The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Debate => Topic started by: SantaClause on February 13, 2018, 09:04:25 PM

So i am new to this forum but i have to say this is some of the most entertaining content i have ever read on the entire internet .... ever LOL!! Ive read all about your experiments and theories, and looked at your maps for a flat earth over the last couple of days. I can go with all of your ideas that the whole world is in a giant conspiracy to keep the truth from the world that its flat, even doctored photos from the shadow government at google etc. lol this would make for one heck of a science fiction novel and would convince some people .... BUT....
I have one flaw in your entire theory... according to the "Flat Earth" model there is an enormous ice wall surrounding the south pole... or well in this case surrounding the planet.... The problem i have with this is a few years ago i actually took a charter sight seeing flight to Antarctica via this company...
http://www.antarcticaflights.com.au/home#expect
So going along with your theory about this ice wall or barrier or edge or something ... We spent a little more than 4 hours flying over the continent ... maybe the windows were blurry...maybe I'm just nearsighted but i never saw this barrier or the edge of the world... i have 20/20 vision so I'm sure i would have seen it if it were there ... Maybe we just didn't fly close enough because the secret societies tricked our GPS location or something ....last i checked the human eye can see pretty far in a clear atmosphere though.... I duno maybe im just a paid brainwashed informant of the world government...
Just to clarify though just HOW CLOSE to the edge of this world do you have to get to see this wall because i really want to see it would be fantastic... Ive seen allot of things and seeing a giant ice wall would make my day..
Cheers

At what altitude did you fly?

So i am new to this forum but i have to say this is some of the most entertaining content i have ever read on the entire internet .... ever LOL!! Ive read all about your experiments and theories, and looked at your maps for a flat earth over the last couple of days. I can go with all of your ideas that the whole world is in a giant conspiracy to keep the truth from the world that its flat, even doctored photos from the shadow government at google etc. lol this would make for one heck of a science fiction novel and would convince some people .... BUT....
I have one flaw in your entire theory... according to the "Flat Earth" model there is an enormous ice wall surrounding the south pole... or well in this case surrounding the planet.... The problem i have with this is a few years ago i actually took a charter sight seeing flight to Antarctica via this company...
http://www.antarcticaflights.com.au/home#expect
So going along with your theory about this ice wall or barrier or edge or something ... We spent a little more than 4 hours flying over the continent ... maybe the windows were blurry...maybe I'm just nearsighted but i never saw this barrier or the edge of the world... i have 20/20 vision so I'm sure i would have seen it if it were there ... Maybe we just didn't fly close enough because the secret societies tricked our GPS location or something ....last i checked the human eye can see pretty far in a clear atmosphere though.... I duno maybe im just a paid brainwashed informant of the world government...
Just to clarify though just HOW CLOSE to the edge of this world do you have to get to see this wall because i really want to see it would be fantastic... Ive seen allot of things and seeing a giant ice wall would make my day..
Cheers
At 35000ft the horizon is only at most ~360km which is not that far in the scheme of things. Add to that you would never see properly that far due to haze and moisture. You could fly over lots of landmass in Antarctica, look over at the horizon and still not see this rumoured ice wall. If this is what your belief in the globe theory hinges on I suggest you keep looking for better answers if you don't like the flat one.
As for how close you have to get to see the edge? Well the flat earth says the sun acts like a spotlight. That means the edge would fall outside its light and look like this
(http://www.webstore.hulsta.co.uk/wpcontent/uploads/rearlacquerglassblack.jpg)

At 35000ft the horizon is only at most ~360km which is not that far in the scheme of things.
Sure, if Earth was round.
If Earth is flat, the horizon should be the edge.
As for how close you have to get to see the edge? Well the flat earth says the sun acts like a spotlight. That means the edge would fall outside its light and look like this
Yet no magic region of darkness is seen either.
Also, that just means smart people will bring a light.

If you shine a light on a void, what will you see?

If you shine a light on a void, what will you see?
If you shine a light on a flat disc, at the edge of the disc, you will see the edge. The void will make it even easier, as even a dim flashlight should be enough to make the edge of Earth distinct from the void.

