Under the clouds?

No, I mean Johannes. I've never seen him before on this forum and suddenly he make posts refuting other peoples explanation, but not saying why it's wrong or explain what the alternative is.

Johannes is a regular who is back from taking some time off.

He's even worse than Parsifal or Tom, in the sense that he doesn't even come up with an explanation. How can you argue against a statement "No, it's rubbish."??

Quote from: The Question1 on May 14, 2010, 02:15:46 PM

Quote from: Catchpa on May 14, 2010, 02:08:20 PM

Who are you anyway? You suddenly just come in and spurt a bunch of stupid statements with absolutely no basis. Stop it.

I think he is the same Johannes from the roundearthsociety.

I LOL'd he means thevoiceofreason, I'd hope. Johannes is hopelessly addicted to TFES.

dude read my post please, and if you don't have the mathematicals to do understand or defend me,
than don't rage

I could try and explain it without the math and physics, but I need time to diagram

Quote from: Thevoiceofreason on May 14, 2010, 10:21:52 AM

Quote from: Johannes on May 13, 2010, 08:50:24 PM

Quote from: Thevoiceofreason on May 13, 2010, 08:34:30 PM

Quote from: Parsifal on May 13, 2010, 06:22:18 PM

This issue has been brought up before, and bendy light provides an explanation.

Bendy light has been debunked over 9000 times.
Beam neutrinos disproves your sun itself.
that one vid of the sun from near earth orbit disproves the disk shaped sun.
the mathematics behind force and acceleration disprove it right off the bat.

Give me another thing

Sources? Maths?

Light bends upwards; that is why the clouds are illuminated from the bottom.

Mathematicals for debunk of bendy light:

in order for the horizon effect to keep parallel in both systems, then Light must follow a circular path for very obvious reasons. the path of a circle is governed by x(t)=rcost y(t)=rsint. If light travels at c,
then c=sqrt(x'(t)²+y'(t)²), or in other words, light is going in the path of a cir circle at uniform speed.

that said, the acceleration vector is therefore centripetal. if this vector is centripetal, then so is force, because F=MA. note, that the magnitude and angle of this force varies upon altitude, latitude, zenith and azimuth angles of the ray, etc.

That is nonsense.

which part? I'll slow it down for your sake.


Do you agree that for RE to parallel FE, that light needs to travel the path of a circle?
because if it didn't, there would be a height above DC, where you could see shanghai,
the earth would appear to be a quadratic if light bent with a quadratic.
this is the only way that experiments can fit both RE and FE.

in order for the horizon effect to keep parallel in both systems, then Light must follow a circular path for very obvious reasons.

They aren't obvious to me, so why don't you explain yourself?

the path of a circle is governed by x(t)=rcost y(t)=rsint. If light travels at c,
then c=sqrt(x'(t)²+y'(t)²), or in other words, light is going in the path of a cir circle at uniform speed.

lol no

You've used the derivatives of x(t) and y(t) in that equation, so you've said nothing about the original functions themselves. Also, protip: c cos t and c sin t aren't the only two expressions whose squares add to c². I'll leave it as an exercise for you to come up with at least one very simple alternative.

Yet unfortunately for bendy light, if it produces this effect it should also produce other effects such as shifting of star positions proportional to altitude, which does not happen.
Bendy light now disproved 9001 times.

You haven't yet explained why this should be the case, so your proof has null value.

It has been explained elsewhere, try the search function.

Quote from: Thevoiceofreason on May 14, 2010, 10:21:52 AM

in order for the horizon effect to keep parallel in both systems, then Light must follow a circular path for very obvious reasons.

They aren't obvious to me, so why don't you explain yourself?

Quote from: Thevoiceofreason on May 14, 2010, 10:21:52 AM

the path of a circle is governed by x(t)=rcost y(t)=rsint. If light travels at c,
then c=sqrt(x'(t)²+y'(t)²), or in other words, light is going in the path of a cir circle at uniform speed.

lol no

You've used the derivatives of x(t) and y(t) in that equation, so you've said nothing about the original functions themselves. Also, protip: c cos t and c sin t aren't the only two expressions whose squares add to c². I'll leave it as an exercise for you to come up with at least one very simple alternative.

