Anisotropy and Mercury (2)

"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:
Max Keon wrote:

How can "every source" possibly say anything relating to a
gravity anisotropy when a gravity anisotropy has never been
represented in mathematics?

Because you have expressed that anisotropy as an
acceleration and the relationship between acceleration,
velocity and location are fixed by their definitions.

You alone have chosen the maths to suit the outcome you want.

There is no possible choce to make Max, acceleration
is DEFINED and the rate of change of velocity

Well why do you have such a problem understanding why you are
wrongly applying the anisotropy?

I am applying as your equation dictates - I have
no choice.

You profess to know all about the consequences of a gravity
anisotropy when you obviously can't. You have never had reason
to address such a thing in the past, and the course you've chosen
is the one that gives the best result for your purpose. But your
method of applying the anisotropic acceleration is totally wrong.
You are adding it to the naturally flowing acceleration in a
naturally occurring eccentric orbit. Which is of course nothing
more than a circular orbit that has been knocked off center.

Can you not see that?

One thing that you have never given a satisfactory explanation
for is why Mercury's fall to its previous perihelion radius is
prematurely halted as a consequence of a gravity anisotropy.
Perhaps if you try demonstrating that mathematically, it will
help you to understand why you are wrong. You will come to
realize that Mercury's orbital speed cannot be fast enough to
generate the required centrifugal force to prematurely halt its
fall and whiz it back out toward the aphelion because the
anisotropy has caused it to be drawn more slowly to the Sun
throughout the inward journey, so by the time Mercury reaches
the point of last perihelion, it has obviously fallen less
distance than normal and is traveling at a lesser speed than
normal. So how on earth does it turn around and go back out
again.

You will open up a whole new universe for me if you can prove
that it does, where nothing is impossible, or predictable.

The rate of change of velocity
in the direction along which the force is applied, being directly
toward the Sun, _and only directly toward the Sun,_ is exactly
what it should be. Your own _personal_ assumption that the force
points in other directions as well is completely false.

I am applying it exactly towards the Sun. A long
time ago I spent several weeks trying to get you
to clarify the direction and eventually I gathered
that was what you meant though you were never
entirely clear on the point.

The program calculates the _magnitude_ of the
anisotropy using (v_r/c) times the Newtonian value
and assumes the direction is the same as the
Newtonian force, i.e. directly towards the Sun.

Yes, and that's probably why you are confused. Your program is
designed _only_ to plot a natural orbit path, _even circular_.
The Newtonian acceleration is no doubt pointing directly at the
Sun throughout the entire orbit, but directly adding the
anisotropy in the way you have done is still wrong.

My method may not yet be precise, but you are millions of meters
away from your claim of precision within 1 meter.
---

All I will suggest is that you break your program at the
one quarter and three quarter orbit points and check
for yourself whether the anisotropy is increasing or
decreasing the Newtonian acceleration. I think you
will find it is the same in both cases when they should
differ. Don't take my word for it, check.

There's no doubt about it George.

Then do the test I suggest and find out.

I recently made the mistake of not doing such a test before
jumping to conclusions and I'm not about to do it again just
yet. But thanks anyway.

The physical sign manipulation
is essential because the required equations always give a
positive result even when the anisotropy becomes negative.

Just in case you've forgotten;

dt = 1
ana = 1
# = (ana^2 * dt)^.5 is always 1.

This is just another example of your problems with
simple maths. You are again taking the root of a
square

# = (ana^2 * dt)^.5

is the same as

# = ana * sqrt(dt)

(ana^2 * dt)^.5 is only part of the equation. The proper
equation is derived from a^2 + b^2 = c^2. It becomes
s = (v^2 + ana^2)^.5 if dt is removed (s is the updated orbital
speed).

But why are you calculating that in the first place?

I'm trying to determine the orbital speed changes according to
the path taken when the anisotropy is included as compared with
the unaffected path. I've used Pythagoras to do that.

Previous orbital speed
______________________
- _ l Anomalous
New orbital speed - l acceleration


I don't need you to tell me that I'm wrong either, only that
it's not the way you would do it. But we already know that.

I've included the program that was designed to demonstrate the
true effect of an anomalous change in gravity, and it does so
exceptionally well. It demonstrates that the orbiting mass
behaves exactly as it should according to the forces acting on
it, with centrifugal force rising to twice the anomalous force
before the falling mass can be halted and sent back out again.

You will notice that I've treated the "previous orbital speed"
as if it was tangential to the gravity source, but that is of
virtually no consequence. The largest error is an adjacent length
of 48000.0000005569 instead of 48000. The hypotenuse length
change due to the anomalous acceleration is not going to change
much if I use the extended length, is it.

This program justifies the method I'm now using to include the
anisotropy in your program.

I've added comments within the program. Feel free to reply to
them.

'-------------
' Control_break halts the program at any time.

' The program plots the natural oscillation of a mass that was
' in a stable circular orbit prior to the application of a sudden
' anomalous change in the pull of gravity. The program ends after
' a complete orbit cycle.

' The same oscillation would not be set up by a gravity
' anisotropy because it's applied in a sine wave fashion.
' The fall distance would follow the same sine wave but would
' always lag behind.

' The inward moving mass overshoots the balance point between
' centrifugal force and the inward force by twice the current
' fall distance. Centrifugal force needs to be, and is, double
' the anomalous inward force that drives it before it has gained
' the required force to halt the moving mass and send it back
' from whence it came. It's an entirely elastic operation.

DEFDBL A-Z
SCREEN 12
CLS : COLOR 7
'-----------
LINE (176, 280)-(261, 280), 8 ' Shows the scale distortion.
LINE (261, 280)-(261, 358), 8
LINE (176, 280)-(261, 358), 8
LINE (150, 240)-(185, 284), 10
LINE (200, 360)-(200, 450), 8
LOCATE 20, 30: PRINT "313000"
LOCATE 18, 26: PRINT "6.5e10"
LOCATE 15, 10: PRINT "2.76e-4 degrees"
'-----------
r = 58000000000# ' Orbit radius.
g = .04# ' Gravity per Newton.
v = 48000#
vv = v ' vv holds the original orbital speed.

an = .0000008# ' Anomalous gravity change (m/sec^2).
' -.0000008# gives the negative result.

dt = 100 ' Set dt as required.

aa:
f = f + 1 ' Program cycle count.
ana = ana + dt * cfx ' Stores the current fall rate.
' cfx is later defined.
anb = anb + dt * ana ' Stores the total fall distance.

'-------The physical sign manipulation is necessary because the
' result of (ana)^2 is always positive, whatever the sign on ana,
' and orbital speed continues to increase regardless of the true
' value stored in ana.

IF anc = anb THEN s = (v ^ 2# - ((ana) ^ 2 * dt)) ^ .5#
IF anc anb THEN s = (v ^ 2# + (ana ^ 2 * dt)) ^ .5#
' s is updated orbital speed.
' The manner in which dt is included gives exactly the same result
' for all values of dt, including dt = 1. Remove dt from the
' equations altogether if it causes confusion.

