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The Motion of the Perihelion of Mercury



 
 
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  #51  
Old January 3rd 09, 10:35 PM posted to sci.physics,sci.physics.relativity,sci.astro
Eric Gisse
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Posts: 1,465
Default The Motion of the Perihelion of Mercury

On Jan 3, 12:21*pm, Raghar wrote:
On Dec 28 2008, 11:49*pm, wrote:

THE MOTION OF THE PERIHELION OF MERCURY
In his general relativity calculation of the motion of the perihelion
of Mercury Albert Einstein had only taken into account the
gravitational actions between the Sun and the Mercury, which he also
assumed as two points.


What will be, according to the theory of general relativity, the value
of the motion of the perihelion of Mercury if the gravitational
actions of all the planets in the solar system are taken into account
and also it is taken into account that the Sun is a little oblate?


Have any done these calculations?


Well van Fandern did this analysis:http://www.metaresearch.org/cosmolog.../PerihelionAdv...


TvF writes crank nonsense, and it doesn't address the original
question.

Also there are flyby anomalies, which are possibly more accuratehttp://www2.phys.canterbury.ac.nz/editorial/Anderson2008.pdf


Completely irrelevant to the original question.


Its latitude dependence suggests
that the Earth’s rotation may be generating an effect much
larger than the frame dragging effect of General Relativity,
the Lense-Thirring effect




  #52  
Old January 4th 09, 04:29 AM posted to sci.physics,sci.physics.relativity,sci.astro
Koobee Wublee
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Posts: 815
Default The Motion of the Perihelion of Mercury

On Jan 3, 3:02 am, Eric Gisse wrote:
On Jan 2, 10:42 pm, Koobee Wublee wrote:
On Jan 2, 12:03 am, Eric Gisse wrote:


ds^2 = dx^2 + dy^2 + dz^2
ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
ds^2 = dr^2 + r^2 d\theta^2 + dz^2
ds^2 = da^2 + db^2 + dz^2


They all describe the same thing, yet "look different". But I can find
a coordinate transformation relating all of them to eachother, so they
are the same thing.


Your notation is screwed up. The r of the 2nd is not the same as r of
the 3rd. The phi of the 2nd is the same as theta of the 3rd.
Otherwise, they can all represent the same geometry. shrug


See? You are a **** hair's distance away from understanding.


What is that again?

THE LABELS DO NOT MATTER.


Wrong! The labels do matter. shrug

The COORDINATES do not matter. They all
represent the _same_ geometry.


Wrong again. To describe a geometry, you need both the labels
(coordinate) and how the labels are connected (metric). shrug

This is not what I have been talking about. It looks like very few
have understood what I am saying. That is why I feel very lonely to
be the few who after Riemann and perhaps Christoffel that have
understood the curvature business very well. shrug


So where did you properly learn this "curvature business"? All I'm
asking is for the book you learned it from. Is that asking so much?


Just my intelligence which you lack. shrug

What I am talking about is that the following geometries are different
given the same dr, dLongitude, dLatitude.


** ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
** ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2


Where


** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2


Except the coordinate labels are not the same in both coordinate
systems.


What do you mean they are not the same? I have chosen them to be the
same. shrug

Calculate the surface area of a sphere - don't just write it
down, CALCULATE it.


I did. Just look up on my older post.

What coordinate transformation? u is any function of r. This is very
basic mathematics. shrug


Yes, and it is called a "coordinate transformation". You start out in
the (t,u,\theta,\phi) coordinate system and end up in the (t,r,\theta,
\phi) coordinate system.


I did not start out in anything. I merely presented these two
independent geometries of spacetime without showing any favor of one
to the other.

You are still dreaming and jacking off in your mathemagical Realm. In
doing so, probably wishing Einstein can join you. shrug


Let me write down the two very different spacetime in a more concise
way.


** ds1^2 = c^2 (1 – K / r) dt^2 – dr^2 / (1 – K / r) – r^2 dO^2
** ds2^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2


THE GEOMETRY ds1 IS NOT THE SAME AS ds2. THAT IS BECAUSE THE METRIC
OF ds1 IS NOT THE SAME AS ds2, BUT THE COORDINATE SYSTEM DESCRIBING
BOTH ds1 and ds2 IS EXACTLY IDENTICAL.


If they are "exactly identical" they should give the same values for
the surface area of the sphere, as #1 is in Schwarzschild isotropic
coordinates which has spheres of constant (r,t) of surface area 4 pi
r^2.


They are not identical. shrug

Let us see you calculate the surface area of #2.