Quote from: Thermal Detonator on May 14, 2010, 11:03:25 AM

Yet unfortunately for bendy light, if it produces this effect it should also produce other effects such as shifting of star positions proportional to altitude, which does not happen.
Bendy light now disproved 9001 times.

You haven't yet explained why this should be the case, so your proof has null value.

*sigh*
http://www.theflatearthsociety.org/forum/index.php?topic=22317.0
remember the curvature vs. bendy light thing?
well the solution is that light bends in a circle:

This ensures that all data fits with the idea that light bends upward on the flat earth.

Lol,yes
the original equation was to show the path.
the derivatives where there to show constant speed.

Which Parsifail,
x(t)=rcos(t) and y(t)=rsin(t), are dummy equations, there to show you the path,
c²=x'(t)²+y'(t)² is there to show CONSTANT speed.

protip:lrn2higschool mathematicals.

now with information of the path
and info of the vel,
we obtain

x(t)=rcos(tc/r)
y(t)=rsin(tc/r)

x'(t)=-c*sin(tc/r)            magnetude of velocity =c
y'(t)=c*cos(tc/r)

x''(t)=-c²/r *cos(tc/r) magnetude of acceleration=c²/r
y''(t)=-c²/r *sin(tc/r)

now dot x'(t),y'(t) with x''(t),y''(t):
-c*sin(tc/r)* -c²/r *cos(tc/r) + c*cos(tc/r))*-c²/r *sin(tc/r)=0
to find that acceleration is perpendicular to velocity, meaning that it is along the radius of the circle.
and since f_x,f_y=m*x"(t),y"(t)
the force to make light bend in a circle is centripetal, explain that plz

You don't need to prove to us that that the acceleration vector of a circle is directed inwards.

I don't get what you are trying to prove by this. What does "in order for the horizon effect to keep parallel in both systems" mean?

What is your drawing trying to illustrate?

I think the idea of bendy light was the only possible solution to this problem on a flat earth. If bendy light doesn't exist, we can prove, beyond all reasonable doubt, that the earth is not flat. The only other possible explanation is magic, unless someone can find a better idea.

Now the flat earth supporters have two honest options:
1- Find a new explanation that won't be disproved.
2- Give up the idea of a flat earth and act like a rational person

And a third, dishonest option:
3- Ignore this post and keep claiming that the earth is flat.

I strongly recommend taking one of the first two, for the sake of your own integrity.

I, a RealEarth 'er, can make some mathematicals that will work for all of bendy light, but it may or may not have plausible forces, aka
you can create a system where forces cause light to bend like magic, and have it sound in the eyes of physics, but it will be very twisted

Quote from: Maximohoundoom on May 14, 2010, 12:48:57 PM

I think the idea of bendy light was the only possible solution to this problem on a flat earth. If bendy light doesn't exist, we can prove, beyond all reasonable doubt, that the earth is not flat. The only other possible explanation is magic, unless someone can find a better idea.

Now the flat earth supporters have two honest options:
1- Find a new explanation that won't be disproved.
2- Give up the idea of a flat earth and act like a rational person

And a third, dishonest option:
3- Ignore this post and keep claiming that the earth is flat.

I strongly recommend taking one of the first two, for the sake of your own integrity.

I, a RealEarth 'er, can make some mathematicals that will work for all of bendy light, but it may or may not have plausible forces, aka
you can create a system where forces cause light to bend like magic, and have it sound in the eyes of physics, but it will be very twisted

i can make light bend its easy just need some thing with LOT of mass the Sun for one will do
WHICH btw if any of you approve of relativity need to stop now becouse thats how it was proved
a solar eclipse + known stars behind the sun were photographed
the photo plates were checked and the stars were not in the right place
this proves a few things 1. the sun is MASSIVE 2.  light can be bent by gravity 3. Mass = warps space to create gravity ie its not gravity pulling on you its the mass of the planet warping space pushing you remember forces can only push not pull
4. for the sun to be that massive it must be far away
and 5. all things have mass

He's even worse than Parsifal or Tom, in the sense that he doesn't even come up with an explanation. How can you argue against a statement "No, it's rubbish."??

