A few questions

Quote from: Silicon on August 28, 2017, 04:25:34 PM

Quote from: JackBlack on August 28, 2017, 02:57:36 PM

Quote from: Silicon on August 28, 2017, 07:25:43 AM

Uh well look at the numbers he used???

Who cares what numbers he used.
Would you prefer for him to use every 5000 km up to 640 000 km to show a point? That will be a very big post.
Zeno's paradox is nothing more than a red herring you are appealing to to try and escape the fact that the FE model does not match reality.
It isn't a case of breaking down the distance into infinitely smaller parts. That is still a finite distance. The issue is that in order for the sun to reach 0 degrees angle of elevation while it remains 5000 km higher than the FE disc, it needs to be infinitely far away.
So unless your sun is speeding up to infinite speeds, that isn't going to happen.
Even if it did, that would make it set for everyone at the same time.

Thats my point.  The numbers grow unrealistically out of proportion at small angles.  The round earth has the exact same problem, the distance between what the observer sees from the sun and the horizon could be halved infinitely. HELLO?!!?

No, that is his point, and a point I have raised before.
The numbers grow unrealistically out of proportion for your model FE at small angles.

The round Earth does not have the same problem.

In the "common" FE model, the sun circles above you, in a plane parallel to Earth.
This means in order to set, it needs to get infinitely far away which is completely unrealistic (which is one reason why the FE model is unrealistic).

In the RE model, the actual model of reality, the apparent position of the sun is based upon the angular difference between your position and the position which the sun is 90 degrees above.
So if you take one location, with the sun directly overhead, and then go 90 degrees around Earth (in any direction), the sun will be at roughly 0 degrees (in reality it is more like 0.5 degrees due to refraction, but then if you go another 0.5 degrees it will typically drop to 0).

So no, there is no problem like this in the RE.

The other way of thinking about it is from a relative point of view, where the sun appears to circle around an axis near you (for all practical purposes it may as well go through you, the difference is that small), aligned with the axis of Earth (which is the axis it would actually appear to circle around), offset some distance from you along this axis with this distance varying through the year.
This means it will go below the horizon.

The simple case for an observer on the equator at the equinox will be the sun effectively circling them in a plane perpendicular to the ground.
So it will set, no problem.

I was simply trying to show you this problem is everywhere, assuming numbers are infinite.  If you were to get a ruler and put it in front of your eyeball while watching the sun go behind the curve on your fake ball earth, it should never happen. Anyway this is pointless conversation...

Quote from: Silicon on August 28, 2017, 04:25:34 PM

Yup it works.  I've really touched a nerve with you on this cause you can't disprove it, and its fun to watch you flip out.

No, you have touched a nerve with me because you keep repeating the same refuted bullshit.

I have disproven it. I have shown that the "correction" is pure bullshit, contradicts any correction an honest person would make and that you are completely unable to justify it.
I have pointed out that even with this correction you still get large variation.
I have shown how to do it the correct way, and the massive variation it produces.
I have shown how to do it in a way based upon the fact Earth is round which produces very little variation.

So no, I have disproven it, you just refuse to admit it.

So how about you stop lying and just admit that it doesn't work.
If you are unwilling to do so then defend your BS.

The burden of proof is on you to show it works. You are yet to do that.

So...you're calling me a liar and at the same time saying you need a proof of how it works.  Ooooook. Anyway there is more proof, and it is devastating to your globe earth model.  Still it's not perfect, and I'm trying to figure out why.  It has to do with the fact that you cannot project a flat surface onto a globe without cutting some corners somewhere.

Quote from: Silicon on August 28, 2017, 04:25:34 PM

The FE sun never "sets"  It goes out of view due to perspective.   I know you've been kinda slow on the uptake here but... I claim it does this at around the mathematical 20 degree mark under the math that supports a 4828km high sun. It would not surprise me in the least if the actual number was 23.5 degrees.

No, I haven't been slow. I am just not accepting your bullshit.

In order for the FE sun to appear to set due to perspective it would need to be infinitely far away, and that would still result in it shrinking to a point, not setting.

The mathematical angle we are discussing is the angle it appears at due to perspective.

So no, it does not set at around the mathematical 20 degree mark. At the mathematical 20 degree mark it would appear at 20 degrees.

Most likely it does.  From mexico in the example don't you think its strange going NESW from 10k distance using the "axis tilt angle of 23.5" the sun is on the horizon with a height close to the claimed 4828km??  I didn't think you would.  All Jackasses are indoctrinated coincidence theorists.

Quote from: Silicon on August 28, 2017, 04:25:34 PM

Actually I am wrong on the tool I mentioned above, its not time and date its
https://www.sunearthtools.com/dp/tools/pos_sun.php ...
Check it out let me know

Seriously?
So rather than just accept that you were full of shit you instead just provide a link to another tool which will also show you are full of shit.
You didn't even have the decency to do it yourself. (However, going to that site I see your issue. You are adjusting for DST, I won't, and instead will list the proper UTC times).

So lets stick in the 2 points I used:
Khatanga, Russia, at 1:46 UTC +7 the following day:
Elevation 6.63 degrees.
So below 30.