If you shine a light on a void, what will you see?
Gods.

I believe bacon is the closest to God one can get.

I believe bacon is the closest to God one can get.
I can think of one cult that disagrees.

"I am new to this site, but I see one flaw..."
Construct straw man.
Destroy straw man.
Yay!!!
Norhing against you personally, but this is the 15th time this week to see this format.
Fun for you, fun for us, fun for all!

HA i love it.... well as to reply
The actual flight wasn't exceedingly high only about 10k feet. It gives you a great view of the landmass and coast lines. I actually had the privilege of speaking with a few scientists that have explored quite a bit of the landmass personally and they never mentioned anything about seeing a giant ice wall or a giant black void anywhere. Ive flown on low altitude flights as well in the states just up a couple thousand feet or so for short hops or connecting flights to a larger airline.
From personal experience i can tell you at just a couple thousand feet high you can see the Appalachian mountains quite soon after takeoff from Texas... if i were to guess i would say at least a couple hundred miles away from low altitude even with pollution and atmospheric distortion, so it would seem to me that something as endless as this void at the edge of the world should be quite visible given the lack of pollution or major atmospheric distortion that far south... unless the edge of the world is thousands of miles of frozen waste between the rest of the world and the void?

... i have to say this is some of the most entertaining content i have ever read on the entire internet .... ever LOL!!
Are you 12?

... i have to say this is some of the most entertaining content i have ever read on the entire internet .... ever LOL!!
Are you 12?
So what if he is? Are you making a proposition?

There can be two poles and the world can still be flat.

There can be two poles and the world can still be flat.
But as every astronomer on the planet knows its spherical.
Show everyone on this site one reputable astronomer who thinks the world is flat?
Sorry...did you mean to say the world is fat? if that was the case then I'm inclined to agree.

There can be two poles and the world can still be flat.
But as every astronomer on the planet knows its spherical.
Show everyone on this site one reputable astronomer who thinks the world is flat?
Sorry...did you mean to say the world is fat? if that was the case then I'm inclined to agree.
Assertion, no evidence.

There can be two poles and the world can still be flat.
But as every astronomer on the planet knows its spherical.
Show everyone on this site one reputable astronomer who thinks the world is flat?
Sorry...did you mean to say the world is fat? if that was the case then I'm inclined to agree.
Assertion, no evidence.
Agree, no evidence.

There can be two poles and the world can still be flat.
Not with them always being 180 degrees apart.

There can be two poles and the world can still be flat.
Not with them always being 180 degrees apart.
You know full well that's not true. DET exists.

You know full well that's not true. DET exists.
No, I know full well it is true.
DE does not exist other than in your mind, and it doesn't have a flat Earth.
It has 2 separate objects joined by magic.
The join between the discs means it isn't flat.

The common AE FE has accepted critiques that cannot be solved because this model is not valid. (The globe model is helpless by the way). But their idea of flatness of the earth is superb.
So I proposed South Pole Centered FE Model.

The common AE FE has accepted critiques that cannot be solved because this model is not valid. (The globe model is helpless by the way). But their idea of flatness of the earth is superb.
So I proposed South Pole Centered FE Model.
Fine, as long as all distances are correct.

The common AE FE has accepted critiques that cannot be solved because this model is not valid. (The globe model is helpless by the way). But their idea of flatness of the earth is superb.
So I proposed South Pole Centered FE Model.
Don't the exact same critiques exist independent of which pole is at the center?

You know full well that's not true. DET exists.
No, I know full well it is true.
DE does not exist other than in your mind, and it doesn't have a flat Earth.
It has 2 separate objects joined by magic.
The join between the discs means it isn't flat.
Again, you are outright lying.
It's pathetic at this stage. You know that nothing you say is true, but you don't care because people are going to believe you regardless. Roundie playbook 101.

The common AE FE has accepted critiques that cannot be solved because this model is not valid. (The globe model is helpless by the way). But their idea of flatness of the earth is superb.
So I proposed South Pole Centered FE Model.
Fine, as long as all distances are correct.
A homework for cartographers. (If they please).
Skip textbooks for a while and go to living realities.

