stop reaching the observer sooner. Instead, assuming you agree the sun is a sphere, we know that less than half the sphere will be visible to the observer. The top point of that visible area is easy to define, as described below. We could say that the "bottom" point is the point directly 90 degree below the top point. This would be more analogous to the archers in the tower. I'll try to draw some more tomorrow.
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I don't agree you have clarified why "sunset happens because light is obscured by the earth". Imagine you've duct taped yourself to a ceiling, and an archer is trying to hit you. If you're far enough away, he will no longer be able to hit you
I'm not sure why you think these situations are analogous. I'm quite convinced they aren't similar at all. Yes, sure, some light will be affected in this way, but that won't happen around sunset.
I haven't focused on the suns centre, I've focused on two points: the top and bottom of the sun. The first image is purely from the centre to describe the trajectories used for the second image.
Fine, I see what you're doing there.
You may have a point (excuse the pun!) that the two points i've used are bad: the bottom point is also clearly further away, so that will ensure the light rays stop reaching the observer sooner.
Actually, I'd say your two points are bad for very different reasons. The bottom point is not as far away as it should have been. At sunset, a significant portion of the light would have made a turn close to 90 degrees. The "farthest" point is the bottom. This is also why the bottom will disappear first. That said, even without fixing your diagram, it can be clearly seen that the green rays (the bottom of the Sun) are obscured by the Earth much sooner than the blue rays.
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So, to my mind, given that it's accepted that "light rays travel away from the sun in all directions", then the sun would disappear top first in the EA model. If light rays are affected by an upward pull, they can be thought of as trajectories. They can be graphed and modelled as trajectories. For example, here's a plot of multiple rays with multiple trajectories, inline with how EA affects light rays. I've only chosen rays emanating from the centre of the sun in this example
(https://i.imgur.com/pi3iKYE.png)
I don't think this is right. The path of the rays with greater verticality wouldn't make sharp u-turns like that were it not for the earth blocking them. They'd continued in more shallow arc before reaching horizontal and then continuing to curve upward.
Not sure if that matters. I haven't read the rest of your post yet, but that first diagram didn't depict what I think EA is postulating about the path of light.
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I don't think this is right. The path of the rays with greater verticality wouldn't make sharp u-turns like that were it not for the earth blocking them. They'd continued in more shallow arc before reaching horizontal and then continuing to curve upward.
Not sure if that matters. I haven't read the rest of your post yet, but that first diagram didn't depict what I think EA is postulating about the path of light.
I was going to point this out as well. Every light ray, being affected the same way by EA, should have exactly the same shape of curve. It will just be in a different position.
Since your entire argument is based on your diagrams, which for whatever reason treat different light rays in different ways based on who knows what criteria, I don't think there's a case here.
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If EA is a force pulling upwards, is it not a trajectory like any other trajectory? It's just like firing an arrow, but upside down? I don't understand what the alternative would be?
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OK so here's a clearer picture of what i'm thinking. If we plot a large number of light rays from a single point, then we can describe an "area" where that point is visible.
(https://i.imgur.com/lO36bM2.png)
So here you can see i've chosen a point at the top of the sun: the observer is outside of the area of visible light, therefore the sun is not visible. As Pete describes it "the light rays have been obscured by the earth", or as I put it "no light rays are reaching the observer".
Now if we imagine the light rays from a point directly below, then we can visualise two areas: in green shows the area where the "top" point is visible, in orange shows the area where the "bottom" point is visible. You can see the problem: Imagine the sun moving away from the observer, and it's clear that the top would disappear first.
(https://i.imgur.com/J1BXrF9.png)
PS please ignore the more "vertical" looking lines: i just couldn't be bothered with the extra complication of getting the Excel formulas to stop plotting if the light hit the ground. They don't affect the story i'm telling. Also I realise there's some conjecture whether the rays can be plotted as a trajectory with a constant velocity (light) and a constant upwards force (EA). Perhaps EA isn't a constant force? Perhaps it behaves like gravity in that the force is weaker for more distant light rays? I'm not sure that's really relevant: even with that alternate method of plotting paths, you still end up with a shape similiar to above: at some point, rays of light stop hitting the earth, and start to be pulled up away from the observer. That "shape", whatever exact shape it is, will have a bottom point. That bottom point will be higher when described from the top of the sun, then from the bottom of the sun, because well, "the top of things is higher than the bottom of things" ? If EA effected light with less force at a greater distance, then the bottom of the bottom shape would actually be LOWER then the top, so would just further exaggerate the issue. Perhaps EA pulls stronger on light rays at a bigger distance away? That would make it even more magical than it already is... I don't believe that's even mathematically possible?
