| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
28 Apr 2007 12:50:25 PM |
| Object: |
Precession of Binary Star PSR 1913+16 |
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, or is
available in some published document?
Using the system parameters found in http://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r = (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of 1.414 solar
masses:
v = sqrt(G*M / r)
The combined orbit circumference is:
C = 2 * (2*pi*r)
The orbit time is:
t = C / v
The precession rate in orbits per orbit, where e is the average orbit
eccentricity (0.617131):
n = 2*pi*(1 - cos(asin(v/c))) / (1 - e*e)
Converted to degrees per Earth year:
d = n * (365/(t/24/60/60)) * 360 = 4.4,
where modern General Relativity provides an approximation of 4.2
degrees per year.
If you are interested, I have put up a short PDF document online,
which includes C++ source code:
http://cavekitty.com/archives/The_Relativistic_Precession_of_Binary_Star_PSR_1913+16.zip
Thank you for any information that you can provide.
- Shawn
.
|
|
| User: "Androcles" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
28 Apr 2007 01:09:17 PM |
|
|
<shalayka@gmail.com> wrote in message =
news:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, or is
available in some published document?
=20
Using the system parameters found in =
http://www.johnstonsarchive.net/relativity/binpulsar.html
:
=20
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r =3D (1576800e3 + 373300e3) / 2.
=20
The average orbit velocity, where M is an average mass of 1.414 solar
masses:
v =3D sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the top.
How fast must it descend the same distance the other side to have an=20
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph most of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
=20
The combined orbit circumference is:
C =3D 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
.
|
|
|
| User: "" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
28 Apr 2007 03:27:28 PM |
|
|
Hi Androcles,
If I hadn't included eccentricity in the equation, the result would
have been extremely different (and wrong).
However, to perform my due diligence, and appease your objections, I
will work on modifying the equation so that the eccentricity is not
abstracted to the highest level possible as done in my original post
to this thread. I am very familiar with Ramanujan's 2nd method for
approximating the circumference of an ellipse, and use it exclusively
in my other endeavours.
What I am wondering now is if this method works for binary stars of
much different mass (but not so much different that it would simplify
back down to a Sun-Mercury type relationship).
I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the equation. I
just can't seem to find a lot of information on binaries at this time.
I appreciate your response, because the points you bring up are very
important to note, ex: eccentricity has a major part to play in this
equation.
It might be a little pugnacious of me to say so, but I'm not going to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your credit,
you are officially more communicative than the local university at
which I am a registered student. So, thank you for your time. I do
truly appreciate it.
- Shawn
On Apr 28, 12:09 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, or is
available in some published document?
Using the system parameters found inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r = (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of 1.414 solar
masses:
v = sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the top.
How fast must it descend the same distance the other side to have an
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph most of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
The combined orbit circumference is:
C = 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
28 Apr 2007 04:53:54 PM |
|
|
<shalayka@gmail.com> wrote in message =
news:1177792048.598466.293320@n59g2000hsh.googlegroups.com...
Hi Androcles,
=20
If I hadn't included eccentricity in the equation, the result would
have been extremely different (and wrong).
=20
However, to perform my due diligence, and appease your objections, I
will work on modifying the equation so that the eccentricity is not
abstracted to the highest level possible as done in my original post
to this thread. I am very familiar with Ramanujan's 2nd method for
approximating the circumference of an ellipse, and use it exclusively
in my other endeavours.
=20
What I am wondering now is if this method works for binary stars of
much different mass (but not so much different that it would simplify
back down to a Sun-Mercury type relationship).
Show that the advance of longitude of perihelion of Mercury is not=20
completely satisfied by Newtonian Mechanics.=20
Show that PSR 1913+16 is binary.=20
I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the equation. I
just can't seem to find a lot of information on binaries at this time.
Guessing is what horse race punters do. The bookie sets the odds and=20
wins every time. Mathematicians do not guess, they prove.
I appreciate your response, because the points you bring up are very
important to note, ex: eccentricity has a major part to play in this
equation.
No circular orbit can show precession.
=20
It might be a little pugnacious of me to say so, but I'm not going to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your credit,
you are officially more communicative than the local university at
which I am a registered student. So, thank you for your time. I do
truly appreciate it.