He's one of the less intelligent members of the forum. His standard technique is to stroll in and back up other flat guys by reposting stuff that's already been dealt with and disproved. He has never, as far as I know, ventured an original idea of his own.

*sigh*
http://www.theflatearthsociety.org/forum/index.php?topic=22317.0
remember the curvature vs. bendy light thing?
well the solution is that light bends in a circle:

This ensures that all data fits with the idea that light bends upward on the flat earth.

No, those diagrams do not make equivalent predictions. Light bending in a circle would predict that if I look straight ahead, I would see the back of my own head.

Lol,yes
the original equation was to show the path.
the derivatives where there to show constant speed.

Which Parsifail,
x(t)=rcos(t) and y(t)=rsin(t), are dummy equations, there to show you the path,
c²=x'(t)²+y'(t)² is there to show CONSTANT speed.

I'm sorry, I thought you were trying to prove that light moves in a circle with that equation, since you hadn't done so anywhere else. We already know that light moves at a constant speed (in a vacuum) thanks to relativity, but what's even more astounding about this is that your equation doesn't prove that at all; in other words, you've drawn a correct conclusion using an incorrect method. You see, the parameter t in your equation is simply the angle on the number plane measured at the origin between the positive x-axis and a straight line joining the origin with a point on the circle; a derivative of a function of t is the rate of change of the function's value as this angle changes, and your constant value c is a measure of the constant arc length for a given angular displacement, an obvious property of a circle. Your calculations don't involve time at all, and therefore don't say anything about speed.

now with information of the path
and info of the vel,
we obtain

x(t)=rcos(tc/r)
y(t)=rsin(tc/r)

x'(t)=-c*sin(tc/r)            magnetude of velocity =c
y'(t)=c*cos(tc/r)

x''(t)=-c²/r *cos(tc/r) magnetude of acceleration=c²/r
y''(t)=-c²/r *sin(tc/r)

now dot x'(t),y'(t) with x''(t),y''(t):
-c*sin(tc/r)* -c²/r *cos(tc/r) + c*cos(tc/r))*-c²/r *sin(tc/r)=0
to find that acceleration is perpendicular to velocity, meaning that it is along the radius of the circle.
and since f_x,f_y=m*x"(t),y"(t)
the force to make light bend in a circle is centripetal, explain that plz

These calculations are based on the misunderstanding that you have calculated anything to do with speed or velocity up until now, which - as I pointed out above - you have not.

Quote from: Thevoiceofreason on May 14, 2010, 11:20:23 PM

*sigh*
http://www.theflatearthsociety.org/forum/index.php?topic=22317.0
remember the curvature vs. bendy light thing?
well the solution is that light bends in a circle:

This ensures that all data fits with the idea that light bends upward on the flat earth.

No, those diagrams do not make equivalent predictions. Light bending in a circle would predict that if I look straight ahead, I would see the back of my own head.

Quote from: Thevoiceofreason on May 14, 2010, 11:20:23 PM

Lol,yes
the original equation was to show the path.
the derivatives where there to show constant speed.

Which Parsifail,
x(t)=rcos(t) and y(t)=rsin(t), are dummy equations, there to show you the path,
c²=x'(t)²+y'(t)² is there to show CONSTANT speed.

I'm sorry, I thought you were trying to prove that light moves in a circle with that equation, since you hadn't done so anywhere else. We already know that light moves at a constant speed (in a vacuum) thanks to relativity, but what's even more astounding about this is that your equation doesn't prove that at all; in other words, you've drawn a correct conclusion using an incorrect method. You see, the parameter t in your equation is simply the angle on the number plane measured at the origin between the positive x-axis and a straight line joining the origin with a point on the circle; a derivative of a function of t is the rate of change of the function's value as this angle changes, and your constant value c is a measure of the constant arc length for a given angular displacement, an obvious property of a circle. Your calculations don't involve time at all, and therefore don't say anything about speed.