I would also like you calculate the Ricci scalars of both. You'll
notice something interesting.


Calculating the area does not make any difference. To any observer
observing the geometry of any curved spacetime, the answer is always
going to be 4 pi r^2. shrug

Can you dig this, punk? I don’t think you possess any minute
intelligence to get after all these years, and I really don’t care.
shrug


With any matheMagical tricks, show me how you can coordinate-transform
from one equation to the other and what justify one solution is more
unique than the others. Remember that the coordinate system is the
same with both equations.


Except the coordinate systems are different


Hey, I am the one presenting the problem. Both ds1 and ds2 share the
same coordinate system. They are all solved from the set of field
equations with the same coordinate system. shrug

- the second one is
displaced from the origin by an amount K, which you can see
immediately by looking at the coefficients of the angular components.
Or by calculating the surface area. Or by calculating the Ricci
scalar. Or by looking at the determinant of the metric.


Nonsense! There is no displacement of anything. There is no
coordinate transformation. shrug

I cannot claim one solution is more unique than the other, which is
the whole point of my explanation. I just prefer Schwarzschild in
isotropic coordinates. I can easily convert from Schwarzschild and
every /OTHER/ "different" solution to yours - just chain the
transformations.


Hold your bullsh*t! Both ds1 and ds2 are spherically symmetric in
isotropic coordinate system. Again, for the n’th time, these
solutions are static, spherically symmetric, and asymptotically flat.
I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC,
ABUT NOT ASYMPTOTICALLY FLAT. That proves Birkhoff’s theorem false by
example. shrug

Your ranting is because you refuse the God you worship is a nitwit, a
plagiarist, and a liar. Yes, Einstein was nobody. He was a nitwit, a
plagiarist, and a liar. Despite the nitwit, the plagiarist, and the
liar did not come up with all these nonsense, whatever is credited to
this nitwit, this plagiarist, and this liar is utter nonsense in the
first place. shrug
  #53  
Old January 4th 09, 04:37 AM posted to sci.physics,sci.physics.relativity,sci.astro
Koobee Wublee
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Posts: 815
Default The Motion of the Perihelion of Mercury

On Jan 3, 9:09 am, George Hammond wrote:

Correct, he admits to "making a typo" which is why there
are two different versions of the Kooby Metric floating
around.


Oh, there are more than that. In the following post, you have been
entertained with a solution that is static, spherically symmetric, but
not asymptotically flat.

http://groups.google.com/group/sci.p...1526793d?hl=en

Of course, allow me to correct my typo. The spacetime I was talking
about is:

ds^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2 – (r + K)^2 dO^2

In the original one, you discovered and I have
independently confirmed that it is simply a radial
coordinate transformation of Schwarzchild given by r=R-K
with K=2m. So the original Kooby Metric is nothing but the
Schwarzchild Metric in disguise.


Oh, it looks like you have been eating the same sh*t with Gisse.
shrug

In the second case his "Typo Version" of the Kooby Metric
is not a coordinate transformation of Schwarzchild and
naturally therefore R_uv does not vanish as you discovered
and hence it is not a valid solution of the EFE's.


shrug

[Rest of whining crap snipped]


So, the great reverend Hammond is also a nitwit. He had no problem to
go to bed with Einstein Dingleberries to save his grace. shrug


  #54  
Old January 4th 09, 06:34 AM posted to sci.physics,sci.physics.relativity,sci.astro
George Hammond[_2_]
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Posts: 22
Default The Motion of the Perihelion of Mercury

On Sat, 3 Jan 2009 19:29:47 -0800 (PST), Koobee Wublee
wrote:


......... That proves Birkhoff’s theorem false by
example. shrug

[Hammond]
"proves Birkhoff's theorem false by example".....aaahahaha
ha hah hah aaaahaha ha ha ha ha......

Your ranting is because you refuse the God you worship is a nitwit, a
plagiarist, and a liar. Yes, Einstein was nobody. He was a nitwit, a
plagiarist, and a liar. Despite the nitwit, the plagiarist, and the
liar did not come up with all these nonsense, whatever is credited to
this nitwit, this plagiarist, and this liar is utter nonsense in the
first place. shrug

[Hammond]
Oh here we go ... turns out you're not just a moron,
you're a jive ass Natzi too....aaahahaha ha hah hah aaaahaha
ha ha ha......
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================


  #55  
Old January 4th 09, 06:55 AM posted to sci.physics,sci.physics.relativity,sci.astro
George Hammond[_2_]
external usenet poster
 