The common AE FE has accepted critiques that cannot be solved because this model is not valid. (The globe model is helpless by the way). But their idea of flatness of the earth is superb.
So I proposed South Pole Centered FE Model.
Don't the exact same critiques exist independent of which pole is at the center?
South centered FE is welcome for critiques. So far so good.

The common AE FE has accepted critiques that cannot be solved because this model is not valid. (The globe model is helpless by the way). But their idea of flatness of the earth is superb.
So I proposed South Pole Centered FE Model.
Don't the exact same critiques exist independent of which pole is at the center?
South centered FE is welcome for critiques. So far so good.
Okay, I'm confused. What critiques did a north centered map fail that pass with a south centered version?
Mike

The common AE FE has accepted critiques that cannot be solved because this model is not valid.
No FE model is valid.
Your model is no better, it just shifts the problems around.
The globe model is helpless by the way
It doesn't need help. It has withstood the test of time with no one finding any significant problem with it.
So I proposed South Pole Centered FE Model.
Which just shifts the problems around.
Instead of the south pole being the problem as it is for the more common FE models, the north pole is a problem for you.
South centered FE is welcome for critiques. So far so good.
No.
So far it has failed repeatedly.
It cannot explain the Arctic midnight sun.
It cannot explain why Polaris is due north of everyone.
Or more generally, it cannot explain why there are 2 poles, always 180 degrees apart.
It cannot explain distances and flight durations and routes near the north pole.
Everywhere the NP centred one fails, your model just fails at the other pole.

Again, you are outright lying.
It's pathetic at this stage. You know that nothing you say is true, but you don't care because people are going to believe you regardless. Roundie playbook 101.
There you go projecting again.
You have provided no evidence to indicate your DE model exists as part of reality, as such I am being quite honest when I say it just exists in your mind.
But more importantly, it has 2 separate flat discs. This means the entirety of Earth is not flat.
At best, if you have 2 perfectly flat discs, the join between them represents the curvature, but that doesn't even match reality.
The simple fact is, on any flat surface you can construct a triangle, anywhere on the surface, and it will have an angle sum of 180 degrees.
This even applies to mathematically "flat" surfaces like the surface of an infinitely long cylinder.
But as soon as the surface is mathematically curved, this is no longer true and the angle sum deviates from 180 degrees.
Which applies for Earth?
Well, as the 2 poles are always 180 degrees apart (and there are subpolar points, i.e. points on the surface of Earth below the poles):
You have one line, directly connecting the 2 poles.
You then have a point, somewhere else (the exact location is irrelevant, as long as it isn't on the earlier line).
Now, you add 2 more lines, connecting this point to either pole.
As it is not on the pole the angle at each pole will be >0.
The angle at this point will be 180 degrees.
This means the angle sum of the triangle will be 180 degrees plus 2 angles greater than 0, making the angle sum greater than 180 degrees.
That means it isn't flat.
The only FE model (i.e. a model which has a single flat Earth) I know of that has 2 poles is the bipolar FE model.
In this model, the poles are not always 180 degrees apart.
So no, I know that what I am saying is true.

There can be two poles and the world can still be flat.
But as every astronomer on the planet knows its spherical.
Show everyone on this site one reputable astronomer who thinks the world is flat?
Sorry...did you mean to say the world is fat? if that was the case then I'm inclined to agree.
Assertion, no evidence.
That's not true, it has been evidenced by every person and machine that have ever took a photograph from space.

There can be two poles and the world can still be flat.
Not with them always being 180 degrees apart.
You know full well that's not true. DET exists.
I have recently read your evidence section again, there is nothing to show why two disks would form under your model.