Quote from: Thevoiceofreason on May 14, 2010, 11:20:23 PM

now with information of the path
and info of the vel,
we obtain

x(t)=rcos(tc/r)
y(t)=rsin(tc/r)

x'(t)=-c*sin(tc/r)            magnetude of velocity =c
y'(t)=c*cos(tc/r)

x''(t)=-c²/r *cos(tc/r) magnetude of acceleration=c²/r
y''(t)=-c²/r *sin(tc/r)

now dot x'(t),y'(t) with x''(t),y''(t):
-c*sin(tc/r)* -c²/r *cos(tc/r) + c*cos(tc/r))*-c²/r *sin(tc/r)=0
to find that acceleration is perpendicular to velocity, meaning that it is along the radius of the circle.
and since f_x,f_y=m*x"(t),y"(t)
the force to make light bend in a circle is centripetal, explain that plz

These calculations are based on the misunderstanding that you have calculated anything to do with speed or velocity up until now, which - as I pointed out above - you have not.

My god sir. no, when light bends in a circle, it makes it so you can only see things 45 degrees (lat and long) from any direction no matter how high you get.

keyword "dummy" equations. I just wrote out mathematically what the path of light should be. now you tell me how the forces work
tc/r is the angle, el vato. plz explain how light bends in a circle.

If light were to bend in a quadratic, the world would look like a rly big hill.

My god sir. no, when light bends in a circle, it makes it so you can only see things 45 degrees (lat and long) from any direction no matter how high you get.

I assume you understand, at the very least, that a circle is round? Yes? Good.

Having established that, if a light ray starts moving horizontally (say, by reflecting off the back of your head) and bends in a circle, then unless it strikes some obstacle it will eventually come all the way back around to where it started from, to meet the front of your head where your eyes are. Your vision, if there are no opaque obstructions along that circular path, should therefore be completely filled by an image of the back of your own head.

Also, your claim that it limits your vision to 45 degrees (which itself is very badly phrased, not least because the actual corresponding surface distance for a degree of longitude varies with latitude) is technically correct if we assume that "45 degrees" means a constant distance from the point on the Earth's surface directly beneath the observer, but it would predict that all such light from below would come from straight down, so the Earth would appear as a point. Ascending further would then reduce the field of vision once more, and the Earth would spread out into a coherent object again - but its image would be mirrored. Finally, ascending to an altitude greater than the diameter of one of these circles would render the Earth invisible.

keyword "dummy" equations. I just wrote out mathematically what the path of light should be. now you tell me how the forces work
tc/r is the angle, el vato. plz explain how light bends in a circle.

It doesn't.

If light were to bend in a quadratic, the world would look like a rly big hill.

Would you care to prove this, or are you just expecting me to take your word as fact?

I assume you understand, at the very least, that a circle is round? Yes? Good.

Do you have any evidence to back up your outrageous claim?

Quote from: Thevoiceofreason on May 16, 2010, 02:53:04 PM

My god sir. no, when light bends in a circle, it makes it so you can only see things 45 degrees (lat and long) from any direction no matter how high you get.

I assume you understand, at the very least, that a circle is round? Yes? Good.

Having established that, if a light ray starts moving horizontally (say, by reflecting off the back of your head) and bends in a circle, then unless it strikes some obstacle it will eventually come all the way back around to where it started from, to meet the front of your head where your eyes are. Your vision, if there are no opaque obstructions along that circular path, should therefore be completely filled by an image of the back of your own head.

