Demented idiot, carl eto wrote:
....
> According to Newton gravity equation (equ 91), a 50 kg
> astronaut would weigh 43.2 kg (equ 93) while in the
> international space station (ISS) yet the described
> astronaut is weightless It is argued that the astronaut
> is weightless because of a free fall but there is no
> sufficient free fall downward velocity that could justify
> the free fall argument.

1. Microgravity is not caused by being far (it's not that
far anyway) from Earth surface

2. Downward velocity is pointless, what matters is centripetal
acceleration while in orbit. Do the maths, idiot "carl eto",
it is equal to gravitational acceleration.

3. Bring your dirty ass, imbecile, into a plane for a parabolic
flight. You will experience microgravity even when the plane
is flying up.

Learn some elementary physics, meanwhile shut the fuck up
(this includes your sock puppets). Report for spam sent to
abuse@humboldt.edu also.

On Sunday, 20 February 2022 at 21:24:01 UTC+1, Python wrote:
> Demented idiot, carl eto wrote:
> ...
> > According to Newton gravity equation (equ 91), a 50 kg
> > astronaut would weigh 43.2 kg (equ 93) while in the
> > international space station (ISS) yet the described
> > astronaut is weightless It is argued that the astronaut
> > is weightless because of a free fall but there is no
> > sufficient free fall downward velocity that could justify
> > the free fall argument.
> 1. Microgravity is not caused by being far (it's not that
> far anyway) from Earth surface
>
> 2. Downward velocity is pointless, what matters is centripetal

And what is your definition of a "theory" written
in the terms of Peano arithmetic, poor stinker?

Stinky Maciej Wozniak wrote:
> On Sunday, 20 February 2022 at 21:24:01 UTC+1, Python wrote:
>> Demented idiot, carl eto wrote:
>> ...
>>> According to Newton gravity equation (equ 91), a 50 kg
>>> astronaut would weigh 43.2 kg (equ 93) while in the
>>> international space station (ISS) yet the described
>>> astronaut is weightless It is argued that the astronaut
>>> is weightless because of a free fall but there is no
>>> sufficient free fall downward velocity that could justify
>>> the free fall argument.
>> 1. Microgravity is not caused by being far (it's not that
>> far anyway) from Earth surface
>>
>> 2. Downward velocity is pointless, what matters is centripetal
>
> And what is your definition of a "theory" written
> in the terms of Peano arithmetic?

42.

> --
> poor stinker

Nice signature Maciej.

Maciej "stinky" Maciej wrote:
> In the meantime in the real world, Peano arithmetics
> is a theory of numbers, not about theories. Godel's
> theorem isn't a part of it. Godel has concocted a brandly
> new theory, but somehow he has forgotten to
> specify its assumptions.
> There are some other reasons why his theorem has no
> value

On Sunday, 20 February 2022 at 22:39:46 UTC+1, Python wrote:
> Stinky Maciej Wozniak wrote:
> > On Sunday, 20 February 2022 at 21:24:01 UTC+1, Python wrote:
> >> Demented idiot, carl eto wrote:
> >> ...
> >>> According to Newton gravity equation (equ 91), a 50 kg
> >>> astronaut would weigh 43.2 kg (equ 93) while in the
> >>> international space station (ISS) yet the described
> >>> astronaut is weightless It is argued that the astronaut
> >>> is weightless because of a free fall but there is no
> >>> sufficient free fall downward velocity that could justify
> >>> the free fall argument.
> >> 1. Microgravity is not caused by being far (it's not that
> >> far anyway) from Earth surface
> >>
> >> 2. Downward velocity is pointless, what matters is centripetal
> >
> > And what is your definition of a "theory" written
> > in the terms of Peano arithmetic?
>
> 42.

And the one you're teaching your *students*,
poor stinker?

> Maciej "stinky" Maciej wrote:
> > In the meantime in the real world, Peano arithmetics
> > is a theory of numbers, not about theories. Godel's
> > theorem isn't a part of it. Godel has concocted a brandly
> > new theory, but somehow he has forgotten to
> > specify its assumptions.
> > There are some other reasons why his theorem has no
> > value

Yes, I did. Your answer was, of course, "UUUU!!!! UUUUUUU!!!
UUUUUUUUUU!!!!! UUUUUUUUU!!!!!!". Python's answer for
everything.

