Anisotropy and Mercury (2)

On 3 Aug, 03:50, "Max Keon" wrote:
"George Dishman" wrote in message ups.com...
On 29 Jul, 13:40, "Max Keon" wrote:
"George Dishman" wrote in message oups.com...
On 26 Jul, 13:27, "Max Keon" wrote:

On the journey inward, orbital speed increases as it falls and
orbit radius shrinks. Since centrifugal force is proportional to
radius and to the square of orbital speed, the end result is that
the mass is simply oscillating back and forth in the middle of an
invisible spring located perpendicular to the direction of
motion.

None of that has anything to do with an anomalous change in the
pull of gravity.
Right, and that's a good way of looking
at how it works.

It certainly is George.

Max, there is no point in my replying if all you
do is snip what I say and go off on a new tack
every time. Your description above goes a long
way towards understanding the situation so follow
it through and you will grasp what your equation
implies. This is how it works:

Right, and that's a good way of looking
at how it works. If you were moving in a
circular orbit following Mercury with the
Sun always off to your left, you would see
the planet swinging to the left and right
of your path like a pendulum.

Your anisotropy then applies a force to the
planet other than when it is at aphelion and
perihelion where the radial velocity is zero.
During the inward leg, the planet moves from
right to left but the anisotropy pushes from
left to right. It is a small force so the
effect is slight but it reduces the length of
the swing just as pushing against a pendulum
would reduce its swing.

On the outward leg, the planet moves from
left to right but the anisotropy pushes from
right to left. It again reduces the length of
the swing so slowly the swing is damped out
until you are left with the planet in a
circular orbit.

Think about that Max.

But you still don't understand the
consequences of an anomalous force being applied to a naturally
stable orbit at all.

The consequences are stated above. Think carefully
about the direction your force acts in and you will
see what I said is exactly what you have been telling
me all along.

The attached program plots the path of a mass in a circular
orbit (plotted in a linear fashion) after an anomalous force is
suddenly applied. If you run the program you will notice that
centrifugal force has increased to twice the anomalous force
before the fall is halted. That's exactly what should be
expected.

This equation s = (v^2 + (ana^2 * dt)^ 2)^.5 extracted from
the program, is used as a means of determining orbital speed
increase according to a^2 + b^2 = c^2.

Pythagoras is only valid if a and b are at right
angles. That is not the case for an elliptical
orbit.

Max, I've said this many times but you are still
not listening - you CANNOT work out the correct
answers without taking the DIRECTION into account.

I attempted using signed fall direction in the same manner as
signed velocity, but I encountered a problem that scuttled the
whole idea. The problem was in the above equation, in that
orbital speed continues to increase when the value stored in
'ana' becomes negative. I had little choice but to go back to
physical sign manipulation. That doesn't make the result wrong
either, it makes it right.

It should have been a warning that something was amiss
somewhere. To solve the problem break both the velocity
and acceleration into x and y components and apply the
same rules. Since you no longer need the square root,
the correct signs are preserved. If you do that, you
will find you have reproduced my program.

... While the anomalous force remains
active, if the oscillations can be elastically diminished to zero
through some external means, the mass will come to rest in an
orbit that has reduced in radius by half the depth of the
oscillation, and there it will stay. ...

Not an "external means", your equation for the
anisotropy is INELASTIC and does the job of
slowly diminishing the eccentricity.

Notice that the process is entirely elastic?

Now that should have set your alarms bells ringing,
your equation is inelastic.

The slightest change to gravity does have an immediate
consequence, as you have been saying all along.

Thank you.

But that
consequence is far removed from what you perceived.

When you correct the errors in your maths you
will find that you end up with the same program
as I wrote and that my results are precisely
what your equation predicts. Keep going though,
you are gradually finding the flaws in your
thinking and correcting them. The main thing you
need to address remains the direction of motion,
you _can_ use the familiar equations from linear
motion but only if you split the motion and
accelerations into x and y components and process
them independently, trying to apply it to the
radial speed and acceleration (effectively using
polar coordinates) doesn't work unless you include
coriolis force as well as centrifugal and that will
be much more complex.

George

"Eric Gisse" wrote in message
ups.com...
On Aug 2, 7:02 pm, "Max Keon" wrote:
I wrote:
This is the result from running the program, along with the
figures generated for the final plot point.
http://members.optusnet.com.au/maxkeon/wave.jpg
Or for the whole story;
http://optusnet.com.au/maxkeon/proven2.html

The link address should be
http://members.optusnet.com.au/maxkeon/proven2.html

You do realize that you are completely wasting everyone's, including
your own, time right?

Thankfully, you're not the smartest person on this planet and
you're not everyone. So I just keep on plodding along.

Don't kid yourself Eric, truth is much bigger than you and me,
or the fear of its cataclysmic consequences. The ooooohhhhh
factor may work in superstitious beliefs, but it has no place
in physics.

-----

Max Keon.

"George Dishman" wrote in message
ps.com...
On 3 Aug, 03:50, "Max Keon" wrote:
"George Dishman" wrote in message
ups.com...
On 29 Jul, 13:40, "Max Keon" wrote:

On the journey inward, orbital speed increases as it falls and
orbit radius shrinks. Since centrifugal force is proportional to
radius and to the square of orbital speed, the end result is that
the mass is simply oscillating back and forth in the middle of an
invisible spring located perpendicular to the direction of
motion.

None of that has anything to do with an anomalous change in the
pull of gravity.

Right, and that's a good way of looking
at how it works.

It certainly is George.

Max, there is no point in my replying if all you
do is snip what I say and go off on a new tack
every time.

That's certainly what I do George, and it's called progress.
Which has been an unusual event in theoretical physics ever since
postulates were introduced into our reality. That's not intended
as humor either.

I snip what you say when I think the rest of your reply has been
addressed, which is quite understandably what you do with my
posts.

Your description above goes a long
way towards understanding the situation so follow
it through and you will grasp what your equation
implies. This is how it works:
---
On the outward leg, the planet moves from
left to right but the anisotropy pushes from
right to left. It again reduces the length of
the swing so slowly the swing is damped out
until you are left with the planet in a
circular orbit.

Think about that Max.

But you still don't understand the
consequences of an anomalous force being applied to a naturally
stable orbit at all.

The consequences are stated above. Think carefully
about the direction your force acts in

Do you think I'm stupid. I can see exactly where the force is
acting and why you are led to believe that it will cause an
orbit eccentricity decay. But I can also see why you are wrong.

The attached program plots the path of a mass in a circular
orbit (plotted in a linear fashion) after an anomalous force is
suddenly applied. If you run the program you will notice that
centrifugal force has increased to twice the anomalous force
before the fall is halted. That's exactly what should be
expected.

This equation s = (v^2 + (ana^2 * dt)^ 2)^.5 extracted from
the program, is used as a means of determining orbital speed
increase according to a^2 + b^2 = c^2.

Pythagoras is only valid if a and b are at right
angles. That is not the case for an elliptical
orbit.

It makes no difference. The true tangential speed can be
obtained using a^2+b^2=c^2. The fall rate for every second is
already known, so the same equation will give an accurate orbital
speed increase per second.

Regardless of how accurate the result may be, nothing changes
the obvious fact that what is lost while the anisotropy is
increasing is reclaimed in full when it has again reduced to
zero at aphelion or perihelion.

I can only suggest that you visit this web page,
http://members.optusnet.com.au/maxkeon/proven2.html
or refer back to my previous reply.

The set of equations in your program describe every possible
naturally occurring orbit, between a circular orbit and a direct
fall. Every one of those orbits are therefore identical. They
are each equivalent to a circular orbit, or to any other orbit
as well. Whatever applies for a circular orbit also applies for
every other orbit, _AND VICE VERSA_.

Is that clear enough?

I think your problem stems from the belief that because those
equations accurately describe nature, they can also be used
to determine the laws of nature.

Max, I've said this many times but you are still
not listening - you CANNOT work out the correct
answers without taking the DIRECTION into account.

I can do EXACTLY as truth dictates.

