How far is Sun in FE model? Definitely more than 5000km.

Hi FES!

I got some interest in this theory recently, I read lots of stuff but it looks like there are few things that are plain wrong based on assumptions.

The topic I'd like to discuss is how far the Sun is.

(you can skip this post and go to http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1729321#msg1729321 for more details).

According to wikipedia, the angular size of the sun is 31'6'' to 32'7''. This is enough to calculate both diameter and the distance to the Sun.

Assumptions:
1. Earth is flat.
2. Sun has fixed physical size (a ball).  Apparent change in size of the Sun is caused by the Sun moving away from us. Assuming disc shape the results are similar to the ones below.
3. Calculations are done for the observer on the equator, during 12h long day.
4. Whenever I write "=", I approximate. I tried minimizng rounding errors.

Notion:
d - the distance to the Sun
R - radius of the Sun

Using simple trigonometry we can find that, during the noon, where the Sun is above us, hence it is the closest and has the largest angular size, we have the relation d=104.131456R.
Same calculations lead us to d=107.7909868R for the case where the Sun is the furthest away, setting (or rising).

In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).

Using Pythagoras theorem we find that R=323.2912km, and so d=33653.7747km.


So the result is far from 5000km.  Such far distance automatically causes massive problems with day/night times, but I leave that off this thread.

How far the Sun really is?

Any comments?

This simple math basically proves the Sun can't be as low as FET claims. Unless you prove I am wrong...

I don't think that anybody even understands what your rant means.  I was tempted to move it to the AR section when I first saw it, but left it just in case someone else can make any sense out of your drivel.  If you want people to actually respond, it would help to make an intelligible post to start with.

If you're going to try to use your math, explain your process rather than just claiming results. You might be right or wrong, probably wrong, but until you actually explain what you're done...
"we have the relation d=104.131456R," makes no sense until you can actually show your calculations, rather than claiming results.

I don't think that anybody even understands what your rant means.  I was tempted to move it to the AR section when I first saw it, but left it just in case someone else can make any sense out of your drivel.  If you want people to actually respond, it would help to make an intelligible post to start with.

Why is it a rant?  Just needs more explanation.

Quote from: jroa on November 06, 2015, 12:51:10 AM

I don't think that anybody even understands what your rant means.  I was tempted to move it to the AR section when I first saw it, but left it just in case someone else can make any sense out of your drivel.  If you want people to actually respond, it would help to make an intelligible post to start with.

Why is it a rant?  Just needs more explanation.

He could actually explain his point, instead of just hitting random letters and numbers. 

I don't think that anybody even understands what your rant means.

If they ever bothered to make a proper picture, what I wrote would be easy to explain.

Let me explain triangles and the math a little more precise.

We have a right triangle with side (d^2+2dR)^(1/2) and legs: R and d+R. If A is the angular diameter, then A/2 is the angle between (d^2+2dR)^(1/2) and d+R. R is the radius of the Sun. Then we have

sin A/2= R/(d+R)

so

d/R=(1-sin A/2)/(sin A/2).

Now take A=31.6' to obtain

d/R=107.7909868

and A=32.7' to obtain

d/R=104.131456

This gives d between 104.131456R and 107.7909868R.


The rest is simple. We have yet another right triangle, with sides 104.131456R, 107.7909868R and 9003.1632km (107.7909868R is the hypotenuse). This triangle is formed by the observator point, the Sun at noon and at set/rise. From Pythagoras we get R, and then d.

We have a right triangle with side (d^2+2dR)^(1/2)

2dR.
Just what.

Hi FES!

I got some interest in this theory recently, I read lots of stuff but it looks like there are few things that are plain wrong based on assumptions.

The topic I'd like to discuss is how far the Sun is.

According to wikipedia, the angular size of the sun is 31'6'' to 32'7''. This is enough to calculate both diameter and the distance to the Sun.

Assumptions:
1. Earth is flat.
2. Sun has fixed physical size (a ball).  Apparent change in size of the Sun is caused by the Sun moving away from us. Assuming disc shape the results are similar to the ones below.
3. Calculations are done for the observer on the equator, during 12h long day.
4. Whenever I write "=", I approximate. I tried minimizng rounding errors.