The common AE FE has accepted critiques that cannot be solved because this model is not valid. (The globe model is helpless by the way). But their idea of flatness of the earth is superb.
So I proposed South Pole Centered FE Model.
Don't the exact same critiques exist independent of which pole is at the center?
South centered FE is welcome for critiques. So far so good.
Okay, I'm confused. What critiques did a north centered map fail that pass with a south centered version?
Mike
They say south pole doesn't exist. In fact so many people have visited it, travel agents offer trip to Antarctica. Circumnavigating was done. etc.
Northern hemisplane distances look narrower. Even wider version of northern distances by RET is not wide enough. As a distance detective, 8) I found odd distances on northern hemisplane presented by RET.
#ScrollDown

The common AE FE has accepted critiques that cannot be solved because this model is not valid.
No FE model is valid.
Your model is no better, it just shifts the problems around.
The globe model is helpless by the way
It doesn't need help. It has withstood the test of time with no one finding any significant problem with it.
So I proposed South Pole Centered FE Model.
Which just shifts the problems around.
Instead of the south pole being the problem as it is for the more common FE models, the north pole is a problem for you.
South centered FE is welcome for critiques. So far so good.
No.
So far it has failed repeatedly.
It cannot explain the Arctic midnight sun.
It cannot explain why Polaris is due north of everyone.
Or more generally, it cannot explain why there are 2 poles, always 180 degrees apart.
It cannot explain distances and flight durations and routes near the north pole.
Everywhere the NP centred one fails, your model just fails at the other pole.
#ScrollDown

They say south pole doesn't exist. In fact so many people have visited it, travel agents offer trip to Antarctica. Circumnavigating was done. etc.
Your model doesn't solve that, it just shifts the problem.
You claim the north pole doesn't exist, even though plenty have visited and navigation is done there all the time.

South centered FE is welcome for critiques. So far so good.
No.
So far it has failed repeatedly.
It cannot explain the Arctic midnight sun.
It cannot explain why Polaris is due north of everyone.
Or more generally, it cannot explain why there are 2 poles, always 180 degrees apart.
It cannot explain distances and flight durations and routes near the north pole.
Everywhere the NP centred one fails, your model just fails at the other pole.
HINT TRY: Spherical RE & the correct π = 3.141592653589793238.....

Oh I thought pi is already dead. (?)
Draw a periference with a compass by yourself. You'll find that 30 degree point is NOT exactly at 0.5 of Y coordinate position.
It is somewhere below 0.5.
Thank for your gut to challenge PHEW ;')

Oh I thought pi is already dead. (?)
No, it's not.
This was shown multiple times.
Draw a periference with a compass by yourself. You'll find that 30 degree point is NOT exactly at 0.5 of Y coordinate position.
If it isn't, it is because you are not drawing it accurately.
You do not need a circle for it.
It is based upon a right angle triangle.
You have the hypotenuse have a length of 2, and the far side have a length of 1.
This is because it is half of an equilateral triangle.
There is no doubt about this.
IT IS 0.5.
If you say it is something else you are wrong.

Oh I thought pi is already dead. (?)
No, it's not.
This was shown multiple times.
Draw a periference with a compass by yourself. You'll find that 30 degree point is NOT exactly at 0.5 of Y coordinate position.
If it isn't, it is because you are not drawing it accurately.
You do not need a circle for it.
It is based upon a right angle triangle.
You have the hypotenuse have a length of 2, and the far side have a length of 1.
This is because it is half of an equilateral triangle.
There is no doubt about this.
IT IS 0.5.
If you say it is something else you are wrong.
You denied what you *see. Case closed. 8)

You denied what you *see. Case closed. 8)
No, I didn't.
I explained quite clearly why it is the case.
Here is an image for you:
(https://i.imgur.com/CC0ghUV.png)
Firstly, it is not to scale.
It is showing an equilateral triangle, which has been cut in half.
Lengths are in red, angles in purple.
Let me know at what you point you disagree with me:
1  It's (the black triangle) an equilateral triangle (remember, not to scale).
2  Thus all three angles are the same
3  Thus a1=a2=a=b1+b2=60 degrees.
4  It also means all sides are equal.
5  Thus r1=r2=r=x1+x2
6  This triangle is cut in half by the grey line.
7  This means x1=x2=x.
8  Thus r=2*x
9  As the smaller triangles have 3 sides each, which are equivalent, i.e. y is common, r1=r2, and x1=x2, these triangles are congruent.
10  As these triangles are congruent the corresponding angles are equal.
11  Thus b1=b2=b.
12  Thus c1=c2=c.
13  As c1 and c2 make a straight line, c1+c2=2*c=180 degrees
14  Thus c=90 degrees
15  Thus a, b, c make a right angle triangle, with angles 60, 30 and 90 respectively with a hypotenuse of 2*x and a side adjacent to the 60 degree angle of x.
16  Cos(a), refers to the cosine of a, which is the ratio of the side in a right angle triangle adjacent to the angle a and its hypotenuse, i.e. cos(a)=x/r.
17  thus cos(60 degrees)=x/r=x/2*x=0.5
Which point of these do you disagree with?
It is quite simple reasoning which shows beyond any doubt that your claim is wrong.