Stinky Maciej Wozniak wrote:
> On Sunday, 20 February 2022 at 22:39:46 UTC+1, Python wrote:
>> Stinky Maciej Wozniak wrote:
>>> On Sunday, 20 February 2022 at 21:24:01 UTC+1, Python wrote:
>>>> Demented idiot, carl eto wrote:
>>>> ...
>>>>> According to Newton gravity equation (equ 91), a 50 kg
>>>>> astronaut would weigh 43.2 kg (equ 93) while in the
>>>>> international space station (ISS) yet the described
>>>>> astronaut is weightless It is argued that the astronaut
>>>>> is weightless because of a free fall but there is no
>>>>> sufficient free fall downward velocity that could justify
>>>>> the free fall argument.
>>>> 1. Microgravity is not caused by being far (it's not that
>>>> far anyway) from Earth surface
>>>>
>>>> 2. Downward velocity is pointless, what matters is centripetal
>>>
>>> And what is your definition of a "theory" written
>>> in the terms of Peano arithmetic?
>>
>> 42.
>
> And the one you're teaching your *students*,
> poor stinker?

Don't ask for what you don't deserve, Maciej.

>> Maciej "stinky" Maciej wrote:
>>> In the meantime in the real world, Peano arithmetics
>>> is a theory of numbers, not about theories. Godel's
>>> theorem isn't a part of it. Godel has concocted a brandly
>>> new theory, but somehow he has forgotten to
>>> specify its assumptions.
>>> There are some other reasons why his theorem has no
>>> value
>
> Yes, I did. Your answer was, of course, "UUUU!!!! UUUUUUU!!!
> UUUUUUUUUU!!!!! UUUUUUUUU!!!!!!". Python's answer for
> everything.

You forgot what my answer was:

What assumption Gödel has forgotten to specify?

Quite different from "UUU!", right?

You answer Maciej was to snip my answer and not a single word.

I can now add this answer:

What are the other reasons for Gödel theorem to have no value?

This is also very different from "UUU!!!".

Python
Feb 19, 2022, 6:45:46 PM (16 hours ago)
to
"carl eto" wrote:
>
>
>
> ... the earth and moon gravitational forces on the CSML
> while propagating to the moon is negligible since the astronauts
> within the CSML are weightless during the voyage to the moon

You really think CSM is massless then? You are THAT idiot?

> which proves the Apollo 11 mission is fictional.

Oh, yes you are.

BTW, abuse sent to ab...@humboldt.edu for spam.

vintage donkie

Python
12:24 PM (5 hours ago)
to
Demented idiot, carl eto wrote:
....
> According to Newton gravity equation (equ 91), a 50 kg
> astronaut would weigh 43.2 kg (equ 93) while in the
> international space station (ISS) yet the described
> astronaut is weightless It is argued that the astronaut
> is weightless because of a free fall but there is no
> sufficient free fall downward velocity that could justify
> the free fall argument.

Downward velocity is pointless, what matters is centripetal
acceleration while in orbit.

The centripetal acceleration would produce a force yet the astronauts are weightless. Do they have Donkies in Russia or just ice and Hockey arenas?

On Sunday, February 20, 2022 at 5:16:33 PM UTC-8, carl eto wrote:

Why is Newton's equation resulting in a 43.2 kg force when the astronauts are weightless? Example, if you take a string and attach it to a ball and rotate the ball, the ball rotates at a constant distance that does not increase yet the astronauts are moving inside the ISS which proves the ISS is a fabrication. According to Newton's law, the astronauts within the ISS would experience a 43 kg force towards the earth and not be experience weightless. Maybe my good friend Vladimir Putin will change your minds tomorrow with a Russian surprise and present made from abstract physics, Ha ha ha ah... Dr. Evil..