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If EA is a force pulling upwards, is it not a trajectory like any other trajectory? It's just like firing an arrow, but upside down? I don't understand what the alternative would be?
The way I understand it is if there is any x-axis component, there is an acceleration in that axis as y-decelerates, as would not be the case in a ballistics trajectory. The light rays should never cross if you were to constructively remove the obstruction of earth. The rays with more verticality would bow out further than the ones already with a horizontal component.
Basically, I think the point is to explain the appearance of terrestrial curvature and other phenomenon that might suggest earth curvature by explaining it with a formula for light behavior that would account for the degree of earth curvature we think we'd see. Your diagram wouldn't produce a round earth curvature appearance like the one we see.
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Hmm I've only heard of an upward pull, not an outward one. It would have to either be a pull from a disc surrounding the sun, or a force pushing outward i.e. accelerating light? What do you think happens to rays of light going directly down if they weren't to hit the ground?
I think my understanding does result in the image of a curved earth ( not that I've drawn it).. the rays of light from the ground will curve up and make the earth appear lower. But that's a different argument
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Aside from the shape of the curve, you're still wrong about where the top and bottom of the sun are.
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If EA is a force pulling upwards, is it not a trajectory like any other trajectory? It's just like firing an arrow, but upside down? I don't understand what the alternative would be?
Have you considered learning about the ways in which light is different from projectiles? That might clear up your confusion, and is not at all specific to EAT.
Even if we did treat these paths as "trajectories", which is not strictly accurate, there is one important property of light which makes your trajectory-plotting grossly inaccurate. Can you work out what it is?
OK so here's a clearer picture of what i'm thinking. If we plot a large number of light rays from a single point, then we can describe an "area" where that point is visible.
(https://i.imgur.com/lO36bM2.png)
Ignoring being told why you're wrong and posting another diagram with exactly the same problems is not going to make any difference to the validity of your "argument".
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Well yes one of the ways light is different from a projectile is it doesn't have any mass, but i figure it's easier to prove EA wrong by using the details of the theory against itself.
The thing is, no one has told me why i'm wrong in a way i can understand. When someone says they don't understand something, i understand it might be frustrating, but i'm asking you to please explain it in a different way. Just like it appears to me that i haven't explained why EA is wrong in a way you can understand. Hey one of us is wrong: could be me? Could be you? Perhaps EA has to be proven wrong in some other way? Perhaps EA isn't wrong at all? With respect, I'm not going to try to google why i'm wrong just because you think i'm wrong... Where would i start? I've put a lot of thought in to whether or not my hypothesis is wrong before i made the post. I've also spent a lot of time drawing graphs to bring some credibility to my argument. I haven't seen a single drawing about EA other than the original one from Tom, which shows exactly the same thing that I've drawn myself [Edit: actually that's not true, there were some diagrams showing the suns apparent position due to light entering the eye at an angle, consistent with the original drawing]. I've also spent a lot of time getting clarification on EA, such that light rays do emanate in a multitude of directions from multiple points on he sun.
I'm going to a lot of time and effort to point out why i think EA is wrong, and when someone seems to not understand what i'm posting, then i put more time in to further explaining myself. I'm not repeating my posts, I'm giving more information. And Pete's right: the original diagram had a clear problem, in that it seemed to show that the bottom WOULD disappear first... I was expecting the viewer to "read between the lines" but it obviously left some ambiguity and conflicting information, so the second post makes it blatantly clear by showing clear areas which the light from the sun would be visible, choosing a different "bottom" point: in the second post it's more of a "lower" point: there would be a point in the sun even lower than that. But the point is, the top would disappear before that lower point in the diagram.
If the replies saying i'm wrong made any sense, I'm sure the rest of the community would be able to point out to me why i'm wrong, perhaps they will in the coming days? Just like some RE'ers don't agree that light acts like a trajectory in EA, even though the EA diagram shows it like that, Pete explains it as an upwards force, EA proponents talk of some light rays continuing to climb in to the sky without hitting the earth... Everything I'm posting, I'm trying to include all the information I have at hand. Everyone's welcome to disagree with me, i'm more than happy to be wrong, but i'm far from an idiot: if i don't understand your response, perhaps i've miss read, or perhaps it just wasn't a clear response? I also understand not everyone has the time to try to prove someone wrong when they are assuming they're right (whats in it for them?), but if you have the time, i would appreciate it if you could please try.