You denied what you *see. Case closed. 8)
No, I didn't.
I explained quite clearly why it is the case.
Here is an image for you:
(https://i.imgur.com/CC0ghUV.png)
Firstly, it is not to scale.
It is showing an equilateral triangle, which has been cut in half.
Lengths are in red, angles in purple.
Let me know at what you point you disagree with me:
1  It's (the black triangle) an equilateral triangle (remember, not to scale).
2  Thus all three angles are the same
3  Thus a1=a2=a=b1+b2=60 degrees.
4  It also means all sides are equal.
5  Thus r1=r2=r=x1+x2
6  This triangle is cut in half by the grey line.
7  This means x1=x2=x.
8  Thus r=2*x
9  As the smaller triangles have 3 sides each, which are equivalent, i.e. y is common, r1=r2, and x1=x2, these triangles are congruent.
10  As these triangles are congruent the corresponding angles are equal.
11  Thus b1=b2=b.
12  Thus c1=c2=c.
13  As c1 and c2 make a straight line, c1+c2=2*c=180 degrees
14  Thus c=90 degrees
15  Thus a, b, c make a right angle triangle, with angles 60, 30 and 90 respectively with a hypotenuse of 2*x and a side adjacent to the 60 degree angle of x.
16  Cos(a), refers to the cosine of a, which is the ratio of the side in a right angle triangle adjacent to the angle a and its hypotenuse, i.e. cos(a)=x/r.
17  thus cos(60 degrees)=x/r=x/2*x=0.5
Which point of these do you disagree with?
It is quite simple reasoning which shows beyond any doubt that your claim is wrong.
Yes but you forgot to mention refraction, the Sagnac effect and the fact that equilateral triangles have obviously been created by NASA to fool you. Thus your proof doesn't hold, general relativity is wrong, and the Sun is 45.6 centimeters away from us.
Seriously though, do you really expect him to know what a cosine is ?

You denied what you *see. Case closed. 8)
No, I didn't.
I explained quite clearly why it is the case.
Here is an image for you:
(https://i.imgur.com/CC0ghUV.png)
Firstly, it is not to scale.
It is showing an equilateral triangle, which has been cut in half.
Lengths are in red, angles in purple.
Let me know at what you point you disagree with me:
1  It's (the black triangle) an equilateral triangle (remember, not to scale).
2  Thus all three angles are the same
3  Thus a1=a2=a=b1+b2=60 degrees.
4  It also means all sides are equal.
5  Thus r1=r2=r=x1+x2
6  This triangle is cut in half by the grey line.
7  This means x1=x2=x.
8  Thus r=2*x
9  As the smaller triangles have 3 sides each, which are equivalent, i.e. y is common, r1=r2, and x1=x2, these triangles are congruent.
10  As these triangles are congruent the corresponding angles are equal.
11  Thus b1=b2=b.
12  Thus c1=c2=c.
13  As c1 and c2 make a straight line, c1+c2=2*c=180 degrees
14  Thus c=90 degrees
15  Thus a, b, c make a right angle triangle, with angles 60, 30 and 90 respectively with a hypotenuse of 2*x and a side adjacent to the 60 degree angle of x.
16  Cos(a), refers to the cosine of a, which is the ratio of the side in a right angle triangle adjacent to the angle a and its hypotenuse, i.e. cos(a)=x/r.
17  thus cos(60 degrees)=x/r=x/2*x=0.5
Which point of these do you disagree with?
It is quite simple reasoning which shows beyond any doubt that your claim is wrong.
The comparison is okay. 2:1:√3. If Y coordinate converted to be 1, the X coordinate becomes 1:√3=0.57735. BUT the point position of 30 degrees projection at coordinate Y is a bit less than 0.5 or 0.49047 due to curving of the periference. Sorry I initially talked about calculating the earth radius based on latitude (phew tangen). We spoke about two different things. Now what's the radius scale of 30 degrees in your pi calculation/equation?