I attempted using signed fall direction in the same manner as
signed velocity, but I encountered a problem that scuttled the
whole idea. The problem was in the above equation, in that
orbital speed continues to increase when the value stored in
'ana' becomes negative. I had little choice but to go back to
physical sign manipulation. That doesn't make the result wrong
either, it makes it right.

It should have been a warning that something was amiss
somewhere. To solve the problem break both the velocity
and acceleration into x and y components and apply the
same rules. Since you no longer need the square root,
the correct signs are preserved. If you do that, you
will find you have reproduced my program.

... While the anomalous force remains
active, if the oscillations can be elastically diminished to zero
through some external means, the mass will come to rest in an
orbit that has reduced in radius by half the depth of the
oscillation, and there it will stay. ...

Not an "external means", your equation for the
anisotropy is INELASTIC and does the job of
slowly diminishing the eccentricity.

No it's doesn't. For some reason, you continue to miss the point
altogether.

I'll snip the rest because it no longer applies.

I've included another progress report in the form of a Qbasic
program. Note that my position has never changed at all. Meaning
that I'm not "off on a new tack", as you put it. I am in fact
still on my original tack because you have not shown me anything
that could sway me one bit toward yours.

The total fall distance is now added to the radius in the same
manner as described at the above address. That is the proper
method of applying an anomalous gravity change to a circular
orbit, _so it must apply for every other natural orbit_,
regardless of its shape.

The program now detects Mercury's aphelion position within 3900
meters and the perihelion position within 5900 meters, in the
direction motion. That's within .1 seconds of travel time.

The results are not exactly reproduced for different
values of dt, but they are always similar, while the average
advance wanders all over the place from one cycle to the next
in a cyclic fashion when the anisotropy is included. It's fairly
evident that an oscillation has been set in motion right at the
start of the first cycle. It's necessarily very dependent on the
precision at the start.

The advance must be positive for both the aphelion or perihelion
detection because the Sun is always to the left of the start point,
and the orbit direction is anticlockwise.

I've tried to keep the program as understandable as possible,
mainly so that I can understand what it's all about next month.
The added clutter is necessary to eliminate the need for me to
stuff around with animations and JPEG images. You can make your
own simply by running the program. It may take some time, but at
least the result is falsifiable.

-----

Max Keon


'-----------Control/Break halts the program at any time.-------

' The only adjustments required are highlighted thus
'//////////////////////

' A perihelion start generates the most obvious orbit
' instability, and that just keeps on resonating around the
' orbit. The greatest change in orbit advance is obviously when
' the oscillation peak coincides with the perihelion.

' Taking the total over a long period will pull the sinewave
' down to show the true advance. Maybe?

DEFDBL A-Z
SCREEN 12: CLS: COLOR 7
LINE (30, 380)-(600, 380) ' y axis zero, for the added graph.
LOCATE 24, 4: PRINT "0"
PRINT "Light blue is current advance position."
PRINT "Red is the total advance."

CIRCLE (250, 250), 4, 14 ' Sun.

c = 299792458#
G = .0000000000667#
M = 1.99D+30
multi = .000000002# ' Multiplier for the graphics.
colr = 3 ' sets the initial graphics color.

LOCATE 12, 1

'////// Swap the 'x switch /////////////////////////////////
x = 69820000000#: vy = 38850#: fl = 0 ' Aphelion start.
'x = 45961000000#: vy = 59017#: fl = 1 ' Perihelion start.
'///////////////////////////////////////////////////////////

IF fl = 0 THEN PRINT " meter aphelion radius detection error."
IF fl = 1 THEN PRINT " meter perihelion radius detection error."

vv = (G * M / x) ^ .5 ' Sets the initial centripetal-centrifugal
' force imbalance to zero.
vz = vv
lastradius = x
rad = x
initrad = x ' Initial start radius is compared with
' the updated aphelion-perihelion radius.

dt = 200 ' dt = 600 is the upper limit for the current setup.
' dt = 200 is the lower limit.

df = dt ' df is used to later reset dt following the dt = .1
' subroutine, where the aphelion detection is within .1 seconds.

aa: '-------- This section is as described at -------------
' http://members.optusnet.com.au/maxkeon/proven2.html
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

cf = vz ^ 2# / vv ^ 2# * newton

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

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

'////////////////////
radius = radius + anb ' Switch off this line for no anisotropy.
'////////////////////

' The total anisotropic fall distance is now added to the radius
' in the same manner as described at the web address. That is the
' way it MUST be applied to a natural orbit, whatever its shape.

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.

CIRCLE (250 + x * multi, 250 - y * multi), 0, colr

IF fl = 0 THEN GOSUB ag ' aphelion start
IF fl = 1 THEN GOSUB ah ' perihelion start

aph = radius ' previous radius compare routine.
ft = ft + 1 ' ft is only included to eliminate startup errors.

GOTO aa
'--------------------------

ad:
CIRCLE (250 + x * multi, 250 - y * multi), 2, 14
' Identifies the aphelion or perihelion.

LOCATE 1, 1: cycl = cycl + 1
PRINT cycl; "cycles"
PRINT x; "x", y; "y"
rr = (radius - rad)
rad = radius: PRINT radius; "radius"
PRINT
fw = fw + rr
PRINT rr; "meter radius change (now - last)"
PRINT fw; "total radius change."
PRINT y; "total advance."
PRINT y - lasty; "position on the y axis for this cycle."

GOSUB ak ' Loops through the graph subroutine.

lasty = y
initrad = radius
colr = colr + 1: IF colr 15 THEN colr = 1
IF cycl = 20 THEN END
fa = 0
RETURN

ag:
IF initrad - radius 10000 THEN dt = df: fa = 0: fb = 0

IF ft 1000 AND fb = 0 AND initrad-radius 10000 THEN dt=.1:fa=1
' 20000 for dt=100. 10000 for higher dt.

LOCATE 11, 1
IF fa = 1 THEN PRINT aph - radius
IF fa = 1 AND aph - radius 0 THEN GOSUB ad: fb = 1
' Detects aphelion.
RETURN

ah:
IF radius - initrad 100000 THEN dt = df: fa = 0: fb = 0
IF ft 1000 AND fb=0 AND radius -initrad 100000 THEN dt=.1:fa=1

LOCATE 11, 1
IF fa = 1 THEN PRINT aph - radius
IF fa = 1 AND aph - radius 0 THEN GOSUB ad: fb = 1
' Detects perihelion.
RETURN

ak:
ub = y - lasty ' This is only included to
ua = y / cycl ' shorten the next two lines.
LINE (50 + incb, 380 - ud / 160000)-(50 + inca, 380-ua/160000),12
LINE (50 + incb, 380 - uc / 160000)-(50 + inca, 380-ub/160000),11
'////////////////// It's necessary to change the 160000 divisor
' to 80 for 2000x graph magnification when the anisotropy is not
' included.
'///////////////////////////////////

incb = inca: uc = ub: ud = ua
inca = inca + 25
RETURN

"Max Keon" wrote in message
u...

"George Dishman" wrote in message
ps.com...
On 3 Aug, 03:50, "Max Keon" wrote:
"George Dishman" wrote in message
ups.com...
On 29 Jul, 13:40, "Max Keon" wrote:

On the journey inward, orbital speed increases as it falls and
orbit radius shrinks. Since centrifugal force is proportional to
radius and to the square of orbital speed, the end result is that
the mass is simply oscillating back and forth in the middle of an
invisible spring located perpendicular to the direction of
motion.

None of that has anything to do with an anomalous change in the
pull of gravity.

Right, and that's a good way of looking
at how it works.

It certainly is George.

Max, there is no point in my replying if all you
do is snip what I say and go off on a new tack
every time.

That's certainly what I do George, and it's called progress.

No, it's called "rude" when you disregard my work
entirely, but ...

Which has been an unusual event in theoretical physics ever since
postulates were introduced into our reality. That's not intended
as humor either.

I snip what you say when I think the rest of your reply has been
addressed, which is quite understandably what you do with my
posts.

Yes, I have no objection to you doing that,
it reduces duplication.

