Under the clouds?

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

You've talked down to me so often about how I don't understand this mathematics that I'd hate to get in the way of the work of a genius like yourself. I'll let you figure it out on your own time.

I know he isn't a flat earth believer. Are there any real flat earth supporters who think my experiment is unreliable? Why have they all ignored this post? I think they are all being dishonest and avoiding questions they can't answer.

What about Lord Wilmore? or Pongo? What do they think? They should do the experiments themselves. Someone who was really interested in knowing the shape of the world would have done everything they could to measure it. These people do not want to know.

Quote from: Thevoiceofreason on May 18, 2010, 08:39:31 AM

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

You've talked down to me so often about how I don't understand this mathematics that I'd hate to get in the way of the work of a genius like yourself. I'll let you figure it out on your own time.

Delivery.



what I did was, find the length from when the beam of light is tangential, and drew a parallel line to the beam I found both the length and the height as functions of theta, the angle the beam of light makes with the vertical. I then solved for height (y) in terms of length from the tangential point (x).

Delivery.



what I did was, find the length from when the beam of light is tangential, and drew a parallel line to the beam I found both the length and the height as functions of theta, the angle the beam of light makes with the vertical. I then solved for height (y) in terms of length from the tangential point (x).

Very good. Although that function can be simplified significantly, it is nonetheless correct.

Quote from: Thevoiceofreason on May 18, 2010, 10:41:42 AM

Delivery.



what I did was, find the length from when the beam of light is tangential, and drew a parallel line to the beam I found both the length and the height as functions of theta, the angle the beam of light makes with the vertical. I then solved for height (y) in terms of length from the tangential point (x).

Very good. Although that function can be simplified significantly, it is nonetheless correct.

And so?
was I write or wrong about the circle thing?

And so?
was I write or wrong about the circle thing?

y = 1/sin(π/2 - x) - 1 (or, in simplified form, y = sec(x) - 1) is not an equation for a circle, so you were wrong.

Quote from: Thevoiceofreason on May 18, 2010, 10:50:52 AM

And so?
was I write or wrong about the circle thing?

y = 1/sin(?/2 - x) - 1 (or, in simplified form, y = sec(x) - 1) is not an equation for a circle, so you were wrong.

is it circular? would the earth not look like a circle?

is it circular? would the earth not look like a circle?

The Earth cannot look like a circle.

Quote from: Thevoiceofreason on May 18, 2010, 11:22:12 AM

is it circular? would the earth not look like a circle?

The Earth cannot look like a circle.

so are you saying that that curve doesn't remotely look similar to a circle, that if light did bend in that fashion,
the earth would atleast appear to be curved like an sphere?

You just proposed a function 1/sin(PI*0.5-x), where x is the horizontal distance of the observer compared to the sun, am I right? I looked at the function in a graph and it seems to be in a U shape. So it starts vertical, goes horizontal and then goes back up vertically into outer space.

One part of the experiment is to propose the model, but now you have to test it. A proposed model doesn't automatically become true just because it's well defined. Have you made a prediction of the angle the sunlight will have when it reaches you? Did you then measure the angle and compare the results?

By the way, you seem to have defined this function for vertical light coming from the sun. Have you thought about how this function will treat the light if it originates from the earth?

Quote from: Parsifal on May 18, 2010, 08:46:02 AM

Quote from: Thevoiceofreason on May 18, 2010, 08:39:31 AM

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

You've talked down to me so often about how I don't understand this mathematics that I'd hate to get in the way of the work of a genius like yourself. I'll let you figure it out on your own time.

Delivery.



what I did was, find the length from when the beam of light is tangential, and drew a parallel line to the beam I found both the length and the height as functions of theta, the angle the beam of light makes with the vertical. I then solved for height (y) in terms of length from the tangential point (x).

Quote from: Parsifal on May 18, 2010, 11:26:10 AM

Quote from: Thevoiceofreason on May 18, 2010, 11:22:12 AM

is it circular? would the earth not look like a circle?

The Earth cannot look like a circle.

so are you saying that that curve doesn't remotely look similar to a circle, that if light did bend in that fashion,
the earth would atleast appear to be curved like an sphere?

A sphere is not a circle.