Correction:
"If X coordinate converted to be 1, the Y coordinate becomes 1:√3=0.57735"

The comparison is okay. 2:1:√3. If Y coordinate converted to be 1, the X coordinate becomes 1:√3=0.57735. BUT the point position of 30 degrees projection at coordinate Y is a bit less than 0.5 or 0.49047 due to curving of the periference. Sorry I initially talked about calculating the earth radius based on latitude (phew tangen). We spoke about two different things. Now what's the radius scale of 30 degrees in your pi calculation/equation?
I asked you a simple question, what point did you disagree with?
Why can't you answer it by pointing out a simple number?
There is no magic curving.
It is a triangle with straight lines.
You are also projecting it completely wrong.
Have the angle marked a2 as the centre (i.e. 0,0).
Scale it such that r=1.
Then you will find the x coordinate (corresponding to cos(60 deg) and sin(30 deg)) of the apex, or the line dividing the 2 triangles) is 0.5.
Now if you can't tell me which of the above 17 points were wrong, you have no justification to say cos(60 deg) is anything other than 0.5.

The comparison is okay. 2:1:√3. If Y coordinate converted to be 1, the X coordinate becomes 1:√3=0.57735. BUT the point position of 30 degrees projection at coordinate Y is a bit less than 0.5 or 0.49047 due to curving of the periference. Sorry I initially talked about calculating the earth radius based on latitude (phew tangen). We spoke about two different things. Now what's the radius scale of 30 degrees in your pi calculation/equation?
I asked you a simple question, what point did you disagree with?
Why can't you answer it by pointing out a simple number?
There is no magic curving.
It is a triangle with straight lines.
You are also projecting it completely wrong.
Have the angle marked a2 as the centre (i.e. 0,0).
Scale it such that r=1.
Then you will find the x coordinate (corresponding to cos(60 deg) and sin(30 deg)) of the apex, or the line dividing the 2 triangles) is 0.5.
Now if you can't tell me which of the above 17 points were wrong, you have no justification to say cos(60 deg) is anything other than 0.5.
Before the pointer line (hypotenuse) touches the Y value of 0.57735, at coordinate (1, 0.57735), it comes on a point at the periference curved line. For 30 degrees, the value of Y=((30:45)*0.7929)  ((30:45)*0.2929)^2 = 0.49047
It is the "phew tangent"  which is used to define distances  while for (RET) earth radius it will be translated into "phew cotangent" 60 degrees =>> 149047=0.50953.
Phew co tangent has a fixed pointer line, not like sin and cos that grows longer/shorter whose value is between 0 & 1. That's why it is like sin and cos, but actually not.
I wonder whether you've got pi math for earth radius based on latitude? Please show me the calculation.

Before the pointer line (hypotenuse) touches the Y value of 0.57735, at coordinate (1, 0.57735), it comes on a point at the periference curved line. For 30 degrees, the value of Y=((30:45)*0.7929)  ((30:45)*0.2929)^2 = 0.49047
It is the "phew tangent"  which is used to define distances  while for (RET) earth radius it will be translated into "phew cotangent" 60 degrees =>> 149047=0.50953.
Again, I gave you a simple argument with 17 points clearly outlined.
Which point don't you agree with?