Your description above goes a long
way towards understanding the situation so follow
it through and you will grasp what your equation
implies. This is how it works:
---
On the outward leg, the planet moves from
left to right but the anisotropy pushes from
right to left. It again reduces the length of
the swing so slowly the swing is damped out
until you are left with the planet in a
circular orbit.

Think about that Max.

But you still don't understand the
consequences of an anomalous force being applied to a naturally
stable orbit at all.

The consequences are stated above. Think carefully
about the direction your force acts in

Do you think I'm stupid.

I can only judge from what you write and based on
that I would say you are innumerate. You did not
know the basic laws of arithmetic and you are still
making a serious mistake with the sign of the
velocity, one you repeat in the code at the bottom
even though you corrected it a few posts back.

Further you don't appear to have any knowledge at
all of calculus which is fundamental to this problem,
there is NO other way to get from the acceleration
to the resulting path. These are basic aspects of
maths which you should have mastered in your early
teens. I cannot say whether you have a learning
disability, or perhaps you were home-schooled by
parents who didn't cover maths, or maybe you are
yet to reach your teens, but your inability to
handle the maths is evident whatever the reason.

I can see exactly where the force is
acting and why you are led to believe that it will cause an
orbit eccentricity decay. But I can also see why you are wrong.

I've said it before but it seems I must repeat it,
acceleration is defined as the derivative of velocity
which in turn is the derivative of the location.
Integrate the acceleration twice and you get the path.
The is what the program I wrote for you does, and the
path is produces is the _consequence_ of your equation.

The attached program plots the path of a mass in a circular
orbit (plotted in a linear fashion) after an anomalous force is
suddenly applied. If you run the program you will notice that
centrifugal force has increased to twice the anomalous force
before the fall is halted. That's exactly what should be
expected.

This equation s = (v^2 + (ana^2 * dt)^ 2)^.5 extracted from
the program, is used as a means of determining orbital speed
increase according to a^2 + b^2 = c^2.

Pythagoras is only valid if a and b are at right
angles. That is not the case for an elliptical
orbit.

It makes no difference.

Yes it does, and that remark certainly qualifies as stupid.
It means you cannot use it in the circumstances.

The true tangential speed can be
obtained using a^2+b^2=c^2. The fall rate for every second is
already known,

The "fall rate" is NOT known, you are calculating it as
if the planet were moving directly towards the Sun which
it is not and a again your results are entirely wrong as
a result.

so the same equation will give an accurate orbital
speed increase per second.

Regardless of how accurate the result may be, nothing changes
the obvious fact that what is lost while the anisotropy is
increasing is reclaimed in full when it has again reduced to
zero at aphelion or perihelion.

I can only suggest that you visit this web page,
http://members.optusnet.com.au/maxkeon/proven2.html
or refer back to my previous reply.

I have looked at the page and it is riddled with the same
basic mathematical errors that were in all your previous
replies. Until you learn basic maths, you have no hope of
correcting these.

The set of equations in your program describe every possible
naturally occurring orbit, between a circular orbit and a direct
fall.

Yes.

Every one of those orbits are therefore identical.

Don't be silly, obviously an ellipse is not identical to
a circle.

They
are each equivalent to a circular orbit, or to any other orbit
as well. Whatever applies for a circular orbit also applies for
every other orbit, _AND VICE VERSA_.

Is that clear enough?

It is very clear, you have no idea what you are talking
about. The various coloured lines on this diagram are
far from identical:

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

You are certainly supplying evidence for anyone to call
you stupid so perhaps you should rethink your statement.

I think your problem stems from the belief that because those
equations accurately describe nature, they can also be used
to determine the laws of nature.

No, my view is trivial actually, it is that the rules
of mathematics (not physics) allow me to calculate the
path which results from your equation and that is what
I have done. If the path predicted by my calculation
doesn't happen in nature (and we know it does not) then
you equation does not "accurately describe nature", the
equation is wrong.

Max, I've said this many times but you are still
not listening - you CANNOT work out the correct
answers without taking the DIRECTION into account.

I can do EXACTLY as truth dictates.

When calculating the path from the equation, you can
only do what the rules of maths dictate.

....
... While the anomalous force remains
active, if the oscillations can be elastically diminished to zero
through some external means, the mass will come to rest in an
orbit that has reduced in radius by half the depth of the
oscillation, and there it will stay. ...

Not an "external means", your equation for the
anisotropy is INELASTIC and does the job of
slowly diminishing the eccentricity.

No it's doesn't. For some reason, you continue to miss the point
altogether.

I'll snip the rest because it no longer applies.

I've included another progress report in the form of a Qbasic
program. Note that my position has never changed at all. Meaning
that I'm not "off on a new tack", as you put it. I am in fact
still on my original tack because you have not shown me anything
that could sway me one bit toward yours.

The total fall distance ...

Stop there. What you call the "total fall distance" is calculated
as if the planet was falling directly towards the planet but it
is not. That formula and everything thereafter is therefore wrong.

is now added to the radius in the same
manner as described at the above address. That is the proper
method of applying an anomalous gravity change to a circular
orbit,

Wrong it is not the proper way to add any acceleration to
any orbit. The only valid way is to add the change of velocity
due to acceleration to the current velocity taking into account
the directions because that is the DEFINITION of acceleration.

Try to get that into your skull Max, acceleration is DEFINED
as the rate of change of velocity.

I'll snip most of the rest, since it is invalidated by that
error in your approach already, but I will pick up on this
again:

IF anc = anb THEN s = (vz ^ 2# - (ana ^ 2 * dt)) ^ .5#
IF anc anb THEN s = (vz ^ 2# + (ana ^ 2 * dt)) ^ .5#

If you swap the sign, you change an increase of gravity into a
decrease or vice versa. Your program _appears_ to restore the
planet to its starting radius after each orbit because you have
reversed the effect of the anisotropy on one leg, either gravity
is increased on both the inward and outward legs or, depending
on other signs, it might be decreased on both. Either way, that
is why you get a different result from me. If you check that it
is decreased on the inward leg and increased on the outward,
then once you get rid of the other errors, you will get the same
result as my program.

George

"George Dishman" wrote in message
...
"Max Keon" wrote in message
. au...
"George Dishman" wrote in message
ps.com...
On 3 Aug, 03:50, "Max Keon" wrote:
"George Dishman" wrote in message
ups.com...

On the outward leg, the planet moves from
left to right but the anisotropy pushes from
right to left. It again reduces the length of
the swing so slowly the swing is damped out
until you are left with the planet in a
circular orbit.

Think about that Max.

But you still don't understand the
consequences of an anomalous force being applied to a naturally
stable orbit at all.

The consequences are stated above. Think carefully
about the direction your force acts in

Do you think I'm stupid.

I can only judge from what you write and based on
that I would say you are innumerate. You did not
know the basic laws of arithmetic and you are still
making a serious mistake with the sign of the
velocity, one you repeat in the code at the bottom
even though you corrected it a few posts back.

I'm not really buying your rather uncharacteristically emotional
reply because, apart from justifying the emotional rant, it
doesn't seem to be you at all. But it's a shame that you wasted
it on a false premise, in that I'm not actually wrong this time.

Do get this straight though George, in the quest for truth, not
getting it right yesterday or today is of no consequence, so long
as I get it right tomorrow.

Further you don't appear to have any knowledge at
all of calculus which is fundamental to this problem,
there is NO other way to get from the acceleration
to the resulting path. These are basic aspects of
maths which you should have mastered in your early
teens. I cannot say whether you have a learning
disability, or perhaps you were home-schooled by
parents who didn't cover maths, or maybe you are
yet to reach your teens, but your inability to
handle the maths is evident whatever the reason.

I can see exactly where the force is
acting and why you are led to believe that it will cause an
orbit eccentricity decay. But I can also see why you are wrong.

I've said it before but it seems I must repeat it,
acceleration is defined as the derivative of velocity
which in turn is the derivative of the location.
Integrate the acceleration twice and you get the path.
The is what the program I wrote for you does, and the
path is produces is the _consequence_ of your equation.