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Pete explains it as an upwards force
If at any point I described it as a force - my apologies. That said, I can't find the post you're referring to.
I'm not repeating my posts, I'm giving more information.
The problem is that you also ignore information you've been provided. Parsifal's post aside, I've informed you twice now that you're misrepresenting the positions of "top" and "bottom". What you did was adjust your diagram to make it even more wrong, while not addressing my point at all.
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Here's a hint: The reason why projectiles fired at a higher angle have a sharper curve where they reach maximum height is because they are moving more slowly. More of their initial velocity was in the vertical dimension, so gravity slows them down more than those fired at a lower angle.
Now, can this apply to light? Does it make sense for EA to slow light rays down?
I eagerly await your thoughts.
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OK sorry, you're right you didn't use the word force... but you explained that EA pulled light upwards. It's my understanding that when something effects something else, or specifically when something disturbs something that would otherwise be travelling in a constant velocity at a constant speed, then it's called a force. But we don't have to call it that if you don't want.
I've informed you twice now that you're misrepresenting the positions of "top" and "bottom"
Saying the same thing twice does not make me understand it any better. I have given you two points: call them whatever you like. One is above the other yes? There is a "higher" point and a "lower" point... the higher point would disappear first. Regardless of exact definitions of exact points and analysing whether the points I've used are "correct" or not or match what you think they should be, my point is one is higher than the other, and the higher one would disappear first. Or are you saying the sun isn't a sphere? I've heard some say it's a disc....
Now, can this apply to light? Does it make sense for EA to slow light rays down?
You tell me? Again, just trying to make sense of EA from the info I've been given. Perhaps I've misunderstood that. Thank you for explaining. So do you agree with this Pete? Are we saying light stays at a constant velocity, but can also be "pulled" away from straight? I'm not entirely sure that makes mathematical sense for something to be present pulling something upwards, yet is incapable of slowing it down, but i'll put some thought in to it. It would really help if one of you could start drawing some diagrams.
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You tell me? Again, just trying to make sense of EA from the info I've been given. Perhaps I've misunderstood that. Thank you for explaining. So do you agree with this Pete? Are we saying light stays at a constant velocity, but can also be "pulled" away from straight? I'm not entirely sure that makes mathematical sense for something to be present pulling something upwards, yet is incapable of slowing it down, but i'll put some thought in to it. It would really help if one of you could start drawing some diagrams.
I'll try.
A projectile will slow or speed up only in the y-axis due to G. It's overall velocity (hypotenuse of x+y) changes because y is changing but x isn't (well, it's slowing due to friction).
Light, on the other hand is constant C. So though y changes due to EA, x must change inversely in order to preserve C. So as velocity of light in the y-axis slows to 0 at the point where it is...er, I mean...the flat earth's surface is tangent to the direction of light, the velocity in the x-axis will have increased to C, and then it reverses, with it increasing in the y-axis and decreasing in x, until, I presume, it is perpendicular to earth (assuming there is enough space).
So those light rays that would continue were it not for the obstruction of earth, they don't return to vertical as sharply as you have depicted them. They'd extend further out beyond those that already had a greater horizontal component.
If that's not right, then I'm totally lost too.
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OK sorry, you're right you didn't use the word force...
I wouldn't shy from using the term "force." The concept is an "accelerator."
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Now, can this apply to light? Does it make sense for EA to slow light rays down?
You tell me?
Well, the speed of light is easily measurable and has never been observed to vary significantly. Not only is it measurable, it is calculable based on the nature of light. Again, this has absolutely nothing to do with EA, and EA cannot violate established laws of nature.
Are we saying light stays at a constant velocity, but can also be "pulled" away from straight?
That cannot be true because velocity is a vector quantity. What you mean is that light travels at a constant speed. You may have heard of the speed of light before, it's a well known constant.
This is high school physics. If you don't grasp these fundamentals then you are going to have a difficult time understanding EAT.
I'm not entirely sure that makes mathematical sense for something to be present pulling something upwards, yet is incapable of slowing it down, but i'll put some thought in to it.
EA operates in an upward direction, but the light is not pulled directly upwards. The acceleration is always perpendicular to the direction of a light ray.