Notion:
d - the distance to the Sun
R - radius of the Sun

Using simple trigonometry we can find that, during the noon, where the Sun is above us, hence it is the closest and has the largest angular size, we have the relation d=104.131456R.
Same calculations lead us to d=107.7909868R for the case where the Sun is the furthest away, setting (or rising).

In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).

Using Pythagoras theorem we find that R=323.2912km, and so d=33653.7747km.


So the result is far from 5000km.  Such far distance automatically causes massive problems with day/night times, but I leave that off this thread.

How far the Sun really is?

If they liked/understood math and geometry, they wouldn't believe the earth is flat in the first place

2dR.
Just what.

Can you be more specific? I do not have mindreading skills.

If they liked/understood math and geometry, they wouldn't believe the earth is flat in the first place

You may assume that people know math (trig) in suffucient level, but it turns out that most of them tend to forget. I can explain things in more details, but I expect readers not to show ignorance.

It is not only understanding, but also proper application showing that there is something wrong.

Also, posting things like that:

I was tempted to move it to the AR section when I first saw it, but left it just in case someone else can make any sense out of your drivel.

just proves one thing: some people here are willing to discuss FE stuff, but they do not even put an effort in reading and checking numbers. It is a shame that a moderator (someone who should present a generic, proper behaviour style on the forum) simpy wanted to throw my posts calling them "drivel". If he spent few minutes, he wouldn't call them "random letters and numbers".

Quote from: TheBigOne on November 06, 2015, 05:42:31 AM

2dR.
Just what.

Can you be more specific? I do not have mindreading skills.

Where the hell did you get that? You're either multiplying the distance to the Sun by its diameter, which is pointless, or you've inexplicably tried to use twice the distance to the Sun.
Try explaining a post rather than handwaving through a series of assertions.

Try explaining a post rather than handwaving through a series of assertions.

We have a right triangle.

Hypotenuse is R+d.
One of legs is R.

Pythagoras theorem says that 2nd leg is sqrt((R+d)^2-R^2).

Did you even bother to draw a triangle? This is completely elementary "math" (actually, it is just calculations).

This lenght actually does not matter, It is here just for completion and can be skipped without ruining the rest of the argument.

brouwer, let us go back to your first message.

There is a difference in applying angular diameter calculations re: spherical bodies, as opposed to applying them to objects in the shape of a disk.

Here is the formal definition: The angle subtended at the observer by a diameter of a distant spherical body which is perpendicular to the line between the observer and the center of the body.

For a disk-shaped body we have this formula, using the graphic http://wpcontent.answers.com/wikipedia/commons/thumb/4/49/Angular_dia_formula.JPG/400px-Angular_dia_formula.JPG :

@ = 2 arctan (1/2 x d/D), if D is much larger than d, then we can approximate by @ = d/D
@ = angular diameter

Let me remind you that indeed the sun has the shape of a disk.

Impossibility of a round Sun shape:

The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

The pressure of light is sometimes referred to as to explain the low atmospheric pressure on the sun. At the surface of the sun, the pressure of light must be 2.75 milligrams per square centimeter; a cubic centimeter of one gram weight at the surface of the earth would weigh 27.47 grams at the surface of the sun. Thus the attraction by the solar mass is 10,000 times greater than the repulsion of the solar light. Recourse is taken to the supposition that if the pull and the pressure are calculated for very small masses, the pressure exceeds the pull, one acting in proportion to the surface, the other in proportion to the volume. But if this is so, why is the lowest pressure of the solar atmosphere observed over the sunspots where the light pressure is least?

Because of its swift rotation, the gaseous sun should have the latitudinal axis greater than the longitudinal, but it does not have it. The sun is one million times larger than the earth, and its day is but twenty-six times longer than the terrestrial day; the swiftness of its rotation at its equator is over 125 km. per minute; at the poles, the velocity approaches zero. Yet the solar disk is not oval but round: the majority of observers even find a small excess in the longitudinal axis of the sun. The planets act in the same manner as the rotation of the sun, imposing a latitudinal pull on the luminary.

Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.

Near the polar regions of the sun, streamers of the corona are observed, which prolong still more the axial length of the sun.

If planets and satellites were once molten masses, as cosmological theories assume, they would not have been able to obtain a spherical form, especially those which do not rotate, as Mercury or the moon (with respect to its primary).



Solar Atmosph. Pressure as a Function of Depth (official science information)

Depth (km) % Light from this Depth Temperature (K) Pressure (bars)

0 99.5 4465 6.8 x 10-3
100 97 4780 1.7 x 10-2
200 89 5180 3.9 x 10-2
250 80 5455 5.8 x 10-2
300 64 5840 8.3 x 10-2
350 37 6420 1.2 x 10-1
375 18 6910 1.4 x 10-1
400 4 7610 1.6 x 10-1

This table indicates that the solar atmosphere changes from being almost completely transparent to being almost opaque over a distance of about 400 km. Notice also that in this region the temperature drops rapidly as we near the surface, and that the pressure (measured in bars, where one bar is the average atmospheric pressure at the surface of the Earth) is very low - generally 1% or less of Earth surface atmospheric pressure.


Now, let us go back to the very subject of your thread.

http://evildrganymede.net/rpg/world/angular_diameters.pdf

To calculate the angular diameter, you need to know the diameter of the object in question, and how far away it is from the observer.


Here is where each and every scientist (from Picard, official chronology of history, to today) makes the mistake: they will use the following data, diameter 1392000 km, distance 149600000 km (for the sun-earth system).


You made the same mistake.

According to wikipedia, the angular size of the sun is 31'6'' to 32'7'', but they used the same wrong data taken from the textbooks on heliocentricity.

Moreover, the assumptions made by the official figures offerred by textbooks on astronomy (including the work attributed to J. Picard), rely on the very wrong ideas about stellar parallax:

http://web.archive.org/web/20150321094726/http://www.realityreviewed.com/Negative%20parallax.htm


Here is a classic work on the angular size of Mars, how the assumptions made by the figures offered by official astronomy, are absolutely wrong:

On the angular size of Mars:

http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4

Let me remind you that indeed the sun has the shape of a disk.

That does not make any difference. 5000km remains too small. I repeated calculations for the disk and obtained ~320km diameter and ~33600km distance. Basically the same result as for ball model.

Unfortunately, disk shape suspended 5000km above the surface would make no sense, as it would be seen as an ellipse near the set or rise time, unless it somehow faces towards us. But a disk shape cannot face toward every observer on the Earth at the same time in such a way each observer sees a disk. This is possible for ball shape.

To calculate the angular diameter, you need to know the diameter of the object in question, and how far away it is from the observer.

Or you can measure this directly on the sky. No diameter and distance required. Measuring gives approximately 32'. Using a better equipment you can increase precision. And find a proper interval. Very small intervals work against 5000km.

In case of the Sun with fixed size, suspended 5000km above the surface, the noticable change of the angular diameter would be significant. At noon it would be 2x larger than at set/rise. Anyone would notice that huge difference.

Quote from: sandokhan on November 08, 2015, 12:38:00 AM

Let me remind you that indeed the sun has the shape of a disk.

That does not make any difference. 5000km remains too small. I repeated calculations for the disk and obtained ~320km diameter and ~33600km distance. Basically the same result as for ball model.

Unfortunately, disk shape suspended 5000km above the surface would make no sense, as it would be seen as an ellipse near the set or rise time, unless it somehow faces towards us. But a disk shape cannot face toward every observer on the Earth at the same time in such a way each observer sees a disk. This is possible for ball shape.

Quote from: sandokhan on November 08, 2015, 12:38:00 AM

To calculate the angular diameter, you need to know the diameter of the object in question, and how far away it is from the observer.

Or you can measure this directly on the sky. No diameter and distance required. Measuring gives approximately 32'. Using a better equipment you can increase precision. And find a proper interval. Very small intervals work against 5000km.

In case of the Sun with fixed size, suspended 5000km above the surface, the noticable change of the angular diameter would be significant. At noon it would be 2x larger than at set/rise. Anyone would notice that huge difference.