The attached program plots the path of a mass in a circular
orbit (plotted in a linear fashion) after an anomalous force is
suddenly applied. If you run the program you will notice that
centrifugal force has increased to twice the anomalous force
before the fall is halted. That's exactly what should be
expected.

This equation s = (v^2 + (ana^2 * dt)^ 2)^.5 extracted from
the program, is used as a means of determining orbital speed
increase according to a^2 + b^2 = c^2.

Pythagoras is only valid if a and b are at right
angles. That is not the case for an elliptical
orbit.

It makes no difference.

Yes it does, and that remark certainly qualifies as stupid.
It means you cannot use it in the circumstances.

The true tangential speed can be
obtained using a^2+b^2=c^2. The fall rate for every second is
already known,

The "fall rate" is NOT known, you are calculating it as
if the planet were moving directly towards the Sun which
it is not and a again your results are entirely wrong as
a result.

The tangential speed at any given time can be calculated using
Pythagoras. i.e. At one point in Mercury's elliptical orbit,
Mercury is heading toward x. (c) length per chosen time duration
is already known from the current orbital speed, while (b) is
(vr) according to your equations. (a) now has a value to which
the anisotropic acceleration can be correctly applied. It cannot
be applied as a component of the natural orbit, as you propose.

a
o_--------.
- _ . b
c -x


0
Sun

so the same equation will give an accurate orbital
speed increase per second.

Regardless of how accurate the result may be, nothing changes
the obvious fact that what is lost while the anisotropy is
increasing is reclaimed in full when it has again reduced to
zero at aphelion or perihelion.

I can only suggest that you visit this web page,
http://members.optusnet.com.au/maxkeon/proven2.html
or refer back to my previous reply.

I have looked at the page and it is riddled with the same
basic mathematical errors that were in all your previous
replies. Until you learn basic maths, you have no hope of
correcting these.

The set of equations in your program describe every possible
naturally occurring orbit, between a circular orbit and a direct
fall.

Yes.

Every one of those orbits are therefore identical.

Don't be silly, obviously an ellipse is not identical to
a circle.

It's time to take off your rose colored glasses and face reality.
An elliptical orbit is nothing more than a circular orbit which
has been kicked off center. And by the way, energy must be
transferred from somewhere to somewhere for that imbalance to be
removed, in a completely closed system George. How do you propose
to do that?
---

... While the anomalous force remains
active, if the oscillations can be elastically diminished to zero
through some external means, the mass will come to rest in an
orbit that has reduced in radius by half the depth of the
oscillation, and there it will stay. ...

Not an "external means", your equation for the
anisotropy is INELASTIC and does the job of
slowly diminishing the eccentricity.

No it's doesn't. For some reason, you continue to miss the point
altogether.

I'll snip the rest because it no longer applies.

I've included another progress report in the form of a Qbasic
program. Note that my position has never changed at all. Meaning
that I'm not "off on a new tack", as you put it. I am in fact
still on my original tack because you have not shown me anything
that could sway me one bit toward yours.

The total fall distance ...

Stop there. What you call the "total fall distance" is calculated
as if the planet was falling directly towards the planet but it
is not. That formula and everything thereafter is therefore wrong.

is now added to the radius in the same
manner as described at the above address. That is the proper
method of applying an anomalous gravity change to a circular
orbit,

Wrong it is not the proper way to add any acceleration to
any orbit. The only valid way is to add the change of velocity
due to acceleration to the current velocity taking into account
the directions because that is the DEFINITION of acceleration.

Exactly. And the force direction determines where that
acceleration is pointing, which is always (near enough to)
directly toward each gravitating body. The RADIAL velocity is
the only one of any consequence, not the tangential "velocity",
or any part thereof. You are still plotting natural orbits.
Get over it!

Try to get that into your skull Max, acceleration is DEFINED
as the rate of change of velocity.

So what's the problem? The anisotropy causes Mercury to
anomalously accelerate _DIRECTLY_ toward or away from the Sun.
It's not pushed sideways by the anisotropy, as you claim.

I'll snip most of the rest, since it is invalidated by that
error in your approach already, but I will pick up on this
again:

IF anc = anb THEN s = (vz ^ 2# - (ana ^ 2 * dt)) ^ .5#
IF anc anb THEN s = (vz ^ 2# + (ana ^ 2 * dt)) ^ .5#

If you swap the sign, you change an increase of gravity into a
decrease or vice versa. Your program _appears_ to restore the
planet to its starting radius after each orbit

No George. My program restores the planet to its true perihelion
and aphelion radii after the anisotropy finally reduces to zero
at each turnaround point in the orbit.

because you have
reversed the effect of the anisotropy on one leg, either gravity
is increased on both the inward and outward legs or, depending
on other signs, it might be decreased on both.

You should know very well why the sign must be physically
changed. If you plug these numbers into your calculator you will
always get the same answer from each one, regardless of the sign
on ana.

dt = 1
ana = 1

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

- ((-ana)^2 * dt)^.5 = -1

In the program, I have left out the set of brackets around
((ana). It still gives the correct result when ana becomes
negative, even if it's not supposed to. Perhaps that has confused
you. Anyway, I've added the brackets to the program, and that has
now been updated to generate a 250 cycle data file, which is used
to plot some quite interesting graphs.

I won't clutter the post with details this time. You seem to
ignore most of them anyway.

http://members.optusnet.com.au/maxkeon/proven2.html

-----

Max Keon

Max Keon wrote:
"George Dishman" wrote in message ...
"Max Keon" wrote in message u...
"George Dishman" wrote in message ps.com...
On 3 Aug, 03:50, "Max Keon" wrote:
"George Dishman" wrote in message ups.com...

On the outward leg, the planet moves from
left to right but the anisotropy pushes from
right to left. It again reduces the length of
the swing so slowly the swing is damped out
until you are left with the planet in a
circular orbit.

Think about that Max.

You still aren't thinking about it.

But you still don't understand the
consequences of an anomalous force being applied to a naturally
stable orbit at all.

The consequences are stated above. Think carefully
about the direction your force acts in

Do you think I'm stupid.

I can only judge from what you write and based on
that I would say you are innumerate. You did not
know the basic laws of arithmetic and you are still
making a serious mistake with the sign of the
velocity, one you repeat in the code at the bottom
even though you corrected it a few posts back.

I'm not really buying your rather uncharacteristically emotional
reply because, apart from justifying the emotional rant, it
doesn't seem to be you at all.

It's not me normally but you specifically asked so
I gave a quiet and measured reply, not a rant. Your
difficulties with maths are at the bottom of all of this
and until you recognise and address them, there is
no way to resolve the disagreement.

But it's a shame that you wasted
it on a false premise, in that I'm not actually wrong this time.

You are wrong in the way you calculate the orbit
from the equation, the methods you are trying to
use don't work. The method I showed you in the
program where the acceleration, velocity and location
are all treated as separate x and y components is
the rigorous and correct method and any alternative
must be equivalent, giving the same answer.

Do get this straight though George, in the quest for truth, not
getting it right yesterday or today is of no consequence, so long
as I get it right tomorrow.

Yes, that's why I have continued. Eventually, if you
get fed up with me pointing out the same errors in
your maths each time, I hope you will go off and try
to find a maths book or website where you can find
the truth in the hope of disproving my statements.
When you do that, you will find every source says
the same thing and that I am right about the maths.


This equation s = (v^2 + (ana^2 * dt)^ 2)^.5 extracted from
the program, is used as a means of determining orbital speed
increase according to a^2 + b^2 = c^2.

Pythagoras is only valid if a and b are at right
angles. That is not the case for an elliptical
orbit.

It makes no difference.

Yes it does, and that remark certainly qualifies as stupid.
It means you cannot use it in the circumstances.

The true tangential speed can be
obtained using a^2+b^2=c^2. The fall rate for every second is
already known,

The "fall rate" is NOT known, you are calculating it as
if the planet were moving directly towards the Sun which
it is not and a again your results are entirely wrong as
a result.

