The distance to the stars in FE theory

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mak3m

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Re: The distance to the stars in FE theory
« Reply #60 on: October 28, 2019, 04:11:51 PM »
Failing to take into account that Sirus is a Binary star, gravitationally unbound from the cluster where it formed. Its moving towards the solar system and will do for tens of thousands of years, it won't be visible in the northern hemisphere in about 4000 years.

Sure.

Please explain why Sirius does keep up so precisely with the exponentially increasing rate of precession?

How can Sirius' proper motion stay synched up so precisely with precession, when the rate of precession itself is changing?


If any local force in here the "heliocentrical" solar system drove up the rate of precession, it would NOT also drive up the proper motion of Sirius across the sky.


In the official theory of astrophysics, Sirius is 8.6 LIGHT YEARS from Earth.

THAT IS 81 TRILLION KILOMETERS.

And yet it keeps up precisely with the exponential increase of the rate of precession.

Its displacement from the ecliptic makes its helical rising appear more regular than other stars. Its period is also almost exactly a solar year.

I have already pointed out that you do not understand the term exponential.

But the Sirius binary system is not static, its changing but at a much reduced rate than other observations, such as the so called zodiac constellations on the ecliptic plane.

Your inability to appreciate long timescales does not make your position correct. Neither does it provide a shred of evidence to calculate the 50km distance of an object of just over 2 solar masses
You have to learn to reply without quoting a long previous answer.

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sandokhan

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Re: The distance to the stars in FE theory
« Reply #61 on: October 28, 2019, 10:42:25 PM »
http://I have already pointed out that you do not understand the term exponential.

I wrote the book on exponentials.

I provided the first explicit formula in the history of mathematics for the exponential function.

GLOBAL COSINE/ARCCOSINE/EXPONENTIAL/LOGARITHM/ARCTANGENT FUNCTIONS



The hypothenuse is labeled as c (which unites points A and C), side a is located on the x axis (which unites points A and B), and we also have side b. Angle θ is located between sides c and a (cos θ = a/c).

Point D will be the intersection of the circle with the positive x axis.

We first calculate the value of segment CD, in terms of a, b and c: (2c2 - 2ac)1/2

We then succesively bisect the chord CD, and each hypothenuse thus obtained (if we divide CD in half the midpoint will be E, and the intersection of the segment AE with the circle will be labeled as F; then we calculate this new hypothenuse CF in terms of the values obtained earlier, and so on, aiming to get as close to the value of s [arc length of CD] as possible], into smaller and smaller equal segments, calculating each succesive value in terms of a, b and c, in order to obtain a very close approximation of s (the arc length between points C and D); since s = rθ, where r = c = 1, by acquiring an exceptional figure for s, we correspondingly then get the value of θ.

Letting c = 1, we finally obtain:


COS θ =  1/2 x (({ [( (2 - θ2/2N)2 - 2)2...]2 - 2}2 - 2))    (n/2 + 1 evaluations)

COS-1 θ =  2n x {2 - ((2 + {2 + [2 + (2 + 2θ)1/2]1/2}1/2...))}1/2   (n + 1 evaluations)


The cosine formula is a GLOBAL formula; by contrast the Maclaurin cosine series is a local formula:



My global cosine formula is the SUM of the Maclaurin cosine expansion.


We know that the Maclaurin hyperbolic cosine expansion is:

cosh x = 1 + x2/2! + x4/4! + ...

Therefore, by just changing the sign in the global cosine formula, we obtain immediately the GLOBAL hyperbolic cosine formula:

COSH V =  1/2 x (([(({[(2 + V2/2n)2 - 2]2} - 2))2...-2]2 - 2))   (n/2 + 1 evalutions)

This is the global hyperbolic cosine formula which is the sum of the corresponding local Maclaurin power series expansion.


We then immediately obtain the GLOBAL natural logarithm formula:

LN V =  2n x ((-2 + {2 + [2 + (2 + 1/V + V)1/2]1/2...}1/2))1/2   (n+1 evaluations)


By summing the nested continued square root function, we finally obtain:


LN V = 2n x (V1/2n+1 - 1/V1/2n+1)

This is the first explicit global formula for the natural logarithm, which can be used immediately to find LN V without resorting to logarithm tables, or calculators which feature the logarithm key: all we need is a calculator which has the four basic operations and the square root key. It links algebraic functions with elementary and higher transcendental functions.