=20
You can have my time but it goes with my skepticism and incredulity.=20
I'm almost willing to accept this:
http://antwrp.gsfc.nasa.gov/apod/ap040513.html
is a close orbit eclipsing binary.=20
I am not willing to accept the forerunner of all binaries is a binary, =
based=20
on the word of an 18-year-old kid with a wooden telescope.
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
If you wish to call a star and planet a binary, then Sirius is a star
and planet.
http://antwrp.gsfc.nasa.gov/apod/ap001006.html
After all, it has a 50 year period placing the orbit comparably a bit=20
futher away than Saturn (29 years) and closer than Uranus (84 years).
You have enough there to keep you going for a few years.
Try not to guess or rely on other people's guesses and maybe you'll
become a real scientist. There are not many of those, but plenty of=20
pseudo-scientists.
- Shawn
=20
=20
=20
=20
=20
=20
On Apr 28, 12:09 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in =
messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, or =
is
available in some published document?
Using the system parameters found =
inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r =3D (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of 1.414 =
solar
masses:
v =3D sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the =
top.
How fast must it descend the same distance the other side to have an
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph most of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
The combined orbit circumference is:
C =3D 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
=20
.
|
|
|
| User: "" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
28 Apr 2007 08:57:44 PM |
|
|
Like I said, I am only guessing because I don't have the data. Call it
a prediction if you like.
As for being a scientist, I'm fine with not fitting the bill. The
methods I'm using are on par with General Relativity, regardless of
what I'm called. In fact, all of the scientists that I've tried to
communicate with have obviously lost the flare that comes with
learning and exploring the world. Otherwise, I would assume that they
would be interested in helping others learn about the universe instead
of berating or ignoring them. I'm sure you can agree that it's
lugubrious to want to associate oneself with that type of person,
regardless of title.
- Shawn
On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in messagenews:1177792048.598466.293320@n59g2000hsh.googlegroups.com...
Hi Androcles,
If I hadn't included eccentricity in the equation, the result would
have been extremely different (and wrong).
However, to perform my due diligence, and appease your objections, I
will work on modifying the equation so that the eccentricity is not
abstracted to the highest level possible as done in my original post
to this thread. I am very familiar with Ramanujan's 2nd method for
approximating the circumference of an ellipse, and use it exclusively
in my other endeavours.
What I am wondering now is if this method works for binary stars of
much different mass (but not so much different that it would simplify
back down to a Sun-Mercury type relationship).
Show that the advance of longitude of perihelion of Mercury is not
completely satisfied by Newtonian Mechanics.
Show that PSR 1913+16 is binary.
I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the equation. I
just can't seem to find a lot of information on binaries at this time.
Guessing is what horse race punters do. The bookie sets the odds and
wins every time. Mathematicians do not guess, they prove.
I appreciate your response, because the points you bring up are very
important to note, ex: eccentricity has a major part to play in this
equation.
No circular orbit can show precession.
It might be a little pugnacious of me to say so, but I'm not going to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your credit,
you are officially more communicative than the local university at
which I am a registered student. So, thank you for your time. I do
truly appreciate it.
You can have my time but it goes with my skepticism and incredulity.
I'm almost willing to accept this:
http://antwrp.gsfc.nasa.gov/apod/ap040513.html
is a close orbit eclipsing binary.
I am not willing to accept the forerunner of all binaries is a binary, based
on the word of an 18-year-old kid with a wooden telescope.
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
If you wish to call a star and planet a binary, then Sirius is a star
and planet.
http://antwrp.gsfc.nasa.gov/apod/ap001006.html
After all, it has a 50 year period placing the orbit comparably a bit
futher away than Saturn (29 years) and closer than Uranus (84 years).
You have enough there to keep you going for a few years.
Try not to guess or rely on other people's guesses and maybe you'll
become a real scientist. There are not many of those, but plenty of
pseudo-scientists.
- Shawn
On Apr 28, 12:09 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, or is
available in some published document?
Using the system parameters found inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r = (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of 1.414 solar
masses:
v = sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the top.
How fast must it descend the same distance the other side to have an
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph most of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
The combined orbit circumference is:
C = 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
29 Apr 2007 01:59:15 AM |
|
|
<shalayka@gmail.com> wrote in message =
news:1177811864.862424.58670@e65g2000hsc.googlegroups.com...