The tangential speed at any given time can be calculated using
Pythagoras. i.e. At one point in Mercury's elliptical orbit,
Mercury is heading toward x. (c) length per chosen time duration
is already known from the current orbital speed, while (b) is
(vr) according to your equations. (a) now has a value to which
the anisotropic acceleration can be correctly applied. It cannot
be applied as a component of the natural orbit, as you propose.

a
o_--------.
- _ . b
c -x


0
Sun

OK, if I follow the diagram, Mercury is at 'o' heading for 'x'.
The orbital speed is the length of ox = 'c' and you break
that into components 'a' and 'b' using Pythagoras where
line 'b' points to the Sun. That is OK but you are missing
two points, firstly, in the next time period, Mercury will be
to the right of this diagram hence the line 'b' must be
slightly rotated to still point at the Sun. That mixes up
the components so some of speed 'b' from this diagram
becomes part of speed 'a' in the next and vice versa, hence
as I said the fall rate is not known. Secondly the speed 'c'
is not constant. For an elliptical orbit, the speed is greater
at perihelion than aphelion and your diagram will not
reproduce that since you are not including the basic
gravitational force in the calculations.

The set of equations in your program describe every possible
naturally occurring orbit, between a circular orbit and a direct
fall.

Yes.

Every one of those orbits are therefore identical.

Don't be silly, obviously an ellipse is not identical to
a circle.

It's time to take off your rose colored glasses and face reality.

I am telling you reality, your maths is wrong and you
nedd to lay aside the physics for a while and study
basic schoolboy calculus as a _pure_ maths topic
before you will resolve that.

An elliptical orbit is nothing more than a circular orbit which
has been kicked off center.

Rubbish, the major axis is longer than the minor axis.

And by the way, energy must be
transferred from somewhere to somewhere for that imbalance to be
removed, in a completely closed system George. How do you propose
to do that?

You should already know that in a standard Keplerian orbit
energy transfers between kinetic and potential to an extent
dependent on the eccentricity. The mechanism is the
Newtonian gravitational force.

I've included another progress report in the form of a Qbasic
program. Note that my position has never changed at all. Meaning
that I'm not "off on a new tack", as you put it. I am in fact
still on my original tack because you have not shown me anything
that could sway me one bit toward yours.

The total fall distance ...

Stop there. What you call the "total fall distance" is calculated
as if the planet was falling directly towards the planet but it
is not. That formula and everything thereafter is therefore wrong.

is now added to the radius in the same
manner as described at the above address. That is the proper
method of applying an anomalous gravity change to a circular
orbit,

Wrong it is not the proper way to add any acceleration to
any orbit. The only valid way is to add the change of velocity
due to acceleration to the current velocity taking into account
the directions because that is the DEFINITION of acceleration.

Exactly. And the force direction determines where that
acceleration is pointing, which is always (near enough to)
directly toward each gravitating body.

The force is always EXACTLY towards the Sun in Newtonian
gravitation, and as I understand your suggestion, the anisotropy
is just a chaneg in the size of that force, not the direction.

The RADIAL velocity is
the only one of any consequence, not the tangential "velocity",
or any part thereof.

Sorry Max, both matter. You cannot just ignore aspects
simply because you don't have the mathematical ability
to include them.

You are still plotting natural orbits.
Get over it!

I am plotting what happens to the orbit if you add an
anisotropic effect to the Newtonian force which is
exactly what you have been describing. My plots are
mathematically correct and accurate to better than a
metre per orbit.

Try to get that into your skull Max, acceleration is DEFINED
as the rate of change of velocity.

So what's the problem?

The problem is that you don't have enough mathematical
ability to do the integration and have tried to use alternative
methods that use too many simplifications and don't work
as a consequence. I have done the integration for you.

The anisotropy causes Mercury to
anomalously accelerate _DIRECTLY_ toward or away from the Sun.
It's not pushed sideways by the anisotropy, as you claim.

I havn't suggested it is pushed sideways by the anisotropy
but even for a normal Keplerian orbit, the orbital speed
varies. It is higher at perihelion than at aphelion and the
direction of motion is purely tangential at both points so
there is clearly a change of tangential speed. Your method
doesn't reproduce that and again that is symptomatic of
your errors in the maths.

I'll snip most of the rest, since it is invalidated by that
error in your approach already, but I will pick up on this
again:

IF anc = anb THEN s = (vz ^ 2# - (ana ^ 2 * dt)) ^ .5#
IF anc anb THEN s = (vz ^ 2# + (ana ^ 2 * dt)) ^ .5#

If you swap the sign, you change an increase of gravity into a
decrease or vice versa. Your program _appears_ to restore the
planet to its starting radius after each orbit

No George. My program restores the planet to its true perihelion
and aphelion radii after the anisotropy finally reduces to zero
at each turnaround point in the orbit.

Yes Max, and that is wrong. the radii at those points are
changed by the anisotropy.

because you have
reversed the effect of the anisotropy on one leg, either gravity
is increased on both the inward and outward legs or, depending
on other signs, it might be decreased on both.

You should know very well why the sign must be physically
changed. If you plug these numbers into your calculator you will
always get the same answer from each one, regardless of the sign
on ana.

dt = 1
ana = 1

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

- ((-ana)^2 * dt)^.5 = -1

In the program, I have left out the set of brackets around
((ana). It still gives the correct result when ana becomes
negative, even if it's not supposed to.

If it results in the aphelion radius reverting to the original
value then you have the sign wrong. Starting from aphelion,
the radius at the first perihelion should be increased
compared to the usual Keplerian value while the next
aphelion will be reduced from the Keplerian value for that
orbit.

Perhaps that has confused
you.

I haven't actually checked your signs because it depends
on a number of other factors as well, but if the perihelion
is restored, you either have an increase of gravity on both
legs or a decrease on both where in fact you should get
a decrease on the inward leg and an increase on the
outward. I'll leave it to you to find out which leg is wrong.

Anyway, I've added the brackets to the program, and that has
now been updated to generate a 250 cycle data file, which is used
to plot some quite interesting graphs.

I won't clutter the post with details this time. You seem to
ignore most of them anyway.

Indeed, I have told you what is wrong with your maths
and studying results I know to be wrong is pointless.

I have a already put plots of what your equation actually
predicts on the web - you saw them some weeks ago,
and although I haven't published it I also ran a slight
variation of the program which added the "rest of the
universe" effect and confirmed the slower decay of the
circular orbit that we derived some months ago.

Until you sort out your problems with calculus, there
isn't any point in discussing my results either because
neither of us accepts the other's method.

George

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

Do get this straight though George, in the quest for truth, not
getting it right yesterday or today is of no consequence, so long
as I get it right tomorrow.

Yes, that's why I have continued. Eventually, if you
get fed up with me pointing out the same errors in
your maths each time, I hope you will go off and try
to find a maths book or website where you can find
the truth in the hope of disproving my statements.
When you do that, you will find every source says
the same thing and that I am right about the maths.

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

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

The true tangential speed can be
obtained using a^2+b^2=c^2. The fall rate for every second is
already known,

The "fall rate" is NOT known, you are calculating it as
if the planet were moving directly towards the Sun which
it is not and a again your results are entirely wrong as
a result.

The tangential speed at any given time can be calculated using
Pythagoras. i.e. At one point in Mercury's elliptical orbit,
Mercury is heading toward x. (c) length per chosen time duration
is already known from the current orbital speed, while (b) is
(vr) according to your equations. (a) now has a value to which
the anisotropic acceleration can be correctly applied. It cannot
be applied as a component of the natural orbit, as you propose.

a
o_--------.
- _ . b
c -x


0
Sun

OK, if I follow the diagram, Mercury is at 'o' heading for 'x'.
The orbital speed is the length of ox = 'c' and you break
that into components 'a' and 'b' using Pythagoras where
line 'b' points to the Sun. That is OK but you are missing
two points, firstly, in the next time period, Mercury will be
to the right of this diagram hence the line 'b' must be
slightly rotated to still point at the Sun. That mixes up
the components so some of speed 'b' from this diagram
becomes part of speed 'a' in the next and vice versa, hence
as I said the fall rate is not known.