For a first approximation,

LN V = 2n x (V1/2n - 1)

First results appear for n = 8 to 12, all the remaining digits for n = 19 and higher...

Example: x = 100,000        LN x = 11.5129255

with n = 20 the first approximation is LN x = 11.512445 (e11.512445 = 100,001.958)


We also can get the corresponding arctangent formula:

ARCTAN V =  2n x ((2- {2+ [2+ (2+ 2{1/(1+ V2)}1/2)1/2]...1/2}))1/2 (n+1 parentheses to be evaluated)


ERROR ANALYSIS

Here is the Maclaurin expansion for ex:



Let us obtain a remainder form for the Maclaurin expansion for ex (Lagrange remainder):

Rn(x) = f(n+1)(c)[xn+1]/(n+1)! , where c is between 0 and x

f(n+1)(c) = ec


An approximation is said to be accurate to n decimal places if the magnitude of the error is less than 0.5 x 10-n.

e1 to four decimal place accuracy:

Rn = ec/(n+1)!

since c<1, then ec < e

since e<3, then Rn < 3/(n+1)!

then, for n=8 we will obtain four decimal place accuracy.


Local formulas are difficult to use because of their very slow convergence.


By contrast, my formula is a GLOBAL formula, which rapidly converges to the result, even for large x.


COSH v = (ev + e-v)/2 =~ 1/2ev = 1/2 x (({[( ( (2 + v2/2n)2) -2)2] -2)2 ...-2}2 -2))       
(n/2 +1 evaluations)

We can turn this formula into an exact formula for ev by simply substituting y for ev, and then solve the quadratic equation for y.

One might ask, could you not use Taylor expansions to obtain cosh 10 (as an example)? No, because you would need some other value to start with, cosh 9.5 or cosh 9.8 or cosh 10.3, to apply Taylor series.

With my global formula, no such approximations are needed, we can start directly with the value v = 10.

My formula also has a built-in remainder approximation estimation: the term v2/2n.

That is, we can estimate the accuracy from the very start: this is the power of a global formula.

The higher the value of n, the better the approximation that we will obtain.


Example:

COSH 10 = 11013.233

102/220 = 0.00009536

Using the global hyperbolic cosine formula with n = 20, we get: 11012.762

« Last Edit: October 28, 2019, 10:45:42 PM by sandokhan »

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sandokhan

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Re: The distance to the stars in FE theory
« Reply #62 on: October 28, 2019, 10:50:05 PM »
Its displacement from the ecliptic makes its helical rising appear more regular than other stars. Its period is also almost exactly a solar year.

But the Sirius binary system is not static, its changing but at a much reduced rate than other observations, such as the so called zodiac constellations on the ecliptic plane.


Is this supposed to be a joke on your part?

You are trolling this thread!

EXPLAIN WHY Sirius does keep up so precisely with the exponentially increasing rate of precession.

How can Sirius' proper motion stay synched up so precisely with precession, when the rate of precession itself is changing?


If any local force in here the "heliocentrical" solar system drove up the rate of precession, it would NOT also drive up the proper motion of Sirius across the sky.


In the official theory of astrophysics, Sirius is 8.6 LIGHT YEARS from Earth.

THAT IS 81 TRILLION KILOMETERS.

And yet it keeps up precisely with the exponential increase of the rate of precession.


You are wrong also on your assessment of the situation.

SIRIUS DOES NOT UNDERGO ANY KIND OF A PRECESSIONAL MOTION AT ALL.

This is what you do not understand.

Please read:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1939662#msg1939662

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mak3m

  • 737
Re: The distance to the stars in FE theory
« Reply #63 on: October 28, 2019, 11:19:50 PM »
http://I have already pointed out that you do not understand the term exponential.

I wrote the book on exponentials.

I provided the first explicit formula in the history of mathematics for the exponential function.

GLOBAL COSINE/ARCCOSINE/EXPONENTIAL/LOGARITHM/ARCTANGENT FUNCTIONS



The hypothenuse is labeled as c (which unites points A and C), side a is located on the x axis (which unites points A and B), and we also have side b. Angle θ is located between sides c and a (cos θ = a/c).

Point D will be the intersection of the circle with the positive x axis.