Before the pointer line (hypotenuse) touches the Y value of 0.57735, at coordinate (1, 0.57735), it comes on a point at the periference curved line. For 30 degrees, the value of Y=((30:45)*0.7929)  ((30:45)*0.2929)^2 = 0.49047
It is the "phew tangent"  which is used to define distances  while for (RET) earth radius it will be translated into "phew cotangent" 60 degrees =>> 149047=0.50953.
Again, I gave you a simple argument with 17 points clearly outlined.
Which point don't you agree with?
You should have concluded that I had no objection with that. Except, I questioned you about phew value for Y=0.4907. Either you agree or disagree, please show me your version of pi calculation for 30 degree angle within a periference, not at Y coordinate value. (0.57735).

The comparison is okay. 2:1:√3. If Y coordinate converted to be 1, the X coordinate becomes 1:√3=0.57735. BUT the point position of 30 degrees projection at coordinate Y is a bit less than 0.5 or 0.49047 due to curving of the periference. Sorry I initially talked about calculating the earth radius based on latitude (phew tangen). We spoke about two different things. Now what's the radius scale of 30 degrees in your pi calculation/equation?
I asked you a simple question, what point did you disagree with?
Why can't you answer it by pointing out a simple number?
There is no magic curving.
It is a triangle with straight lines.
You are also projecting it completely wrong.
Have the angle marked a2 as the centre (i.e. 0,0).
Scale it such that r=1.
Then you will find the x coordinate (corresponding to cos(60 deg) and sin(30 deg)) of the apex, or the line dividing the 2 triangles) is 0.5.
Now if you can't tell me which of the above 17 points were wrong, you have no justification to say cos(60 deg) is anything other than 0.5.
Before the pointer line (hypotenuse) touches the Y value of 0.57735, at coordinate (1, 0.57735), it comes on a point at the periference curved line. For 30 degrees, the value of Y=((30:45)*0.7929)  ((30:45)*0.2929)^2 = 0.49047
It is the "phew tangent"  which is used to define distances  while for (RET) earth radius it will be translated into "phew cotangent" 60 degrees =>> 149047=0.50953.
Phew co tangent has a fixed pointer line, not like sin and cos that grows longer/shorter whose value is between 0 & 1. That's why it is like sin and cos, but actually not.
I wonder whether you've got pi math for earth radius based on latitude? Please show me the calculation.
periference  no such word, you have made it up.

You should have concluded that I had no objection with that.
No. If you had no objection you would have admitted cos(60 deg)=0.5, you would have accepted that you were wrong.
Instead you continued to falsely assert that cos(60) is not 0.5.
So, do you accept cos(60 deg)=0.5?
Yes or no?
If no, what point do you disagree with?

It's a tangent or cotangent stuff. Not sin/cosin.
If you wouldn't present the calulation in pi version, fine, see you on other threads. ~

The comparison is okay. 2:1:√3. If Y coordinate converted to be 1, the X coordinate becomes 1:√3=0.57735. BUT the point position of 30 degrees projection at coordinate Y is a bit less than 0.5 or 0.49047 due to curving of the periference. Sorry I initially talked about calculating the earth radius based on latitude (phew tangen). We spoke about two different things. Now what's the radius scale of 30 degrees in your pi calculation/equation?
I asked you a simple question, what point did you disagree with?
Why can't you answer it by pointing out a simple number?
There is no magic curving.
It is a triangle with straight lines.
You are also projecting it completely wrong.
Have the angle marked a2 as the centre (i.e. 0,0).
Scale it such that r=1.
Then you will find the x coordinate (corresponding to cos(60 deg) and sin(30 deg)) of the apex, or the line dividing the 2 triangles) is 0.5.
Now if you can't tell me which of the above 17 points were wrong, you have no justification to say cos(60 deg) is anything other than 0.5.
Before the pointer line (hypotenuse) touches the Y value of 0.57735, at coordinate (1, 0.57735), it comes on a point at the periference curved line. For 30 degrees, the value of Y=((30:45)*0.7929)  ((30:45)*0.2929)^2 = 0.49047
It is the "phew tangent"  which is used to define distances  while for (RET) earth radius it will be translated into "phew cotangent" 60 degrees =>> 149047=0.50953.
Phew co tangent has a fixed pointer line, not like sin and cos that grows longer/shorter whose value is between 0 & 1. That's why it is like sin and cos, but actually not.
I wonder whether you've got pi math for earth radius based on latitude? Please show me the calculation.
periference  no such word, you have made it up.
Haha sorry for my English.
"Circumference" ~