But what about the point south?
Well, going to Punta Arenas at 12:43:
Elevation 13.58 degrees.
So again below 30.

So that site works out the same.

Going to admit you are full of shit now?
If not, how many more times will I need to refute you before you do?

Yes that was the issue, but as far as the sun setting, the theory on the 'mathematical' 23-25 degree range still holds.

Quote from: Silicon on August 28, 2017, 07:38:23 AM

Isn't it funny how my formula doesn't work on your fake example?  It shows big differences, so because of this you know it doesn't "force" anything.  This could only mean your numbers would never occur in reality. Why can't you find this example at all in past present or future numbers???

The only reason it is a fake example is because the FE is fake.
Yes, your BS doesn't "force" anything and as such is still left with large variability. It just matches much better than the correct math for FE, so you act like it has solved the problem. It hasn't. Your BS still doesn't work.

Quote from: Silicon on August 28, 2017, 04:25:34 PM

If my claim is true, it would certainly explain why its been hidden for so long

It isn't being hidden. You are just being paranoid and delusional.

BS this, BS that ok Jackass.  Only a puppet in a lab would make such a response.

Quote from: Silicon on August 28, 2017, 04:25:34 PM

Well there is that video on the FE forum link I posted where the guy does use a solar filter...  Any explanations for that?

Do you mean the one I already responded to, pointing out he isn't using a solar filter and still suffers from serious glare?

Is it the one where he is claiming he is using a solar filter?  If it doesn't meet your standards, why don't you tell us what a jackass would use.

But the fact remains, the sun does set below the horizon.  You can see it.  It does not change shape or size, again you can see it.
Just go look.

How does magnetism work?

**Grabs popcorn.

Quote from: Silicon on August 28, 2017, 04:25:34 PM

Thats my point.  The numbers grow unrealistically out of proportion at small angles.  The round earth has the exact same problem, the distance between what the observer sees from the sun and the horizon could be halved infinitely. HELLO?!!?

I was simply trying to show you this problem is everywhere, assuming numbers are infinite.  If you were to get a ruler and put it in front of your eyeball while watching the sun go behind the curve on your fake ball earth, it should never happen. Anyway this is pointless conversation...

Cut the crap. You claimed the RE has the exact same problem. IT DOES NOT!!!
In the RE, the sun does not circle above us in a plane parallel to the ground, so it doesn't have the same problem.

I have no idea what you are trying to suggest with the ruler.

The only reason it is a pointless conversation is because you have no interest in the truth, are happy to continually repeat BS, making up whatever BS you want to try and pretend your delusional fantasy world is real.

So...you're calling me a liar and at the same time saying you need a proof of how it works.  Ooooook. Anyway there is more proof, and it is devastating to your globe earth model.  Still it's not perfect, and I'm trying to figure out why.  It has to do with the fact that you cannot project a flat surface onto a globe without cutting some corners somewhere.

I am calling you a liar, because that is what you are. You are repeatedly lying, claiming it works for the FE and that it hasn't been disproven.
There is no proof at all.
If you had any you would have provided it by now.
That is why I am demanding it.

If you are suggesting I shouldn't need proof because you have already provided it then you are just further backing up my claims of you being a liar as you have not shown that it works. Instead you showed that even with your BS, it still doesn't work.

Your BS isn't perfect because it doesn't match reality at all.

Yes, you cannot project a flat surface onto a globe (or vice versa) without some kind of distortion. This is further proof that Earth is round, as you cannot map Earth's surface onto a flat plane without serious distortion.

Most likely it does.  From mexico in the example don't you think its strange going NESW from 10k distance using the "axis tilt angle of 23.5" the sun is on the horizon with a height close to the claimed 4828km??  I didn't think you would.  All Jackasses are indoctrinated coincidence theorists.

No, the height isn't close to the claimed 4828 km.
You don't go 10 000 km away from Mexico. The furthest you go is 5 800 km.

When you go 10 000 km away from where the sun is directly overhead, it ends up on the horizon.
For example, using your site, going to the same time (June 13, 2016, 13:46 UTC-5 (or 14:46 GMT-5+1 for DST), and then looking around a bit, I found these coordinates:
23.2658021, -101.434021

That is where the sun is at 90 degrees.

Now then, lets go 90 degrees away from that position, which would correspond to roughly 10000 km on the flat Earth (closer to 10002 km), due north or due south. Lets start with south. So that puts us at 23.26...-90=-66.7341979.
So the location we want is -66.7341979, -101.434021

It is being mean and not letting me search there (it seems there is some error with how it handles both numbers being negative).
So I used this location instead:
-66.5787143, -101.4587402
It's close enough.

Now then, it has an elevation angle of 0.16 degrees.
So first, doing it the honest way:
tan(0.16 degrees)=h/10 000
h=10000 *tan(0.16 degrees)=27.9 km.

Last time I checked, 27.9 km is not 4828 km.