' The point is that it can't be wrong because it always gives the
' same result as with no dt at all.

anc = anb
'-------------------

v = s
' v holds the updated orbital speed for the next cycle.

cf = v ^ 2# / vv ^ 2# * g ' cf is centrifugal force.

cfx = g + an - cf ' Compares the inward force including
' the anomalous acceleration, with the current centrifugal force
' value. The result gives the true acceleration rate, which is
' added to ana, above.

CIRCLE (10 + f * dt / 16000, 280 + anb / 4000), 0, 14
' 16000 and 4000 are multipliers for the graphics.

IF f * dt 7603200 THEN GOSUB ab: END
fa = fa + 1: IF fa * dt 10000 THEN GOSUB ab: fa = 0

GOTO aa

ab: LOCATE 4, 1
PRINT f * dt; "seconds elapsed time. "
PRINT ana; "meter fall per"; dt; "second batch. "
PRINT anb; "meter total fall so far. "
PRINT s - vv; "m/sec orbital speed change from the normal. "
PRINT cf; "m/sec^2 centrifugal force. "
LOCATE 24, 18
PRINT r - anb; "radius. "
RETURN

'---------------------

http://members.optusnet.com.au/maxkeon/proven2.html
has been updated slightly.
---

Dark matter is non-baryonic so new physics would not
be surprising, it is an exciting prospect indeed. The best
handle we have so far seems to be the Bullet Cluster
where the two clumps of dark matter associatated with
the galaxies have passed through each other.

And the latest evidence doesn't comply with that.

Cite the paper please.

http://www.eurekalert.org/pub_releas...-dmm081607.php
seems to indicate that the Abell 520 system "train wreck" is not
as previously noted, as in the Bullet Cluster where the "dark
matter" and galaxies have stayed together.

Einstein's theories cannot explain what nature is clearly
demonstrating, so they break down if some reason for the anomaly
is not forthcoming.

You seem confused. It is the gravitational lensing of
Einsein's theory that provides the tool we use to
investigate dark matter. Lensing of more distant
objects allows us the map the distribution of the
dark matter.

How can a fact of nature be taken aside and labeled "Einstein's
theory" ?

What I labelled "Einstein's theory" gives us the maths
that describes that aspect of Nature.

Such lensing was obviously going to be present long
before Einstein came along. It _is_ a fact of nature you know.

Of course.

Anyway, what makes you think the lensing is causes by dark
matter?

Dark matter just means something that has mass but
doesn't interact with light, it is a 'catch all'
generic term and deliberately vague.

Why not something that nature predicts, like a black
hole(s)?

Because microlensing surveys should detect them.
The find some events but not enough.

How do you differentiate between dark matter and black holes?

-----

Max Keon

Max Keon wrote:
"George Dishman" wrote in message ...
Max Keon wrote:
George Dishman wrote:
Max Keon wrote:

How can "every source" possibly say anything relating to a
gravity anisotropy when a gravity anisotropy has never been
represented in mathematics?

Because you have expressed that anisotropy as an
acceleration and the relationship between acceleration,
velocity and location are fixed by their definitions.

You alone have chosen the maths to suit the outcome you want.

There is no possible choce to make Max, acceleration
is DEFINED and the rate of change of velocity

Well why do you have such a problem understanding why you are
wrongly applying the anisotropy?

I am applying as your equation dictates - I have
no choice.

You profess to know all about the consequences of a gravity
anisotropy when you obviously can't.

I have no such need, _you_ have provided the equation
that supposedly describes the anisotropy, all _I_ need
to know is how to integrate that. It is simple maths that
I learnt in secondary school about forty years ago. You
should know it too but apparently you haven't done that.

You have never had reason
to address such a thing in the past, and the course you've chosen
is the one that gives the best result for your purpose. But your
method of applying the anisotropic acceleration is totally wrong.

I'm sorry Max, there is only one method permitted
by the definition of acceleration. If you don't like the
results, the only thing you can do is change the
equation you have given me.

You are adding it to the naturally flowing acceleration in a
naturally occurring eccentric orbit. Which is of course nothing
more than a circular orbit that has been knocked off center.

Can you not see that?

Your statement is factually untrue, the minor axis
of an ellipse is smaller than the major axis.

One thing that you have never given a satisfactory explanation
for is why Mercury's fall to its previous perihelion radius is
prematurely halted as a consequence of a gravity anisotropy.
Perhaps if you try demonstrating that mathematically, it will
help you to understand why you are wrong.

I have told you over and over again why that happens,
and we discussed it in detail two months ago, see
message number 29 in this thread on Google.

You will come to
realize that Mercury's orbital speed cannot be fast enough to
generate the required centrifugal force to prematurely halt its
fall and whiz it back out toward the aphelion because the
anisotropy has caused it to be drawn more slowly to the Sun
throughout the inward journey, so by the time Mercury reaches
the point of last perihelion, it has obviously fallen less
distance than normal and is traveling at a lesser speed than
normal. So how on earth does it turn around and go back out
again.

You will open up a whole new universe for me if you can prove
that it does, where nothing is impossible, or predictable.

It is fully predicaable Max, just integrate the acceleration
as the maths requires. It may well open up a whole new
universe for you if you learn calculus and vectors.

The rate of change of velocity
in the direction along which the force is applied, being directly
toward the Sun, _and only directly toward the Sun,_ is exactly
what it should be. Your own _personal_ assumption that the force
points in other directions as well is completely false.

I am applying it exactly towards the Sun. A long
time ago I spent several weeks trying to get you
to clarify the direction and eventually I gathered
that was what you meant though you were never
entirely clear on the point.

The program calculates the _magnitude_ of the
anisotropy using (v_r/c) times the Newtonian value
and assumes the direction is the same as the
Newtonian force, i.e. directly towards the Sun.

Yes, and that's probably why you are confused. Your program is
designed _only_ to plot a natural orbit path, _even circular_.

Not true, the program integrates your equation for the
acceleration wherever that may lead.

The Newtonian acceleration is no doubt pointing directly at the
Sun throughout the entire orbit, but directly adding the
anisotropy in the way you have done is still wrong.

The Newtonian acceleration points at the Sun, we
both know that. I have asked you many times what
the direction of the anisotropy is and yo never gave
a clear answer but my understanding is that it also
points along the line betwen the un and the planet
but towards the Sun on the outward leg and away
from it on the inward leg. Please either confirm that
or correct it.

The rules of vector addition in cartesian coordinates
say we must add the two x components to get the
x component of the sum and similarly for y. Since
both vectors point along the same Sun-Mercury line
those rules mean we get the same answer if we add
the signed magnitudes and the result points in the
same direction as the Newtonian force which is slightly
easier for coding so that's what the program does.

My method may not yet be precise, ..

Your method is fundamentally wrong. An alternative
which would be similar to what you are trying to do
would be to work in polar coordinates but it is much
more complex and you need to accomodate coriolis
effects. That's not the real reason your results are
wrong thugh, you are making a few assumptions
which are incorrect. An ellipse is _not_ an offset
circle, the speed in an elliptical orbit is _not_ the
same as a cicular orbit and the path is _not_
perpendicular to the Sun-Mercury line so you cannot
use Pythagoras.

but you are millions of meters
away from your claim of precision within 1 meter.