We first calculate the value of segment CD, in terms of a, b and c: (2c2 - 2ac)1/2

We then succesively bisect the chord CD, and each hypothenuse thus obtained (if we divide CD in half the midpoint will be E, and the intersection of the segment AE with the circle will be labeled as F; then we calculate this new hypothenuse CF in terms of the values obtained earlier, and so on, aiming to get as close to the value of s [arc length of CD] as possible], into smaller and smaller equal segments, calculating each succesive value in terms of a, b and c, in order to obtain a very close approximation of s (the arc length between points C and D); since s = rθ, where r = c = 1, by acquiring an exceptional figure for s, we correspondingly then get the value of θ.

Letting c = 1, we finally obtain:


COS θ =  1/2 x (({ [( (2 - θ2/2N)2 - 2)2...]2 - 2}2 - 2))    (n/2 + 1 evaluations)

COS-1 θ =  2n x {2 - ((2 + {2 + [2 + (2 + 2θ)1/2]1/2}1/2...))}1/2   (n + 1 evaluations)


The cosine formula is a GLOBAL formula; by contrast the Maclaurin cosine series is a local formula:



My global cosine formula is the SUM of the Maclaurin cosine expansion.


We know that the Maclaurin hyperbolic cosine expansion is:

cosh x = 1 + x2/2! + x4/4! + ...

Therefore, by just changing the sign in the global cosine formula, we obtain immediately the GLOBAL hyperbolic cosine formula:

COSH V =  1/2 x (([(({[(2 + V2/2n)2 - 2]2} - 2))2...-2]2 - 2))   (n/2 + 1 evalutions)

This is the global hyperbolic cosine formula which is the sum of the corresponding local Maclaurin power series expansion.


We then immediately obtain the GLOBAL natural logarithm formula:

LN V =  2n x ((-2 + {2 + [2 + (2 + 1/V + V)1/2]1/2...}1/2))1/2   (n+1 evaluations)


By summing the nested continued square root function, we finally obtain:


LN V = 2n x (V1/2n+1 - 1/V1/2n+1)

This is the first explicit global formula for the natural logarithm, which can be used immediately to find LN V without resorting to logarithm tables, or calculators which feature the logarithm key: all we need is a calculator which has the four basic operations and the square root key. It links algebraic functions with elementary and higher transcendental functions.

For a first approximation,

LN V = 2n x (V1/2n - 1)

First results appear for n = 8 to 12, all the remaining digits for n = 19 and higher...

Example: x = 100,000        LN x = 11.5129255

with n = 20 the first approximation is LN x = 11.512445 (e11.512445 = 100,001.958)


We also can get the corresponding arctangent formula:

ARCTAN V =  2n x ((2- {2+ [2+ (2+ 2{1/(1+ V2)}1/2)1/2]...1/2}))1/2 (n+1 parentheses to be evaluated)


ERROR ANALYSIS

Here is the Maclaurin expansion for ex:



Let us obtain a remainder form for the Maclaurin expansion for ex (Lagrange remainder):

Rn(x) = f(n+1)(c)[xn+1]/(n+1)! , where c is between 0 and x

f(n+1)(c) = ec


An approximation is said to be accurate to n decimal places if the magnitude of the error is less than 0.5 x 10-n.

e1 to four decimal place accuracy:

Rn = ec/(n+1)!

since c<1, then ec < e

since e<3, then Rn < 3/(n+1)!

then, for n=8 we will obtain four decimal place accuracy.


Local formulas are difficult to use because of their very slow convergence.


By contrast, my formula is a GLOBAL formula, which rapidly converges to the result, even for large x.


COSH v = (ev + e-v)/2 =~ 1/2ev = 1/2 x (({[( ( (2 + v2/2n)2) -2)2] -2)2 ...-2}2 -2))       
(n/2 +1 evaluations)

We can turn this formula into an exact formula for ev by simply substituting y for ev, and then solve the quadratic equation for y.

One might ask, could you not use Taylor expansions to obtain cosh 10 (as an example)? No, because you would need some other value to start with, cosh 9.5 or cosh 9.8 or cosh 10.3, to apply Taylor series.

With my global formula, no such approximations are needed, we can start directly with the value v = 10.

My formula also has a built-in remainder approximation estimation: the term v2/2n.

That is, we can estimate the accuracy from the very start: this is the power of a global formula.

The higher the value of n, the better the approximation that we will obtain.