Doing it your BS way, welll, now we end up with a problem.
Remember your formula?
(((ATAN(DEGREES((elevation))))/height)*distance)
Well, the height is 27.9 km. So dividing by that wont do much. The distance is 10 000 km, so that will do a lot.
The elevation angle is 0.16 degrees, so "converting" that to degrees makes it 9.16, the inverse tan of that is 83.8 degrees or 1.46 radians (the default unit excel works in)
So ballbark figures, we have 15000/30, and thus get a "correction" of 500 degrees (523.5899786 to be more precise and accurate).
This ends up giving us a height of -2914.738256 km. So not only is it the wrong sign, it is also no where near your BS 3000 miles.

If instead you go north, you go to the other side of the world, ending up at 23.2658021+90 N, or 90-23.2658021 N on the other side (90-((23.2658021+90)-90) =66.7341979
And the latitude is -101.434021+180=78.565979

So the location we want is:
66.7341979, 78.565979
This gives an elevation angle of 0.
So you get a height of 0, and a div0 error for your calculation.
If we go a bit closer (to the north), then we are back to where we were before.

Or by 10 k did you mean 10 km?
10 km would be roughly 90/1000 or .09 degrees, so going north for simplicity that puts you at 23.3558021, -101.434021, with an elevation of 89.91 degrees.
So putting that in to the real formula you get 6366.192488 km height for the sun. Using your BS, you get close to 5000 km.

So no, it still doesn't work, and once again you are spouting pure garbage.

And what does the tilt angle have to do with this other than being why the sun is above that location?

Yes that was the issue, but as far as the sun setting, the theory on the 'mathematical' 23-25 degree range still holds.

No it doesn't.
The mathematical angle still is what you would expect to observe in reality if Earth was flat.

BS this, BS that ok Jackass.  Only a puppet in a lab would make such a response.

So is that your problem? You are just a pathetic puppet?

I have pointed out why it is BS. Something you are yet to refute.

Is it the one where he is claiming he is using a solar filter?  If it doesn't meet your standards, why don't you tell us what a jackass would use.

Whatever the jackass that made the video used.

I'm not a jackass so that isn't what I would use.

I would use a proper solar filter which is made for direct viewing of the sun. One in which only the sun is visible through (or other extremely bright lights), which you are capable of zooming in on the sun and seeing it clearly as a sharp circle, typically with various solar activity visible to some extent if the camera is good enough.

I would not use some piece of shit that still has the sky lit up with massive glare.

Quote from: JackBlack on August 29, 2017, 04:19:56 AM

Quote from: Silicon on August 28, 2017, 04:25:34 PM

Well there is that video on the FE forum link I posted where the guy does use a solar filter...  Any explanations for that?

Do you mean the one I already responded to, pointing out he isn't using a solar filter and still suffers from serious glare?

Is it the one where he is claiming he is using a solar filter?  If it doesn't meet your standards, why don't you tell us what a jackass would use.

I can't find the link to your magic video,  so would you mind posting it again?

Now as to filter standards, if it doesn't score the sharpness of image and freedom from glare if the following video, then your video is useless as evidence.
Sun size stays constant from the flat earther,  Matrix Decode:.

And the moon also stays very close to the same size, no filter needed:
And while you are at it, you seem to totally ignore the simply fact that the uncorrected,  unfudged  sun angles do not give consistent sun heights on the flat earth, but fit the Globe with a very distant sun almost perfectly.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

So this time, I will present the sun elevations and azimuth from five locations all close to longitude 70°W. These locations are shown on the Google Earth map on the right. The sun azimuth and elevations have been found from: Sun Earth Tools. If you have any doubts as the accuracy of this site, I suggest that a good test would be to check its accuracy where you live. I think if it is accurate at a lot of random locations is could be relied on for these locations.

Locations for Sun Height Calculations

The following table gives the data for each location. All sun elevation was obtained from Sun Earth Tools as close as possible to the local midday on the last equinox. The time was UTC 20/Mar/2016  16:48.

Location	Latitude	Longitude	Sun Elev	Dist from    Vaupes	Flat Sun Ht	Lat Diff from    Vaupes	Calc Circum
Kimmirut, Canada	62.847°	-69.869°	27.36°	7,034 km	3,609 km	63.58°	39,828 km
Santo Domingo	18.486°	-69.931°	71.72°	2,107 km	6,077 km	19.22°	39,465 km
Vaupes, Colombia	-0.565°	-69.634°	89.06°	0 km	------
Chupa District, Peru	-15.109°	-69.998°	74.69°	1,610 km	6,256 km	14.37°	40,334 km
Punta Arenas, Chile	-53.164°	-70.917°	36.63°	5,830 km	4,388 km	52.43°	40,031 km

These locations and the directions to the sun on a flat earth are shown in the left hand  diagram below:
Once we have the angles from two sites the height of the sun can be calculated from: h = d/(1/tan(A₁) + 1/tan(A₂)).

Sun Height on Flat Earth along 70°W Long

Sun Height on Globe Earth along 70°W Long

Using this method to find the height of the sun on the Flat earth gives measurements from 3609 km (for Kimmirut and Vaupes) to 6256 km (for Chupa District to Vaupes) depending on the spacing of the measurement sites.

In other words, claiming that the Flat Earth sun is at about 5,000 km altitude has no foundation whatever.