https://www.google.com/search?q=dr+evil&tbm=isch&chips=q:dr+evil,g_1:austin:9G1J8h9BVJc%3D&rlz=1C1GCEA_enUS993US993&hl=en&sa=X&ved=2ahUKEwizwZfX14_2AhUDIX0KHat_AaIQ4lYoAHoECAEQGw&biw=1903&bih=969#imgrc=wr9CQHGzko1iwM

On Sunday, February 20, 2022 at 5:53:12 PM UTC-8, carleto4...@gmail.com wrote:
> On Sunday, February 20, 2022 at 5:16:33 PM UTC-8, carl eto wrote:
>
>
> Why is Newton's equation resulting in a 43.2 kg force when the astronauts are weightless? Example, if you take a string and attach it to a ball and rotate the ball, the ball rotates at a constant distance that does not increase yet the astronauts are moving inside the ISS which proves the ISS is a fabrication. According to Newton's law, the astronauts within the ISS would experience a 43 kg force towards the earth and not be experience weightless. Maybe my good friend Vladimir Putin will change your minds tomorrow with a Russian surprise and present made from abstract physics, Ha ha ha ah... Dr. Evil..
>
>
> https://www.google.com/search?q=dr+evil&tbm=isch&chips=q:dr+evil,g_1:austin:9G1J8h9BVJc%3D&rlz=1C1GCEA_enUS993US993&hl=en&sa=X&ved=2ahUKEwizwZfX14_2AhUDIX0KHat_AaIQ4lYoAHoECAEQGw&biw=1903&bih=969#imgrc=wr9CQHGzko1iwM

You are apparently quite proud of your complete ignorance of the subject matter...

On Sunday, 20 February 2022 at 23:41:42 UTC+1, Python wrote:
> Stinky Maciej Wozniak wrote:
> > On Sunday, 20 February 2022 at 22:39:46 UTC+1, Python wrote:
> >> Stinky Maciej Wozniak wrote:
> >>> On Sunday, 20 February 2022 at 21:24:01 UTC+1, Python wrote:
> >>>> Demented idiot, carl eto wrote:
> >>>> ...
> >>>>> According to Newton gravity equation (equ 91), a 50 kg
> >>>>> astronaut would weigh 43.2 kg (equ 93) while in the
> >>>>> international space station (ISS) yet the described
> >>>>> astronaut is weightless It is argued that the astronaut
> >>>>> is weightless because of a free fall but there is no
> >>>>> sufficient free fall downward velocity that could justify
> >>>>> the free fall argument.
> >>>> 1. Microgravity is not caused by being far (it's not that
> >>>> far anyway) from Earth surface
> >>>>
> >>>> 2. Downward velocity is pointless, what matters is centripetal
> >>>
> >>> And what is your definition of a "theory" written
> >>> in the terms of Peano arithmetic?
> >>
> >> 42.
> >
> > And the one you're teaching your *students*,
> > poor stinker?
> Don't ask for what you don't deserve, Maciej.
> >> Maciej "stinky" Maciej wrote:
> >>> In the meantime in the real world, Peano arithmetics
> >>> is a theory of numbers, not about theories. Godel's
> >>> theorem isn't a part of it. Godel has concocted a brandly
> >>> new theory, but somehow he has forgotten to
> >>> specify its assumptions.
> >>> There are some other reasons why his theorem has no
> >>> value
> >
> > Yes, I did. Your answer was, of course, "UUUU!!!! UUUUUUU!!!
> > UUUUUUUUUU!!!!! UUUUUUUUU!!!!!!". Python's answer for
> > everything.
> You forgot what my answer was:
>
> What assumption Gödel has forgotten to specify?
>
> Quite different from "UUU!", right?

Not especially, poor stinker, since it's clearly
written above. He has forgotten to specify any
assumptions of the theory his theorems belong to.
And what is your definition of a "theory" in the terms
of AP, poor stinker?

..