Like I said, I am only guessing because I don't have the data. Call it
a prediction if you like.
As for being a scientist, I'm fine with not fitting the bill. The
methods I'm using are on par with General Relativity, regardless of
what I'm called. In fact, all of the scientists that I've tried to
communicate with have obviously lost the flare that comes with
learning and exploring the world. Otherwise, I would assume that they
would be interested in helping others learn about the universe instead
of berating or ignoring them. I'm sure you can agree that it's
lugubrious to want to associate oneself with that type of person,
regardless of title.
I am interested in helping homo sapiens sapiens to think about
Nature as it is, I am not interested in training homo neanderthalensis=20
to become a shaman pushing buttons on a calculator, dividing by zero
and claiming he's now a mathematician.
GR is on a par with palmistry, reading horoscopes and tea leaves.=20
Learning about Nature does not mean inventing it.
You choose to be a shaman, attempting to fool others as you succeed
in fooling yourself.=20
Whether you agree or not, it's idiotic to want to associate oneself with
that type of imbecile, regardless of title. Better to ignore them.
Please go away, start a new newsgroup, "shaman.physics", you are fine
by not fitting the bill, and you don't.
=20
=20
=20
On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in =
messagenews:1177792048.598466.293320@n59g2000hsh.googlegroups.com...
Hi Androcles,
If I hadn't included eccentricity in the equation, the result would
have been extremely different (and wrong).
However, to perform my due diligence, and appease your objections, =
I
will work on modifying the equation so that the eccentricity is not
abstracted to the highest level possible as done in my original =
post
to this thread. I am very familiar with Ramanujan's 2nd method for
approximating the circumference of an ellipse, and use it =
exclusively
in my other endeavours.
What I am wondering now is if this method works for binary stars of
much different mass (but not so much different that it would =
simplify
back down to a Sun-Mercury type relationship).
Show that the advance of longitude of perihelion of Mercury is not
completely satisfied by Newtonian Mechanics.
Show that PSR 1913+16 is binary.
I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the =
equation. I
just can't seem to find a lot of information on binaries at this =
time.
Guessing is what horse race punters do. The bookie sets the odds and
wins every time. Mathematicians do not guess, they prove.
I appreciate your response, because the points you bring up are =
very
important to note, ex: eccentricity has a major part to play in =
this
equation.
No circular orbit can show precession.
It might be a little pugnacious of me to say so, but I'm not going =
to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your credit,
you are officially more communicative than the local university at
which I am a registered student. So, thank you for your time. I do
truly appreciate it.
You can have my time but it goes with my skepticism and incredulity.
I'm almost willing to accept this:
http://antwrp.gsfc.nasa.gov/apod/ap040513.html
is a close orbit eclipsing binary.
I am not willing to accept the forerunner of all binaries is a =
binary, based
on the word of an 18-year-old kid with a wooden telescope.
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
If you wish to call a star and planet a binary, then Sirius is a star
and planet.
http://antwrp.gsfc.nasa.gov/apod/ap001006.html
After all, it has a 50 year period placing the orbit comparably a bit
futher away than Saturn (29 years) and closer than Uranus (84 years).
You have enough there to keep you going for a few years.
Try not to guess or rely on other people's guesses and maybe you'll
become a real scientist. There are not many of those, but plenty of
pseudo-scientists.
- Shawn
On Apr 28, 12:09 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in =
messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, =
or is
available in some published document?
Using the system parameters found =
inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r =3D (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of 1.414 =
solar
masses:
v =3D sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the =
top.
How fast must it descend the same distance the other side to have =
an
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph most =
of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
The combined orbit circumference is:
C =3D 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
=20
.
|
|
|
|
| User: "" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
28 Apr 2007 09:09:59 PM |
|
|
I should be more specific here. I am only commenting on the behaviour
of those scientists at the university which I have paid good money
toward. I wouldn't have wasted my time and effort if I had known they
were going to be so completely useless to me in my desire to learn. I
know it's final examinations right now, but taking a month to reply to
an email message is stretching the limits of my good patience -- and
I'm a parent, so please believe me when I say that I have a near-
infinite supply of patience when it's appropriate.
- Shawn
On Apr 28, 7:57 pm, wrote:
Like I said, I am only guessing because I don't have the data. Call it
a prediction if you like.