It is very telling when we note that when we plot these angles on a spherical earth the directions to the sun are all parallel.

Explain that!

Now, if instead of using these measurements to determine the Flat Earth sun height, we use them as Eratosthenes did, assuming a distant sun and use this data to calculate the circumference of the earth.

The circumference can be calculated from (distance from Vaupes) * 360°/(angle difference of sun from Vaupes)

This time, we get far better consistency.
              The estimated figures for the circumference of the earth range from 39,465 km to 40,334 km.

Certainly these figures would indicate that the earth is a globe with a distant sun.

And this time don't try any stupid crap about the 89.06° angle - there is no numerical problem caused by that.

Mr Silicon, just face the facts!
The sun elevation angles do not fit you flat earth, without your fudge factors that try  to force the measurements to you preconceived sun height.
Some scientific method - guess the answer first then fudge the measurements til they fit your guess!

<< edit: had left JackBlack out and we can'r have that! >>

Maybe Silicon doesn't know what an angle is...?

Silicon, an angle is the measure of how spread apart two lines are when they intersect in the same plane. A good example of this is a pair of scissors. The two sides meet at the joint of the scissors, but they can be spread open different amounts (closed, wide, or in between). A degree is a unit for measuring angles. One degree is 1/360 of a circle. It's a little like the tiny second marks on an analog clock, but there are 360 marks instead of just 60.

One fourth of a circle is 90 degrees. This is the measurement of a "square" corner. Most buildings have many 90-degree corners: where the floor meets the wall is a good example.

We can use this "floor meets wall" example to think about the angle of the Sun in the sky. Imagine standing on a tiny island in the Pacific at solar noon on the equator on the equinox. The ocean will be like the floor, and the Sun will be straight up the wall. The Sun will be at 90 degrees from all directions. If you continue to watch the Sun, it will appear to move 15 degrees per hour to the west. In six hours (6x15=90), it will reach the horizon at zero degrees. After a few more minutes, it will appear to move below the horizon (below zero degrees) and disappear.

Quote from: Sponge Guy on August 17, 2017, 10:26:31 AM

How does magnetism work?

**Grabs popcorn.

In my model, the force that accellerates the Earth also moves the electons that are bound to the solid mass.  When ions are moved, electric charge is produced.  As electricity moves, current causes a magnetic field.  This magnetic field is what you witness. 

Quote from: disputeone on August 29, 2017, 11:47:35 PM

Quote from: Sponge Guy on August 17, 2017, 10:26:31 AM

How does magnetism work?

**Grabs popcorn.

In my model, the force that accellerates the Earth also moves the electons that are bound to the solid mass.  When ions are moved, electric charge is produced.  As electricity moves, current causes a magnetic field.  This magnetic field is what you witness.

There are a few issues with that. Firstly, Earth isn't charged, so all the positive ions would produce a magnetic field which counteracts the negative ions.
Secondly, ignoring that and pretending Earth was just negative ions, the field produced from various locations around the disk would result in a weaker field the further from the edge you go, with no field in the middle.
Thirdly, ignoring both of those, that would produce a field parallel to the surface of Earth. There would be no dip.

Quote from: jroa on August 30, 2017, 04:49:51 AM

Quote from: disputeone on August 29, 2017, 11:47:35 PM

Quote from: Sponge Guy on August 17, 2017, 10:26:31 AM

How does magnetism work?

Idiot.  Ions have a charge.  Perhaps you ate going to rewrite all of physics? 

**Grabs popcorn.

In my model, the force that accellerates the Earth also moves the electons that are bound to the solid mass.  When ions are moved, electric charge is produced.  As electricity moves, current causes a magnetic field.  This magnetic field is what you witness.

There are a few issues with that. Firstly, Earth isn't charged, so all the positive ions would produce a magnetic field which counteracts the negative ions.
Secondly, ignoring that and pretending Earth was just negative ions, the field produced from various locations around the disk would result in a weaker field the further from the edge you go, with no field in the middle.
Thirdly, ignoring both of those, that would produce a field parallel to the surface of Earth. There would be no dip.

Idiot.  Ions have a charge.  Please stop lying. 

Quote from: disputeone on August 29, 2017, 11:47:35 PM

Quote from: Sponge Guy on August 17, 2017, 10:26:31 AM

How does magnetism work?

**Grabs popcorn.

In my model, the force that accellerates the Earth also moves the electons that are bound to the solid mass.  When ions are moved, electric charge is produced.  As electricity moves, current causes a magnetic field.  This magnetic field is what you witness.

My spelling is too bad for me to criticize others but with you i must ask this question:

Is "electons" a typo or have you again invented  another magical substance to force observations to fit your model?

Quote from: jroa on August 30, 2017, 04:49:51 AM

Quote from: disputeone on August 29, 2017, 11:47:35 PM

Quote from: Sponge Guy on August 17, 2017, 10:26:31 AM

How does magnetism work?

**Grabs popcorn.

In my model, the force that accellerates the Earth also moves the electons that are bound to the solid mass.  When ions are moved, electric charge is produced.  As electricity moves, current causes a magnetic field.  This magnetic field is what you witness.