On Sunday, February 20, 2022 at 6:38:46 PM UTC-8, Paul Alsing wrote:
> On Sunday, February 20, 2022 at 5:53:12 PM UTC-8, carleto4...@gmail.com wrote:
> > On Sunday, February 20, 2022 at 5:16:33 PM UTC-8, carl eto wrote:
> >
> >
> > Why is Newton's equation resulting in a 43.2 kg force when the astronauts are weightless? Example, if you take a string and attach it to a ball and rotate the ball, the ball rotates at a constant distance that does not increase yet the astronauts are moving inside the ISS which proves the ISS is a fabrication. According to Newton's law, the astronauts within the ISS would experience a 43 kg force towards the earth and not be experience weightless. Maybe my good friend Vladimir Putin will change your minds tomorrow with a Russian surprise and present made from abstract physics, Ha ha ha ah.... Dr. Evil..
> >
> >
> > https://www.google.com/search?q=dr+evil&tbm=isch&chips=q:dr+evil,g_1:austin:9G1J8h9BVJc%3D&rlz=1C1GCEA_enUS993US993&hl=en&sa=X&ved=2ahUKEwizwZfX14_2AhUDIX0KHat_AaIQ4lYoAHoECAEQGw&biw=1903&bih=969#imgrc=wr9CQHGzko1iwM
> You are apparently quite proud of your complete ignorance of the subject matter...

Yes I am completely ignorant of the matter. Please do tell. That is why I am asking you.

On Wednesday, February 23, 2022 at 1:26:25 PM UTC-8, carleto4...@gmail.com wrote:
> On Sunday, February 20, 2022 at 6:38:46 PM UTC-8, Paul Alsing wrote:
> > On Sunday, February 20, 2022 at 5:53:12 PM UTC-8, carleto4...@gmail.com wrote:
> > > On Sunday, February 20, 2022 at 5:16:33 PM UTC-8, carl eto wrote:
> > >
> > >
> > > Why is Newton's equation resulting in a 43.2 kg force when the astronauts are weightless? Example, if you take a string and attach it to a ball and rotate the ball, the ball rotates at a constant distance that does not increase yet the astronauts are moving inside the ISS which proves the ISS is a fabrication. According to Newton's law, the astronauts within the ISS would experience a 43 kg force towards the earth and not be experience weightless. Maybe my good friend Vladimir Putin will change your minds tomorrow with a Russian surprise and present made from abstract physics, Ha ha ha ah... Dr. Evil..
> > >
> > >
> > > https://www.google.com/search?q=dr+evil&tbm=isch&chips=q:dr+evil,g_1:austin:9G1J8h9BVJc%3D&rlz=1C1GCEA_enUS993US993&hl=en&sa=X&ved=2ahUKEwizwZfX14_2AhUDIX0KHat_AaIQ4lYoAHoECAEQGw&biw=1903&bih=969#imgrc=wr9CQHGzko1iwM

> > You are apparently quite proud of your complete ignorance of the subject matter...

> Yes I am completely ignorant of the matter.

I am glad to see that you are finally admitting that you are as stupid as a dead tree...

Astronomers use parallax to determine the distance to a star using the earth's orbital diameter as the parallax reference distance but the distance to a 4.22 light year star (4 x 1016 m) is more than 105 times larger than the earth's orbital diameter (2.99 x 1011 m). The earth's orbital diameter is too short of a distance to produce a change in the angular position θ that can be measured using a telescope in the measurement of the distance to the closest star. The telescopic resolution θ required to determine the distance to a 4.22 ly star is calculated using,

A/B = cos θ.......................................................................................................99

when A/B → 0, equation 99 becomes,

A/B = θ................................................................................................................100

Using A as the earth's orbital diameter, B is the distance to a 4.22 ly star, and θ represents the change in the angular displacement of the star, the telescopic resolution (power) required to determine the distance to a 4.22 ly (4 x 1016 meters) star is calculated,

θ = A/B = (2.99 x 1011 m) / (4 x 1016 meters) = 7.475 x 10-6 degrees or 0.027 arcsec...101

To measure the distance to a 4.22 ly star using the earth's orbital diameter as the parallax reference distance requires a telescopic resolution of 0.027 arcsec (equ 83) which is 3.7 times more power than the Hubble (.1 arcsec). Using equation 100, the maximum parallax distance to a star calculated using the Hubble is calculated,

B = A/θ = (2.99 x 1011 m) x (3600o) / arcec) x (.1 arcsec)-1 = 1..0764 x 1015 m = 0.114 ly....102

"Hey, wait a sec! Hubble’s resolution is only 0.1 arcseconds, so the lander is way too small to be seen as anything more than a dot, even by Hubble. It would have to be a lot bigger to be seen at all. In fact, if you do the math (set Hubble’s resolution to 0.1 arcseconds and the distance to 400,000 kilometers) you see that Hubble’s resolution on the Moon is about 200 meters! In other words, even a football stadium on the Moon would look like a dot to Hubble." (By Phil Plait | August 12, 2008, Discover online).