When you learn how to do the maths, you will get the
same result as me.

All I will suggest is that you break your program at the
one quarter and three quarter orbit points and check
for yourself whether the anisotropy is increasing or
decreasing the Newtonian acceleration. I think you
will find it is the same in both cases when they should
differ. Don't take my word for it, check.

There's no doubt about it George.

Then do the test I suggest and find out.

I recently made the mistake of not doing such a test before
jumping to conclusions and I'm not about to do it again just
yet. But thanks anyway.

? Don't make the same mistake, do the test and
find out which way the anisotropy is pointing. If
you get the sign right you should at least see that
the path doesn't repeat but there are several other
errors listed above that will also make your result
somewhat inaccurate.

The physical sign manipulation
is essential because the required equations always give a
positive result even when the anisotropy becomes negative.

Just in case you've forgotten;

dt = 1
ana = 1
# = (ana^2 * dt)^.5 is always 1.

This is just another example of your problems with
simple maths. You are again taking the root of a
square

# = (ana^2 * dt)^.5

is the same as

# = ana * sqrt(dt)

(ana^2 * dt)^.5 is only part of the equation. The proper
equation is derived from a^2 + b^2 = c^2. It becomes
s = (v^2 + ana^2)^.5 if dt is removed (s is the updated orbital
speed).

That's quite different, but v is a speed while ana is an
acceleration so you cannot add them. You could correct
that error using

s = (v^2 + (ana*dt)^2)^.5

but as I said before, v and (ana*dt) are not perpendicular
so you cannot use Pythagoras. Instead you might try
breaking v into x and y components and likewise (ana*dt)
keeping both signed. Then add the components

s_x = v_x + dt * ana_x
s_y = v_y + dt * ana_y

then you could write:

s = sqrt(s_x^2 + s_y^2)

but you don't need that since you already have the x and y
components that you use for the change of location.

But why are you calculating that in the first place?

I'm trying to determine the orbital speed changes according to
the path taken when the anisotropy is included as compared with
the unaffected path. I've used Pythagoras to do that.

OK, see above for how to do that properly, Pythagoras
does not apply since v and ana are not perpendicular.


Previous orbital speed
______________________
- _ l Anomalous
New orbital speed - l acceleration


I don't need you to tell me that I'm wrong either, only that
it's not the way you would do it. But we already know that.

You have to follow the rules of maths, Pythagoras only
applies when the adjacent and opposite are perfectly
perpendicular.

I've included the program that was designed to demonstrate the
true effect of an anomalous change in gravity, and it does so
exceptionally well.

You need to fix the problems listed above before it
will be worth looking at any code. The bottom line
is that my code is written such that it follows the
only valid method _assuming_ you really mean
that the anisotropic acceleration points directly
towards or away from the Sun depending on which
leg is being calculated. The results are accurate
enough for our needs as I have demonstrated.

It demonstrates that the orbiting mass
behaves exactly as it should according to the forces acting on
it, with centrifugal force rising to twice the anomalous force
before the falling mass can be halted and sent back out again.

You will notice that I've treated the "previous orbital speed"
as if it was tangential to the gravity source, but that is of
virtually no consequence.

Sorry Max, it is critical. Nobody will treat your results
as anything other than worthless until you correct that
and the other errors.

This program justifies the method I'm now using to include the
anisotropy in your program.

The problem is that you have always expected a
particular outcome and you are writing the code
to meet your expectations. I had a rough idea what
would happen but I have always been prepared to
be surprised by what the maths produced. The
speed with which the eccentricity decays is such
a surprise, I thought it would take longer, but the
maths has shown me that it would be quite rapid.

I've added comments within the program. Feel free to reply to
them.

I'll skip it until you remove the use of Pythagoras,
deal with the x and y components separately and
account for the varying orbital speed.

snip code

Dark matter is non-baryonic so new physics would not
be surprising, it is an exciting prospect indeed. The best
handle we have so far seems to be the Bullet Cluster
where the two clumps of dark matter associatated with
the galaxies have passed through each other.

And the latest evidence doesn't comply with that.

Cite the paper please.

http://www.eurekalert.org/pub_releas...-dmm081607.php
seems to indicate that the Abell 520 system "train wreck" is not
as previously noted, as in the Bullet Cluster where the "dark
matter" and galaxies have stayed together.

You have misunderstood the problem what I said
remains true but the galaxies have been displaced
as well as the gas so there is an additional mystery.
Dark Matter is called 'dark' not just because it doesn't
interact with EM but slightly tongue in cheek because
we are 'in the dark' about its true nature.

Dark matter just means something that has mass but
doesn't interact with light, it is a 'catch all'
generic term and deliberately vague.

Why not something that nature predicts, like a black
hole(s)?

Because microlensing surveys should detect them.
The find some events but not enough.

How do you differentiate between dark matter and black holes?

Black holes have a mass greater than the Sun in a very
small volume while a background of basic particles is
thin and spread out. The dark matter spread round a
galaxy will magnify background objects and the effect
will stay the same for centuries, black holes passing
in front of a more distant star produce lensing events
that may only last a few minutes and then be gone.

George

"George Dishman" wrote in message
oups.com...
Max Keon wrote:
George Dishman wrote:

I am applying as your equation dictates - I have
no choice.

You profess to know all about the consequences of a gravity
anisotropy when you obviously can't.

I have no such need, _you_ have provided the equation
that supposedly describes the anisotropy, all _I_ need
to know is how to integrate that. It is simple maths that
I learnt in secondary school about forty years ago.

That's where the problem apparently lies George. You are trying
to apply a very simple logic that was never intended to describe
the consequences of a gravity anisotropy. The anisotropic
acceleration cannot be simply added to the Newtonian component
of an elliptical orbit, as you have done. It doesn't work.

You have never had reason
to address such a thing in the past, and the course you've chosen
is the one that gives the best result for your purpose. But your
method of applying the anisotropic acceleration is totally wrong.

I'm sorry Max, there is only one method permitted
by the definition of acceleration. If you don't like the
results, the only thing you can do is change the
equation you have given me.

You are adding it to the naturally flowing acceleration in a
naturally occurring eccentric orbit. Which is of course nothing
more than a circular orbit that has been knocked off center.

Can you not see that?

Your statement is factually untrue, the minor axis
of an ellipse is smaller than the major axis.

What are you trying to say? Are you suggesting that a circular
orbit that has been deflected off course won't fall into a
natural eccentric orbit, and have a minor axis that's smaller
than the major axis? Or what?

One thing that you have never given a satisfactory explanation
for is why Mercury's fall to its previous perihelion radius is
prematurely halted as a consequence of a gravity anisotropy.
Perhaps if you try demonstrating that mathematically, it will
help you to understand why you are wrong.

I have told you over and over again why that happens,
and we discussed it in detail two months ago, see
message number 29 in this thread on Google.