As for being a scientist, I'm fine with not fitting the bill. The
methods I'm using are on par with General Relativity, regardless of
what I'm called. In fact, all of the scientists that I've tried to
communicate with have obviously lost the flare that comes with
learning and exploring the world. Otherwise, I would assume that they
would be interested in helping others learn about the universe instead
of berating or ignoring them. I'm sure you can agree that it's
lugubrious to want to associate oneself with that type of person,
regardless of title.
- Shawn
On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
< > wrote in messagenews:1177792048.598466.293320@n59g2000hsh.googlegroups.com...
Hi Androcles,
If I hadn't included eccentricity in the equation, the result would
have been extremely different (and wrong).
However, to perform my due diligence, and appease your objections, I
will work on modifying the equation so that the eccentricity is not
abstracted to the highest level possible as done in my original post
to this thread. I am very familiar with Ramanujan's 2nd method for
approximating the circumference of an ellipse, and use it exclusively
in my other endeavours.
What I am wondering now is if this method works for binary stars of
much different mass (but not so much different that it would simplify
back down to a Sun-Mercury type relationship).
Show that the advance of longitude of perihelion of Mercury is not
completely satisfied by Newtonian Mechanics.
Show that PSR 1913+16 is binary.
I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the equation. I
just can't seem to find a lot of information on binaries at this time.
Guessing is what horse race punters do. The bookie sets the odds and
wins every time. Mathematicians do not guess, they prove.
I appreciate your response, because the points you bring up are very
important to note, ex: eccentricity has a major part to play in this
equation.
No circular orbit can show precession.
It might be a little pugnacious of me to say so, but I'm not going to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your credit,
you are officially more communicative than the local university at
which I am a registered student. So, thank you for your time. I do
truly appreciate it.
You can have my time but it goes with my skepticism and incredulity.
I'm almost willing to accept this:
http://antwrp.gsfc.nasa.gov/apod/ap040513.html
is a close orbit eclipsing binary.
I am not willing to accept the forerunner of all binaries is a binary, based
on the word of an 18-year-old kid with a wooden telescope.
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
If you wish to call a star and planet a binary, then Sirius is a star
and planet.
http://antwrp.gsfc.nasa.gov/apod/ap001006.html
After all, it has a 50 year period placing the orbit comparably a bit
futher away than Saturn (29 years) and closer than Uranus (84 years).
You have enough there to keep you going for a few years.
Try not to guess or rely on other people's guesses and maybe you'll
become a real scientist. There are not many of those, but plenty of
pseudo-scientists.
- Shawn
On Apr 28, 12:09 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
< > wrote in messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, or is
available in some published document?
Using the system parameters found inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r = (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of 1.414 solar
masses:
v = sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the top.
How fast must it descend the same distance the other side to have an
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph most of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
The combined orbit circumference is:
C = 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
29 Apr 2007 02:09:16 AM |
|
|
<shalayka@gmail.com> wrote in message =
news:1177812599.034859.148750@y80g2000hsf.googlegroups.com...
I should be more specific here. I am only commenting on the behaviour
of those scientists at the university which I have paid good money
toward. I wouldn't have wasted my time and effort if I had known they
were going to be so completely useless to me in my desire to learn. I
know it's final examinations right now, but taking a month to reply to
an email message is stretching the limits of my good patience -- and
I'm a parent, so please believe me when I say that I have a near-
infinite supply of patience when it's appropriate.
=20
Under discussion is "Precession of Binary Star PSR 1913+16", not exam =
time, parenting,=20
cost of tuition or tardiness of faculty.=20
I'm not interested in your personal problems of everyday life no matter
how specific you are or what excuses you make, to me they are mere=20
gossip, and nor can I render you any assistance in resolving them.
Show that the advance of longitude of perihelion of Mercury is not
completely satisfied by Newtonian Mechanics.
Show that PSR 1913+16 is binary.
=20
=20
On Apr 28, 7:57 pm, wrote:
Like I said, I am only guessing because I don't have the data. Call =
it
a prediction if you like.