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Saying the same thing twice does not make me understand it any better.
Right, but just complaining that you don't understand doesn't advance us, either. You need to tell me what you don't understand and/or disagree with. In the meantime, I'll try saying the same thing for a third time, just to see if it sticks this time.
You already drew a diagram which is OK for this purpose, even if not otherwise accurate. Have a look at it and look at the angle the light, and conversely the image, has been rotated by, around the time/place where sunset can be seen. It's fairly close to 90 degrees.
If you agree that the image has rotated, then it naturally follows that... well, the image is rotated. I really don't know how to describe this differently This means that the point you're describing as the "top" is actually the back of the rotated image (and consequently it's obstructed by the rest of the Sun). In your side view, the sun would appear to the observer to be rotated like so:
(https://i.imgur.com/ZZfZ0iG.png)
The orange and green points are remarkably irrelevant when sunset is concerned. If there's something about this you don't understand, or with which you disagree, you have to actually say it. If you're just gonna sit here complaining, I doubt any of us will be particularly charitable.
Or are you saying the sun isn't a sphere? I've heard some say it's a disc...
For the very same reason (the ~90-degree rotation), it doesn't matter in this case whether or not the Sun is a sphere.
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This means that the point you're describing as the "top" is actually the back of the rotated image (and consequently it's obstructed by the rest of the Sun).
I think i see what you mean, and you're rotating my sun to match what the observer sees: i'm saying leave the sun and my points where they are: I accept that the observer views that sun at an angle. We do not disagree with each other. I'm picturing that "roughly" at sunset set the person would be seeing an image of the sun upwards at 45 degrees or so, i.e. tracing the light rays back from the observer to the sun. As you can see in my diagram, the angles of those rays are roughly 45 degrees: it's diagrammatic. So the observer sees an image of the sun, with a "top" and a "bottom" yeah?... If the top is 0 degrees, and the bottom is 180 degrees (for arguments sake), my points are at 0 degrees (green) and 90 degrees (orange). The lower point is not the bottom of the sun, it's just lower than the top point, and disappears after the top point. Nothing you've said seems to address the obvious issue of light towards the top of the sun failing to reach the observer before the lower point of the sun. All you seem to be doing is picking unrelated holes in how I've described the problem, without addressing the problem.
This is high school physics. If you don't grasp these fundamentals then you are going to have a difficult time understanding EAT
lol seriously? Flat earth believes disregard so much of high school physics, how am i meant to know what bits you accept or not? But yes i misspoke and used velocity instead of speed. I realise velocity is a vector. Please don't insult me, it makes me want to do the same and i'm trying reeeeally hard not to.
But yeah i get you're point, light rays are pulled in a perpendicular direction and maintaining the same speed. That still seems a bit "odd" to me but i accept it, and I don't believe it effects what i'm saying in the slightest: the paths of light that i assumed were bending up sharply hit the ground anyway... they're irrelevant to my point just because they hit the ground at a slightly different angle. Roughly speaking: light rays are bending upwards. My point still holds as far as i can tell.
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i'm saying leave the sun and my points where they are: I accept that the observer views that sun at an angle.
That makes no sense. The figure you're looking for is (very close to) 90 degrees, not 45, and if we leave your points where they are, then you're proposing that the back of the Sun would disappear before the front. You can't even see the back of the Sun because it's obscured by the rest of the Sun.
The reason I'm attacking your understanding of the problem is that if you manage to finally draw a correct diagram, you'll realise that there is no problem. I can't address something that doesn't exist, especially when all you have done to make your case is present diagrams with glaring holes. Currently, your argument is "if light were a projectile, a person with x-ray vision could see the back of the Sun fade away sooner than the front of the Sun". It shouldn't be shocking that we're trying to get you to say something that makes at least some sense prior to addressing it.
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It's hardly "at the back" of the sun... unless you're starting to propose again that all light from the sun starts off by travelling downwards from the centre like the original EA diagram from Tom? I thought we had established that you agree that light from the sun emanates in all directions and isn't an array of lasers?
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Virtually all light that any observer on the Earth will see will have originated downwards. That much should be obvious. Other light tends not to be illustrated because it's frankly irrelevant.
I'm sorry, but if you can't see how your own diagrams (even before fixing your curves and the fact that your observer is hundreds of kilometres tall) result in a 90-degree rotation of the Sun for the observer at sunset, then there's very little I can do to help you. You already drew an adequate explanation of this, and if you fail to comprehend your own explanations, then I sure as hell won't be able to do any better.