The distance to the sun is 149Mkm.

Hypotenuse is R+d.

Just what. Did you draw the triangle? The only way R+d is going to give a relevant straight line, let alone a hypotenuse, is if you measure to the core of the Sun, which is nonsenical.
I'm not going to trust your calculations if you can't get a basic step correct.

Unfortunately, disk shape suspended 5000km above the surface would make no sense...

Agreed. The sun cannot possibly orbit at an orbit of 3000 miles (4800 km) above the Earth.

Or you can measure this directly on the sky. No diameter and distance required. Measuring gives approximately 32'.

Do not forget that in each such measurement certain assumptions are made (just as J. Picard, official chronology of history, had to make re: stellar parallax). And if no diameter and distance are required, then you might get a similar reading for a DIFFERENT set of figures: a smaller Sun, and a much lower orbit.

I can dismiss the 149million km Sun figure immediately using the faint young sun paradox, not to mention the double forces of attractive gravitation paradox...

Again, read how the angular size of Mars is completely wrong...

http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4

Unfortunately, disk shape suspended 5000km above the surface would make no sense...

Agreed. The sun cannot possibly orbit at an orbit of 3000 miles (4800 km) above the Earth.

Or you can measure this directly on the sky. No diameter and distance required. Measuring gives approximately 32'.

Do not forget that in each such measurement certain assumptions are made (just as J. Picard, official chronology of history, had to make re: stellar parallax). And if no diameter and distance are required, then you might get a similar reading for a DIFFERENT set of figures: a smaller Sun, and a much lower orbit.

I can dismiss the 149million km Sun figure immediately using the faint young sun paradox, not to mention the double forces of attractive gravitation paradox...

Again, read how the angular size of Mars is completely wrong...

http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4

I think your link, although it probably has some applications, probably does not apply to astronomy. If so, please answer the following question.

How can I see the light from a small flashlight (1" in diameter) a mile away?

Given:

• 1 mi = 5280' = 63,360"
• 24 arcsec = 0.006667°

Max height = tan(0.006667°) x 63,360" = 7.37"

You should not see the light from the flashlight. Max distance I should see it is 8594" = 716'

Please explain...

Quote from: sandokhan on November 08, 2015, 02:33:31 AM

...

...
Please explain...

  I think there should somewhere be permanent post which encourages users to not start debating with sandokhan. It really lives in its own world and never really listens anyone but goes on and on and on with its half baked stories. Just my 2 cents.

Agreed. The sun cannot possibly orbit at an orbit of 3000 miles (4800 km) above the Earth.

So how high is it?

The point of this thread is to answer this question, possibly giving a final and correct answer. It is clear that 5000km is wrong which is what, based on personal observation, most FET give and believe.

Do not forget that in each such measurement certain assumptions are made (just as J. Picard, official chronology of history, had to make re: stellar parallax).

I know that. I assumed that the change in size is due to the Sun moving away during the day, nothing more.

And if no diameter and distance are required, then you might get a similar reading for a DIFFERENT set of figures: a smaller Sun, and a much lower orbit.

You do not. Actually, all you need is any round object of known size that you can suspend in air and a little bit of math to figure, how far you need to be from this object to see it with proper angular diameter.


The main question remains open. How far is the Sun in FE model? I hope you can present more than just a number, because explaining the value is the most important part.

brouwer, you were asked several questions by thebigone to which you provided no answers, so far. Basic questions about your calculations.

Actually, all you need is any round object of known size that you can suspend in air and a little bit of math to figure, how far you need to be from this object to see it with proper angular diameter.

You should do some homework before you come here to post your messages.

Let us go to the textbook on angular size/diameter actual calculations.


The angular sizes of objects show how much of the sky an object appears to cover. Angular size does not, however, say anything about the actual size of an object.

How large an object appears depends not only on its size, but also on its distance.


Therefore, contrary to your armchair opinion, DISTANCE IS CRUCIAL TO CALCULATE THE ACTUAL SIZE OF AN OBJECT.

You are correct. You can not determine size or distance from just the angular size of the sun. You need its actual diameter or its actual distance to calculate the other.