That is nonsense.

This is the error per second for the circular orbit specified
below. It will be much the same for the tangential speed of any
part of an eccentric orbit.

__b____
l .
l
a l .
l c
l .
l.
l
a = 5.8e10 meters
b = 48000
c = sqr(a^2 + b^2)
c = sqr(5.8e10^2 + 48000^2) = 58000000000.01986
c - a = .01986


And then:
a
-----------------l
- _ l b
c - l
a = 48000
b = .01986
c = sqr(a^2 + b^2)
c = sqr(48000^2 + .01986^2) = 48000.00000000411
c - a = 4.11e-9

The orbital speed error is 4.11e-9 m/sec or 0.0312 meters per
orbit. The fall rate is known to a high enough degree of
accuracy for the purpose.

Secondly the speed 'c'
is not constant. For an elliptical orbit, the speed is greater
at perihelion than aphelion and your diagram will not
reproduce that since you are not including the basic
gravitational force in the calculations.

The set of equations in your program describe every possible
naturally occurring orbit, between a circular orbit and a direct
fall.

Yes.

Every one of those orbits are therefore identical.

Don't be silly, obviously an ellipse is not identical to
a circle.

It's time to take off your rose colored glasses and face reality.

I am telling you reality, your maths is wrong and you
nedd to lay aside the physics for a while and study
basic schoolboy calculus as a _pure_ maths topic
before you will resolve that.

An elliptical orbit is nothing more than a circular orbit which
has been kicked off center.

Rubbish, the major axis is longer than the minor axis.

How could it not be? You really are missing the point here.

And by the way, energy must be
transferred from somewhere to somewhere for that imbalance to be
removed, in a completely closed system George. How do you propose
to do that?

You should already know that in a standard Keplerian orbit
energy transfers between kinetic and potential to an extent
dependent on the eccentricity. The mechanism is the
Newtonian gravitational force.

What are you trying to say? Are you suggesting that if an
anomalous change in gravity occurs enroute to the aphelion and
is removed prior to arriving, the rise to the previous aphelion
radius will somehow be prematurely halted? Why on earth would
you think that? Or were you not really saying anything at all?
---

You are still plotting natural orbits.
Get over it!

I am plotting what happens to the orbit if you add an
anisotropic effect to the Newtonian force which is
exactly what you have been describing. My plots are
mathematically correct and accurate to better than a
metre per orbit.

Very impressive George. It's a shame that you can't apply the
gravity anisotropy the way you do.

Your plots are mathematically correct according to a program
which can plot the coordinates for any degree of eccentricity,
to a perfect circle. That's all it does. You have CHOSEN to add
the anisotropy in such a manner as to generate the result that
best suits your purpose. And you alone have made that choice
because there cannot be anything already established to support
it. A gravity anisotropy has not previously been known to exist.

This is how it works
http://members.optusnet.com.au/maxkeon/proven2.html
---

You should know very well why the sign must be physically
changed. If you plug these numbers into your calculator you will
always get the same answer from each one, regardless of the sign
on ana.

dt = 1
ana = 1

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

- ((-ana)^2 * dt)^.5 = -1

In the program, I have left out the set of brackets around
((ana). It still gives the correct result when ana becomes
negative, even if it's not supposed to.

If it results in the aphelion radius reverting to the original
value then you have the sign wrong. Starting from aphelion,
the radius at the first perihelion should be increased
compared to the usual Keplerian value while the next
aphelion will be reduced from the Keplerian value for that
orbit.

Perhaps that has confused
you.

I haven't actually checked your signs because it depends
on a number of other factors as well, but if the perihelion
is restored, you either have an increase of gravity on both
legs or a decrease on both where in fact you should get
a decrease on the inward leg and an increase on the
outward. I'll leave it to you to find out which leg is wrong.

You are still on the wrong tram. The sign change is correct.
But your reasoning is not.

Anyway, I've added the brackets to the program, and that has
now been updated to generate a 250 cycle data file, which is used
to plot some quite interesting graphs.

I won't clutter the post with details this time. You seem to
ignore most of them anyway.

Indeed, I have told you what is wrong with your maths
and studying results I know to be wrong is pointless.

I have a already put plots of what your equation actually
predicts on the web - you saw them some weeks ago,
and although I haven't published it I also ran a slight
variation of the program which added the "rest of the
universe" effect and confirmed the slower decay of the
circular orbit that we derived some months ago.

You've had your little rant, now it must be my turn.

If you've applied the same logic as you have in the past, you
are wrong again. But I think I understand why at this time you
feel the need to denounce a gravity anisotropy which is acting
on anything that moves relative to the local frame of the
universe. I saw that post on sci.astro.research too. It seems
that your dark matter faces yet another dilemma.

I was quite taken by one of the possible causes, that dark matter
is possibly affected, not only by gravity but also by some as yet
unknown interaction between dark matter particles. That amazingly
postulated speculation was then labeled "an exciting
alternative". An exciting alternative that would require new
physics.

Einstein's theories cannot explain what nature is clearly
demonstrating, so they break down if some reason for the anomaly
is not forthcoming. So we dream up anything that will fill the
void, even if there is no prior evidence of its existence
whatever. So long as it fills the void, that's all that matters.
Let's just say that it's invisible. Hey, we're the scientists,
so we can say whatever we like.

Do you really think that the world is going along for the ride
with you on this? Do you not realize that the world is rapidly
turning its back on physics, that the credibility of physics is
rapidly eroding away. Do you not know that you are just jabbering
amongst yourselves? Do you not know that the future of humankind
is in crisis?

As Richard Dawkins put it, militant faith is again on the march
despite the progress of science.

Can you guess why? Throwing in more bull**** just makes the pile
bigger doesn't it.

The only thing that will neutralize bull**** is stark reality,
and it's about time you all started doing the job the world is
paying you to do.

Until you sort out your problems with calculus, there
isn't any point in discussing my results either because
neither of us accepts the other's method.

Maths can't make your result any less wrong because your basic
principles are wrong. Sort out _your_ problems George.

We will never agree to disagree either because YOU ARE WRONG.

-----

Max Keon

Max Keon wrote:

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

Do get this straight though George, in the quest for truth, not
getting it right yesterday or today is of no consequence, so long
as I get it right tomorrow.

Yes, that's why I have continued. Eventually, if you
get fed up with me pointing out the same errors in
your maths each time, I hope you will go off and try
to find a maths book or website where you can find
the truth in the hope of disproving my statements.
When you do that, you will find every source says
the same thing and that I am right about the maths.

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 so you
get the velocity by integrating the total acceleration.
That is the ONLY method permitted by the definition
and other techniques can only be equivalent to that
or wrong.

The true tangential speed can be
obtained using a^2+b^2=c^2. The fall rate for every second is
already known,

The "fall rate" is NOT known, you are calculating it as
if the planet were moving directly towards the Sun which
it is not and a again your results are entirely wrong as
a result.

The tangential speed at any given time can be calculated using
Pythagoras. i.e. At one point in Mercury's elliptical orbit,
Mercury is heading toward x. (c) length per chosen time duration
is already known from the current orbital speed, while (b) is
(vr) according to your equations. (a) now has a value to which
the anisotropic acceleration can be correctly applied. It cannot
be applied as a component of the natural orbit, as you propose.

a
o_--------.
- _ . b
c -x


0
Sun

OK, if I follow the diagram, Mercury is at 'o' heading for 'x'.
The orbital speed is the length of ox = 'c' and you break
that into components 'a' and 'b' using Pythagoras where
line 'b' points to the Sun. That is OK but you are missing
two points, firstly, in the next time period, Mercury will be
to the right of this diagram hence the line 'b' must be
slightly rotated to still point at the Sun. That mixes up
the components so some of speed 'b' from this diagram
becomes part of speed 'a' in the next and vice versa, hence
as I said the fall rate is not known.

That is nonsense.

It is complex admittedly but it is correct. It leads to
the apparent centrifugal and coriolis pseudo-forces.