Also, your claim that it limits your vision to 45 degrees (which itself is very badly phrased, not least because the actual corresponding surface distance for a degree of longitude varies with latitude) is technically correct if we assume that "45 degrees" means a constant distance from the point on the Earth's surface directly beneath the observer, but it would predict that all such light from below would come from straight down, so the Earth would appear as a point. Ascending further would then reduce the field of vision once more, and the Earth would spread out into a coherent object again - but its image would be mirrored. Finally, ascending to an altitude greater than the diameter of one of these circles would render the Earth invisible.

Quote from: Thevoiceofreason on May 16, 2010, 02:53:04 PM

keyword "dummy" equations. I just wrote out mathematically what the path of light should be. now you tell me how the forces work
tc/r is the angle, el vato. plz explain how light bends in a circle.

It doesn't.

Quote from: Thevoiceofreason on May 16, 2010, 02:53:04 PM

If light were to bend in a quadratic, the world would look like a rly big hill.

Would you care to prove this, or are you just expecting me to take your word as fact?

Take the circle as a vector field, light starts out parallel to the ground. starts curving with circle. exits out of Earth's atmosphere at about 1/4th of what would be a great circle. this ensures that you cannot see at or past 45 degrees latitude or longitude. aka you can only see half of the world. (I know there is variance, but stick with the simple case) no matter how high you travel.

Now lets take the case of light traveling a parabola. if you were on a flat plain, and light bent in the path Ax2, then things 1 meter away would appear to be A meters lower than your current point. If things were 2 meter's away, then it would be 4A meters lower. if things were 3 meters away, they would be 9A meters lower...etc do you understand? It would be like standing on top a quadratic, where at your point of observation is the top of said quadratic. As a consequence, for any given point x on the quadratic earth (-ax², you could stand at some height h above the vertex, where you can see that point.



Now If light were to bend in a circle on the other hand... you get the idea.

Take the circle as a vector field, light starts out parallel to the ground. starts curving with circle. exits out of Earth's atmosphere at about 1/4th of what would be a great circle. this ensures that you cannot see at or past 45 degrees latitude or longitude. aka you can only see half of the world. (I know there is variance, but stick with the simple case) no matter how high you travel.

Now continue following that circle all the way around, and it will become obvious why that model doesn't work.

Now lets take the case of light traveling a parabola. if you were on a flat plain, and light bent in the path Ax2, then things 1 meter away would appear to be A meters lower than your current point. If things were 2 meter's away, then it would be 4A meters lower. if things were 3 meters away, they would be 9A meters lower...etc do you understand? It would be like standing on top a quadratic, where at your point of observation is the top of said quadratic. As a consequence, for any given point x on the quadratic earth (-ax², you could stand at some height h above the vertex, where you can see that point.

I might understand if you were using clear language. As it is, you're talking about things which "appear" some distance "lower", without explaining what you mean by that, and it certainly isn't the intuitive meaning (the path length of the light ray) because your numbers don't work out that way. You need to use clear, precise language in mathematics, or you lose people simply because your meaning is too ambiguous.

Now If light were to bend in a circle on the other hand... you get the idea.

Yes, I get the idea that light bending in a circle doesn't work at all.

Quote from: Thevoiceofreason on May 17, 2010, 12:31:42 PM

Take the circle as a vector field, light starts out parallel to the ground. starts curving with circle. exits out of Earth's atmosphere at about 1/4th of what would be a great circle. this ensures that you cannot see at or past 45 degrees latitude or longitude. aka you can only see half of the world. (I know there is variance, but stick with the simple case) no matter how high you travel.

Now continue following that circle all the way around, and it will become obvious why that model doesn't work.

Quote from: Thevoiceofreason on May 17, 2010, 12:31:42 PM

Now lets take the case of light traveling a parabola. if you were on a flat plain, and light bent in the path Ax2, then things 1 meter away would appear to be A meters lower than your current point. If things were 2 meter's away, then it would be 4A meters lower. if things were 3 meters away, they would be 9A meters lower...etc do you understand? It would be like standing on top a quadratic, where at your point of observation is the top of said quadratic. As a consequence, for any given point x on the quadratic earth (-ax², you could stand at some height h above the vertex, where you can see that point.