As for being a scientist, I'm fine with not fitting the bill. The
methods I'm using are on par with General Relativity, regardless of
what I'm called. In fact, all of the scientists that I've tried to
communicate with have obviously lost the flare that comes with
learning and exploring the world. Otherwise, I would assume that they
would be interested in helping others learn about the universe =
instead
of berating or ignoring them. I'm sure you can agree that it's
lugubrious to want to associate oneself with that type of person,
regardless of title.
- Shawn
On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
< > wrote in =
messagenews:1177792048.598466.293320@n59g2000hsh.googlegroups.com...
Hi Androcles,
If I hadn't included eccentricity in the equation, the result =
would
have been extremely different (and wrong).
However, to perform my due diligence, and appease your =
objections, I
will work on modifying the equation so that the eccentricity is =
not
abstracted to the highest level possible as done in my original =
post
to this thread. I am very familiar with Ramanujan's 2nd method =
for
approximating the circumference of an ellipse, and use it =
exclusively
in my other endeavours.
What I am wondering now is if this method works for binary stars =
of
much different mass (but not so much different that it would =
simplify
back down to a Sun-Mercury type relationship).
Show that the advance of longitude of perihelion of Mercury is not
completely satisfied by Newtonian Mechanics.
Show that PSR 1913+16 is binary.
I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the =
equation. I
just can't seem to find a lot of information on binaries at this =
time.
Guessing is what horse race punters do. The bookie sets the odds =
and
wins every time. Mathematicians do not guess, they prove.
I appreciate your response, because the points you bring up are =
very
important to note, ex: eccentricity has a major part to play in =
this
equation.
No circular orbit can show precession.
It might be a little pugnacious of me to say so, but I'm not =
going to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your =
credit,
you are officially more communicative than the local university =
at
which I am a registered student. So, thank you for your time. I =
do
truly appreciate it.
You can have my time but it goes with my skepticism and =
incredulity.
I'm almost willing to accept this:
http://antwrp.gsfc.nasa.gov/apod/ap040513.html
is a close orbit eclipsing binary.
I am not willing to accept the forerunner of all binaries is a =
binary, based
on the word of an 18-year-old kid with a wooden telescope.
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
If you wish to call a star and planet a binary, then Sirius is a =
star
and planet.
http://antwrp.gsfc.nasa.gov/apod/ap001006.html
After all, it has a 50 year period placing the orbit comparably a =
bit
futher away than Saturn (29 years) and closer than Uranus (84 =
years).
You have enough there to keep you going for a few years.
Try not to guess or rely on other people's guesses and maybe you'll
become a real scientist. There are not many of those, but plenty of
pseudo-scientists.
- Shawn
On Apr 28, 12:09 pm, "Androcles" =
<Engin...@hogwarts.physics.co.uk>
wrote:
< > wrote in =
messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a =
name, or is
available in some published document?
Using the system parameters found =
inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r =3D (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of =
1.414 solar
masses:
v =3D sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to =
the top.
How fast must it descend the same distance the other side to =
have an
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph =
most of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
The combined orbit circumference is:
C =3D 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
=20
.
|
|
|
|
|
| User: "Eric Gisse" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
28 Apr 2007 09:21:47 PM |
|
|
On Apr 28, 5:57 pm, wrote:
[...]
Don't listen to Androcles. He is a spewing idiot with delusions of
competence.
.
|
|
|
|
|
| User: "" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
28 Apr 2007 09:01:57 PM |
|
|
P.S. Thank you again for the links. My main desire at the moment is
for actual system data like eccentricity, apastron/periastron
separation distances, and precession estimates/observations. However,
I do find the pictures of these systems on NASA's page to be
fascinating.
On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in messagenews:1177792048.598466.293320@n59g2000hsh.googlegroups.com...
Hi Androcles,
If I hadn't included eccentricity in the equation, the result would
have been extremely different (and wrong).
However, to perform my due diligence, and appease your objections, I
will work on modifying the equation so that the eccentricity is not
abstracted to the highest level possible as done in my original post
to this thread. I am very familiar with Ramanujan's 2nd method for
approximating the circumference of an ellipse, and use it exclusively
in my other endeavours.
What I am wondering now is if this method works for binary stars of
much different mass (but not so much different that it would simplify
back down to a Sun-Mercury type relationship).
Show that the advance of longitude of perihelion of Mercury is not
completely satisfied by Newtonian Mechanics.
Show that PSR 1913+16 is binary.