You will probably need an "Eratosthenes experiment" but with at least 3 and preferably 5-10 locations. This will determine the shape of the Earth (Flat or Round) based on where the measurements converge, then based on that you can find the distance to the Sun, then the diameter of the Sun - probably with a good deal of accuracy (like <1%). This can probably also be applied to the Moon as well using a Full Moon.

If they ever bothered to make a proper picture, what I wrote would be easy to explain.

Let me explain triangles and the math a little more precise.

We have a right triangle with side (d^2+2dR)^(1/2) and legs: R and d+R. If A is the angular diameter, then A/2 is the angle between (d^2+2dR)^(1/2) and d+R. R is the radius of the Sun. Then we have

mistaken assumptions. 
Your initial premises are wrong. 

The sun does not rotate exactly parallel to the surface of the earth nor does it stay exactly equidistant from the North Pole. 

The sun moves up and down as well as in and out. 

Quote from: Brouwer on November 07, 2015, 10:36:56 PM

Hypotenuse is R+d.

Just what. Did you draw the triangle? The only way R+d is going to give a relevant straight line, let alone a hypotenuse, is if you measure to the core of the Sun, which is nonsenical.
I'm not going to trust your calculations if you can't get a basic step correct.

Sorry that I missed your post. I measure d to be the distance to the surface to the Sun, not to the center of the Sun. The triangle I am talking about has legs R and (d^2+2dR)^(1/2) and hypotenuse is R+d. I hope that should clarify the stuff.

The way I measure d (to the core or to the surface) will not change the final result at all (the value of R will remain the same).

The angular sizes of objects show how much of the sky an object appears to cover. Angular size does not, however, say anything about the actual size of an object.

Which is all what I need - how much of the sky is covered.

Therefore, contrary to your armchair opinion, DISTANCE IS CRUCIAL TO CALCULATE THE ACTUAL SIZE OF AN OBJECT.

It is crutial, but we are do not need the distance and size to start with. They are found using simple trig.

mistaken assumptions. 
Your initial premises are wrong. 

The sun does not rotate exactly parallel to the surface of the earth nor does it stay exactly equidistant from the North Pole. 

The sun moves up and down as well as in and out. 

To keep the angular size in measurable range, objects moving up and down/in and out should change their size to perfectly match obesrvable angular size.

Can you present your numbers/math then? Mine is based on 6h time shift (from noon to set) and assumptions that seemed natural. I would like too see a different math that would help me understanding, how it is possible that the fixed size object can keep its angular diameter from such low distance.

For instance, assuming the sun is really 5000 km above, it moves 9000km away and say 1000km up/down, it should look significantly smaller (about twice as smaller angular size). But it does not.

I cannot seem to understand the distance, but I am the only one who presented any numbers and calculations. Handwaving and saying stuff without supporting it with citations/math/other sources is not what I call a reliable argument.

There are more paradoxes with the distance, but I figured I will stick to the observable angular size. If you want, I can present the other one.

Quote from: TheBigOne on November 08, 2015, 02:09:45 AM

Quote from: Brouwer on November 07, 2015, 10:36:56 PM

Hypotenuse is R+d.

Just what. Did you draw the triangle? The only way R+d is going to give a relevant straight line, let alone a hypotenuse, is if you measure to the core of the Sun, which is nonsenical.
I'm not going to trust your calculations if you can't get a basic step correct.

Sorry that I missed your post. I measure d to be the distance to the surface to the Sun, not to the center of the Sun. The triangle I am talking about has legs R and (d^2+2dR)^(1/2) and hypotenuse is R+d. I hope that should clarify the stuff.

The way I measure d (to the core or to the surface) will not change the final result at all (the value of R will remain the same).

Quote from: sandokhan on November 09, 2015, 01:00:22 PM

The angular sizes of objects show how much of the sky an object appears to cover. Angular size does not, however, say anything about the actual size of an object.

Which is all what I need - how much of the sky is covered.

Quote from: sandokhan on November 09, 2015, 01:00:22 PM

Therefore, contrary to your armchair opinion, DISTANCE IS CRUCIAL TO CALCULATE THE ACTUAL SIZE OF AN OBJECT.