Example:

COSH 10 = 11013.233

102/220 = 0.00009536

Using the global hyperbolic cosine formula with n = 20, we get: 11012.762

Utter dribble

Power series expansions are littered all over the internet, copy and paste again. But that's the point isnt it, would take a while to work through.

But even if you did write it, it's nothing special. First glance no limitation to the variables so you can take the series pretty much anywhere you want.

More importantly context, what had exponentials of a log series got to do with a claimed exponential increase elliptical procession, which your own citations DOES NOT DEMONSTRATE
You have to learn to reply without quoting a long previous answer.

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sandokhan

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Re: The distance to the stars in FE theory
« Reply #64 on: October 28, 2019, 11:24:50 PM »
Your knowledge of advanced mathematics is lacking.

My formulas ARE NOT power series expansions at all: a power series expansion is a LOCAL formula.

I published the GLOBAL FORMULAS for the logarithm, hyperbolic cosine, cosine and arccosine functions, a first in mathematics.

You have just shown to everyone here that you are just trolling the upper forums.

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mak3m

  • 737
Re: The distance to the stars in FE theory
« Reply #65 on: October 28, 2019, 11:28:45 PM »
Its displacement from the ecliptic makes its helical rising appear more regular than other stars. Its period is also almost exactly a solar year.

But the Sirius binary system is not static, its changing but at a much reduced rate than other observations, such as the so called zodiac constellations on the ecliptic plane.


Is this supposed to be a joke on your part?

You are trolling this thread!

EXPLAIN WHY Sirius does keep up so precisely with the exponentially increasing rate of precession.

How can Sirius' proper motion stay synched up so precisely with precession, when the rate of precession itself is changing?


If any local force in here the "heliocentrical" solar system drove up the rate of precession, it would NOT also drive up the proper motion of Sirius across the sky.


In the official theory of astrophysics, Sirius is 8.6 LIGHT YEARS from Earth.

THAT IS 81 TRILLION KILOMETERS.

And yet it keeps up precisely with the exponential increase of the rate of precession.


You are wrong also on your assessment of the situation.

SIRIUS DOES NOT UNDERGO ANY KIND OF A PRECESSIONAL MOTION AT ALL.

This is what you do not understand.

Please read:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1939662#msg1939662

Erm no.

I have provided an explanation.

You made the claim that Sirius is 50km and cannot provide any proof.

Throw up as many paradoxes as you want, as most of your threads do, you will word spam, I will repeat question, then you report me to Mods for asking you questions.

You have to learn to reply without quoting a long previous answer.

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mak3m

  • 737
Re: The distance to the stars in FE theory
« Reply #66 on: October 28, 2019, 11:30:15 PM »
Your knowledge of advanced mathematics is lacking.

My formulas ARE NOT power series expansions at all: a power series expansion is a LOCAL formula.

I published the GLOBAL FORMULAS for the logarithm, hyperbolic cosine, cosine and arccosine functions, a first in mathematics.

You have just shown to everyone here that you are just trolling the upper forums.

Semantics doest change my point

Link to published paper please

You have to learn to reply without quoting a long previous answer.

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mak3m

  • 737
Re: The distance to the stars in FE theory
« Reply #67 on: October 28, 2019, 11:31:32 PM »
Two claims of trolling already   >:(

Just supply substantive proof of your 50km claim
You have to learn to reply without quoting a long previous answer.

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Stash

  • 3826
Re: The distance to the stars in FE theory
« Reply #68 on: October 29, 2019, 01:50:59 AM »
Your knowledge of advanced mathematics is lacking.

My formulas ARE NOT power series expansions at all: a power series expansion is a LOCAL formula.

I published the GLOBAL FORMULAS for the logarithm, hyperbolic cosine, cosine and arccosine functions, a first in mathematics.

You have just shown to everyone here that you are just trolling the upper forums.

Still confused - How did you arrive at 50 km, specifically? Nothing you've shown so far even remotely addresses where that number came from; 50 km, specifically.
No. That sudden lurch forwards is the atmospheric slosh effect.