This is how you explained it.
_You misundrstand what I said. Compare a circular
_orbit with a slightly elliptical one of the same
_energy. The elliptical path is inside the circular
_path for about half the orbit. The centrifugal
_force becomes sufficient to throw the planet out
_roughly where the paths cross but momentum keeps
_the planet moving inward to perihelion in the
_elliptical case. You say the same yourself later,
_I just expressed it from a different point of view.

You have described the difference between a circular and an
elliptical orbit. I've been trying to tell you all along that
they are identical, that an eccentric orbit is only a circular
orbit that has been deflected off course.

I've also been trying to tell you all along that your program
is only capable of doing what it was designed to do, and that is
of course to plot a natural orbit path, whether it be circular
or elliptical. Adding the anisotropy directly to Newtonian
gravity certainly causes the perihelion radius to increase and
the aphelion radius to reduce, but the only reason for that is
that your program sets the perihelion and aphelion at the points
where the y axis crosses zero. There can never be a perihelion
advance using your method.

Mercury has been drawn more slowly than normal to the Sun on the
fall to perihelion and will be at a greater radius from the Sun
when the y axis crosses through zero. But gravity conditions have
now returned to normal and Mercury is now drawn to the Sun as
normal, so it's orbital speed is now too slow to halt the inward
fall. _It is not going to stop there and turn around._

Include the anisotropy correctly and your program is fine.
http://members.optusnet.com.au/maxkeon/proven2.html

You will come to
realize that Mercury's orbital speed cannot be fast enough to
generate the required centrifugal force to prematurely halt its
fall and whiz it back out toward the aphelion because the
anisotropy has caused it to be drawn more slowly to the Sun
throughout the inward journey, so by the time Mercury reaches
the point of last perihelion, it has obviously fallen less
distance than normal and is traveling at a lesser speed than
normal. So how on earth does it turn around and go back out
again.

You will open up a whole new universe for me if you can prove
that it does, where nothing is impossible, or predictable.

It is fully predicaable Max, just integrate the acceleration
as the maths requires. It may well open up a whole new
universe for you if you learn calculus and vectors.

A gravity anisotropy is yet another inconvenient truth that you
are just going to have to get used to.
---

The Newtonian acceleration is no doubt pointing directly at the
Sun throughout the entire orbit, but directly adding the
anisotropy in the way you have done is still wrong.

The Newtonian acceleration points at the Sun, we
both know that. I have asked you many times what
the direction of the anisotropy is and yo never gave
a clear answer but my understanding is that it also
points along the line betwen the un and the planet
but towards the Sun on the outward leg and away
from it on the inward leg. Please either confirm that
or correct it.

The pull of gravity is reduced for motion toward a gravity
source, and is increased for outward motion. It can't get much
clearer than that.
---

This is just another example of your problems with
simple maths. You are again taking the root of a
square

# = (ana^2 * dt)^.5

is the same as

# = ana * sqrt(dt)

(ana^2 * dt)^.5 is only part of the equation. The proper
equation is derived from a^2 + b^2 = c^2. It becomes
s = (v^2 + ana^2)^.5 if dt is removed (s is the updated orbital
speed).

That's quite different, but v is a speed while ana is an
acceleration so you cannot add them.

That is ridiculous. If the distance added in 1 second of
acceleration is 1 meter, then the distance is 1 meter. What do
you think it would be. The acceleration rate at the Earth's
surface adds an additional 9.8 meter fall in ever second. That's
a very specific quantity you know. How can that be a problem to
you?

You could correct
that error using

s = (v^2 + (ana*dt)^2)^.5

The real test of dt is to run the program with different dt
values and note the results. If they are not all the same as when
dt = 1 then it's wrongly applied. Your modification does not
work. My method does work by the way.

but as I said before, v and (ana*dt) are not perpendicular
so you cannot use Pythagoras. Instead you might try
breaking v into x and y components and likewise (ana*dt)
keeping both signed. Then add the components

s_x = v_x + dt * ana_x
s_y = v_y + dt * ana_y

then you could write:

s = sqrt(s_x^2 + s_y^2)

but you don't need that since you already have the x and y
components that you use for the change of location.

I don't need to do any of that because it's all wrong.

But why are you calculating that in the first place?

I'm trying to determine the orbital speed changes according to
the path taken when the anisotropy is included as compared with
the unaffected path. I've used Pythagoras to do that.

OK, see above for how to do that properly, Pythagoras
does not apply since v and ana are not perpendicular.

Previous orbital speed
______________________
- _ l Anomalous
New orbital speed - l acceleration


I don't need you to tell me that I'm wrong either, only that
it's not the way you would do it. But we already know that.

You have to follow the rules of maths, Pythagoras only
applies when the adjacent and opposite are perfectly
perpendicular.

If I choose to use Pythagoras I automatically define those
conditions whether they be true or not. Since the degree of error
was of little consequence in demonstrating my point in the case
of the circular orbit described in my previous post, I have done
no wrong at all. I'm certainly not breaking any laws of
mathematics by tolerating an acceptable error.

-----

Max Keon

"Max Keon" wrote in message
u...
"George Dishman" wrote in message
oups.com...
Max Keon wrote:
George Dishman wrote:

I am applying as your equation dictates - I have
no choice.

You profess to know all about the consequences of a gravity
anisotropy when you obviously can't.

I have no such need, _you_ have provided the equation
that supposedly describes the anisotropy, all _I_ need
to know is how to integrate that. It is simple maths that
I learnt in secondary school about forty years ago.

That's where the problem apparently lies George. You are trying
to apply a very simple logic that was never intended to describe
the consequences of a gravity anisotropy.

Maths is universal whatever it is used to describe.

The anisotropic
acceleration cannot be simply added to the Newtonian component
of an elliptical orbit, as you have done. It doesn't work.

In that case you equation is invalid because what you
have given me is nothing more than a way to calculate
the amount of acceleration that I have to add to the
Newtonian acceleration.

You have never had reason
to address such a thing in the past, and the course you've chosen
is the one that gives the best result for your purpose. But your
method of applying the anisotropic acceleration is totally wrong.

I'm sorry Max, there is only one method permitted
by the definition of acceleration. If you don't like the
results, the only thing you can do is change the
equation you have given me.

You are adding it to the naturally flowing acceleration in a
naturally occurring eccentric orbit. Which is of course nothing
more than a circular orbit that has been knocked off center.

Can you not see that?

Your statement is factually untrue, the minor axis
of an ellipse is smaller than the major axis.

What are you trying to say? Are you suggesting that a circular
orbit that has been deflected off course won't fall into a
natural eccentric orbit, and have a minor axis that's smaller
than the major axis? Or what?

To put it in crude terms, I am saying that an ellipse is
longer than it is wide, a statement that is not true of
an off-centre circle. Look at the red path he

http://en.wikipedia.org/wiki/Image:S...d_ellipses.svg

It is obviously not a circle.

One thing that you have never given a satisfactory explanation
for is why Mercury's fall to its previous perihelion radius is
prematurely halted as a consequence of a gravity anisotropy.
Perhaps if you try demonstrating that mathematically, it will
help you to understand why you are wrong.

I have told you over and over again why that happens,
and we discussed it in detail two months ago, see
message number 29 in this thread on Google.