I might understand if you were using clear language. As it is, you're talking about things which "appear" some distance "lower", without explaining what you mean by that, and it certainly isn't the intuitive meaning (the path length of the light ray) because your numbers don't work out that way. You need to use clear, precise language in mathematics, or you lose people simply because your meaning is too ambiguous.

Quote from: Thevoiceofreason on May 17, 2010, 12:31:42 PM

Now If light were to bend in a circle on the other hand... you get the idea.

Yes, I get the idea that light bending in a circle doesn't work at all.

who said it would have to continue on forever in that circle?
As for the parabolic, Imagine. someone takes a laser at some point y=0  and x=0 going along the x axis. on a parabolic path, when the light is at x=3, it is 9A meters tall, with A being the coefficient of the parabola.

I'm going to show a flaw in the reasoning of bendy light.

1- The earth looks flat, therefore it is most likely flat.
2- The morning and evening sunlight can hit the underside of the clouds.
3- The sun is 32000 miles up in the sky, the clouds are about 1km up.
4- On a flat earth, the sunlight could not be under the clouds if light moved in a straight path.
The earth doesn't look flat, it looks round. Point number 1 has just been invalidated.
5- Because the earth is flat, light must bend in order to make it look like it does.

As you see, in order to propose bendy light, you first have to assume that the earth looks flat and then assume that it doesn't. But let's forget about the reason why you need to propose bendy light and focus on the implications. The problem with bendy light is that you forgot to establish a well defined formula for the path it takes. Not just a vague guess, it should be written down with all the constants defined so we can test it. Until you come up with the formula, the flat earth is just as improbable as a torus, a triangular prism, a sphere or even an irregular blob.

Experiment: How to test the validity of your bendy light equation

Before the experiment:

1- Propose a mathematical function that describes bendy light.
2- Propose a model of the earth that predicts where the sun will be at a certain time of the day.
3- Choose a time of the day and your location on earth and predict what angle the sunlight will have in that area.

The experiment:

1- Get a thin straight tube connected to an angle measurer(A circular object with angle marks on it).
2- Then hang a string with a weight on it.
3- Point the tube at the sun (you don't have to look into it) and measure the angle marked by the string.

After the experiment:

Compare the measured value of the light angle with the ones predicted by your model.

I have confidence that, if you actually do the experiment, a linear light model (Although light does curve slightly around gravitational force) on a round earth will work just fine. It's easier to calculate, too.

I'm going to show a flaw in the reasoning of bendy light.

1- The earth looks flat, therefore it is most likely flat.
2- The morning and evening sunlight can hit the underside of the clouds.
3- The sun is 32000 miles up in the sky, the clouds are about 1km up.
4- On a flat earth, the sunlight could not be under the clouds if light moved in a straight path.
The earth doesn't look flat, it looks round. Point number 1 has just been invalidated.
5- Because the earth is flat, light must bend in order to make it look like it does.

As you see, in order to propose bendy light, you first have to assume that the earth looks flat and then assume that it doesn't. But let's forget about the reason why you need to propose bendy light and focus on the implications. The problem with bendy light is that you forgot to establish a well defined formula for the path it takes. Not just a vague guess, it should be written down with all the constants defined so we can test it. Until you come up with the formula, the flat earth is just as improbable as a torus, a triangular prism, a sphere or even an irregular blob.

Experiment: How to test the validity of your bendy light equation

Before the experiment:

1- Propose a mathematical function that describes bendy light.
2- Propose a model of the earth that predicts where the sun will be at a certain time of the day.
3- Choose a time of the day and your location on earth and predict what angle the sunlight will have in that area.

The experiment:

1- Get a thin straight tube connected to an angle measurer(A circular object with angle marks on it).
2- Then hang a string with a weight on it.
3- Point the tube at the sun (you don't have to look into it) and measure the angle marked by the string.

After the experiment:

Compare the measured value of the light angle with the ones predicted by your model.