The Hubble (.1 arcsec) has a maximum parallax range of .114 ly (1015 m) using the earth's orbital diameter as the parallax reference distance but distance to the closest star is 4.22 ly which is four times larger then the range of the Hubble using the earth's orbital diameter as the parallax reference distance which is experimental proof modern astronomy is fictional. In addition, to determine the distance to a star using parallax requires a straight path reference distance; the curvature of the earth's orbit changes the direction of the observer on the earth by 360o after 180 days (six months); consequently, the earth's orbital diameter cannot be used as the parallel reference distance since the observe must be facing the same direction after propagating the distance of the parallax reference distance. The maximum parallax reference distance that can be obtain using the earth's orbit is approximately 18 days to form a semi straight reference distance (1.65 x 1010 m) resulting in the maximum parallax range of .0114 ly. As a result, astronomer cannot resolve the distances to any of the stars. The maximum parallax distance that can be resolved with the Hubble (.1 arcsec) is 0.0114 light years yet the Hubble is viewing the Eagle Nebula that is said to be more than 7,000 ly from the earth and composed of more than 1,000 stars which would require a power three orders of magnitude greater than the Hubble. The velocity of the Sun (220 km/s) is used to increase the maximum parallax distance but the stars are stationary (celestial maps) which includes the Sun. If the Sun were in motion celestial maps would have to be up dated every year. A dimming method of the intensity of the stars (celestial ladder) is used to measure the distance to a star that is more than 350 ly from the earth but the variation of the intensity of a star is caused by the earth's atmosphere. Astronomers cannot determine if a star is far away or dim. Chilean astronomers use the dimming of a 4.22 ly star Proxima Centauri caused by a planet passing in front of the star Centauri viewed using the La Silla telescope that dimming of the intensity of the star Centauri is used to justify the existence of a planet outside our solar system (Escude, p. 408) but the variation of the intensity (dimming) of a 4.22 light year star is caused by the earth's atmosphere since the La Silla telescope cannot possibly resolve the effect of a planet decreasing the intensity of a 4.22 ly star..

carl eto <carleto4157990662@gmail.com> wrote:

> Astronomers use parallax to determine the distance to a star using the
> earth's orbital diameter as the parallax reference distance but the
> distance to a 4.22 light year star (4 x 1016 m) is more than 105 times
> larger than the earth's orbital diameter (2.99 x 1011 m). The earth's
> orbital diameter is too short of a distance to produce a change in the
> angular position ? that can be measured using a telescope in the
> measurement of the distance to the closest star. The telescopic resolution
> ? required to determine the distance to a 4.22 ly star is calculated
> using,

That's another thing you misunderstand completely.
The yearly displacement of nearby stars (the parallax)
is measured by comparing their positions
with those of far away reference stars,
First done by Bessel in 1838.
Resolving anything is not necessary.

Using satellite measurements, stellar distances from parallax
can be measured out to about 10 000 light years nowadays,
so for billions of stars,

Jan

<https://nl.wikipedia.org/wiki/Hipparcos>
<https://en.wikipedia.org/wiki/Gaia_(spacecraft)>
<https://en.wikipedia.org/wiki/Hubble_Space_Telescope>

> That's another thing you misunderstand completely.
> The yearly displacement of nearby stars (the parallax)
> is measured by comparing their positions
> with those of far away reference stars,
> First done by Bessel in 1838.
> Resolving anything is not necessary.
>
> Using satellite measurements, stellar distances from parallax
> can be measured out to about 10 000 light years nowadays,
> so for billions of stars,
>
> Jan
>
> <https://nl.wikipedia.org/wiki/Hipparcos>
> <https://en.wikipedia.org/wiki/Gaia_(spacecraft)>
> <https://en.wikipedia.org/wiki/Hubble_Space_Telescope>

First, to use parallax requires a parallax reference distance which is usually the earth's orbital diameter. What are you using as the parallax reference distance? The problem using the earth's orbital diameter is that the dark side of the earth (night) would not be facing the star, after a 6 month time interval, that is used to determine the parallax angular displacement that you are measuring to determine the distance to the said star.