This is how you explained it.
_You misundrstand what I said. Compare a circular
_orbit with a slightly elliptical one of the same
_energy. The elliptical path is inside the circular
_path for about half the orbit. The centrifugal
_force becomes sufficient to throw the planet out
_roughly where the paths cross but momentum keeps
_the planet moving inward to perihelion in the
_elliptical case. You say the same yourself later,
_I just expressed it from a different point of view.

Yes, so far that is OK.

You have described the difference between a circular and an
elliptical orbit. I've been trying to tell you all along that
they are identical, that an eccentric orbit is only a circular
orbit that has been deflected off course.

That is NOT true, think of a cometary orbit and it
is obvious that an ellipse can be long and narrow
such as the red path in the earlier diagram.

I've also been trying to tell you all along that your program
is only capable of doing what it was designed to do, and that is
of course to plot a natural orbit path, whether it be circular
or elliptical.

Again that is not true, it integrates the acceleration
and will plot any possible path, it is not restricted
to orbits at all.

Adding the anisotropy directly to Newtonian
gravity certainly causes the perihelion radius to increase and
the aphelion radius to reduce, but the only reason for that is
that your program sets the perihelion and aphelion at the points
where the y axis crosses zero. There can never be a perihelion
advance using your method.

Untrue again, the program determines the perihelion
when the radius stops decreasing and starts to increase
and vice versa for aphelion. It inserts a short cross
line at both points and you can see they are advanced.

Mercury has been drawn more slowly than normal to the Sun on the
fall to perihelion and will be at a greater radius from the Sun
when the y axis crosses through zero. But gravity conditions have
now returned to normal and Mercury is now drawn to the Sun as
normal, so it's orbital speed is now too slow to halt the inward
fall. _It is not going to stop there and turn around._

Include the anisotropy correctly and your program is fine.
http://members.optusnet.com.au/maxkeon/proven2.html

The anisotropy is included as your equation demands. Change
your equation and I will change the program to match.

You will come to
realize that Mercury's orbital speed cannot be fast enough to
generate the required centrifugal force to prematurely halt its
fall and whiz it back out toward the aphelion because the
anisotropy has caused it to be drawn more slowly to the Sun
throughout the inward journey, so by the time Mercury reaches
the point of last perihelion, it has obviously fallen less
distance than normal and is traveling at a lesser speed than
normal. So how on earth does it turn around and go back out
again.

You will open up a whole new universe for me if you can prove
that it does, where nothing is impossible, or predictable.

It is fully predicaable Max, just integrate the acceleration
as the maths requires. It may well open up a whole new
universe for you if you learn calculus and vectors.

A gravity anisotropy is yet another inconvenient truth that you
are just going to have to get used to.

Not at the level of your equation. There ARE effects
similar to what you describe in GR which matches what
really happens but your equation gives an effect that
is far too large.

The Newtonian acceleration is no doubt pointing directly at the
Sun throughout the entire orbit, but directly adding the
anisotropy in the way you have done is still wrong.

The Newtonian acceleration points at the Sun, we
both know that. I have asked you many times what
the direction of the anisotropy is and yo never gave
a clear answer but my understanding is that it also
points along the line betwen the un and the planet
but towards the Sun on the outward leg and away
from it on the inward leg. Please either confirm that
or correct it.

The pull of gravity is reduced for motion toward a gravity
source, and is increased for outward motion. It can't get much
clearer than that.

OK, that's what I have been working on. Now stop
your program at the 1/4 and 3/4 orbit points and
test if it does that. I think you'll find it doesn't.

This is just another example of your problems with
simple maths. You are again taking the root of a
square

# = (ana^2 * dt)^.5

is the same as

# = ana * sqrt(dt)

(ana^2 * dt)^.5 is only part of the equation. The proper
equation is derived from a^2 + b^2 = c^2. It becomes
s = (v^2 + ana^2)^.5 if dt is removed (s is the updated orbital
speed).

That's quite different, but v is a speed while ana is an
acceleration so you cannot add them.

That is ridiculous. If the distance added in 1 second of
acceleration is 1 meter, then the distance is 1 meter. What do
you think it would be. The acceleration rate at the Earth's
surface adds an additional 9.8 meter fall in ever second.

Wrong, the acceleration adds 9.8 m/s to the speed in
every second, so in 2 seconds it adds 19.6m/s and
in dt seconds it adds (9.8 * dt) m/s to the speed.

That's
a very specific quantity you know. How can that be a problem to
you?

You got it wrong again. If it was always in the same
direction (which ist isn't) it would add 4.9 m to
the distance in the first second, 14.7 m to the distance
in the next second and so on.

You could correct
that error using

s = (v^2 + (ana*dt)^2)^.5

The real test of dt is to run the program with different dt
values and note the results. If they are not all the same as when
dt = 1 then it's wrongly applied. Your modification does not
work. My method does work by the way.

Sorry, yours is clearly wrong.

but as I said before, v and (ana*dt) are not perpendicular
so you cannot use Pythagoras. Instead you might try
breaking v into x and y components and likewise (ana*dt)
keeping both signed. Then add the components

s_x = v_x + dt * ana_x
s_y = v_y + dt * ana_y

then you could write:

s = sqrt(s_x^2 + s_y^2)

but you don't need that since you already have the x and y
components that you use for the change of location.

I don't need to do any of that because it's all wrong.

Get a textbook Max, learn some schoolboy maths.

But why are you calculating that in the first place?

I'm trying to determine the orbital speed changes according to
the path taken when the anisotropy is included as compared with
the unaffected path. I've used Pythagoras to do that.

OK, see above for how to do that properly, Pythagoras
does not apply since v and ana are not perpendicular.

Previous orbital speed
______________________
- _ l Anomalous
New orbital speed - l acceleration


I don't need you to tell me that I'm wrong either, only that
it's not the way you would do it. But we already know that.

You have to follow the rules of maths, Pythagoras only
applies when the adjacent and opposite are perfectly
perpendicular.

If I choose to use Pythagoras I automatically define those
conditions whether they be true or not.

You automatically _assume_ those conditions and
since they do NOT apply, your result is wrong.

Since the degree of error
was of little consequence in demonstrating my point in the case
of the circular orbit described in my previous post, I have done
no wrong at all. I'm certainly not breaking any laws of
mathematics by tolerating an acceptable error.

Learn some basic maths Max, there are plenty of
K12 resources on the web that will teach you how
to do this. I've shown you how to do it correctly,
the rest is up to you.

George

"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:
Max Keon wrote:
George Dishman wrote:
---

I've also been trying to tell you all along that your program
is only capable of doing what it was designed to do, and that is
of course to plot a natural orbit path, whether it be circular
or elliptical.

Again that is not true, it integrates the acceleration
and will plot any possible path, it is not restricted
to orbits at all.

It's certainly not restricted to orbits because it doesn't work
in that application.

This is obvious.
If a constant force is suddenly applied perpendicular to a mass
traveling a straight line path, the mass will immediately shift
course to follow a constant curve that will eventually form a
complete circle. If the force is suddenly removed, the mass will
immediately take off along a straight line trajectory again.