The answer to your question is "this doesn't happen under EAT because your premise directly contradicts EAT."
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Why do you have to talk to me like im an idiot? To me, i don't get what you don't understand about the diagram... the rays exit the sun at roughly 45 degrees, the sun hasn't "rotated 90 degrees", nothings rotated at all, just the view that the observer would see would have the green point at the top, and the bottom point 180 degrees below that. If you fail to understand that after careful explanation there's very little i can do to help you.
But I think we can both have a win here... if you're saying the majority of the light rays visible by the observer are from those emanating downwards, then yes, it makes total sense that my hypothesis of the top disappearing first would probably be wrong. What you could've said at the start is that "light rays exiting the sun at anything greater than a few degrees to vertical will never reach the observer" but hey... whatevs.
So we can at least conclude that "light rays that reach the observer must emanate a 'half sun' described by a line parallel to the surface of the earth. If light rays reaching the observer were to eminate from a point ABOVE that line, then that point would become the effective 'top' of the sun to the observer, and because that point will have a corresponding point 90 degrees below it, then the apparent 'top' of the sun will disappear before that lower point". Agreed? Note I've said "lower point", not bottom. Or to be more precise, because it wouldn't be the perfect half sphere (i.e. you'd have to work out the angle towards the eye and reduce the half sphere visible) then you could say "once the top most point rotates such that there is a second point directly below it", or once that first point rotates past the true half way parallel line... something like that...
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Why do you have to talk to me like im an idiot?
You keep saying this. I don't think you're an idiot. I mean it sincerely that, if you still don't see the 90-degree rotation despite producing a diagram that illustrates it yourself, then I do not know how to explain it better. This isn't a case of "I think you're an idiot", but rather one of "this is an unfortunate lose-lose scenario".
What you could've said at the start is that "light rays exiting the sun at anything greater than a few degrees to vertical will never reach the observer" but hey... whatevs.
I did not realise that this was unclear until you've made it explicit in your previous post. I really can't just guess what is and isn't confusing to you. That's why I previously asked you to state your contentions.
So we can at least conclude that "light rays that reach the observer must emanate a 'half sun' described by a line parallel to the surface of the earth. If light rays reaching the observer were to eminate from a point ABOVE that line, then that point would become the effective 'top' of the sun to the observer, and because that point will have a corresponding point 90 degrees below it, then the apparent 'top' of the sun will disappear before that lower point". Agreed? Note I've said "lower point", not bottom.
No, we can't agree on that. What you call the "lower point" will not be visible to the observer around that time, since it will literally be obstructed by the front of the Sun. The concept of something you can't see "disappearing" is not accessible to me. For something to disappear, it must have first appeared.
Similarly, if you look at my avatar, you can't see the back of my head. This is not due to any property of perspective. It's simply a case of my beautiful face being in the way.
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But yeah i get you're point, light rays are pulled in a perpendicular direction and maintaining the same speed. That still seems a bit "odd" to me but i accept it, and I don't believe it effects what i'm saying in the slightest: the paths of light that i assumed were bending up sharply hit the ground anyway... they're irrelevant to my point just because they hit the ground at a slightly different angle. Roughly speaking: light rays are bending upwards. My point still holds as far as i can tell.
No, you never had a point to begin with. As a consequence of different rays having different curves in your diagram, they cross over each other all the time. This means that, for most observers, the same point on the Sun's edge will be observable in multiple directions at once.
The technical term for this is "blur", and you're going to have a tough time making the case that the Sun is not observable as a coherent circular image in the sky. If your diagram cannot even model a single point on the Sun correctly, how do you hope to produce a meaningful description of what the entire Sun looks like?
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But yeah i get you're point, light rays are pulled in a perpendicular direction and maintaining the same speed. That still seems a bit "odd" to me but i accept it, and I don't believe it effects what i'm saying in the slightest: the paths of light that i assumed were bending up sharply hit the ground anyway... they're irrelevant to my point just because they hit the ground at a slightly different angle. Roughly speaking: light rays are bending upwards. My point still holds as far as i can tell.
No, you never had a point to begin with. As a consequence of different rays having different curves in your diagram, they cross over each other all the time. This means that, for most observers, the same point on the Sun's edge will be observable in multiple directions at once.