It is crutial, but we are do not need the distance and size to start with. They are found using simple trig.

Quote from: Charming Anarchist on November 09, 2015, 08:19:58 PM

mistaken assumptions. 
Your initial premises are wrong. 

The sun does not rotate exactly parallel to the surface of the earth nor does it stay exactly equidistant from the North Pole. 

The sun moves up and down as well as in and out. 

To keep the angular size in measurable range, objects moving up and down/in and out should change their size to perfectly match obesrvable angular size.

Can you present your numbers/math then? Mine is based on 6h time shift (from noon to set) and assumptions that seemed natural. I would like too see a different math that would help me understanding, how it is possible that the fixed size object can keep its angular diameter from such low distance.

For instance, assuming the sun is really 5000 km above, it moves 9000km away and say 1000km up/down, it should look significantly smaller (about twice as smaller angular size). But it does not.

I cannot seem to understand the distance, but I am the only one who presented any numbers and calculations. Handwaving and saying stuff without supporting it with citations/math/other sources is not what I call a reliable argument.

There are more paradoxes with the distance, but I figured I will stick to the observable angular size. If you want, I can present the other one.

The distance has been proven, why are you trying to disagree when you are guessing at numbers.

and hypotenuse is R+d.

Here's a hint. When I directly question where on earth you get a ridiculous measure of the hypotenuse from, you need to do a bit more than just repeat it. If you are not measuring to the center of the Sun then R+d as you've defined them mean nothing.

The distance has been proven, why are you trying to disagree when you are guessing at numbers.

I am not guessing any numbers. I am making assumptions and use simple math to derive the outcome. It does not matter whether the Sun changs its distance daily by a small amount, it does not matter if it does that when chaning seasons, it does not matter if the orbit is ellipctic. You repeat the math and obtain the same thing - 5000km is definitely not enough.

Here's a hint. When I directly question where on earth you get a ridiculous measure of the hypotenuse from, you need to do a bit more than just repeat it. If you are not measuring to the center of the Sun then R+d as you've defined them mean nothing.

There is no difference in measuring to the center and to the surface. It is just setting a convention for this calculation only...

tangent^2+radius^2=(d+radius)^2

There the mysterious d+R comes, and so (d^2+2dR)^(1/2). Got it?

If I take d to be measured to the point B, I will have sin A/2 =R/d and then I can repeat the entire argument as I did in 1st post to obtain the same value of R.

To show how wrong the argument from wiki is:

Thomas Winship, author of Zetetic Cosmogony, provides a calculation demonstrating that the sun can be computed to be relatively close to the earth's surface if one assumes that the earth is flat --
    On March 21-22 the sun is directly overhead at the equator and appears
    45 degrees above the horizon at 45 degrees north and south latitude. As
    the angle of sun above the earth at the equator is 90 degrees while it is
    45 degrees at 45 degrees north or south latitude, it follows that the angle
    at the sun between the vertical from the horizon and the line from the
    observers at 45 degrees north and south must also be 45 degrees. The result
    is two right angled triangles with legs of equal length. The distance between
    the equator and the points at 45 degrees north or south is approximately 3,000
    miles. Ergo, the sun would be an equal distance above the equator.

We can repeat the same experiment (same date). The Sun remains overhead at the equator. But this time instead of 45 degrees we take 5 degrees (this can be seen close to the North Pole). The math says that if one of legs is 3000 miles and the angle between hypotenuse and the other leg is 5 degrees, then the other leg has almost 34290 miles. This is how far the observer would have to be from the equator to see the Sun at 5 degree angle. Which is nonsense as the distance from the equator to the NP is significantly smaller.

I see no more replies.

Why?

The distance is one of FET's foundation and noone is willing to discuss it? Noone is willing to provide anything that could eventually  solve the issue I have noticed?

I see no more replies.

Why?

The distance is one of FET's foundation and noone is willing to discuss it? Noone is willing to provide anything that could eventually  solve the issue I have noticed?

If you would please post your point in a single, concise question, I will be happy to provide you with an answer.