"Henri Wilson" HW@.... wrote in message
...
On Sun, 18 Mar 2007 12:55:45 -0000, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
. ..
....
OK, I should have also said "very disparate masses". There is
an upper limit of a mass ratio of ~24:1 for the Lagrange point
stability.

http://en.wikipedia.org/wiki/Lagrangian_point#Stability

yes yes, theories theories.

Not this time Henry, pure maths and not open to argument.

Three body problems are not easily solved generally....let alone four of
five
body problem...

But special cases can be solved and the Lagrange configuration
is one of them. If the system is stable then the bodies have
a fixed relationship. In general three body systems are chaotic.

Yes I realise that.
That's why I used the 60 degree lag.

Yes, you got the angle rght but there is also
an upper mass limit that you missed.

I am not worrying about speed unification at the moment.

Don't forget you still need it to avoid the problem of
multiple images. I would explain more but it will be hard
to find words you can follow (no insult intended, it's
just tricky to describe unless you already know the answer).

I'm not worrying about it because I don't need it and cannot see any way
to
include it. That doesn't mean I think it doesn't happen. It certainly
does...but not on anything like the scale that I previously required to
explain
the distance discrepancy.

There is no discrepancy in the distance, the extinction
distance is the key and the observer distance can be very
much larger. Basically once the speeds have equalised it
doesn't matter how much farther away we are, the patterns
stay the same as the light travels. The key though is that
the extinction distance has to be less than that at which
the multiple images first appear.

I did this some time ago, the vertical scale of the waveform
is voltage, horizontal scale is distance, no extinction.

http://www.georgedishman.f2s.com/Henri/RitzSine.html

The ADoppler is obvious but the VDoppler is harder to see.
Just after launch the wavelength is constant but the speed
varies hence frequency varies. I'm going to try to add a
sample locations and have a graph showing the frequency
measured at that point as a function of time but it isn't
easy to see how to illustrate the relation to orbital phase.

I can't see any 'VDoppler' effect.

I told you it would be hard to explain :-(

Normally we are accustomed to the speed being c and Doppler
shift producing a change of wavelength. In ballistic theory
the wavelength can be constant and Doppler can result from
a change in speed. You have to watch very carefully to see
that the speed at the 90 and 270 emissions is different.

Your program does exactly what mine does.
It shows that maximum compression occurs in pulses emitted at 270 (the
furthest
point). The TRUE maximum velocity occurs at zero phase.

? Have you got that right? your diagram showed ADoppler
compression being a maximum at zero phase and the maximum
velocity at 90 degrees. I agree with your diagram if it is
wavelength rather than frequency.

http://www.users.bigpond.com/hewn/bunching.jpg

I gather you are refering to the fact that the phase of the compression
maximum
does change slightly as the distance increases. It asymptotes towards 90
wrt
the true velocity maximum.

No, what I am referring to is that the maximum _frequency_
does not correspond to the minimum _wavelength_ because the
speed is also changing. Your static diagram cannot show the
variation in speed.

You can see now why astronomy has been completely wrong for a century. All
'doppler determined' velocity curves are likely to be about 90 degrees
out.

Nope, the phase of the Shapiro delay tells me it is exactly
as the conventional theory predicts.

This version shows "pulses". I fact I cheated, the waveform
is the 11th power of a sine wave but it looks like a pulse
and means the shape changes correctly:

http://www.georgedishman.f2s.com/Henri/RitzPulse.html

that's OK.
Do you now agree with what I said? If pulse arrival rate is used in
conventional doppler, then the calculated velocity curve will be 90 out
wrt the
true one.

No, the problem is your diagram illustrates wavelength
(I guess), not arrival rate. The arrival rate would be
most compressed at 90 degrees on the top line and
move towards zero degrees as the distance increases. It
would be asymptotic to some intermediate value.

The chancess are its
effect is much less than I thought it was. Rather, my 'distance
discrepancies'
are largely due to orbit pitch.

Your model should include all the parameters, you may not
think them important now but in some future discussion
they may become important.

Pitch can be varied in my variables.exe program.

I don't think you realise the constraints Keplerian orbits
place on you Henry.

George, there are probably 10 billion stars in our galaxy, most with
companions
and orbiting planets.
Do you really think we know every possible configuration just by
investigating
our own solar system?

No, I think we can eliminate unstable configurations by
applying Newton's Laws (relativistic effects are small).

That would be nice..

For the Lagrange it has been done.

I read somewhere that a conglomerate of asteroids might possibly exists
around
a Lagrange point. ...maybe from an exploded star or planet.

Sure, look up "Trojans". However, asteroids are much less
massive than the planets. There is an upper mass limit for
the orbit to be stable.

However my program IS strictly limited to Keplerian orbits. ..

Not if you use such a high mass.


..but the dip can be explained with an object rotating in the same orbit
but
with 60 degree lag.

see S Cas in:
www.users.bigpond.com/hewn/group1.jpg

Mine is the yellow curve....a perfect fit...

Unfortunately, however, I cannot explain the claimed magnitude change of
about
9. In fact I don't believe it. According to the britastro website, there
is a
group of stars that appears to have very large changes in brightness.
How do
YOU explain those? I think somebody forgot to convert to a log scale.

That's unlikely but they are unlikely to be cepheids, the
nromal range for them goes up to about 2.0 IIRC. I would
need to s bit of research to find out what would cause
such a large range. From your point of view it is trivial,
the extinction distance is 99.9498% of the critical distance
compared to 72.6% for a mag 2.0 change. You might think that
having the extinction at 99.9498% of critical when we are
certain it never _exceeds_ critical (because we never see
multiple images) is a remarkable coincidence but that's
what the whole Cepheid variation idea relies on. I doubt
Sekerin even fully understood that. Oh and note that a tiny
change in inclination would put it over that limit, the radial
component of acceleration depends on inclination for a given
orbit while the extinction distance depends on the "quality
of space". That's why you suspected extinction depended on
the orbit, for 9 mag it has to be 99.9498% regardless of your
pitch factor.

But George, you are completely ignoring the fact that the calculated
radial
velocities are much higher than the real ones...due to the fact that
bunching
is used as a measure.

Nope, what I said above is independent of the velocity, the
distance ratio comes directly from the brightness ratio.

In the case of your pulsar, astronomers have used the maximum rate of
pulse
arrival as an indicator of maximum doppler shift. As you are now aware,
this is
way out in both magnitude and phase.

What mass ratio?

You can get an estimate from the relative sizes of the dip and main
curve. I
would say about 4:1 .

Then it is not possible. Try putting it into your orbital
simulation and if your software is accurate the system will
be unstable and become chaotic or possibly eject one of the
bodies leaving a tighter binary.

The velocities are the same.
The rest would be too speculative.

There's nothing speculative about it, just the same
inverse square law that you use for two-body Keplerian
orbits which you say your model already uses.

George