This is the error per second for the circular orbit specified
below. It will be much the same for the tangential speed of any
part of an eccentric orbit.

__b____
l .
l
a l .
l c
l .
l.
l
a = 5.8e10 meters
b = 48000
c = sqr(a^2 + b^2)
c = sqr(5.8e10^2 + 48000^2) = 58000000000.01986
c - a = .01986


And then:
a
-----------------l
- _ l b
c - l
a = 48000
b = .01986
c = sqr(a^2 + b^2)
c = sqr(48000^2 + .01986^2) = 48000.00000000411
c - a = 4.11e-9

The orbital speed error is 4.11e-9 m/sec or 0.0312 meters per
orbit.

Sorry, you are missing Kepler's Law, the speed varies
in an eliptical orbit.

The fall rate is known to a high enough degree of
accuracy for the purpose.

Nothing like it, you have now missed two major
effects. What you have said above is badly flawed.

Secondly the speed 'c'
is not constant. For an elliptical orbit, the speed is greater
at perihelion than aphelion and your diagram will not
reproduce that since you are not including the basic
gravitational force in the calculations.

The set of equations in your program describe every possible
naturally occurring orbit, between a circular orbit and a direct
fall.

Yes.

Every one of those orbits are therefore identical.

Don't be silly, obviously an ellipse is not identical to
a circle.

It's time to take off your rose colored glasses and face reality.

I am telling you reality, your maths is wrong and you
nedd to lay aside the physics for a while and study
basic schoolboy calculus as a _pure_ maths topic
before you will resolve that.

An elliptical orbit is nothing more than a circular orbit which
has been kicked off center.

Rubbish, the major axis is longer than the minor axis.

How could it not be?

Look at the diagram of the ellipses I cited a few posts
back, it is obvious that the high eccentricity values
have a long, narrow shape, they are nowhere near
being circles.

You really are missing the point here.

You are talking complete rubbish here.

And by the way, energy must be
transferred from somewhere to somewhere for that imbalance to be
removed, in a completely closed system George. How do you propose
to do that?

You should already know that in a standard Keplerian orbit
energy transfers between kinetic and potential to an extent
dependent on the eccentricity. The mechanism is the
Newtonian gravitational force.

What are you trying to say?

You asked how energy was "transferred from somewhere
to somewhere" in a "completely closed system" when I
previously talked about the change of speed between
perihelion and aphelion so I guess you are asking how
the kinetic energy is transferred to permit that speed
change. If not, make your question clearer.

Are you suggesting that if an
anomalous change in gravity occurs enroute to the aphelion and
is removed prior to arriving, the rise to the previous aphelion
radius will somehow be prematurely halted? Why on earth would
you think that? Or were you not really saying anything at all?

I was trying to answer the question you asked which
relates to the newtonian part of gravity, nothing to do
with the anisotropy.

You are still plotting natural orbits.
Get over it!

I am plotting what happens to the orbit if you add an
anisotropic effect to the Newtonian force which is
exactly what you have been describing. My plots are
mathematically correct and accurate to better than a
metre per orbit.

Very impressive George. It's a shame that you can't apply the
gravity anisotropy the way you do.

That is the ONLY way permitted by the definition of
acceleration. You need to learn the basics of calculus
before we cann talk sensibly.

Your plots are mathematically correct according to a program
which can plot the coordinates for any degree of eccentricity,
to a perfect circle. That's all it does. You have CHOSEN to add
the anisotropy in such a manner as to generate the result that
best suits your purpose.

No, I have NO choice. I have added the anisotropic
acceleration in the only way permitted by the rules
of mathematics.

And you alone have made that choice
because there cannot be anything already established to support
it. A gravity anisotropy has not previously been known to exist.

Acceleration is a uniquely defined quantity, there is
no option in how you use it. Your equation defines
an acceleration and my maths handles that in the
only way posible.

This is how it works
http://members.optusnet.com.au/maxkeon/proven2.html

Sorry Max, until you learn calculus, you are not going
to be able to work this out.

....
I haven't actually checked your signs because it depends
on a number of other factors as well, but if the perihelion
is restored, you either have an increase of gravity on both
legs or a decrease on both where in fact you should get
a decrease on the inward leg and an increase on the
outward. I'll leave it to you to find out which leg is wrong.

You are still on the wrong tram. The sign change is correct.
But your reasoning is not.

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.

Anyway, I've added the brackets to the program, and that has
now been updated to generate a 250 cycle data file, which is used
to plot some quite interesting graphs.

I won't clutter the post with details this time. You seem to
ignore most of them anyway.

Indeed, I have told you what is wrong with your maths
and studying results I know to be wrong is pointless.

I have a already put plots of what your equation actually
predicts on the web - you saw them some weeks ago,
and although I haven't published it I also ran a slight
variation of the program which added the "rest of the
universe" effect and confirmed the slower decay of the
circular orbit that we derived some months ago.

You've had your little rant, now it must be my turn.

Sorry Max, a 'rant' is an emotional outburst, what I said
is simply a factual statement that I have run the
simulation and posted some of the results.

If you've applied the same logic as you have in the past, you
are wrong again. But I think I understand why at this time you
feel the need to denounce a gravity anisotropy which is acting
on anything that moves relative to the local frame of the
universe. I saw that post on sci.astro.research too. It seems
that your dark matter faces yet another dilemma.

I was quite taken by one of the possible causes, that dark matter
is possibly affected, not only by gravity but also by some as yet
unknown interaction between dark matter particles. That amazingly
postulated speculation was then labeled "an exciting
alternative". An exciting alternative that would require new
physics.

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.

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.

So we dream up anything that will fill the
void, even if there is no prior evidence of its existence
whatever. So long as it fills the void, that's all that matters.
Let's just say that it's invisible. Hey, we're the scientists,
so we can say whatever we like.

Again you are confused, dark matter has similar
charateristics to neutrinos but seems to have a
slightly higher mass.

snip rant

Until you sort out your problems with calculus, there
isn't any point in discussing my results either because
neither of us accepts the other's method.

Maths can't make your result any less wrong because your basic
principles are wrong. Sort out _your_ problems George.

There are no problems in my maths Max, get
yourself a book on calculus and find out what
the definition of acceleration is.

We will never agree to disagree either because YOU ARE WRONG.

Sorry Max, this is not something that is open to
debate. I am right about the maths, end of story.
You can confirm that by reading any high school
text books covering calculus and basic mechanics.

Your equation predicts eccentricity will decay to
zero in a two body situation and therafter the orbit
will be stable. With three or more bodies, it says
even the circular orbit will decay though more
slowly. For Mercury, it is still fast enough to crash
it into the Sun in a million years.

Your emotional outbusts may make you feel better
but they will not changes the rules of maths and
they won't change the consequences of your
equation, it is undoubtedly wrong.

George

"George Dishman" wrote in message
ups.com...
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? 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.

When the anisotropy is correctly applied, it still has
consequences of course, but nothing like what you have proposed.
Those consequences are exactly as described at this address,
http://members.optusnet.com.au/maxkeon/proven2.html
See if you can get it right this time.

so you
get the velocity by integrating the total acceleration.

No. You only get it right if you apply the anisotropy correctly,
along the line of the force and nowhere else, then note the
consequences that it has on the natural eccentricity of the
orbit.

That is the ONLY method permitted by the definition
and other techniques can only be equivalent to that
or wrong.

????
---

You are still plotting natural orbits.
Get over it!

I am plotting what happens to the orbit if you add an
anisotropic effect to the Newtonian force which is
exactly what you have been describing. My plots are
mathematically correct and accurate to better than a
metre per orbit.

Very impressive George. It's a shame that you can't apply the
gravity anisotropy the way you do.

That is the ONLY way permitted by the definition of
acceleration. You need to learn the basics of calculus
before we cann talk sensibly.

Your plots are mathematically correct according to a program
which can plot the coordinates for any degree of eccentricity,
to a perfect circle. That's all it does. You have CHOSEN to add
the anisotropy in such a manner as to generate the result that
best suits your purpose.