Construct an algorithm that can track the projectile motion of an artillery shell to the source using GPS then use a smart bomb to attack of the source---->.fire the retaliatory projectile while the initial projectile is still in the air since the source may be mobile. You are basically counteracting the mobility of the artillery system. ROC main threat is artillery and tanks ----------->>M829.

carl eto <carleto4157990662@gmail.com> wrote:

> > That's another thing you misunderstand completely.
> > The yearly displacement of nearby stars (the parallax)
> > is measured by comparing their positions
> > with those of far away reference stars,
> > First done by Bessel in 1838.
> > Resolving anything is not necessary.
> >
> > Using satellite measurements, stellar distances from parallax
> > can be measured out to about 10 000 light years nowadays,
> > so for billions of stars,
> >
> > Jan
> >
> > <https://nl.wikipedia.org/wiki/Hipparcos>
> > <https://en.wikipedia.org/wiki/Gaia_(spacecraft)>
> > <https://en.wikipedia.org/wiki/Hubble_Space_Telescope>
>
> First, to use parallax requires a parallax reference distance which is
> usually the earth's orbital diameter. What are you using as the parallax
> reference distance? The problem using the earth's orbital diameter is that
> the dark side of the earth (night) would not be facing the star, after a 6
> month time interval, that is used to determine the parallax angular
> displacement that you are measuring to determine the distance to the said
> star.

Do you believe in the 'Dark Side of the Moon' too?

Jan
(just curious)

On Saturday, February 26, 2022 at 11:29:40 AM UTC-8, carleto4...@gmail.com wrote:
> > That's another thing you misunderstand completely.
> > The yearly displacement of nearby stars (the parallax)
> > is measured by comparing their positions
> > with those of far away reference stars,
> > First done by Bessel in 1838.
> > Resolving anything is not necessary.
> >
> > Using satellite measurements, stellar distances from parallax
> > can be measured out to about 10 000 light years nowadays,
> > so for billions of stars,
> >
> > Jan
> >
> > <https://nl.wikipedia.org/wiki/Hipparcos>
> > <https://en.wikipedia.org/wiki/Gaia_(spacecraft)>
> > <https://en.wikipedia.org/wiki/Hubble_Space_Telescope>

> First, to use parallax requires a parallax reference distance which is usually the earth's orbital diameter. What are you using as the parallax reference distance? The problem using the earth's orbital diameter is that the dark side of the earth (night) would not be facing the star, after a 6 month time interval, that is used to determine the parallax angular displacement that you are measuring to determine the distance to the said star.

You are unbelievably clueless regarding parallax. Clueless!

This hobby is way beyond your capabilities, perhaps basket-weaving would be more productive for you...

Paul Alsing <pnalsing@gmail.com> wrote:

> On Saturday, February 26, 2022 at 11:29:40 AM UTC-8, carleto4...@gmail.com:
> > > That's another thing you misunderstand completely.
> > > The yearly displacement of nearby stars (the parallax)
> > > is measured by comparing their positions
> > > with those of far away reference stars,
> > > First done by Bessel in 1838.
> > > Resolving anything is not necessary.
> > >
> > > Using satellite measurements, stellar distances from parallax
> > > can be measured out to about 10 000 light years nowadays,
> > > so for billions of stars,
> > >
> > > Jan
> > >
> > > <https://nl.wikipedia.org/wiki/Hipparcos>
> > > <https://en.wikipedia.org/wiki/Gaia_(spacecraft)>
> > > <https://en.wikipedia.org/wiki/Hubble_Space_Telescope>
>
> > First, to use parallax requires a parallax reference distance which is
> > usually the earth's orbital diameter. What are you using as the parallax
> > reference distance? The problem using the earth's orbital diameter is
> > that the dark side of the earth (night) would not be facing the star,
> > after a 6 month time interval, that is used to determine the parallax
> > angular displacement that you are measuring to determine the distance to
> > the said star.
> >
> You are unbelievably clueless regarding parallax. Clueless!
>
> This hobby is way beyond your capabilities, perhaps basket-weaving would
> be more productive for you...