I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the equation. I
just can't seem to find a lot of information on binaries at this time.
Guessing is what horse race punters do. The bookie sets the odds and
wins every time. Mathematicians do not guess, they prove.
I appreciate your response, because the points you bring up are very
important to note, ex: eccentricity has a major part to play in this
equation.
No circular orbit can show precession.
It might be a little pugnacious of me to say so, but I'm not going to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your credit,
you are officially more communicative than the local university at
which I am a registered student. So, thank you for your time. I do
truly appreciate it.
You can have my time but it goes with my skepticism and incredulity.
I'm almost willing to accept this:
http://antwrp.gsfc.nasa.gov/apod/ap040513.html
is a close orbit eclipsing binary.
I am not willing to accept the forerunner of all binaries is a binary, based
on the word of an 18-year-old kid with a wooden telescope.
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
If you wish to call a star and planet a binary, then Sirius is a star
and planet.
http://antwrp.gsfc.nasa.gov/apod/ap001006.html
After all, it has a 50 year period placing the orbit comparably a bit
futher away than Saturn (29 years) and closer than Uranus (84 years).
You have enough there to keep you going for a few years.
Try not to guess or rely on other people's guesses and maybe you'll
become a real scientist. There are not many of those, but plenty of
pseudo-scientists.
- Shawn
On Apr 28, 12:09 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, or is
available in some published document?
Using the system parameters found inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r = (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of 1.414 solar
masses:
v = sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the top.
How fast must it descend the same distance the other side to have an
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph most of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
The combined orbit circumference is:
C = 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
29 Apr 2007 02:19:16 AM |
|
|
<shalayka@gmail.com> wrote in message =
news:1177812117.896871.53830@u30g2000hsc.googlegroups.com...
P.S. Thank you again for the links. My main desire at the moment is
for actual system data like eccentricity, apastron/periastron
separation distances, and precession estimates/observations.=20
Totally impossible to obtain. Anything you think you have is idle
guesswork.
However,
I do find the pictures of these systems on NASA's page to be
fascinating.
Where your heart line reaches and crosses your life line there'll be a =
crux
in your life, you'll be at a crossroads. =20
http://www.handanalysis.co.uk/
How fascinating...<yawn>
"That fellow seems to me to possess but one idea, and that is a wrong =
one."--Dr. Samuel Johnson
"There are well-dressed foolish ideas, just as there are well-dressed =
fools."--Nicolas Chamfort
=20
=20
=20
On Apr 28, 3:53 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in =
messagenews:1177792048.598466.293320@n59g2000hsh.googlegroups.com...
Hi Androcles,
If I hadn't included eccentricity in the equation, the result would
have been extremely different (and wrong).
However, to perform my due diligence, and appease your objections, =
I
will work on modifying the equation so that the eccentricity is not
abstracted to the highest level possible as done in my original =
post
to this thread. I am very familiar with Ramanujan's 2nd method for
approximating the circumference of an ellipse, and use it =
exclusively
in my other endeavours.
What I am wondering now is if this method works for binary stars of
much different mass (but not so much different that it would =
simplify
back down to a Sun-Mercury type relationship).
Show that the advance of longitude of perihelion of Mercury is not
completely satisfied by Newtonian Mechanics.
Show that PSR 1913+16 is binary.
I'm guessing at this point that it will, but that I would have to
average their eccentricities before plugging them into the =
equation. I
just can't seem to find a lot of information on binaries at this =
time.
Guessing is what horse race punters do. The bookie sets the odds and
wins every time. Mathematicians do not guess, they prove.
I appreciate your response, because the points you bring up are =
very
important to note, ex: eccentricity has a major part to play in =
this
equation.
No circular orbit can show precession.
It might be a little pugnacious of me to say so, but I'm not going =
to
answer your questions. I have asked my questions due to an honest
dilemma, where it seems that you have not. However, to your credit,
you are officially more communicative than the local university at
which I am a registered student. So, thank you for your time. I do
truly appreciate it.
You can have my time but it goes with my skepticism and incredulity.
I'm almost willing to accept this:
http://antwrp.gsfc.nasa.gov/apod/ap040513.html
is a close orbit eclipsing binary.