No, I have NO choice. I have added the anisotropic
acceleration in the only way permitted by the rules
of mathematics.

You have done no such thing.
---

I haven't actually checked your signs because it depends
on a number of other factors as well, but if the perihelion
is restored, you either have an increase of gravity on both
legs or a decrease on both where in fact you should get
a decrease on the inward leg and an increase on the
outward. I'll leave it to you to find out which leg is wrong.

You are still on the wrong tram. The sign change is correct.
But your reasoning is not.

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

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

---

If you've applied the same logic as you have in the past, you
are wrong again. But I think I understand why at this time you
feel the need to denounce a gravity anisotropy which is acting
on anything that moves relative to the local frame of the
universe. I saw that post on sci.astro.research too. It seems
that your dark matter faces yet another dilemma.

I was quite taken by one of the possible causes, that dark matter
is possibly affected, not only by gravity but also by some as yet
unknown interaction between dark matter particles. That amazingly
postulated speculation was then labeled "an exciting
alternative". An exciting alternative that would require new
physics.

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.

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

Anyway, what makes you think the lensing is causes by dark
matter? Why not something that nature predicts, like a black
hole(s)? They are almost certainly a component of a galaxy
center and can be scattered according to how they collide, or
don't collide. There's no reason why the deflection directions
can't be along the line to the Earth.

So we dream up anything that will fill the
void, even if there is no prior evidence of its existence
whatever. So long as it fills the void, that's all that matters.
Let's just say that it's invisible. Hey, we're the scientists,
so we can say whatever we like.

Again you are confused, dark matter has similar
charateristics to neutrinos but seems to have a
slightly higher mass.

So what's the mandatory speed at which these dark matter beasties
travel? The so called massive neutrino is compelled by some man
made law of nature to travel at a speed which is vaguely less
than the speed of E/M radiation. If the speed of dark matter is
anything like that, it could not be constrained to cycle about a
galaxy unless the galaxy was almost dense enough to form a black
hole. The massive neutrino would be trapped only just prior to
light. Dark matter would become entrapped before that.

But the speed of dark matter can be no more than the orbital
speed required to hold a mass in the outer region of a galaxy,
otherwise it will fly off into the universe.

Has a dark matter speed been established?

What god is telling you all of this stuff George.


Here's a Universe that's not an alternative to the big
bang theory or any other theory. It's the real Universe,
and it clearly predicts a gravity anisotropy.
http://members.optusnet.com.au/maxkeon/the1-1a.html


snip rant

You've snipped the wrong part.

Until you sort out your problems with calculus, there
isn't any point in discussing my results either because
neither of us accepts the other's method.

Maths can't make your result any less wrong because your basic
principles are wrong. Sort out _your_ problems George.

There are no problems in my maths Max,

The problem has never been with your maths, only with the way
you apply it.

get
yourself a book on calculus and find out what
the definition of acceleration is.

We will never agree to disagree either because YOU ARE WRONG.

Sorry Max, this is not something that is open to
debate. I am right about the maths, end of story.
You can confirm that by reading any high school
text books covering calculus and basic mechanics.

Your equation predicts eccentricity will decay to
zero in a two body situation and therafter the orbit
will be stable. With three or more bodies, it says
even the circular orbit will decay though more
slowly. For Mercury, it is still fast enough to crash
it into the Sun in a million years.

Your emotional outbusts may make you feel better

They are designed as a wake up call George.
But you're not yet ready to be unplugged.

-----

Max Keon

"Max Keon" wrote in message
u...

"George Dishman" wrote in message
ups.com...
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.

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.

When the anisotropy is correctly applied, it still has
consequences of course, but nothing like what you have proposed.

The consequences of the equation as you have written
it and an assumed direction of towards the Sun are
as I have told you.

Those consequences are exactly as described at this address,
http://members.optusnet.com.au/maxkeon/proven2.html

Your methods are mathematically incorrect, I have
explained repeatedly how to do it properly but if
you continue to refuse to get out your maths books
and learn how to handle vectors and calculus, you
are not going to correct your errors

See if you can get it right this time.

so you
get the velocity by integrating the total acceleration.

No. You only get it right if you apply the anisotropy correctly,
along the line of the force and nowhere else,

That is precisely what my program does.

then note the
consequences that it has on the natural eccentricity of the
orbit.

Yes, it decays rapidly.

That is the ONLY method permitted by the definition
and other techniques can only be equivalent to that
or wrong.

????

Acceleration is defined as the derivative of the
velocity. The inverse operation to differentiation
is integration. To get the velocity therefore, you
must integrate the acceleration. Since your equation
fully defines the acceleration and the results of
integration are unique, it follows that the result
of that process is unique. The consequence my program
calculates is the _only_ correct result.

No, I have NO choice. I have added the anisotropic
acceleration in the only way permitted by the rules
of mathematics.

You have done no such thing.

Yes I have, open up a maths textbook and find out.

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.

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)

But why are you calculating that in the first place?
If you are trying to find the distance moved due to
acceleration it should be s = 1/2 * ana^2 * dt and
if you are finding the change of speed it is
dv = ana * dt, neither involves the square root of
the time.

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.

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.

They are almost certainly a component of a galaxy
center and can be scattered according to how they collide, or
don't collide. There's no reason why the deflection directions
can't be along the line to the Earth.

So we dream up anything that will fill the
void, even if there is no prior evidence of its existence
whatever. So long as it fills the void, that's all that matters.
Let's just say that it's invisible. Hey, we're the scientists,
so we can say whatever we like.

Again you are confused, dark matter has similar
charateristics to neutrinos but seems to have a
slightly higher mass.

So what's the mandatory speed at which these dark matter beasties
travel?

There are a number of fairly complex measurements
that suggest they move slowly while neutrinos move
near the speed of light. You will hear the term
"cold dark matter" often abbreviated to "CDM" and
that refers to the mean speed being much less than
the speed of light.

The so called massive neutrino is compelled by some man
made law of nature to travel at a speed which is vaguely less
than the speed of E/M radiation. If the speed of dark matter is
anything like that, it could not be constrained to cycle about a
galaxy unless the galaxy was almost dense enough to form a black
hole.

Even then it would either fall in or escape, the
stable region is negligibly thin. You are exactly
right, that is a key piece of evidence that DM is
cold, it has less than the escape velocity of the
galaxies with which it is associated.

The massive neutrino would be trapped only just prior to
light. Dark matter would become entrapped before that.

But the speed of dark matter can be no more than the orbital
speed required to hold a mass in the outer region of a galaxy,
otherwise it will fly off into the universe.

Has a dark matter speed been established?

What god is telling you all of this stuff George.

No god Max, you have a good understanding of how
that aspect is worked out.

Here's a Universe that's not an alternative to the big
bang theory or any other theory. It's the real Universe,
and it clearly predicts a gravity anisotropy.
http://members.optusnet.com.au/maxkeon/the1-1a.html


snip rant

You've snipped the wrong part.

It had no more scientific content.

Until you sort out your problems with calculus, there
isn't any point in discussing my results either because
neither of us accepts the other's method.

Maths can't make your result any less wrong because your basic
principles are wrong. Sort out _your_ problems George.

There are no problems in my maths Max,

The problem has never been with your maths, only with the way
you apply it.

It is the only method permitted by the rules of
mathematics Max, you need to get your schoolbooks
out again and do some revision.

get
yourself a book on calculus and find out what
the definition of acceleration is.

We will never agree to disagree either because YOU ARE WRONG.

Sorry Max, this is not something that is open to
debate. I am right about the maths, end of story.
You can confirm that by reading any high school
text books covering calculus and basic mechanics.

Your equation predicts eccentricity will decay to
zero in a two body situation and therafter the orbit
will be stable. With three or more bodies, it says
even the circular orbit will decay though more
slowly. For Mercury, it is still fast enough to crash
it into the Sun in a million years.

Your emotional outbusts may make you feel better

They are designed as a wake up call George.
But you're not yet ready to be unplugged.

Unplugged from the rules of maths? No I don't
think that would help.

George