Other than terrestrial water and tidal flex heating

On Jul 28, 4:01 pm, "Warren G. Harding" wrote:
BradGuth wrote:
On Jul 27, 10:39 pm, Timberwoof
wrote:
In article
,

BradGuth wrote:
On Jul 27, 8:22 pm, "Warren G. Harding" wrote:
Timberwoof wrote:
In article ,
Odysseus wrote:

In article
,
Timberwoof wrote:

snip

An AU is 15.813*10^6 LY.

Say, what? At first I thought you might have just switched the units
--
but even when inverted the figure is wrong by orders of magnitude.

Whoops. Brain fart. I guess I should redo my math and show it.

AU = 149.60 x 10^9 m
ly = 9.461 x 1^15 m
ly = 63.24 x 10^3 AU

There we go. An LY is ~ 10^5 AU. And since gravitational force decreases
with the square of the distance, Brad's favorite star Sirius, which is
about half a million light years away, has about 1 x 10^-11[1] the
gravitational influence on the Earth that the sun has. In other words,
unmeasurable.

Sorry about the error; thanks for spotting it and giving me the chance
to correct it. Brad, shut up. I never claimed to be infallible, and I've
given you dozens and dozens of chances to correct your errors.

[1] The mass of Sirius is 3 solar masses; it's 8.6 LY away, which equals
5.44 x 10^5 au. One over the square of the distance, times the mass of
Sirius, is 1.0 x 10^-11.

I'm still waiting to see the formal definition of "tidal radius."

Better yet is the tidal elliptic radius, as an extended orbital radius
variable in 3D interactive space, whereas everything is continually in
3D motion. To say the least, this is highly complex.

Meaningless drivel.

In 2D space, a basic tidal radius of a purely circular orbit is the
same as specified by the usual internet posted definitions.

Post some URLs please. I am unable to find any that make sense in this
context.

If both items are of the exact same mass and always of equal velocity
throughout each of their nearly circular treks, and with having no
other gravitational influences would obviously make this relatively
easy to spreadsheet this one out.

Not without the definition of "tidal radius".

But are you going to help redefine this "tidal radius", or are you
only going to stalk and bash everything to death, as is your usual MO?

You are the person who uses the phrase in a technical manner, therefore
it is your responsibility to define what it means if you expect others
to understand. Otherwise you might as well be posting in Dzongkha.



- Brad Guth Brad_Guth Brad.Guth BradGuth

Now you're insulting the Dzongkha.

Clearly you are not the least interested in helping one damn bit. So,
why are you here?

- Brad Guth Brad_Guth Brad.Guth BradGuth

On Jul 28, 4:03 pm, "Warren G. Harding" wrote:
BradGuth wrote:
You do realize that my "train of thought" is working this analogy from
the reverse conclusion method, that's based upon a few rogue planets
and icy proto-moons having arrived into our solar system. Therefore
we need to start off by assuming there's a fundamental logic and
method within the known laws of physics as to how that scope of
interstellar trek could have transpired if everything went exactly
according to plan.

Will nothing dissuade you from this assumption?

A full blown do-everything imaginable via supercomputer simulation, or
rather millions upon million of simulations, if that's what it takes.
Isn't that dark and scary, or what?

- Brad Guth Brad_Guth Brad.Guth BradGuth

BradGuth wrote:

On Jul 28, 3:57 pm, "Warren G. Harding" wrote:

What is "3D interactive space"?

Everything is in orbit around something, and mostly of the elliptical
orbit format.

Then why not use common language and say "orbit"?


What is "an extended orbital radius variable"?

When there's something of gravity other than the primary two or three
items, such as other significant stellar groups, and of course those
pesky black holes that should account for something.

In physics, these are called "forces". See Newton, Isaac.


How is "tidal elliptic radius" defined?

I'm still working on that one. Would you care to help?

It is your term, I can't read your mind. If you mean "elliptic orbit",
then say so. By the way, Newton described all orbits as ellipses.


In 2D space, a basic tidal radius of a purely circular orbit is the
same as specified by the usual internet posted definitions.