The technical term for this is "blur", and you're going to have a tough time making the case that the Sun is not observable as a coherent circular image in the sky. If your diagram cannot even model a single point on the Sun correctly, how do you hope to produce a meaningful description of what the entire Sun looks like?
er, no, i don't think you know how an eye works... only the paths of light that enter the eye will be visible. The diagram just shows all of the possible paths of light: something is visible if there is a path of light entering the eye. The eye will "focus" a number of rays entering the lens at certain angles from the same point by converging them to a single point. If you drew all of the rays of light in the room you're in right now it might freak you out. It doesn't look "blurry" though does it? There's light bouncing around all of the place, they just don't enter the eye, or they enter the eye at the wrong angle
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I mean it sincerely that, if you still don't see the 90-degree rotation despite producing a diagram that illustrates it yourself, then I do not know how to explain it better
OK i'll try this one more time, but i can concede my entire point is moot if light rays start downwards. I can concede that with what I now understand to be the correct EAT theory of the the suns light rays, then my problem doesn't occur. I'm completely happy with this outcome as it gives additional information to EA and provides a path for further discussion. But anyway, one final time for the folks back home:
(https://i.imgur.com/JAIH4Dh.png)
Curves not shown, just "starting angle" and "ending angle"
(https://i.imgur.com/P0qbQCc.png)
There's no "90 degree rotation", there's a 45 degree angle of light, so the "visible portion" of the sun describes a perpendicular 45 degree angle...
(https://i.imgur.com/WiuyEjW.png)
I imagine what's happened here is you've taken your understanding of the sun "pointing downwards", and then seen my two points on a 45 degree sun, and my two points DO describe a 90 degree angle to your version of the sun... but in of themselves, they are not 90 degrees to MY understanding of the sun: they're just an additional 45 degree angle (45 + 45 = 90!)
So on my apparently totally incorrect version of the angles of light from the sun, then the top point would disappear first. This is now made almost entirely irrelevant, as you've explained that in EAT the light from the sun is almost completely downwards. Although we still have some points higher than others, they're also "further away" so you don't end up with the same problem of the area of visible light ending at the top before the bottom.
I think you'd still find if you modelled what the sun would look like as it disappears in your version, it would not be as perfectly uniform as observed
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I understand it this way:
(http://oi65.tinypic.com/2dkxvev.jpg)
Not all light that reaches an observer on earth starts straight down. It depends on where you are.
The sun is radiating light in all directions, but only that light emanating within a certain angular range reaches earth, and where you are on earth relative to the sun determines the angle from vertical of those light rays leaving the sun that reach you. As the sun moves westward, those angles of light leaving the sun increase slightly, but result in a greater angle incident to earth due to EA.
If there was no earth and you were suspended in space on a plane below the sun, the sun wouldn't "set." You'd just see it appear to get lower below horizontal. (Edit: have to think on this more. Might not be true. checks out.) The sole reason for the sun "setting" in either standard RE mechanics or EA theory is the obstruction of earth. With EA, to see the rays that emanated at angles such that they are curving upward before they reach earth, you have to climb in elevation to intercept them.
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It doesn't start straight down. Just close enough to it for practical purposes. Sure, if it makes you feel any better, we can make the angle 80-90 degrees. 45 is a no-go.
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er, no, i don't think you know how an eye works... only the paths of light that enter the eye will be visible. The diagram just shows all of the possible paths of light: something is visible if there is a path of light entering the eye. The eye will "focus" a number of rays entering the lens at certain angles from the same point by converging them to a single point. If you drew all of the rays of light in the room you're in right now it might freak you out. It doesn't look "blurry" though does it? There's light bouncing around all of the place, they just don't enter the eye, or they enter the eye at the wrong angle
If two light rays intersect at a point, and you place an eye at that point, both of those rays are going to enter the eye. The eye can't magically "focus" two light rays that originate from the same point but different directions. If it were somehow able to do that, then mirrors would appear as black objects because all of their light would be "focused" onto the original. Can you see how absurd this claim is?