My spelling is too bad for me to criticize others but with you i must ask this question:

Is "electons" a typo or have you again invented  another magical substance to force observations to fit your model?

Wow, because missing a letter completly changes the message, right, a-hole?

Not at all, merely checking if you invented something else like your magic temperature inversion, magical aether, etc.
Don't  get your panties in a knot, it is your practice to invent things to force observations to fit your model not mine.
I was simply curious if this was another instance  of that or a typo.

Not at all, merely checking if you invented something else like your magic temperature inversion, magical aether, etc.
Don't  get your panties in a knot, it is your practice to invent things to force observations to fit your model not mine.
I was simply curious if this was another instance  of that or a typo.

And, your practice is to come onto a controversial science forum in order to stir up trouble. 

Quote from: Silicon on August 29, 2017, 09:56:02 PM

Quote from: Silicon on August 28, 2017, 04:25:34 PM

Thats my point.  The numbers grow unrealistically out of proportion at small angles.  The round earth has the exact same problem, the distance between what the observer sees from the sun and the horizon could be halved infinitely. HELLO?!!?

I was simply trying to show you this problem is everywhere, assuming numbers are infinite.  If you were to get a ruler and put it in front of your eyeball while watching the sun go behind the curve on your fake ball earth, it should never happen. Anyway this is pointless conversation...

I have no idea what you are trying to suggest with the ruler.

I know.  If you understood even half of what I was trying to say we would be having a very different conversation.  But I don't think you're paid to understand.

Quote from: Silicon on August 29, 2017, 09:56:02 PM

So...you're calling me a liar and at the same time saying you need a proof of how it works.  Ooooook. Anyway there is more proof, and it is devastating to your globe earth model.  Still it's not perfect, and I'm trying to figure out why.  It has to do with the fact that you cannot project a flat surface onto a globe without cutting some corners somewhere.

Yes, you cannot project a flat surface onto a globe (or vice versa) without some kind of distortion.

Well at least we agree on something.  I'll just leave this one here for now.

Quote from: Silicon on August 29, 2017, 09:56:02 PM

Most likely it does.  From mexico in the example don't you think its strange going NESW from 10k distance using the "axis tilt angle of 23.5" the sun is on the horizon with a height close to the claimed 4828km??  I didn't think you would.  All Jackasses are indoctrinated coincidence theorists.

No, the height isn't close to the claimed 4828 km.
You don't go 10 000 km away from Mexico. The furthest you go is 5 800 km.

When you go 10 000 km away from where the sun is directly overhead, it ends up on the horizon.
For example, using your site, going to the same time (June 13, 2016, 13:46 UTC-5 (or 14:46 GMT-5+1 for DST), and then looking around a bit, I found these coordinates:
23.2658021, -101.434021

That is where the sun is at 90 degrees.

Now then, lets go 90 degrees away from that position, which would correspond to roughly 10000 km on the flat Earth (closer to 10002 km), due north or due south. Lets start with south. So that puts us at 23.26...-90=-66.7341979.
So the location we want is -66.7341979, -101.434021

It is being mean and not letting me search there (it seems there is some error with how it handles both numbers being negative).
So I used this location instead:
-66.5787143, -101.4587402
It's close enough.

Now then, it has an elevation angle of 0.16 degrees.
So first, doing it the honest way:
tan(0.16 degrees)=h/10 000
h=10000 *tan(0.16 degrees)=27.9 km.

Last time I checked, 27.9 km is not 4828 km.

Doing it your BS way, welll, now we end up with a problem.
Remember your formula?
(((ATAN(DEGREES((elevation))))/height)*distance)
Well, the height is 27.9 km. So dividing by that wont do much. The distance is 10 000 km, so that will do a lot.
The elevation angle is 0.16 degrees, so "converting" that to degrees makes it 9.16, the inverse tan of that is 83.8 degrees or 1.46 radians (the default unit excel works in)
So ballbark figures, we have 15000/30, and thus get a "correction" of 500 degrees (523.5899786 to be more precise and accurate).
This ends up giving us a height of -2914.738256 km. So not only is it the wrong sign, it is also no where near your BS 3000 miles.

If instead you go north, you go to the other side of the world, ending up at 23.2658021+90 N, or 90-23.2658021 N on the other side (90-((23.2658021+90)-90) =66.7341979
And the latitude is -101.434021+180=78.565979

So the location we want is:
66.7341979, 78.565979
This gives an elevation angle of 0.
So you get a height of 0, and a div0 error for your calculation.
If we go a bit closer (to the north), then we are back to where we were before.

Or by 10 k did you mean 10 km?
10 km would be roughly 90/1000 or .09 degrees, so going north for simplicity that puts you at 23.3558021, -101.434021, with an elevation of 89.91 degrees.
So putting that in to the real formula you get 6366.192488 km height for the sun. Using your BS, you get close to 5000 km.

So no, it still doesn't work, and once again you are spouting pure garbage.

And what does the tilt angle have to do with this other than being why the sun is above that location?