However, if a force is suddenly applied to a mass in a circular
orbit around i.e. the Sun, the mass will initially be directed
to follow a tighter radius trajectory that would also eventually
form a smaller complete circle. But the accelerating force
increases as the mass is drawn inward, thus increasing orbital
speed, and since centrifugal force increases at a squaring rate
per velocity increase, the fall will be rapidly halted.
Centripetal force is initially increased by the added force, and
that rises at a linear rate as the fall progresses, but
centrifugal forces quickly rise to counter the fall.

The fall is halted and the mass is turned back outward when
centrifugal force has increased to twice the initial force
increase, as is demonstrated in the first program listed at the
link provided later in the post. Study it properly because that
scenario applies for every orbiting mass in the universe.

I've added this introduction to the web page as well.
-------------
The trajectory of a circular orbit is a naturally flowing
transition between the two dimensions of its orbit plane. The
orbit trajectory of an eccentric orbit is also a naturally
flowing transition between those two dimensions. All sustainable
orbits are therefore exactly the same.

An anomalous gravity change cannot be treated as part of that
natural transition. It has nothing to do with it. Any change to
that totally balanced system is equivalent to a force acting to
shift an object from a position where zero forces are being
applied. Any anomalous force is going to be subject to the rules
that apply if the orbiting mass is stationary. THE ANOMALOUS
FORCE CANNOT BE ADDED DIRECTLY TO THE EXISTING FORCES.
-------------

Adding the anisotropy directly to Newtonian
gravity certainly causes the perihelion radius to increase and
the aphelion radius to reduce, but the only reason for that is
that your program sets the perihelion and aphelion at the points
where the y axis crosses zero. There can never be a perihelion
advance using your method.

Untrue again, the program determines the perihelion
when the radius stops decreasing and starts to increase
and vice versa for aphelion. It inserts a short cross
line at both points and you can see they are advanced.

I can see that the orbit eccentricity decays, but I don't see
any perihelion advance.

These sets of figures were generated in accordance with the rules
of your program for the initial fall to perihelion, with the
anisotropy added directly to Newtonian gravity, as you require.
Notice that Mercury's position on the y axis has shifted by
around 41 million meters, which according to the orbit rotation
is in the retarded direction, and that occurs with or without
the anisotropy. So the basic cause is only software.

Excluding anisotropy.
41025996.52243679 current position on the y axis.
45961145477.91186 radius.
3797180.400002688 1 seconds steps.
59017.39331765785 orbital speed.

Including anisotropy.
41030998.40955561 current position on the y axis.
45963158216.10093 radius.
3797291.400001362 1 second steps.
59014.80893124959 orbital speed.

Mercury has shifted from the normal by 5002 meters on the y axis
at turnaround, which is less than the .1 seconds increments
within which the perihelion position was being detected. So the
turnaround point is much the same with or without the anisotropy.

It's actually not in the direction of advance either, it's
further retarded, in this case.
---

You could correct
that error using

s = (v^2 + (ana*dt)^2)^.5

The real test of dt is to run the program with different dt
values and note the results. If they are not all the same as when
dt = 1 then it's wrongly applied. Your modification does not
work. My method does work by the way.

Sorry, yours is clearly wrong.

Apart from the relatively inconsequential orbital speed error,
which I later address, the first of the "wrongs" was in the
program line "radius = radius + anb". That should read
"radius = radius + ana". I was adding the total compounding
change to the radius in each second. ana holds the updated
current fall distance for the current second, and that's what
should have been added to the radius.

"Wrong 2" was in this equation "cf = vz^2# / vv^2# * newton",
which now reads "cf = vz^2# / vv^2# * newt". "newt" now
represents the Newtonian gravity value applicable to a compare
orbital speed "vv" which was established as the speed required
to balance the centripetal and centrifugal forces for the initial
orbit radius from the Sun for a circular orbit at that radius.
"vv = (G * M / x) ^ .5" and then "vz = vv" . vz becomes the
variable. That combination gives the correct balance for all
orbital speed per radius scenarios.

The consequences of the anisotropy are now barely noticeable.
http://members.optusnet.com.au/maxkeon/proven2.html

But dt is functioning perfectly, isn't it.
---

But why are you calculating that in the first place?

I'm trying to determine the orbital speed changes according to
the path taken when the anisotropy is included as compared with
the unaffected path. I've used Pythagoras to do that.

OK, see above for how to do that properly, Pythagoras
does not apply since v and ana are not perpendicular.

Previous orbital speed
______________________
- _ l Anomalous
New orbital speed - l acceleration


I don't need you to tell me that I'm wrong either, only that
it's not the way you would do it. But we already know that.

You have to follow the rules of maths, Pythagoras only
applies when the adjacent and opposite are perfectly
perpendicular.

If I choose to use Pythagoras I automatically define those
conditions whether they be true or not.

You automatically _assume_ those conditions and
since they do NOT apply, your result is wrong.

The dotted line in the first diagram is tangential to a force
applied from the bottom of the page, and it's assumed to be the
same length as the first of the tilted lines. The added fall
distance during 1 second of acceleration is 1 meter.

sqr(48000^2 + 1^2) = 48000.00001042 is the hypotenuse length of
the triangle. The orbital speed increase during that second is
48000.00001042 - 48000 = .00001042 m/sec.

-. . . _ . . . . 48000 m/sec . . . . . . . .
- _ - _
- _ - l ----
- _ l 1 meter
-l ____

The correct way to address the problem is of course to first
establish a true length for the tangential leg.

If the tilt angle of the orbit trajectory is 10 degrees and the
orbital speed is 48000 m/sec, sin10 * 48000 = 8335 meters for
the opposite leg. So the triangle so far is, hypotenuse = 48000
and opposite = 8335.

sqr(48000^2 - 8335^2) = 47270.79 meters is then the true
tangential speed for the triangle adjacent.

Now that wasn't so hard was it.

The orbital speed increase is now according to this diagram.

47270.79 m/sec
(b)---_----------------------------------l
- _ l 1 m/sec
- _ l
47270.790010577 m/sec


47270.790010577 - 47270.79 = .000010577 meter orbital speed
increase.

Comparing that with the previous result of .00001042 meter
increase, .000010577 - .00001042 = 0.000000157 meter error.

I can't see that being a problem just yet. My point can be
demonstrated without further cluttering the program by adding
the extra step.

It's really no problem to add it though.

-----

Max Keon

"Max Keon" wrote in message
. ..

I was going to redo my code to interpolate between
steps for the perihelion advance but it will take
some effort and it is pointless if you don't accept
the path is valid anyway so I'll just correct this:

I've added this introduction to the web page as well.
....
THE ANOMALOUS
FORCE CANNOT BE ADDED DIRECTLY TO THE EXISTING FORCES.

If an object moves 3m and then another 5m in the
same direction, it has moved 8m in that direction.
We call that displacement. Displacements add.

Velocity is the rate at which displacement happens,
it is just the displacement divided by the time
taken so velocities add.