I am not willing to accept the forerunner of all binaries is a =
binary, based
on the word of an 18-year-old kid with a wooden telescope.
http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm
If you wish to call a star and planet a binary, then Sirius is a star
and planet.
http://antwrp.gsfc.nasa.gov/apod/ap001006.html
After all, it has a 50 year period placing the orbit comparably a bit
futher away than Saturn (29 years) and closer than Uranus (84 years).
You have enough there to keep you going for a few years.
Try not to guess or rely on other people's guesses and maybe you'll
become a real scientist. There are not many of those, but plenty of
pseudo-scientists.
- Shawn
On Apr 28, 12:09 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in =
messagenews:1177782625.497884.294940@n59g2000hsh.googlegroups.com...
I am wondering if the following method for calculating the
relativistic precession of a highly eccentric orbit has a name, =
or is
available in some published document?
Using the system parameters found =
inhttp://www.johnstonsarchive.net/relativity/binpulsar.html
:
The orbit radius is taken to be the average of one half of the
separation distances at apastron and periastron:
r =3D (1576800e3 + 373300e3) / 2.
The average orbit velocity, where M is an average mass of 1.414 =
solar
masses:
v =3D sqrt(G*M / r)
That is a schoolboy blunder.
A car drives up a hill for one mile taking 2 minutes to get to the =
top.
How fast must it descend the same distance the other side to have =
an
average velocity of 60 mph?
I drive from home to London (30 miles for 1 hour), at 80 mph most =
of
the way (I really do).
What is my average velocity, my maximum velocity and my minimum
velocity?
The combined orbit circumference is:
C =3D 2 * (2*pi*r)
Wrong. What is the circumference of an ellipse?
http://mathworld.wolfram.com/images/gifs/EllipticGears.gif
=20
.
|
|
|
| User: "" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
30 Apr 2007 01:35:28 PM |
|
|
The precession data is not impossible to obtain. I'm sure there's been
one or two binaries found since 1973.
As for my "problems", they are not personal, they're professional. I
just got a very detailed reply from my university, which means that
they were not ignoring me. I guess that means the problem is theirs,
since they are overworked.
By the way, where did you get the precession data that you used to
verify your methods? Or have you only explained the precession of
Mercury?
- Shawn
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
30 Apr 2007 02:22:35 PM |
|
|
<shalayka@gmail.com> wrote in message =
news:1177958128.065520.238070@y80g2000hsf.googlegroups.com...
The precession data is not impossible to obtain. I'm sure there's been
one or two binaries found since 1973.
Then do it.
.
|
|
|
| User: "" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
30 Apr 2007 03:43:17 PM |
|
|
Do what? Contact the local universities to see if I can get data on
binary stars? I've already done that.
How do you not get dizzy from the circles you're constantly running
in?
Don't get me wrong, I enjoy reading your posts. However, I'm confident
at this point that you are constructing a fantasy conversation in your
head that has nothing to do with what we're actually talking about.
Not exactly frustrating, just bewildering.
- Shawn
On Apr 30, 1:22 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in messagenews:1177958128.065520.238070@y80g2000hsf.googlegroups.com...
The precession data is not impossible to obtain. I'm sure there's been
one or two binaries found since 1973.
Then do it.
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Precession of Binary Star PSR 1913+16 |
30 Apr 2007 06:20:39 PM |
|
|
<shalayka@gmail.com> wrote in message =
news:1177965797.274560.227310@u30g2000hsc.googlegroups.com...
Do what? Contact the local universities to see if I can get data on
binary stars? I've already done that.
It was you that said it was possible to obtain the data, not I. Run =
around
in circles until you have it.
How do you not get dizzy from the circles you're constantly running
in?
It's your circle. I'm sitting back waiting for you to obtain some data. =
<shrug>
=20
Don't get me wrong, I enjoy reading your posts. However, I'm confident
at this point that you are constructing a fantasy conversation in your
head that has nothing to do with what we're actually talking about.
Not exactly frustrating, just bewildering.
=20
- Shawn
Are you?
=20
=20
=20
On Apr 30, 1:22 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
wrote:
<shala...@gmail.com> wrote in =
messagenews:1177958128.065520.238070@y80g2000hsf.googlegroups.com...
The precession data is not impossible to obtain. I'm sure there's =
been
one or two binaries found since 1973.
Then do it.
=20
.
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