The sun is on the horizon roughly 10,030 km in all directions from the example in Mexico.  If you were to take the RE axial tilt number of 23.5 degrees as your angle:

 =TAN(RADIANS(23.5))*10030 = 4,361

using my formula it adds roughly 3.61 which gives a height of about 5,135.01

if you average the two you get 4,748.

Now, this is measured over a distance of 10k km of your distorted globe so I feel this is pretty f'ing accurate considering.

So in short, yes I believe it does.

Quote from: JackBlack on August 29, 2017, 04:19:56 AM

Is it the one where he is claiming he is using a solar filter?  If it doesn't meet your standards, why don't you tell us what a jackass would use.

Whatever the jackass that made the video used.

I'm not a jackass so that isn't what I would use.

I would use a proper solar filter which is made for direct viewing of the sun. One in which only the sun is visible through (or other extremely bright lights), which you are capable of zooming in on the sun and seeing it clearly as a sharp circle, typically with various solar activity visible to some extent if the camera is good enough.

I would not use some piece of shit that still has the sky lit up with massive glare.

Well tell us what you would use.

Maybe Silicon doesn't know what an angle is...?

Silicon, an angle is the measure of how spread apart two lines are when they intersect in the same plane. A good example of this is a pair of scissors. The two sides meet at the joint of the scissors, but they can be spread open different amounts (closed, wide, or in between). A degree is a unit for measuring angles. One degree is 1/360 of a circle. It's a little like the tiny second marks on an analog clock, but there are 360 marks instead of just 60.

One fourth of a circle is 90 degrees. This is the measurement of a "square" corner. Most buildings have many 90-degree corners: where the floor meets the wall is a good example.

We can use this "floor meets wall" example to think about the angle of the Sun in the sky. Imagine standing on a tiny island in the Pacific at solar noon on the equator on the equinox. The ocean will be like the floor, and the Sun will be straight up the wall. The Sun will be at 90 degrees from all directions. If you continue to watch the Sun, it will appear to move 15 degrees per hour to the west. In six hours (6x15=90), it will reach the horizon at zero degrees. After a few more minutes, it will appear to move below the horizon (below zero degrees) and disappear.

I still don't think you fully understand what I'm presenting here.  Think about it some more.

And this time don't try any stupid crap about the 89.06° angle - there is no numerical problem caused by that.

Mr Silicon, just face the facts!
The sun elevation angles do not fit you flat earth, without your fudge factors that try  to force the measurements to you preconceived sun height.
Some scientific method - guess the answer first then fudge the measurements til they fit your guess!

Have you even looked at the differences between the tangent angles between our examples?  Yes there is a difference, and that difference is compounded over distance with the method we're using

The sun is on the horizon roughly 10,030 km in all directions from the example in Mexico.  If you were to take the RE axial tilt number of 23.5 degrees as your angle:

 =TAN(RADIANS(23.5))*10030 = 4,361

using my formula it adds roughly 3.61 which gives a height of about 5,135.01

if you average the two you get 4,748.

Now, this is measured over a distance of 10k km of your distorted globe so I feel this is pretty f'ing accurate considering.

First of all, this isn't measured, its calculated and for calculation the error is very big. Secondly I don't understand why you put round earth axial tilt to your equation when you calculate height for flat earth?

Quote from: jroa on August 30, 2017, 04:49:51 AM

Quote from: disputeone on August 29, 2017, 11:47:35 PM

Quote from: Sponge Guy on August 17, 2017, 10:26:31 AM

How does magnetism work?

**Grabs popcorn.

In my model, the force that accellerates the Earth also moves the electons that are bound to the solid mass.  When ions are moved, electric charge is produced.  As electricity moves, current causes a magnetic field.  This magnetic field is what you witness.

My spelling is too bad for me to criticize others but with you i must ask this question:

Is "electons" a typo or have you again invented  another magical substance to force observations to fit your model?

JROA!!!!!! WHAT DO YOU MEAN "YOUR" MODEL?

You are the forefront of Flat Earth Science and are the only one figuring this our?

Sounds to me like L. Ron Hubbard Junior is working on sequel to Dianetics.... There is only 1 model, and that is the truth... not what ever you come up with... I just hope you are testing these hypotheses and prepping your findings for peer review.

Quote from: NAZA on August 30, 2017, 08:24:49 AM

Not at all, merely checking if you invented something else like your magic temperature inversion, magical aether, etc.
Don't  get your panties in a knot, it is your practice to invent things to force observations to fit your model not mine.
I was simply curious if this was another instance  of that or a typo.

And, your practice is to come onto a controversial science forum in order to stir up trouble.

Science?   LMAO!
Science is making observations and determining a model that fits those observations and can make verifiable  predictions.

Science is not picking a model from a book of fables and twisting laws of nature with magic to make observations fit the model.

Coming soon:
Jroa's multidimensional  earth theory also known as The Beanstalk Model.

RE: science logic science science measurement science logic science logic test measurement science...

FE: fantasy speculation guess "science word" guess legend fantasy "science word" nonsense babble speculation...

RE: Nu'uh because my teacher said so.

FE: "I saw it on Youtube!"