Now you've introduced another term: "basic tidal radius" without any
explanation. Are there no links to these "usual internet posted
definitions"?

Remove the word "basic", and just go with the mainstream accepted
flow.

Then this sentence becomes:

"In 2D space, a(n) /orbit/ of a purely circular orbit is the same as
specified by the usual internet posted definitions."

Or:

"An orbit is an orbit."


If both items are of the exact same mass and always of equal velocity
throughout each of their nearly circular treks, and with having no
other gravitational influences would obviously make this relatively
easy to spreadsheet this one out.

I have no idea what you are trying to communicate here -- "spreadsheet"
is a noun, not a verb. Where and what are these two identical "items"
you refer to? Obviously this has nothing to with the Sun.

You didn't try to tell me what the identical items are.

You still have no definition of "tidal radius."

You obviously like to play word games, and to deliver as much topic/
author ****ology as you can muster, just like your old boss Hitler and
all of those fellow Zionist/Nazi types with their brown-nosed minions
and clowns.

No need to froth and foam, Adolf Hitler was a pile of ash long before I
was born.

If you want other people to comprehend what you are writing, you have
to explicitly explain what the terms you are using mean. I cannot
define them for you, as you asked me above concerning "tidal elliptic
radius".

Physics uses precise language in which the technical terms are defined
or approximated with mathematical relationships. To arbitrarily change
these terms to something else is pointless, no one will understand you.

Can you be trusted to deductively interpret a pile of dog poop before
stepping in it, anyway.

I was trying to understand your words. I hope you understand the
implication of this association with "dog poop".

- Brad Guth Brad_Guth Brad.Guth BradGuth

BradGuth wrote:

On Jul 28, 4:01 pm, "Warren G. Harding" wrote:
BradGuth wrote:
On Jul 27, 10:39 pm, Timberwoof
wrote:
In article
,

BradGuth wrote:
On Jul 27, 8:22 pm, "Warren G. Harding" wrote:
Timberwoof wrote:
In article ,
Odysseus wrote:

In article
,
Timberwoof wrote:

snip

An AU is 15.813*10^6 LY.

Say, what? At first I thought you might have just switched the
units
--
but even when inverted the figure is wrong by orders of magnitude.

Whoops. Brain fart. I guess I should redo my math and show it.

AU = 149.60 x 10^9 m
ly = 9.461 x 1^15 m
ly = 63.24 x 10^3 AU

There we go. An LY is ~ 10^5 AU. And since gravitational force
decreases
with the square of the distance, Brad's favorite star Sirius, which
is
about half a million light years away, has about 1 x 10^-11[1] the
gravitational influence on the Earth that the sun has. In other
words,
unmeasurable.

Sorry about the error; thanks for spotting it and giving me the
chance
to correct it. Brad, shut up. I never claimed to be infallible, and
I've
given you dozens and dozens of chances to correct your errors.

[1] The mass of Sirius is 3 solar masses; it's 8.6 LY away, which
equals
5.44 x 10^5 au. One over the square of the distance, times the mass
of
Sirius, is 1.0 x 10^-11.

I'm still waiting to see the formal definition of "tidal radius."

Better yet is the tidal elliptic radius, as an extended orbital radius
variable in 3D interactive space, whereas everything is continually in
3D motion. To say the least, this is highly complex.

Meaningless drivel.

In 2D space, a basic tidal radius of a purely circular orbit is the
same as specified by the usual internet posted definitions.

Post some URLs please. I am unable to find any that make sense in this
context.

If both items are of the exact same mass and always of equal velocity
throughout each of their nearly circular treks, and with having no
other gravitational influences would obviously make this relatively
easy to spreadsheet this one out.

Not without the definition of "tidal radius".

But are you going to help redefine this "tidal radius", or are you
only going to stalk and bash everything to death, as is your usual MO?

You are the person who uses the phrase in a technical manner, therefore
it is your responsibility to define what it means if you expect others
to understand. Otherwise you might as well be posting in Dzongkha.



- Brad Guth Brad_Guth Brad.Guth BradGuth

Now you're insulting the Dzongkha.