It's a tangent or cotangent stuff. Not sin/cosin.
If you wouldn't present the calulation in pi version, fine, see you on other threads. ~
No, we have been discussing cos and sin before, i.e. cosine and sine.
You claimed the cosine of 60 degrees is not 0.5.
Do you now admit that you were wrong and that the cosine of 60 degrees is in fact 0.5?
Also, sine and cosine are what is used to determine the x and y position of a point on a circle, based upon the radius and angle, not tangent and cotangent.

It's a tangent or cotangent stuff. Not sin/cosin.
If you wouldn't present the calulation in pi version, fine, see you on other threads. ~

It's a tangent or cotangent stuff. Not sin/cosin.
If you wouldn't present the calulation in pi version, fine, see you on other threads. ~
You’re just overly complicating the whole thing.
This is a simple as it gets:
r1 = r2 = (x1+x2) = 2 all sides of the isosceles triangle have a length of 2.
Drop a perpendicular Y:
r2 = 2, x2 = 1, y = 1.73205
Cosine of b2:
y/r2 = 1.73205/2 = 0.866025
Sine of b2:
x2/r2 = 1/2 = 0. 5
Tangent of b2:
x2/y = 1/1.73205 = 0.57735 = Sin(b2)/Cos(b2) = 0.5/.866025 = 0.57735
It’s as simple as that. The use of pi, or phew, should get the above. There are multiple was to calculate the number but I prefer the Taylor series. I’ve laid out how that series works for you before but I’ll do it again.
Tan(b2) = Sin(b2)/Cos(b2)
Using the Taylor series to find the values above for pi:
Tan(b2) = 0.5/0.866025404 = 0.57735027
Using the Taylor series to find the values above for phew:
Tan(b2) = 0.4913206/0.8635163 = 0.56897671
Phew converges at the incorrect values. Therefore, phew is wrong and pi is correct.
Mike

If a pointer line meets the 30° circumference, the ratio Y:X will always be 0.57735:1.
Taylor series is a presetting method. It ain't deserve to change what phew math has presented with real figures & real circumference.
I always talked about the projection of 30° point into Y value.
0.57735 is out off context. It goes without saying.
It's about "phew tangent", not mere 'tangent'.
0.49047 is observable.

If a pointer line meets the 30° circumference, the ratio Y:X will always be 0.57735:1.
That was not the ratio we were discussing.
If you want that, then depending on which is Y and which is X you will have sqrt(3)/1 or 1/sqrt(3).
You seem to have taken the 1/sqrt(3) option.
That is 0.57735...
But that wasn't being discussed.
What was being discuss is cos(60 deg) and sin(30 deg) which you repeatedly claimed was not 0.5.
Again, do you accept that cos(60 deg)=sin(30 deg)=0.5?
Taylor series is a presetting method.
No it isn't.
It is based upon the derivatives of the sin or cos function.
It ain't deserve to change what phew math has presented with real figures & real circumference.
Your phew math has always been a load of bullshit.
I always talked about the projection of 30° point into Y value.
Yes, the projection of it onto a unit circle, i.e. take a circle with radius 1, what is the y value of the point?
In that case, it is 0.5.
You are now trying to scale it to pretend you haven't made a massive mistake.
0.49047 is observable.
Stop lying.
I have just shown that 0.5 is observable beyond any shadow of a doubt.
If you think I am wrong, tell me which of the 17 points are wrong.

If a pointer line meets the 30° circumference, the ratio Y:X will always be 0.57735:1.
Taylor series is a presetting method. It ain't deserve to change what phew math has presented with real figures & real circumference.
I always talked about the projection of 30° point into Y value.
0.57735 is out off context. It goes without saying.
It's about "phew tangent", not mere 'tangent'.
0.49047 is observable.
Observable only if you have no idea what you are talking about. The real value is 0.5. It is indisputable.