RE:  I want to discuss such and such with FE'ers, so why do you delieve such and such, you bunch of inbred idiots?

Jroa...

SHAME *BONG*


SHAME *BONG*


SHAME *BONG*

1: if the Earth is flat, then how does Kim Jong Un, leader of North Korea, located on the absolute eastern side of Asia, expect to bomb California, located on the absolute western side of America, via the Pacific Ocean?

He won't. That's how deep the governments go to make the public think the Earth is spherical.

2: if the Earth is flat, then how does a gravitational pull function?  Even if the world was flat a few million years ago, it would have been pulled into a spherical shape (because a sphere shape is the easiest to form) as a result of its immense size (See http://www.qrg.northwestern.edu/projects/vss/docs/space-environment/zoom-grav.html
for more detail).

That is because gravity is acceleration based on the fact that the Earth goes upwards. That's why we're not floating and that's why the Earth isn't flat.

3: if the Earth is flat, why doesn't the law of parallax show?  Wouldn't the moon and sun shrink as it rises and sets since they would be moving farther away from the center?

They're moving further away from you according to the globe Earth model too since the moon is apparently in an elliptical orbit and so is the Earth itself. You don't see parallax then do you?

4: if the Earth is flat, why is the Earth's shadow a perfect circle upon the moon during a total lunar eclipse?  Wouldn't the shadow be flattened only a small bit at the very least?  Why does the moon have phases?

These questions have both been answered to death recently.

5: if the Earth is flat, then Antarctica would surround the circle, correct?  Then how does, for at the most, 8 months at a time, Antarctica have the sun shining constantly, but when you move North (or inward, towards the center), the sun starts to rise and set normally?

Correct. Antarctica is an ice wall surrounding the Earth. I'd love to answer this question and many more, but it's time for me to get some sleep. In the mean time, probably worth it to ask NASA. Actually, nevermind.

Idiot.  Ions have a charge.  Please stop lying.

Yes, you are an idiot.
When did I say they didn't?
I said Earth as a whole does not.
It has both positive ions and negative ions. These produce opposing magnetic fields.

I know.  If you understood even half of what I was trying to say we would be having a very different conversation.  But I don't think you're paid to understand.

No, I understand almost everything you say. I just realise it is pure bullshit.
Perhaps if you understood and acted like an honest, rational human being we would be having a very different conversation, with you admitting the FE model doesn't match reality.

And you are right, I'm not paid to understand your BS. I'm here on my own volition. No one is paying me.

Quote from: JackBlack on August 30, 2017, 04:03:23 AM

Yes, you cannot project a flat surface onto a globe (or vice versa) without some kind of distortion.

Well at least we agree on something.  I'll just leave this one here for now.

So you accept one of the proofs for Earth being round. Good.

If you wish to object, please provide a map of Earth which doesn't have any distortions.

The sun is on the horizon roughly 10,030 km in all directions from the example in Mexico.  If you were to take the RE axial tilt number of 23.5 degrees as your angle:

Right, so you are just making up pure garbage which has nothing at all to do with axial tilt and nothing at all to do with mexico.

=TAN(RADIANS(23.5))*10030 = 4,361

This formula means nothing.
The elevation angle of the sun is not 23.5 degrees.
And this is not the height you claimed.

using my formula it adds roughly 3.61 which gives a height of about 5,135.01
if you average the two you get 4,748.

Your formula is pure BS, and again, that isn't the height you claimed.

Also, why would you average them?

You seem to just be doing whatever BS you can with the numbers to say "hey look, it's what I want".

Now, this is measured over a distance of 10k km of your distorted globe so I feel this is pretty f'ing accurate considering.

What is measured over a distance?
You stuck numbers in a formula to produce a result that means nothing.

And again, the globe is not the one that is distorted. Measurements on a globe match reality.

So in short, yes I believe it does.

And just like always, you are full of shit.
You are just fucking with numbers to get the number you want.
It would be no better than this:
Hey, the axial tilt is 23.5 degrees.
If I add 4804.5 km to that, I get the height of 4828 km I want.

So no, it doesn't.

Quote from: JackBlack on August 30, 2017, 04:03:23 AM

I would use a proper solar filter which is made for direct viewing of the sun. One in which only the sun is visible through (or other extremely bright lights), which you are capable of zooming in on the sun and seeing it clearly as a sharp circle, typically with various solar activity visible to some extent if the camera is good enough.

Well tell us what you would use.

I already did. Learn to read what has been said.

I still don't think you fully understand what I'm presenting here.  Think about it some more.

No, we understand, we just know it is bullshit.

You are trying to suggest that when all the math indicates an object should appear at a roughly 25 degrees angle of elevation, it magically drops to 0.

It is pure bullshit not based in reality at all.
We can measure it ourselves. It also makes no sense at all.
At the same time it would require an angle offset of 0 to magically change to -25, but only when looking towards where the sun should be, an angle of 65 to magically change to 90 when looking straight up, but at other times not have any effect at all, and for some reason it magically only effects the sun.

It is pure bullshit,

You can try it yourself.
Find a nice long hallway, with a nice flat roof.
Measure the height from your eye level to the roof (feel free to sit down or lie down or otherwise get your eyes closer to the floor).

