[SiDawg]

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



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.



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...

[Pete Svarrior]

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.

[SiDawg]

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.

[Pete Svarrior]

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.

[Bobby Shafto]

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

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.

[xasop]

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.

[SiDawg]

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?

[SiDawg]

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.



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.



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?

[Bobby Shafto]

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. 

[SiDawg]

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

[Pete Svarrior]

Aside from the shape of the curve, you're still wrong about where the top and bottom of the sun are.

[xasop]

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.

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".

[SiDawg]

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.

[Pete Svarrior]

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.

[xasop]

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.

[SiDawg]

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.

[Bobby Shafto]

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.

[Bobby Shafto]

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."

[xasop]

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.

[Pete Svarrior]

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:



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.