Re: The distance to the stars in FE theory
« Reply #69 on: October 29, 2019, 01:54:55 AM »
Sandokhan can be as disparaging about mainstream science as he wishes, but what science aims to do is to provide concise and clear explanations based on real, observational data. Hence the claim about 8.5 lightyears is based on the observed annual parallax of Sirius. Sandokhan so far has not provided an equally clear and concise explanation for his claim about 50km.  We are not interested in pages and pages of copied and pasted equations, as attractive as they might look. All you need is one simple equation. d(distance)= 1/p (parallax)

As I have said before, the distances of the nearest stars are determined using parallax and in the case of Sirius that amounts to 0.38 arc seconds.  That is an annual back and forth change in the position of Sirius relative to the background stars. The distance is inversely proportional so 1/0.38" gives 2.6pc. The wobble of the Earths axis due to precession takes 26,000 years and so it doesn't take a genius to work out that precession has nothing to do with calculating star distances. 

So Sandokhan can call himself a Flat Earth Sultan, Flat Earth Scientist or whatever other labels he wishes to use but that doesn't make him right.  If you seriously believe a star which is physically larger than the Sun (claimed FE Wiki distance 3000 miles or 4,800km) yet appears as a point source of light in the sky from Earth is actually just 50km away (60 times closer than the Sun?!?), then even my 2 year old grand-daughter will detect something is a bit amiss about that claim! 

Also as noted yesterday, Sirius is the fifth nearest star to the Sun.  So what does that mean for the distance of Alpha Centauri, the Suns closes neighbour?

Please correct me on any of the above if you believe it to be wrong.  But if you do then please explain (clearly) why.
« Last Edit: October 29, 2019, 02:00:15 AM by Nucleosynthesis »

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kopfverderber

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Re: The distance to the stars in FE theory
« Reply #70 on: October 29, 2019, 02:17:28 AM »
If stars were 50km away weather balloons would almost reach them. Stars would look bigger when seen from airplanes or mountain peaks.
You must gather your party before venturing forth

Re: The distance to the stars in FE theory
« Reply #71 on: October 29, 2019, 02:29:36 AM »
FE Wiki* states that the stars are 'luminous elements' which apparently move in a layer above the Sun and the Moon.  It doesn't specifically say anything about a size as such.

If the stars are above the Sun and Moon then that would mean further away from our point of view. FE theory states the Sun is 4,800km (3000 miles) above the Earth but Sandokhan claims Sirius is only 50km from the Earth. So on those figures, how can Sirius be above the Sun and Moon?

*I should add that the term luminous elements comes from the 'other' FE Wiki. I acknowledge that there is no similar term in the FE Wiki on this site.  There doesn't appear to be any mention in this FE Wiki about the distance of the stars or indeed what mechanism or process is responsible for making them visible.
« Last Edit: October 29, 2019, 02:40:39 AM by Nucleosynthesis »

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mak3m

  • 737
Re: The distance to the stars in FE theory
« Reply #72 on: October 29, 2019, 02:57:41 AM »
Ahh morning coffee

OK bit of a retraction on the maths, it appears to work but im not sure what the purpose is.

It looks like a function to take a sequence of square roots, Im assuming you are creating a logarithm to to exponent the 2nd variable.

Looks overly complicated to me and not sure how that links to the presumed exponential increase in precession.



You have to learn to reply without quoting a long previous answer.

Re: The distance to the stars in FE theory
« Reply #73 on: October 29, 2019, 03:15:37 AM »
Since both the Sun and Sirius must be connected by the same field, we have two choices:
No, we don't have these choices.
You have provided absolutely no basis for this.

Even if we agree that they are connected by the same field, why does this give the distances you claim?

You are a literally pulling a number from no where.

According to the site you reference, what is keeping it "so perfectly in sync" is that it is held together by gravity over those 8.6 light years.
There is absolutely no justification for your claim of 50 km.

Re: The distance to the stars in FE theory
« Reply #74 on: October 29, 2019, 06:31:13 AM »
In one sentence.
State why the the crux is the crux.

Re: The distance to the stars in FE theory
« Reply #75 on: October 29, 2019, 07:40:20 AM »
How can Sirius' proper motion stay synched up so precisely with precession, when the rate of precession itself is changing?
1. It doesn't stay synced precisely. Your source says "... despite precession, Sirius and the solstice must remain about the same distance in time from one another during most of Egyptian history." [Buchwald, “Egyptian Stars under Paris Skies” (Caltech, Engineering &
Science No. 4, 2003]. You made "precisely" up.
2. The formulas for rate of precession are valid for about the current centuries, it cannot be used to calculate precession rates for ancient times. This is the part you conveniently "missed".