Acceleration is the rate at which velocity changes,
it is just the change of velocity divided by the
time taken so accelerations add.

Displacement, velocity and acceleration are all
vectors

http://www.physchem.co.za/Vectors/Addition.htm

The method used in my code uses x and y components:

http://www.physchem.co.za/Vectors/Ad...tm#Rectangular

Learn the basics Max, this is trivial schoolboy maths.

George

"George Dishman" wrote in message
...
"Max Keon" wrote in message
. ..

I was going to redo my code to interpolate between
steps for the perihelion advance but it will take
some effort and it is pointless if you don't accept
the path is valid anyway

Your program plots the coordinates of a natural eccentric orbit
with great precision, but that's where it ends. It cannot
accommodate a gravity anisotropy the way you are trying to do
it.

The graphs shown on this web page
http://members.optusnet.com.au/maxkeon/proven2.html
were generated using the wrong initial orbital speed, which
didn't matter at the time, but now it does. I've updated the web
page for the correct initial speed. Nothing has really changed
much though.

This program requires the correct orbital speed to demonstrate
my point. It does exactly as is stated below.


'-----------------
' The program is designed to demonstrate that orbital speed can
' also be determined using Pythagoras. It provides an extremely
' fluid method of depicting how a mass in orbit would behave when
' it has been deflected off course by any minor collision. That's
' where part (2) of this program fails to deliver.

' The oscillations that are generated about the mean path have
' been initiated at the start and they continue to resonate
' throughout the orbit, oscillating back and forth within the
' confines of the centripital-centrifugal force imbalance, in
' the same manner as demonstrated on the web page linked below,
' and depending on where the trajectory may be pointing at
' aphelion and perihelion turnaround the oscillation distance
' can be increased or reduced. That's the realm of the gravity
' anisotropy and it can't be treated as part of the natural flow
' of an eccentric orbit.

DEFDBL A-Z
SCREEN 12: COLOR 7
CLS

GOSUB ab

c = 299792458#
G = .0000000000667#
M = 1.99D+30
multi = .000000002#
' Multiplier for the graphics.

x = 69820000000#: vy = 38850#
' Aphelion start.

yx = 57890570000#

vv = vy
' Corrected initial orbital speed.

newt = G * M / x ^ 2
' Centrifugal force is established for
' all cases from this result.

vz = vv
lastradius = x
rad = 45961140000#

dt = 200

lasty = 41073216#
' This is roughly the initial negative advance that's generated
' as a consequence of the aphelion start. And that occurs with
' or without the anisotropy.

aa: ' -------- Part (1) ----------------------------------
' This section is as described in program (1) at
' http://members.optusnet.com.au/maxkeon/proven2.html

' Something that has become very obvious is that radial velocity
' has nothing whatever to do with generating centrifugal force.
' It's the tangential speed alone that generates the force, even
' though the force varies according to radius. The point is, a
' radius change most certainly cannot generate centrifugal force.

' Pythagoras is therefore a valid tool for determining orbital
' speed change caused by a fall or rise from a gravity source.
' The tangent leg of the triangle is always at 90 degrees to the
' force direction, or opposite leg of the triangle. The
' hypotenuse carries the orbital speed change. The true orbital
' speed is derived from that result.

ana = ana + dt * cfx
' ana stores the current fall distance.
anb = anb + ana
' anb stores the total fall distance.

IF anc = anb THEN s = (vz ^ 2# - ((ana) ^ 2 * dt)) ^ .5#
IF anc anb THEN s = (vz ^ 2# + (ana ^ 2 * dt)) ^ .5#
anc = anb
' anc determines which of the two equations is active.

vz = s
' vz carries the current tangential speed.

vzvr = SQR(vz ^ 2 + vr ^ 2)
' vzvr is the true orbital speed.
' vr is extracted from part (2) of the program which plots the
' natural transition of the orbiting body between the two planes
' of its orbit. That has nothing to do with anomalous changes.

LOCATE 8, 45: PRINT INT(vzvr)
' Orbital speed for part (1).

CIRCLE (10 + inc, -180 + vzvr / 100), 0, 9

cf = vz ^ 2# / vv ^ 2# * newt
' (Current speed^2 / original speed^2) gives the centrifugal
' force change ratio. Multiplied by the original centrifugal
' force gives the actual force.

cfx = newton + anisotropy - cf
' cfx holds the true fall rate, which is the difference
' between centripetal and centrifugal forces.
'---------------------------------------------------

' Part (2) -----------------

ryx = x * x + y * y
radius = SQR(ryx)

' radius = radius + ana
' The anisotropy is excluded so that, apart from calculating
' radial velocity, there is no link whatever between the two
' methods for determining orbital speed.

vr = (radius - lastradius) / dt

lastradius = radius

newton = G * M / ryx
anisotropy = vr / c * -newton
acceleration = -newton

ax = acceleration * (x / radius)
ay = acceleration * (y / radius)

vx = vx + dt * ax
vy = vy + dt * ay

x = x + dt * vx
y = y + dt * vy

v = (vx ^ 2# + vy ^ 2#) ^ .5#
' Orbital speed for part (2).

LOCATE 9, 45: PRINT INT(v)
CIRCLE (10 + inc, -180 + v / 100), 0, 12
inc = inc + .015

ft = ft + 1
IF ft * dt 7603200 THEN GOSUB ab

GOTO aa

ab:
IF cycles = 5 THEN END
inc = 0: ft = 0: CLS
COLOR 9: LOCATE 8, 17: PRINT "Orbital speed per Pythagoras"
COLOR 12: LOCATE 9, 19: PRINT "Orbital speed per Part (2)"
COLOR 7
LOCATE 13, 18: PRINT "Aphelion"
LOCATE 27, 18: PRINT "Perihelion"
cycles = cycles + 1
LOCATE 21, 8: PRINT "cycle"; cycles
RETURN
'-----------------


so I'll just correct this:

I've added this introduction to the web page as well.
...
THE ANOMALOUS
FORCE CANNOT BE ADDED DIRECTLY TO THE EXISTING FORCES.

If an object moves 3m and then another 5m in the
same direction, it has moved 8m in that direction.
We call that displacement. Displacements add.

Velocity is the rate at which displacement happens,
it is just the displacement divided by the time
taken so velocities add.

Acceleration is the rate at which velocity changes,
it is just the change of velocity divided by the
time taken so accelerations add.

Displacement, velocity and acceleration are all
vectors

http://www.physchem.co.za/Vectors/Addition.htm

The method used in my code uses x and y components:

http://www.physchem.co.za/Vectors/Ad...tm#Rectangular

Learn the basics Max, this is trivial schoolboy maths.

Even if the anisotropy is inelastic the orbit eccentricity will
only decay by around 41 meters per cycle. And it only requires
less than .1 meters of that amount to be elastic to explain
Mercury's perihelion advance.

Even if I wasn't aware of the basics, my errors would be trivial,
while yours are absolutely enormous.