Clearly you are not the least interested in helping one damn bit. So,
why are you here?

Clearly you missed my point. If you cannot express your meaning in
language that other people understand, they will not understand what
you are attempting to communicate. Others (i.e. me) cannot help you
here, that is your task.

- Brad Guth Brad_Guth Brad.Guth BradGuth

BradGuth wrote:

On Jul 28, 4:03 pm, "Warren G. Harding" wrote:
BradGuth wrote:
You do realize that my "train of thought" is working this analogy from
the reverse conclusion method, that's based upon a few rogue planets
and icy proto-moons having arrived into our solar system. Therefore
we need to start off by assuming there's a fundamental logic and
method within the known laws of physics as to how that scope of
interstellar trek could have transpired if everything went exactly
according to plan.

Will nothing dissuade you from this assumption?

A full blown do-everything imaginable via supercomputer simulation, or
rather millions upon million of simulations, if that's what it takes.
Isn't that dark and scary, or what?

Why must it be in the form of a visual simulation? And why does the
computer hardware make a difference?

- Brad Guth Brad_Guth Brad.Guth BradGuth

On Jul 28, 6:21 pm, "Warren G. Harding" wrote:
BradGuth wrote:
On Jul 28, 4:03 pm, "Warren G. Harding" wrote:
BradGuth wrote:
You do realize that my "train of thought" is working this analogy from
the reverse conclusion method, that's based upon a few rogue planets
and icy proto-moons having arrived into our solar system. Therefore
we need to start off by assuming there's a fundamental logic and
method within the known laws of physics as to how that scope of
interstellar trek could have transpired if everything went exactly
according to plan.

Will nothing dissuade you from this assumption?

A full blown do-everything imaginable via supercomputer simulation, or
rather millions upon million of simulations, if that's what it takes.
Isn't that dark and scary, or what?

Why must it be in the form of a visual simulation? And why does the
computer hardware make a difference?



- Brad Guth Brad_Guth Brad.Guth BradGuth

Because that's what the mainstream status quo gets to play with, and
to eye-candy publish and hype on behalf of whatever their heart
desires, and usually it's in one way or another entirely at public
expense. BTW, I too want to get paid (at least eventually before I
die) for doing this.

- Brad Guth Brad_Guth Brad.Guth BradGuth

On Jul 28, 6:19 pm, "Warren G. Harding" wrote:
BradGuth wrote:
On Jul 28, 4:01 pm, "Warren G. Harding" wrote:
BradGuth wrote:
On Jul 27, 10:39 pm, Timberwoof
wrote:
In article
,

BradGuth wrote:
On Jul 27, 8:22 pm, "Warren G. Harding" wrote:
Timberwoof wrote:
In article ,
Odysseus wrote:

In article
,
Timberwoof wrote:

snip

An AU is 15.813*10^6 LY.

Say, what? At first I thought you might have just switched the
units
--
but even when inverted the figure is wrong by orders of magnitude.

Whoops. Brain fart. I guess I should redo my math and show it.

AU = 149.60 x 10^9 m
ly = 9.461 x 1^15 m
ly = 63.24 x 10^3 AU

There we go. An LY is ~ 10^5 AU. And since gravitational force
decreases
with the square of the distance, Brad's favorite star Sirius, which
is
about half a million light years away, has about 1 x 10^-11[1] the
gravitational influence on the Earth that the sun has. In other
words,
unmeasurable.

Sorry about the error; thanks for spotting it and giving me the
chance
to correct it. Brad, shut up. I never claimed to be infallible, and
I've
given you dozens and dozens of chances to correct your errors.

[1] The mass of Sirius is 3 solar masses; it's 8.6 LY away, which
equals
5.44 x 10^5 au. One over the square of the distance, times the mass
of
Sirius, is 1.0 x 10^-11.

I'm still waiting to see the formal definition of "tidal radius."

Better yet is the tidal elliptic radius, as an extended orbital radius
variable in 3D interactive space, whereas everything is continually in
3D motion. To say the least, this is highly complex.

Meaningless drivel.