Or perhaps making flat earth globes?

Jan
<https://i.guim.co.uk/img/media/cfafb822f6d1a899f191e76770c42cdb14b4d0db/0_0_2362_2835/master/2362.jpg?width=620&quality=85&auto=format&fit=max&s=b0a4ae25adb7481afbfac23048d9d390>

On Saturday, February 26, 2022 at 8:06:40 PM UTC-8, Paul Alsing wrote:

> > First, to use parallax requires a parallax reference distance which is usually the earth's orbital diameter. What are you using as the parallax reference distance? The problem using the earth's orbital diameter is that the dark side of the earth (night) would not be facing the star, after a 6 month time interval, that is used to determine the parallax angular displacement that you are measuring to determine the distance to the said star.
> You are unbelievably clueless regarding parallax. Clueless!
>
> This hobby is way beyond your capabilities, perhaps basket-weaving would be more productive for you...

Please tell us what the parallax reference distance is or how parallax is measured?

carl eto <carleto4157990662@gmail.com> wrote:

> On Saturday, February 26, 2022 at 8:06:40 PM UTC-8, Paul Alsing wrote:
>
> > > First, to use parallax requires a parallax reference distance which is
> > > usually the earth's orbital diameter. What are you using as the
> > > parallax reference distance? The problem using the earth's orbital
> > > diameter is that the dark side of the earth (night) would not be
> > > facing the star, after a 6 month time interval, that is used to
> > > determine the parallax angular displacement that you are measuring to
> > > determine the distance to the said star.
> > >
> > You are unbelievably clueless regarding parallax. Clueless!
> >
> > This hobby is way beyond your capabilities, perhaps basket-weaving would
> > be more productive for you...
>
> Please tell us what the parallax reference distance is or how parallax is
> measured?

Isn't it time that you find out for yourself,
-before- you spout lots of nonsense about it?

Jan

On Sunday, February 27, 2022 at 11:44:15 AM UTC-8, J. J. Lodder wrote:

> Isn't it time that you find out for yourself,
> -before- you spout lots of nonsense about it?
>
> Jan

This includes how Parallax is done. Please Jan don't get sick or bite your dick.

Astronomers use parallax to determine the distance to a star using the earth's orbital diameter as the parallax reference distance but the distance to a 4.22 light year star (4 x 1016 m) is more than 105 times larger than the earth's orbital diameter (2.99 x 1011 m). The earth's orbital diameter is too short of a distance to produce a change in the angular position θ that can be measured using a telescope in the measurement of the distance to the closest star. The telescopic resolution θ required to determine the distance to a 4.22 ly star is calculated using,

A/B = cos θ.............. .........................................................................................99

when A/B → 0, equation 99 becomes,

A/B = θ............. ...................................................................................................100

Using A as the earth's orbital diameter, B is the distance to a 4.22 ly star, and θ represents the change in the angular displacement of the star, the telescopic resolution (power) required to determine the distance to a 4.22 ly (4 x 1016 meters) star is calculated,

θ = A/B = (2.99 x 1011 m) / (4 x 1016 meters) = 7.475 x 10-6 degrees or 0.027 arcsec...101

To measure the distance to a 4.22 ly star using the earth's orbital diameter as the parallax reference distance requires a telescopic resolution of 0.027 arcsec (equ 83) which is 3.7 times more power than the Hubble (.1 arcsec). Using equation 100, the maximum parallax distance to a star calculated using the Hubble is calculated,

B = A/θ = (2.99 x 1011 m) x (3600o) / arcec) x (.1 arcsec)-1 = 1.0764 x 1015 m = 0.114 ly....102

"Hey, wait a sec! Hubble’s resolution is only 0.1 arcseconds, so the lander is way too small to be seen as anything more than a dot, even by Hubble. It would have to be a lot bigger to be seen at all. In fact, if you do the math (set Hubble’s resolution to 0.1 arcseconds and the distance to 400,000 kilometers) you see that Hubble’s resolution on the Moon is about 200 meters! In other words, even a football stadium on the Moon would look like a dot to Hubble." (By Phil Plait | August 12, 2008, Discover online).