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Stash

  • 3826
Re: The distance to the stars in FE theory
« Reply #76 on: October 29, 2019, 04:58:10 PM »
How can Sirius' proper motion stay synched up so precisely with precession, when the rate of precession itself is changing?
1. It doesn't stay synced precisely. Your source says "... despite precession, Sirius and the solstice must remain about the same distance in time from one another during most of Egyptian history." [Buchwald, “Egyptian Stars under Paris Skies” (Caltech, Engineering &
Science No. 4, 2003]. You made "precisely" up.
2. The formulas for rate of precession are valid for about the current centuries, it cannot be used to calculate precession rates for ancient times. This is the part you conveniently "missed".

Regardless of precession and the near synching of Sirius, there is still no explanation as to where a 50 km altitude of Sirius over earth's surface comes into play. The question remains, how is the 50 km claim derived? Without some sort of derivation, the claim seems like an arbitrary number asserted without evidence.
No. That sudden lurch forwards is the atmospheric slosh effect.

Re: The distance to the stars in FE theory
« Reply #77 on: October 30, 2019, 02:07:09 PM »
So sando has still to answer why the crux is the crux or why the crux results in a 50km measurement.

On a side note, johnD has been talking about a samsung weather balloon.
Apparently these things can go up to 30-40km up.
Thus pretty much meeting the star halfway.
Are you, sando, seriously refuting that no ones seen the star from a weather balloon?
An easily ("easily") reproducible experiment?

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Stash

  • 3826
Re: The distance to the stars in FE theory
« Reply #78 on: October 30, 2019, 07:49:48 PM »
So sando has still to answer why the crux is the crux or why the crux results in a 50km measurement.

I'm guessing because there is no calculation/derivation. I think the notion is that something so far away (Sirius) couldn't be in such close procession to earth. Ergo, it must be close, I guess. But then there is a massive gap in logic/facts/calculations. Why 50 km? Why not 75? Why not 25? There is no answer to those questions. Hence the radio silence as such calculations/logic doesn't exist.

Whenever a refutation is brought up it is countered with, "RE has to answer this (random) question..."

RE does not (as they have been answered for 100's of years) but aside from that, if one wants to make a claim the that the 5th closest star to earth is in a sub-weather balloon orbit altitude, one has to justify said claim. And show why and how. No other random questions need to be fielded. Why 50 km? Simple as that.
No. That sudden lurch forwards is the atmospheric slosh effect.

Re: The distance to the stars in FE theory
« Reply #79 on: November 03, 2019, 01:31:49 AM »
The star Sirius that we see with the naked eye is the primary member of a binary system. That means it is possible to calculate that the mass of Sirius 'A' is a tad over 2 solar masses or about 4 x 10^30kg.

Now if Sirius is just 50km or so away then given that we cannot detect any physical disk for a star which has twice as much mass as the Sun, that would mean it would have to be incredibly compact, and hence incredibly dense as well. The Schwarzschild radius for such a mass (radius at which the escape velocity becomes equal to the speed of light and hence it would become a black hole) is a touch under 6km. So if Sirius was only 50km away then I think its reasonable to suggest that we on Earth would feel something of a (very significant) gravitational pull from Sirius.

FE Theory (or at least one version of it) places the Sun only 3000km away above the flat Earth. So there would certainly be a marked mutual gravitational effect between the Sun and Sirius that we don't experience in the real world. That effect would be more than enough to have sucked the Sun and the Earth into Sirius many millions of years ago. So the fact that the Sun, the Earth and Sirius seem to manage to exist without any significant mutual effects suggests that something is definitely amiss somewhere with the FE claims.

Something else which doesn't add up is that FE Wiki claims that the Sun is just 32 miles in diameter. Now we can see a definite disk (32' across on the sky or 1/2 a degree) so if the Sun is 60 times further away from us than Sirius, based on Sandokhans claim of 50 km distance for Sirius then that must make Sirius every small indeed since we cannot see the physical disk of Sirius.  Geometrically that would make Sirius considerably smaller than 6km across and hence it would have a radius less than its Schwarzschild radius which is physically impossible.   
« Last Edit: November 03, 2019, 03:53:56 AM by Nucleosynthesis »


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sokarul

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Re: The distance to the stars in FE theory
« Reply #81 on: November 03, 2019, 02:54:57 PM »
He can’t copy paste anything so he is long gone.
Sokarul

ANNIHILATOR OF  SHIFTER

Run Sandokhan run