George, get used to the idea that I will only ever be proven
wrong if I have misinterpreted the consequences of the zero
origin universe. To the best of my knowledge, I base my
arguments on the truth as defined by the zero origin concept
http://members.optusnet.com.au/maxkeon/the1-1a.html
That universe is akin to an infinitely faceted Rubix "Cube",
where the very first twist sets the ball rolling, and the more
you twist it the worse it gets. It can never be put back the
way it was.

-----

Max Keon

"Max Keon" wrote in message
...

"George Dishman" wrote in message
...
"Max Keon" wrote in message
. ..

I was going to redo my code to interpolate between
steps for the perihelion advance but it will take
some effort and it is pointless if you don't accept
the path is valid anyway

Your program plots the coordinates of a natural eccentric orbit
with great precision, but that's where it ends. It cannot
accommodate a gravity anisotropy the way you are trying to do
it.

You have provided me with an equation for the extra
acceleration which is to be added to take account of
the anisotropy. Acceleration is a vector and I have
added it in the correct manner for adding vectors, so
what I have done is the _only_ valid method that can
make use of what you have provided.

....

I've added this introduction to the web page as well.
...
THE ANOMALOUS
FORCE CANNOT BE ADDED DIRECTLY TO THE EXISTING FORCES.
....
Displacement, velocity and acceleration are all
vectors

http://www.physchem.co.za/Vectors/Addition.htm

The method used in my code uses x and y components:

http://www.physchem.co.za/Vectors/Ad...tm#Rectangular

Learn the basics Max, this is trivial schoolboy maths.

Even if the anisotropy is inelastic the orbit eccentricity will
only decay by around 41 meters per cycle. ...

Sorry Max, it is thousands of km per orbit at the
start, my code gives you the correct numbers, your
calculations are flawed.

Even if I wasn't aware of the basics, my errors would be trivial,

You are not aware of basics and as a result you have
written code that gives you what you expected but not
what the anisotropy would actually produce.

while yours are absolutely enormous.

There are no errors in my method, and I have shown
you the errors that result from the numerical
integrations, you simply need to learn schoolboy
maths. Study the two web site links above and start
again once you have mastered their content, or read
through my code and study how I have done it and
compare that with the reference pages. When you
understand what a vector is, you will find that my
code is correct. I'm not going to waste any more
time teaching you stuff you should have learned as
a teenager Max, it is time for you to do some work
for yourself.

George

"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:

I was going to redo my code to interpolate between
steps for the perihelion advance but it will take
some effort and it is pointless if you don't accept
the path is valid anyway

Your program plots the coordinates of a natural eccentric orbit
with great precision, but that's where it ends. It cannot
accommodate a gravity anisotropy the way you are trying to do
it.

You have provided me with an equation for the extra
acceleration which is to be added to take account of
the anisotropy. Acceleration is a vector and I have
added it in the correct manner for adding vectors, so
what I have done is the _only_ valid method that can
make use of what you have provided.

You are still missing the point completely here George.

A circular orbit is equivalent to a straight line through the
two dimensions in the plane of the orbit. Likewise is a stable
eccentric orbit. It's the line where all forces are in balance.

In the diagram, a force is applied for the duration of 1 second
and it deflects "@" to travel a different straight line
trajectory, where it will continue to point if no other forces
are applied.

(1) on off
\/ ^
@------------- _1 sec
- _
- _

A force is now applied to @'s trajectory when it's traveling a
circular or elliptical orbit path.

(2) on off
\/
@------------- _ ^
---------------

Do you now see the difference?

Your program does not account for the centripetal-centrifugal
force imbalance. It assumes that the trajectory is affected by
the anisotropy as in diagram (1).
---

Even if the anisotropy is inelastic the orbit eccentricity will
only decay by around 41 meters per cycle. ...

Sorry Max, it is thousands of km per orbit at the
start, my code gives you the correct numbers, your
calculations are flawed.

Try studying this page properly.
http://members.optusnet.com.au/maxkeon/proven2.html

Now that your mission has failed, perhaps you should put some
effort into understanding the zero origin universe, the universe
we actaully live in.
http://members.optusnet.com.au/maxkeon/the1-1a.html

-----

Max Keon

"Max Keon" wrote in message
...

"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:

I was going to redo my code to interpolate between
steps for the perihelion advance but it will take
some effort and it is pointless if you don't accept
the path is valid anyway

Your program plots the coordinates of a natural eccentric orbit
with great precision, but that's where it ends. It cannot
accommodate a gravity anisotropy the way you are trying to do
it.

You have provided me with an equation for the extra
acceleration which is to be added to take account of
the anisotropy. Acceleration is a vector and I have
added it in the correct manner for adding vectors, so
what I have done is the _only_ valid method that can
make use of what you have provided.

You are still missing the point completely here George.

The point Max is that the maths I have explained to
you is the _only_ valid method for incorporating your
equation. You have told me the acceleration and I have
used it. If you don't like the answer, tell me a
different equation, one which _can_ be added in the
manner my program applies.

A circular orbit is equivalent to a straight line through the
two dimensions in the plane of the orbit. Likewise is a stable
eccentric orbit. It's the line where all forces are in balance.

The forces are not balanced, the only force acting is
Newtonian gravity which is why the planet doesn't move
in a straight line but instead travels in a circle. If
you work in polar coordinates, there is a centrifugal
"pseudo-force" which appears to balance gravity and
keep the radius constant but we weren't using polar
coordinates.

In the diagram, a force is applied for the duration of 1 second
and it deflects "@" to travel a different straight line
trajectory, where it will continue to point if no other forces
are applied.

(1) on off
\/ ^
@------------- _1 sec
- _
- _

A force is now applied to @'s trajectory when it's traveling a
circular or elliptical orbit path.

(2) on off
\/
@------------- _ ^
---------------

Do you now see the difference?

Yes, the first is correct, the second is wrong. You
have shown an instantaneous change of the velocity
when the force goes off which of course cannot happen.

What has happened in the first case is that the planet
has been pushed off its circular orbit. The segment
after the 1s application of the transverse force is
now part of an elliptical orbit. One orbit later the
planet will return on a course that arrives a little
above the left hand section, pass through the "off"
point and then retrace the part to the lower right.
If the force is exactly perpendicular to the motion
throughout the 1s, then the energy of the orbit
remains the same but the eccentricity has changed.

Your program does not account for the centripetal-centrifugal
force imbalance. It assumes that the trajectory is affected by
the anisotropy as in diagram (1).

(1) is correct, your second diagram is wrong. In an
elliptical orbit the "force imbalance" you mention is
what causes the variation of the radius. In an
elliptical orbit, in polar coordinates, generally the
gravitational and centrifugal forces are _not_ balanced
whereas they are for a circular orbit.

The bottom line here Max is that the maths I use is
correct, it predicts the elliptical orbit correctly
when there is no anisotropy so you know that it is
correctly modelling the continually changing force
imbalance that creates the path, and incorporating
your extra acceleration is done in the correct manner
so the results are accurately what your speculation
would produce. The orbital eccentricity would decay
to near zero over a few thousand orbits and your idea
is unquestionably wrong.

If you decreased the value significantly, something
like using (v/c)^3 instead of (v/c), then you would
get close to the correct values but your equation as
it stands produces far too much effect.

George