You just proposed a function 1/sin(PI*0.5-x), where x is the horizontal distance of the observer compared to the sun, am I right? I looked at the function in a graph and it seems to be in a U shape. So it starts vertical, goes horizontal and then goes back up vertically into outer space.

One part of the experiment is to propose the model, but now you have to test it. A proposed model doesn't automatically become true just because it's well defined. Have you made a prediction of the angle the sunlight will have when it reaches you? Did you then measure the angle and compare the results?

By the way, you seem to have defined this function for vertical light coming from the sun. Have you thought about how this function will treat the light if it originates from the earth?

I'm a Real Earth'er,
I made this function, from a thought experiment of what light would have to bend like if the earth was flat. you can treat it like a vector field, if light starts traveling tangential to the curve. but also, the function is rotatable, but you must adjust for change in latitude. And the angle for light traveling in this path will be the same experimentally regardless, as that is how the function was made. this is the problem with the disproof of bendy light, it forces the results to be parallel between flat earth and round earth. now as for the angle of light from the sun, according to the equation, it should bend like that, the only problem is, is that it doesn't. because last time I checked light that moves nearly parallel doesn't suddenly go tangential to the earth. End of BL
.

Quote from: Thevoiceofreason on May 18, 2010, 10:41:42 AM

Quote from: Parsifal on May 18, 2010, 08:46:02 AM

Quote from: Thevoiceofreason on May 18, 2010, 08:39:31 AM

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

You've talked down to me so often about how I don't understand this mathematics that I'd hate to get in the way of the work of a genius like yourself. I'll let you figure it out on your own time.

Delivery.



what I did was, find the length from when the beam of light is tangential, and drew a parallel line to the beam I found both the length and the height as functions of theta, the angle the beam of light makes with the vertical. I then solved for height (y) in terms of length from the tangential point (x).

Quote from: Thevoiceofreason on May 18, 2010, 11:32:38 AM

Quote from: Parsifal on May 18, 2010, 11:26:10 AM

Quote from: Thevoiceofreason on May 18, 2010, 11:22:12 AM

is it circular? would the earth not look like a circle?

The Earth cannot look like a circle.

so are you saying that that curve doesn't remotely look similar to a circle, that if light did bend in that fashion,
the earth would atleast appear to be curved like an sphere?

A sphere is not a circle.

That's what I meant. it was a mistake, now argue the actual point

RoundEarth isn't a sphere.

Also, prove that the arc in your diagram isn't from an oval.

RoundEarth isn't a sphere.

Also, prove that the arc in your diagram isn't from an oval.

Eh... I was looking that up the other day its damn close.
Better then tolerances for billiard ball any way so for most mathematical stuff it might as well be.

RoundEarth isn't a sphere.

Also, prove that the arc in your diagram isn't from an oval.

I said looked LIKE a sphere.
it actuality, the curve is actually the graph of secant, which LOOKS circular.

What about Lord Wilmore? or Pongo? What do they think?

Personally, I think that it is light bouncing off of the earth and back to the clouds.  The earth is by no means flat as a mirror and everything including tiny dust particles will reflect the light from the spotlight-sun in all different directions.  So, the light you see illuminating the bottoms of the clouds is light from the spotlight that has bounced off the surface of the earth and been reflected (in this case you are seeing the light that is reflected in your direction... obviously). 

This is why the bottoms on the clouds light up before the tops.

There's a problem with the reflection hypothesis, it requires the sun to be very far.

Imagine that the clouds are at 1km height, and they extend from right ontop of the observer all the way up to 3km towards the sun. For the sunlight to hit the clouds right ontop of the observer, the sunlight must hit the ground at a maximum of 1.5km away from him. This means that, for every kilometer of height, the sun must be 1.5 times further away than its height.

A sun that's 32000km high would have to be 48000 km away. That's more than the perimeter of the equator, whether you measure it on a round earth or a flat earth. And this is supposing the clouds only extend 3km away, what if it's 5 or 6?(Which is still possible) The sun would have to be outside of the radius of the earth in order to produce the reflection effect.

There's a problem with the reflection hypothesis, it requires the sun to be very far.