The Hubble (.1 arcsec) has a maximum parallax range of .114 ly (1015 m) using the earth's orbital diameter as the parallax reference distance but distance to the closest star is 4.22 ly which is four times larger then the range of the Hubble using the earth's orbital diameter as the parallax reference distance which is experimental proof modern astronomy is fictional. In addition, to determine the distance to a star using parallax requires a straight path reference distance; the curvature of the earth's orbit changes the direction of the observer on the earth by 360o after 180 days (six months); consequently, the earth's orbital diameter cannot be used as the parallel reference distance since the observe must be facing the same direction after propagating the distance of the parallax reference distance. The maximum parallax reference distance that can be obtain using the earth's orbit is approximately 18 days to form a semi straight reference distance (1.65 x 1010 m) resulting in the maximum parallax range of .0114 ly. As a result, astronomer cannot resolve the distances to any of the stars. The maximum parallax distance that can be resolved with the Hubble (.1 arcsec) is 0..0114 light years yet the Hubble is viewing the Eagle Nebula that is said to be more than 7,000 ly from the earth and composed of more than 1,000 stars which would require a power three orders of magnitude greater than the Hubble. The velocity of the Sun (220 km/s) is used to increase the maximum parallax distance but the stars are stationary (celestial maps) which includes the Sun. If the Sun were in motion celestial maps would have to be up dated every year. A dimming method of the intensity of the stars (celestial ladder) is used to measure the distance to a star that is more than 350 ly from the earth but the variation of the intensity of a star is caused by the earth's atmosphere. Astronomers cannot determine if a star is far away or dim. Chilean astronomers use the dimming of a 4.22 ly star Proxima Centauri caused by a planet passing in front of the star Centauri viewed using the La Silla telescope that dimming of the intensity of the star Centauri is used to justify the existence of a planet outside our solar system (Escude, p. 408) but the variation of the intensity (dimming) of a 4.22 light year star is caused by the earth's atmosphere since the La Silla telescope cannot possibly resolve the effect of a planet decreasing the intensity of a 4.22 ly star.

carl eto <carleto4157990662@gmail.com> wrote:

> On Sunday, February 27, 2022 at 11:44:15 AM UTC-8, J. J. Lodder wrote:
>
> > Isn't it time that you find out for yourself,
> > -before- you spout lots of nonsense about it?
> >
> > Jan
>
> This includes how Parallax is done. Please Jan don't get sick or bite your
> dick.
[snip repeats of nonsense]

So you haven't understood it at all.
Hint, the resolution of the telescope is irrelevant
for measuring parallax,
(which is about measuring -stellar positions-)

Jan

On Sunday, February 27, 2022 at 2:25:23 PM UTC-8, J. J. Lodder wrote:
> carl eto <carleto4...@gmail.com> wrote:
>
> > On Sunday, February 27, 2022 at 11:44:15 AM UTC-8, J. J. Lodder wrote:
> >
> > > Isn't it time that you find out for yourself,
> > > -before- you spout lots of nonsense about it?
> > >
> > > Jan
> >
> > This includes how Parallax is done. Please Jan don't get sick or bite your
> > dick.
> [snip repeats of nonsense]
>
> So you haven't understood it at all.
> Hint, the resolution of the telescope is irrelevant
> for measuring parallax,
> (which is about measuring -stellar positions-)
>
> Jan

The position of the stars forms an angle that is limited to the power (resolution) of the telescope. You cannot determine the change in the position of the star without proper power of the telescope. Also, what is the reference parallax distance that you are using to determining the parallax distance? Also, the closest star is 4.22 ly. So the reference distance has to be in the same order. Do you have a reference distance.? No you do not so modern astronomy is based on your nonsense.