Posts: 22
Default The Motion of the Perihelion of Mercury

On Sat, 3 Jan 2009 19:37:34 -0800 (PST), Koobee Wublee
wrote:

snip

[Koobee Wublee]
......... That proves Birkhoff’s theorem false by
example. shrug

[Hammond]
"proves Birkhoff's theorem false by example".....aaahahaha
ha hah hah aaaahaha ha ha ha ha......
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================
  #56  
Old January 4th 09, 09:29 AM posted to sci.physics,sci.physics.relativity,sci.astro
Eric Gisse
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Posts: 1,465
Default The Motion of the Perihelion of Mercury

On Jan 3, 6:29*pm, Koobee Wublee wrote:
On Jan 3, 3:02 am, Eric Gisse wrote:



On Jan 2, 10:42 pm, Koobee Wublee wrote:
On Jan 2, 12:03 am, Eric Gisse wrote:
ds^2 = dx^2 + dy^2 + dz^2
ds^2 = dr^2 + r^2(d\theta^2 + sin^2(\theta)d\phi^2)
ds^2 = dr^2 + r^2 d\theta^2 + dz^2
ds^2 = da^2 + db^2 + dz^2


They all describe the same thing, yet "look different". But I can find
a coordinate transformation relating all of them to eachother, so they
are the same thing.


Your notation is screwed up. *The r of the 2nd is not the same as r of
the 3rd. *The phi of the 2nd is the same as theta of the 3rd.
Otherwise, they can all represent the same geometry. *shrug


See? You are a **** hair's distance away from understanding.


What is that again?

THE LABELS DO NOT MATTER.


Wrong! *The labels do matter. *shrug


Prove it. Write something with two different labels, and PROVE that
you can measure something to be different.

I have an idea - let's see your proof that "other solutions" that
satisfy the conditions of Birkhoff's theorem produce something other
than 43 arcsec/centry for Mercury's perihelion precession. You did,
after all, assert that to be the case.


The COORDINATES do not matter. They all
represent the _same_ geometry.


Wrong again. *To describe a geometry, you need both the labels
(coordinate) and how the labels are connected (metric). *shrug


True - to write the metric down, you need coordinates. Except the
coordinates do not matter as I can easily transform from one system to
the next and write the metric down in the new coordinate system.

As has been done repeatedly. As has the explanation to you.

Consider that you are just too stupid to understand, like Androcles
and rbwinn before you.


This is not what I have been talking about. *It looks like very few
have understood what I am saying. *That is why I feel very lonely to
be the few who after Riemann and perhaps Christoffel that have
understood the curvature business very well. *shrug


So where did you properly learn this "curvature business"? All I'm
asking is for the book you learned it from. Is that asking so much?


Just my intelligence which you lack. *shrug


So you just expect me to believe you woke up with this knowledge one
morning?

I find it more likely that you arrived at this particular level of
understanding after trying to study modern general relativity
textbooks. Except you failed, and invented all these arguments to
protect your ego.

Why else would you get so defensive and insulting when I ask a simple
question like "where did you learn what you know?".


What I am talking about is that the following geometries are different
given the same dr, dLongitude, dLatitude.


** *ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
** *ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2


Where


** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2


Except the coordinate labels are not the same in both coordinate
systems.


What do you mean they are not the same? *I have chosen them to be the
same. *shrug


Then calculate the surface area of a sphere using the metric.


Calculate the surface area of a sphere - don't just write it
down, CALCULATE it.


I did. *Just look up on my older post.


Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it.

In fact, you seem quite incapable of proving something as simple as
this. Why is the only person who has "truly understood" differential
geometry incapable of deriving the surface area of a sphere?

[...same idiocy as in 2006...]

http://groups.google.com/group/sci.p...c?dmode=source

JanPB explained this to you before with much greater patience than I
can muster. You used the same arguments then as you use now.

I cannot claim one solution is more unique than the other, which is
the whole point of my explanation. I just prefer Schwarzschild in
isotropic coordinates. I can easily convert from Schwarzschild and
every /OTHER/ "different" solution to yours - just chain the
transformations.


Hold your bullsh*t! *Both ds1 and ds2 are spherically symmetric in
isotropic coordinate system. *Again, for the n’th time, these
solutions are static, spherically symmetric, and asymptotically flat.


....and the same. How quickly you forget that this has all been
explained to you in detail before. With the same metrics no less.

http://groups.google.com/group/sci.p...7?dmode=source

I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC,
ABUT NOT ASYMPTOTICALLY FLAT. *That proves Birkhoff’s theorem false by
example. *shrug


Except you didn't prove the solution wasn't asymptotically flat
because you never calculated something like CURVATURE which would
actually prove your point. Plus I showed you the coordinate
transformation to navigate between your "different solution" and
Schwarzschild.