Imagine that the clouds are at 1km height, and they extend from right ontop of the observer all the way up to 3km towards the sun. For the sunlight to hit the clouds right ontop of the observer, the sunlight must hit the ground at a maximum of 1.5km away from him. This means that, for every kilometer of height, the sun must be 1.5 times further away than its height.

A sun that's 32000km high would have to be 48000 km away. That's more than the perimeter of the equator, whether you measure it on a round earth or a flat earth. And this is supposing the clouds only extend 3km away, what if it's 5 or 6?(Which is still possible) The sun would have to be outside of the radius of the earth in order to produce the reflection effect.

this.

the problem is also bendy light. how could light be made to reflect/refract its way under the clouds at that angle, with using the graph above as a vector field??

Imagine that the clouds are at 1km height, and they extend from right ontop of the observer all the way up to 3km towards the sun. For the sunlight to hit the clouds right ontop of the observer, the sunlight must hit the ground at a maximum of 1.5km away from him. This means that, for every kilometer of height, the sun must be 1.5 times further away than its height.

Could you please rephrase this in a way that makes some kind of sense?

Quote from: Maximohoundoom on May 19, 2010, 05:15:58 AM

Imagine that the clouds are at 1km height, and they extend from right ontop of the observer all the way up to 3km towards the sun. For the sunlight to hit the clouds right ontop of the observer, the sunlight must hit the ground at a maximum of 1.5km away from him. This means that, for every kilometer of height, the sun must be 1.5 times further away than its height.

Could you please rephrase this in a way that makes some kind of sense?

If the sun had to shine on the bottom of 3km of clouds using reflection, it would have to be 48000km away from the place it's shining on. This is done using a simple triangle rule, I don't think I need to explain the math to anyone because it's so clear.

48000 kilometers is much larger than the perimeter of the equator. So this would only happen if the sun was somewhere outside the edge of the earth.

Quote from: Parsifal on May 19, 2010, 10:05:42 AM

Quote from: Maximohoundoom on May 19, 2010, 05:15:58 AM

Imagine that the clouds are at 1km height, and they extend from right ontop of the observer all the way up to 3km towards the sun. For the sunlight to hit the clouds right ontop of the observer, the sunlight must hit the ground at a maximum of 1.5km away from him. This means that, for every kilometer of height, the sun must be 1.5 times further away than its height.

Could you please rephrase this in a way that makes some kind of sense?

If the sun had to shine on the bottom of 3km of clouds using reflection, it would have to be 48000km away from the place it's shining on. This is done using a simple triangle rule, I don't think I need to explain the math to anyone because it's so clear.

48000 kilometers is much larger than the perimeter of the equator. So this would only happen if the sun was somewhere outside the edge of the earth.

And even if it was above the earth, that would really screw up the seasons

If the sun had to shine on the bottom of 3km of clouds using reflection, it would have to be 48000km away from the place it's shining on. This is done using a simple triangle rule, I don't think I need to explain the math to anyone because it's so clear.

48000 kilometers is much larger than the perimeter of the equator. So this would only happen if the sun was somewhere outside the edge of the earth.

Ah, your earlier post suggested you meant the cloud layer extended to 3 km vertically, or in other words was 2 km thick. Thanks for the clarification.

Remember when I said that the earth wasn't as flat as a mirror (being that there are things on the ground that reflect the light in all different directions)?  Do you remember that part?  If you don't, you can scroll back up and read it.  Also, did you notice how flat the earth looks in your pic?  I did.  In fact, your measurements seem to rely heavily on that right angle.

Remember when I said that the earth wasn't as flat as a mirror (being that there are things on the ground that reflect the light in all different directions)?  Do you remember that part?  If you don't, you can scroll back up and read it.  Also, did you notice how flat the earth looks in your pic?  I did.  In fact, your measurements seem to rely heavily on that right angle.

Then I guess my graph isn't very reliable for that purpose. But to disprove the idea of straight light on a flat earth, we only need to look at the sun rise on a clear morning. It's completely clear, the atmosphere is thick enough to protect our eyes from most of the harmful rays. The sun rises, top first. It starts at 0º and begins to rise higher into the sky.