In 2D space, a basic tidal radius of a purely circular orbit is the
same as specified by the usual internet posted definitions.

Post some URLs please. I am unable to find any that make sense in this
context.

If both items are of the exact same mass and always of equal velocity
throughout each of their nearly circular treks, and with having no
other gravitational influences would obviously make this relatively
easy to spreadsheet this one out.

Not without the definition of "tidal radius".

But are you going to help redefine this "tidal radius", or are you
only going to stalk and bash everything to death, as is your usual MO?

You are the person who uses the phrase in a technical manner, therefore
it is your responsibility to define what it means if you expect others
to understand. Otherwise you might as well be posting in Dzongkha.

- Brad Guth Brad_Guth Brad.Guth BradGuth

Now you're insulting the Dzongkha.

Clearly you are not the least interested in helping one damn bit. So,
why are you here?

Clearly you missed my point. If you cannot express your meaning in
language that other people understand, they will not understand what
you are attempting to communicate. Others (i.e. me) cannot help you
here, that is your task.



- Brad Guth Brad_Guth Brad.Guth BradGuth

Thanks, I'll try harder to make myself a little more understandable to
those that can't hardly or wouldn't dare figure anything out for
themselves.

If I should manage to redefine "tidal radius", or better to create the
"tidal elliptic" version, do I get to have my name as tagged to that
formula?

- Brad Guth Brad_Guth Brad.Guth BradGuth

Deductive thinking isn't for the innocent, the too young or the too
old. Do you even know what it's like to deductively think or ponder
about much of anything?

- Brad Guth Brad_Guth Brad.Guth BradGuth


On Jul 28, 6:16 pm, "Warren G. Harding" wrote:
BradGuth wrote:
On Jul 28, 3:57 pm, "Warren G. Harding" wrote:

What is "3D interactive space"?

Everything is in orbit around something, and mostly of the elliptical
orbit format.

Then why not use common language and say "orbit"?



What is "an extended orbital radius variable"?

When there's something of gravity other than the primary two or three
items, such as other significant stellar groups, and of course those
pesky black holes that should account for something.

In physics, these are called "forces". See Newton, Isaac.



How is "tidal elliptic radius" defined?

I'm still working on that one. Would you care to help?

It is your term, I can't read your mind. If you mean "elliptic orbit",
then say so. By the way, Newton described all orbits as ellipses.



In 2D space, a basic tidal radius of a purely circular orbit is the
same as specified by the usual internet posted definitions.

Now you've introduced another term: "basic tidal radius" without any
explanation. Are there no links to these "usual internet posted
definitions"?

Remove the word "basic", and just go with the mainstream accepted
flow.

Then this sentence becomes:

"In 2D space, a(n) /orbit/ of a purely circular orbit is the same as
specified by the usual internet posted definitions."

Or:

"An orbit is an orbit."



If both items are of the exact same mass and always of equal velocity
throughout each of their nearly circular treks, and with having no
other gravitational influences would obviously make this relatively
easy to spreadsheet this one out.

I have no idea what you are trying to communicate here -- "spreadsheet"
is a noun, not a verb. Where and what are these two identical "items"
you refer to? Obviously this has nothing to with the Sun.

You didn't try to tell me what the identical items are.



You still have no definition of "tidal radius."

You obviously like to play word games, and to deliver as much topic/
author ****ology as you can muster, just like your old boss Hitler and
all of those fellow Zionist/Nazi types with their brown-nosed minions
and clowns.

No need to froth and foam, Adolf Hitler was a pile of ash long before I
was born.

If you want other people to comprehend what you are writing, you have
to explicitly explain what the terms you are using mean. I cannot
define them for you, as you asked me above concerning "tidal elliptic
radius".

Physics uses precise language in which the technical terms are defined
or approximated with mathematical relationships. To arbitrarily change
these terms to something else is pointless, no one will understand you.



Can you be trusted to deductively interpret a pile of dog poop before
stepping in it, anyway.

I was trying to understand your words. I hope you understand the
implication of this association with "dog poop".



- Brad Guth Brad_Guth Brad.Guth BradGuth