I have confidence that, if you actually do the experiment, a linear light model (Although light does curve slightly around gravitational force) on a round earth will work just fine. It's easier to calculate, too.

If you account for refraction (also very well described mathematically) then you've got a very sound experiment there.

I'm going to show a flaw in the reasoning of bendy light.

1- The earth looks flat, therefore it is most likely flat.
2- The morning and evening sunlight can hit the underside of the clouds.
3- The sun is 32000 miles up in the sky, the clouds are about 1km up.
4- On a flat earth, the sunlight could not be under the clouds if light moved in a straight path.
The earth doesn't look flat, it looks round. Point number 1 has just been invalidated.
5- Because the earth is flat, light must bend in order to make it look like it does.

As you see, in order to propose bendy light, you first have to assume that the earth looks flat and then assume that it doesn't. But let's forget about the reason why you need to propose bendy light and focus on the implications. The problem with bendy light is that you forgot to establish a well defined formula for the path it takes. Not just a vague guess, it should be written down with all the constants defined so we can test it. Until you come up with the formula, the flat earth is just as improbable as a torus, a triangular prism, a sphere or even an irregular blob.

Experiment: How to test the validity of your bendy light equation

Before the experiment:

1- Propose a mathematical function that describes bendy light.
2- Propose a model of the earth that predicts where the sun will be at a certain time of the day.
3- Choose a time of the day and your location on earth and predict what angle the sunlight will have in that area.

The experiment:

1- Get a thin straight tube connected to an angle measurer(A circular object with angle marks on it).
2- Then hang a string with a weight on it.
3- Point the tube at the sun (you don't have to look into it) and measure the angle marked by the string.

After the experiment:

Compare the measured value of the light angle with the ones predicted by your model.

I have confidence that, if you actually do the experiment, a linear light model (Although light does curve slightly around gravitational force) on a round earth will work just fine. It's easier to calculate, too.

here's a diagram. the red line is a light beam. the blue is the earth. the black is there to represent heights. the scale is the same in both sides. note the angle and height of the laser in reference to the ground. Light therefor travels in a circle relative to the earth.
Checkmate Robo.

You seem to have forgotten the diagram.

This haven't had an answer from a single FE'er. Would be pretty funny to get one though.

You seem to have forgotten the diagram.

here it is

who said it would have to continue on forever in that circle?

The diagram you posted earlier would suggest that. If it stops bending after it is moving vertically, then above that point the entire visible Earth would appear as a point.

As for the parabolic, Imagine. someone takes a laser at some point y=0  and x=0 going along the x axis. on a parabolic path, when the light is at x=3, it is 9A meters tall, with A being the coefficient of the parabola.

I've asked you to use more precise language, not to continue spewing incoherently. You haven't even defined your coordinate system.

Maximohoundoom's experiment is not a valid test for bendy light, because it is also testing the validity of a model of the Sun's position in the sky. A negative result doesn't necessarily mean that light doesn't bend, it could mean that our understanding of the Sun's position in the sky is flawed.

Thevoiceofreason, please actually perform the mathematical transformation from a RE model to a FE model along those black lines, and you will see that the path of the light ray does not come out as a circle. I know this, because I have performed that very transformation myself when I was first considering the bendy light hypothesis. If you know as much about mathematics as you like to think you do, this calculation should be no problem at all, as it can be easily done using elementary trigonometry.

Of course, this experiment can only work if you have a working model of the bendy light AND a working model of the position of the sun. But I thought all these flat earth people had a model of the position of the sun for 50 years already.

If you don't happen to have a model, you have two options: "Go get one!" or "Give up already!". Don't keep arguing for an idea you don't even have yet. This isn't how science works.

Parsifal is not a FE'er. Stop taking him serious.

Parsifal is not a FE'er. Stop taking him serious.

I completely agree that non-FEers should never be taken seriously. They obviously have no sense of logic or reason, and only will to continue living in a fantasy universe full of balls whizzing around in empty space.

So what path does light bend in if I'm wrong?