Let's calculate the angle of the sun on a morning according to the flat earth. The perimeter of the equator is about 40.000km around, am I wrong? So the sun would be 1/4 of this distance around the circle: 10.000km away, we'll just consider this the horizontal distance despite the curve of the equator. Let's see what angle of the sun would be at sunrise:

hDist=10.000: Horizontal distance of the sun
sunHeight=32.000: Vertical distance of the sun
totalDist: Shortest distance between earth and sun
Alpha: Angle of the sun compared to the ground, as seen from the observer.

totalDist = sqrt(sunHeight*sunHeight+hDist*hDist) -->
totalDist = 33526.

sin(Alpha)*totalDist = sunHeight -->
sin(Alpha) = sunHeight/totalDist -->
Alpha = arcsin(sunHeight/totalDist) -->
Alpha = 72.64º

This means the morning sun starts way up high in the sky. It's not even close enough to the earth in order to make the illusion of being behind the horizon.

Quote from: Pongo on May 19, 2010, 02:26:00 PM

Remember when I said that the earth wasn't as flat as a mirror (being that there are things on the ground that reflect the light in all different directions)?  Do you remember that part?  If you don't, you can scroll back up and read it.  Also, did you notice how flat the earth looks in your pic?  I did.  In fact, your measurements seem to rely heavily on that right angle.

Then I guess my graph isn't very reliable for that purpose. But to disprove the idea of straight light on a flat earth, we only need to look at the sun rise on a clear morning. It's completely clear, the atmosphere is thick enough to protect our eyes from most of the harmful rays. The sun rises, top first. It starts at 0º and begins to rise higher into the sky.

Let's calculate the angle of the sun on a morning according to the flat earth. The perimeter of the equator is about 40.000km around, am I wrong? So the sun would be 1/4 of this distance around the circle: 10.000km away, we'll just consider this the horizontal distance despite the curve of the equator. Let's see what angle of the sun would be at sunrise:

hDist=10.000: Horizontal distance of the sun
sunHeight=32.000: Vertical distance of the sun
totalDist: Shortest distance between earth and sun
Alpha: Angle of the sun compared to the ground, as seen from the observer.

totalDist = sqrt(sunHeight*sunHeight+hDist*hDist) -->
totalDist = 33526.

sin(Alpha)*totalDist = sunHeight -->
sin(Alpha) = sunHeight/totalDist -->
Alpha = arcsin(sunHeight/totalDist) -->
Alpha = 72.64º

This means the morning sun starts way up high in the sky. It's not even close enough to the earth in order to make the illusion of being behind the horizon.

And this is partly why Parsifal invented Bendy Light. see pic above for vector path

Quote from: Maximohoundoom on May 18, 2010, 08:46:59 AM

What about Lord Wilmore? or Pongo? What do they think?

Personally, I think that it is light bouncing off of the earth and back to the clouds.  The earth is by no means flat as a mirror and everything including tiny dust particles will reflect the light from the spotlight-sun in all different directions.  So, the light you see illuminating the bottoms of the clouds is light from the spotlight that has bounced off the surface of the earth and been reflected (in this case you are seeing the light that is reflected in your direction... obviously). 

This is why the bottoms on the clouds light up before the tops.

See my earlier post about colors.

So does this mean we can finally conclude that a flat earth requires bendy light and get back to testing bendy light? I think the last place we left from was when somebody said that they didn't have a working model of our sun and bendy light. I seem to remember telling them to make one already if they actually want to prove their point.

So does this mean we can finally conclude that a flat earth requires bendy light and get back to testing bendy light? I think the last place we left from was when somebody said that they didn't have a working model of our sun and bendy light. I seem to remember telling them to make one already if they actually want to prove their point.

*sigh*
Testing beeny light is a dead end because we already know that it doesn't exist. I've lost count of the number of times I've said this: bendy light predicts effects on star positions which are not observed happening. If a theory predicts an effect and that effect isn't happening, then the theory cannot be correct.

*sigh*
Testing beeny light is a dead end because we already know that it doesn't exist. I've lost count of the number of times I've said this: bendy light predicts effects on star positions which are not observed happening. If a theory predicts an effect and that effect isn't happening, then the theory cannot be correct.

We should collect all the information about this light problem and present it as a well structured, step-by-step explanation as to why the earth cannot possibly be flat. Then, the only ones who could deny it are the religious flat earth supporters. A religious flat earth supporter would be going against his own principle of searching for the truth through observable facts.