Your ranting is because you refuse the God you worship [...]


Funny how you are always the one to bring up God in these little
conversations.
  #57  
Old January 4th 09, 09:31 AM posted to sci.physics,sci.physics.relativity,sci.astro
Eric Gisse
external usenet poster
 
Posts: 1,465
Default The Motion of the Perihelion of Mercury

On Jan 3, 8:09*am, George Hammond wrote:
[snip]

* *Bear in mind, as a mere 25 year old you still think he's
a sincere Physics buff....


NO, I DO NOT. He is an arrogant idiot who has been repeatedly shown he
can't do something as simple as calculate the surface area of a
sphere, all the way to being unable to understand a derivation of the
field equations.

[snip]
  #58  
Old January 4th 09, 09:32 AM posted to sci.physics,sci.physics.relativity,sci.astro
Eric Gisse
external usenet poster
 
Posts: 1,465
Default The Motion of the Perihelion of Mercury

On Jan 3, 6:37*pm, Koobee Wublee wrote:
On Jan 3, 9:09 am, George Hammond wrote:

* *Correct, he admits to "making a typo" which is why there
are two different versions of the Kooby Metric floating
around.


Oh, there are more than that. *In the following post, you have been
entertained with a solution that is static, spherically symmetric, but
not asymptotically flat.


How do you know it is not asymptotically flat when you did not even
consider calculating curvature components?

[snip]
  #59  
Old January 4th 09, 08:25 PM posted to sci.physics,sci.physics.relativity,sci.astro
George Hammond[_2_]
external usenet poster
 
Posts: 22
Default The Motion of the Perihelion of Mercury

On Sun, 4 Jan 2009 00:29:58 -0800 (PST), Eric Gisse
wrote:




What I am talking about is that the following geometries are different
given the same dr, dLongitude, dLatitude.


** *ds1^2 = g1_11 dr^2 + g1_22 r^2 dO^2
** *ds2^2 = g2_11 dr^2 + g2_22 r^2 dO^2


Where


** *dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2


Except the coordinate labels are not the same in both coordinate
systems.


What do you mean they are not the same? *I have chosen them to be the
same. *shrug


Then calculate the surface area of a sphere using the metric.


Calculate the surface area of a sphere - don't just write it
down, CALCULATE it.


I did. *Just look up on my older post.


Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it.

[Hammond]
Spherical surface area in flat space is obtained by
integrating the surface element cos(theta)d(theta)d(phi) in
sperical coordinates over phi and theta for constant r which
results in the ancient formula Area=4 pi r^2.
In curved space of course the metric must be used if the
phi and theta basis vectors are not orthogonal. However in
the Schwarzchild and Kooby metrics there is only radial
space warping and no angularspace warping (radially
symmetric metric), therefore the surface element is still a
simple rectangle and not a parallogram as in the general
case.
Since integrating phi from 0-2pi and theta from 0-pi
over the angular part of the flat space metric or even the
Schwarzchild Metric:

r^2 cos(theta)d(theta)d(phi)

results in 4 pi r^2i,

we see that the area of an ordinary sphere in flat space, or
even a sphere in Schwarzchild space is 4 pi r^2 for a fixed
value of r.
In the case of the Kooby Metric however, the angular part
is not multiplied by r^2, it is multiplied by (R+K)^2
therefore the same surface integration results in an area
equal to:

4 pi (R+K)^2

for a fixed value of R; the new Kooby radial coordinate.
Notice of course that substituting the coordinate
transformation R=r-K transforms thisresult back to the well
known Schwarzchild result area=4 pi r^2 again.
=====================================
HAMMOND'S PROOF OF GOD WEBSITE
http://geocities.com/scientific_proof_of_god
mirror site:
http://proof-of-god.freewebsitehosting.com
GOD=G_uv (a folk song on mp3)
http://interrobang.jwgh.org/songs/hammond.mp3
=====================================
  #60  
Old January 5th 09, 08:03 AM posted to sci.physics,sci.physics.relativity,sci.astro
Koobee Wublee
external usenet poster
 
Posts: 815
Default The Motion of the Perihelion of Mercury

On Jan 4, 12:29 am, Eric Gisse wrote:
On Jan 3, 6:29 pm, Koobee Wublee wrote:
On Jan 3, 3:02 am, Eric Gisse wrote:


THE LABELS DO NOT MATTER.