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If two light rays intersect at a point, and you place an eye at that point, both of those rays are going to enter the eye. The eye can't magically "focus" two light rays that originate from the same point but different directions
*shrug* don't tell it to me, tell it to science
(https://kaiserscience.files.wordpress.com/2015/02/eye-focusing-rays-of-light-figure_10_24_labeled.jpg?w=1000&h=472)
When things are "in focus" the different direction of light coming from a single point all go to a single point. When things are "out of focus" they go to different points and appear blury. If two paths of light enter the eye at "the same spot" on the front of the lens but originate from different spots, the shape of the lens will "direct" those rays of light to separate spots on the retina. So the eye's dealing with two things: the "location" of something i.e. wheres the top of the sign, the bottom of the sign, the word 'stop' etc, and the eye is also dealing with "focus"... i.e. for these rays all coming from the same spot, am i going to focus and converge them back to a single point, or am i converging other things in the picture... am i focusing on the bird or the building. If it was only single rays of light from each point in an image that reached the retina, then EVERYTHING would be in focus in your image plane.
In fact your iris does control this: it's an "aperture": just like a camera: when the aperture is tiny, then the depth of field is huge: "everything" is in focus but the amount of light entering is less so the image is darker. When the aperture is wide open i.e. big hole, then the depth of field is small: you need to "focus" on certain items at a certain distance, and everything else is blurry. Your eye doesn't change the shape of the iris becaue it gives a crap about depth of field, it's just trying to control the amount of light on your sensitive retina. Depth of field is a side effect.
Or put another way: different rays from the same point enter the eye while "diverging" (spreading out)), different rays from DIFFERENT points enter the eye while "converging" (joining together)... The lens will deform to bend diverging rays in to a single point (focus) but the converging rays will enter the eye, cross over, and hit the retina at different spots to tell the brain that those points are in different spots. Similarly, if something is out of focus, the diverging rays fail to be bent back to converge perfectly, end up hitting "different spots" and our brain tells us this is one object out of focus rather than different objects or a blurry object (and gives us an option to try to bring those things in to focus). Staring at a blurry photo of something can strain our eyes, because our brain tells us it's something is out of focus but our lens fails to deform to bring it back in to focus.
https://kaiserscience.wordpress.com/biology-the-living-environment/physiology/vision-how-do-our-eyes-work/
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If two light rays intersect at a point, and you place an eye at that point, both of those rays are going to enter the eye. The eye can't magically "focus" two light rays that originate from the same point but different directions
*shrug* don't tell it to me, tell it to science
(https://kaiserscience.files.wordpress.com/2015/02/eye-focusing-rays-of-light-figure_10_24_labeled.jpg?w=1000&h=472)
Yes, I understand how focus works. This isn't at all the scenario we're talking about. If you really can't see that there is a huge difference between light rays diverging from a point and getting focused back to another point, and light rays getting bent so that they appear to come from completely different directions, then I'm sorry, but I don't know how to explain the self-evident.
I don't know why we're even having this conversation. You already conceded the point that your diagrams show the curvature of light incorrectly. Why is the burden of proof still on me to show that an incorrect model would make incorrect predictions?
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The eye can't magically "focus" two light rays that originate from the same point but different directions
So you DON'T understand how focus works...
Yes, I understand how focus works. This isn't at all the scenario we're talking about
So the scenario is NOT two light rays that originate from the same point?
If you really can't see that there is a huge difference between light rays diverging from a point and getting focused back to another point, and light rays getting bent so that they appear to come from completely different directions, then I'm sorry, but I don't know how to explain the self-evident.
If you don't understand that "divergence" implies "different directions" then I don't know how to explain the self evident.
But I think I get what you're trying to say, I apologise if i misunderstood. The way I've drawn/calculated the curved light causes an issue because diverging rays will become converging rays. And yes, this would result in blur. i.e. top two pairs of diverging rays in image below = no problem.... bottom where diverging then becomes converging = huge problem: i.e. the eye will think they're rays from two different points. I agree this causes a problem, and provides an additional piece of useful information for EA: if rays of light starting in different directions must not converge, then the effect of the "pull" must be relative to the distance from that light ray. You said earlier the curve for each light ray would be more or less the same: I think it's an important distinction to make. If they were EXACTLY the same curve, the they would eventually converge. If you imagine two light rays that have turned so they're now travelling parallel to the earth (one above the other) then the upwards pull must effect the the higher ray less than the lower ray, otherwise they would converge.
(https://i.imgur.com/5pc00M9.png)
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There is no problem if the rays converge at some point high above the Earth where nobody is ever going to see them. What's important is that there is no noticeable convergence at altitudes people actually visit.
None of this is getting us any closer to a foundation for your claims in the OP.
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True. Let's put this to bed. We've learnt new information about a) the nature of the "force" curving the light rays and b) the initial direction of those light rays. The OP conjecture doesn't hold under those conditions.