Then draw a circle on the roof, or put a ball there.
Now, standing under it, measure the angle of elevation. It will be 90 degrees.
Now move away horizontally the distance between your eye level and the roof. Now it will be roughly 45 degrees.
Move that distance again (in the same direction), it will have dropped to roughly 26 degrees.
Keep moving various distances and measuring the angles.

If you notice, the REAL OBSERVED angles match those predicted by the math.

So it is just your FE BS which doesn't match.

Have you even looked at the differences between the tangent angles between our examples?  Yes there is a difference, and that difference is compounded over distance with the method we're using

Yes, I have seen the difference between them.
Guess what? For a RE it doesn't matter. It all matches and produces a consistent result.

It is only your FE BS where it isn't consistent, and his angle doesn't matter. Even without it, your results are still inconsistent with massive variation and still quite large variation even with your unsubstantiated, refuted BS.

RE:  I want to discuss such and such with FE'ers, so why do you delieve such and such, you bunch of inbred idiots?

Thats sounds quite a lot what some inbread idiot would say. Typos included.

They're moving further away from you according to the globe Earth model too since the moon is apparently in an elliptical orbit and so is the Earth itself. You don't see parallax then do you?

The difference is in scale.
For the RE model, the moon is roughly 400 000 km away. The sun is 150 000 000 km away. Due to these vast distances, the change in length is negligible (but still detectable with the right equipment, parallax will show, just not to the naked eye).
With the FE model, the sun and moon are roughly 5000 km above Earth. That puts them 5000 km away when directly overhead.
At sunset at the north pole, it is above a point roughly 10 000 km away. That makes it roughly 11200 km away.
At the equator on the equinox sets when it is above a point 14 142 km away, to the north west, making it 15 000 km away. Also, it is observed to set due west.

So in the FE model you have significant variations in distances, with one example having the sun's distance change by a factor of 3. You also get quite large variations in angles, with one example showing a difference of 45 degrees between expected and observed.

Quote from: Sponge Guy on August 17, 2017, 10:26:31 AM

4: if the Earth is flat, why is the Earth's shadow a perfect circle upon the moon during a total lunar eclipse?  Wouldn't the shadow be flattened only a small bit at the very least?  Why does the moon have phases?

These questions have both been answered to death recently.

Yep, to the death of the FE, being completely unable to provide any rational answer.

Quote from: jroa on August 30, 2017, 07:40:23 AM

Idiot.  Ions have a charge.  Please stop lying.

Yes, you are an idiot.
When did I say they didn't?
I said Earth as a whole does not.
It has both positive ions and negative ions. These produce opposing magnetic fields.

So, are you saying that ions have no charge, idiot?

Quote from: JackBlack on August 30, 2017, 02:19:38 PM

Quote from: jroa on August 30, 2017, 07:40:23 AM

Idiot.  Ions have a charge.  Please stop lying.

Yes, you are an idiot.
When did I say they didn't?
I said Earth as a whole does not.
It has both positive ions and negative ions. These produce opposing magnetic fields.

So, are you saying that ions have no charge, idiot?

Again, Yes, you are an idiot.
When did I say they didn't?
I said Earth as a whole does not.
It has both positive ions and negative ions. These produce opposing magnetic fields.
Ions have charge, but if you have a neutral object (like Earth), these charges cancel.

Do you understand that, idiot?

The field produced by a moving positive charge will be opposite the field produced by a negative charge moving the same way.
Do you understand that, idiot?

As they produce opposite fields, they cancel.
Do you understand that, idiot?

As they cancel, it means no net field.
Do you understand that, idiot?

Even simpler, DO YOU UNDERSTAND HOW TO READ, IDIOT?

Do you understand what words mean, idiot?

Can you comprehend English, idiot?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
I still don't think you fully understand what I'm presenting here.  Think about it some more.

Quote from: rabinoz on August 30, 2017, 04:15:58 AM

And this time don't try any stupid crap about the 89.06° angle - there is no numerical problem caused by that.

Mr Silicon, just face the facts!
The sun elevation angles do not fit your flat earth, without your fudge factors that try to force the measurements to you preconceived sun height.
Some scientific method - guess the answer first then fudge the measurements till they fit your guess!

Have you even looked at the differences between the tangent angles between our examples?  Yes there is a difference, and that difference is compounded over distance with the method we're using

• I could not care less about your fudge factors!
• You have never responded to the very simple straightforward claims that I have made.
• I still can't find any reference to your magic video that shows the sun's size changing,

So please Mr Silicon, just face the facts!

• Unless you claim that the light from the sun bends through impossible angles your model of the sun over the flat earth is impossible.
• The size of the sun does not change measurably from rising to setting, except for possible small changes near the horizon

Here, I've put the North Pole into the mix, I'll add it to the diagrams when I get time.
The following table gives the data for each location.
All sun elevation was obtained from Sun Earth Tools as close as possible to the local midday on the equinox of UTC 20/Mar/2016  16:48.

Now you try to fit a sun elevation of 0.53° and flat earth sun height of 93 km into your fudged equations.