Wrong! The labels do matter. shrug


Prove it. Write something with two different labels, and PROVE that
you can measure something to be different.


I have said that many times over. To describe the geometry, one must
choose a coordinate system first and then define how the set of
coordinates is related to the geometry. shrug

I have an idea - let's see your proof that "other solutions" that
satisfy the conditions of Birkhoff's theorem produce something other
than 43 arcsec/centry for Mercury's perihelion precession. You did,
after all, assert that to be the case.


Well, it is time consuming. I have lacked any motivation to do so
after discovering GR is a total nonsense based on voodoo mathematics.
shrug

The COORDINATES do not matter. They all
represent the _same_ geometry.


Wrong again. To describe a geometry, you need both the labels
(coordinate) and how the labels are connected (metric). shrug


True - to write the metric down, you need coordinates.


Oh, wow! That is an improvement after all these years. shrug

Except the
coordinates do not matter as I can easily transform from one system to
the next and write the metric down in the new coordinate system.


Again, you are stuck in your matheMagical wonderland of coordinate
transformation. There is no such coordinate transformation. You are
schizophrenic. shrug

As has been done repeatedly. As has the explanation to you.


There is no credible coordinate transformation done besides total
nonsense. shrug

Consider that you are just too stupid to understand, like Androcles
and rbwinn before you.


The Einstein Dingleberries have no problems bouncing these crackpots
up and down the tennis courts, but Koobee Wublee remains
unchallenged. Your whining crap does not count. shrug

So where did you properly learn this "curvature business"? All I'm
asking is for the book you learned it from. Is that asking so much?


Just my intelligence which you lack. shrug


So you just expect me to believe you woke up with this knowledge one
morning?


Well, it is up to you. I have also studies a lot about the nonsense
which you eagerly swallowed. shrug

I find it more likely that you arrived at this particular level of
understanding after trying to study modern general relativity
textbooks. Except you failed, and invented all these arguments to
protect your ego.


There are flaws in GR. I have discovered them, and I have advanced my
studies. In the meantime, you are stuck in what they have told you.
That is why you remain a multi-year super-senior. shrug

Why else would you get so defensive and insulting when I ask a simple
question like "where did you learn what you know?".


I am not getting defensive. I just don’t think my educational
background is any of your business. shrug

Except the coordinate labels are not the same in both coordinate
systems.


What do you mean they are not the same? I have chosen them to be the
same. shrug


Then calculate the surface area of a sphere using the metric.


I have told you the answer more than once. To an observer, the
surface area of a sphere is always (4 pi r^2) regardless how curved up
space is. shrug

Calculate the surface area of a sphere - don't just write it
down, CALCULATE it.


I did. Just look up on my older post.


Nope - you _asserted_ it was 4 pi r^2. You never _derived_ it.

[Remaining whining crap snipped]


You are just trolling me. shrug

JanPB explained this to you before with much greater patience than I
can muster. You used the same arguments then as you use now.


Speaking of the one who calls me mentally unstable, and yet he would
have no problem of swimming naked in a public lake. Mr. Bielawski is
an Einstein Dingleberry who swallows everything the collective group
of Einstein Dingleberries decides what physics should be without any
scientific methodology. You are also one of them. You have also
become a henchman trying to force that crap down on anyone’s throat.
shrug

I cannot claim one solution is more unique than the other, which is
the whole point of my explanation. I just prefer Schwarzschild in
isotropic coordinates. I can easily convert from Schwarzschild and
every /OTHER/ "different" solution to yours - just chain the
transformations.


Hold your bullsh*t! Both ds1 and ds2 are spherically symmetric in
isotropic coordinate system. Again, for the n’th time, these
solutions are static, spherically symmetric, and asymptotically flat.


[snipped whining crap]


So, you have no answers as usual besides whining nonsense.

I HAVE SHOWN YOU SOLUTIONS THAT ARE STATIC, SPHERICALLY SYMMETRIC,
ABUT NOT ASYMPTOTICALLY FLAT. That proves Birkhoff’s theorem false by
example. shrug


Except you didn't prove the solution wasn't asymptotically flat
because you never calculated something like CURVATURE which would
actually prove your point. Plus I showed you the coordinate
transformation to navigate between your "different solution" and
Schwarzschild.


There is no need to talk about this further. It is obvious that you
do not understand what asymptotically flat means. shrug

Your ranting is because you refuse the God you worship [...]


Funny how you are always the one to bring up God in these little
conversations.


That is because you have decided Einstein as your God. shrug

Your education is totally based on faith. shrug
 




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