Why are the 'Fixed Stars' so FIXED?

On 29 Mar 2007 00:09:21 -0700, "George Dishman"
wrote:

On 29 Mar, 01:32, HW@....(Henri Wilson) wrote:
On 28 Mar 2007 06:27:41 -0700, "George Dishman" wrote:
On 28 Mar, 11:40, HW@....(Henri Wilson) wrote:
On 28 Mar 2007 02:16:59 -0700, "George Dishman" wrote:
On 28 Mar, 08:10, HW@....(Henri Wilson) wrote:

The diagram would be like this:

g h --- O

+
B

The pulsar sends one pulse from g and the next from h,
it is orbiting round the barycentre B and the observer
is at O. Obviously there is a v*cos(theta) term for
other parts of the orbit, it is the distance change
in the direction of the line of sight that matters.

I have
incorporated that by adding an Rsin(x) term to the star distance. It is
generally negligible.

It will certainly be small but it is not negligible, it
will produce a 45 degree phase shift when the ADoppler
is about 93 parts per million too and in fact we know
that the VDoppler is probably larger than the ADoppler
_except_that_ the phase can be changed by the effect you
describe at the top of the post regarding an elliptical
orbit looking circular.

I think I had it right before.
The distance for 45 deg phase difference is about 0.0007 LY.
It is independent of velocity.

OK, that is the sort of value I would expect. Now
the general gist of my argument is this: you get
a 45 degree phase shift at 0.0007 LY so you would
expect to get of the order of 5 degrees at a 1/10th
of that distance where the ADoppler only adds a
small fraction to the VDoppler.

You made the point that an elliptical orbit could
look circular provided the periastron was on the
line of sight because the distortion of the sine
wave from the variable speed is cancelled by the
distortion caused by the c+v effect.

A slight change in your yaw factor could then
change the relative phase of those factors to
give a net phase change of a few degrees. That
could cancel the phase shift due to ADoppler
and again make the orbit look circular.

The distortion of the brightness curve for circular orbits looks quite
symmetrical. I tried varying the yaw angle very slightly but it skewed the
curve away from a sine wave.
I think the major axis has to be aligned witrh teh LOS. However, we don;t know
how acccurate the published curves are....so you are probably right.

The bottom line then is that knowing we see what
looks like a circular orbit (or at least very low
eccentricity) there is a relationship between the
extinction distance, the true eccentricity and the
yaw.

Well I can telll you one thing. The extinction distance is directly
proportional to period.
The 0.0007 value is for a period of 0.0042 years.
It becomes 0.007 for 0.042 years, 0.07 for 0.042 years..etc.
....always independent of peripheral velocity.

How can you explain THAT?


From your other reply:

That is what I was alluding to a couple of weeks ago. For
small values you can probably get a match by eye but the
equation for an ellipse and those for Kepler's Laws are
quite different from the effect of ballistic theory. It
would be a curious though unimportant coincidence if they
exactly matched. Just as Ptolemy was able to get a good but
imperfect match with combined circles, I think if you did
the analytical investigation, you would find there was a
small difference but perhaps third or fourth order. That
is what would show up as the shape of a pattern in your
residuals.

I think it is quite likely that there is an exact match. It isn't unreasonable.

Given the form of the equations, I disagree but if
you do the calculation, you might prove me wrong.

The c+v factor might effectively 'squash' the ellipse back into a circle. That
seems plausible.

For different eccentricitiers, the curve becomes a sinewave at diffferent
distances (for he same maximum velocity)

Possibly, but I think the ADoppler distortion continues
to increase with distance and eventually causes multiple
images while the Keplerian distortion will be asymptotic
to some curve as the yaw approaches 90 degrees. The
question is how much the cancellation degrades as higher
order terms become more important. Your simulation is
the easiest way to investigate that.

Yes..and I agree there has to be a degree of extinction...even though the
distance anomaly can now be explained by the fact that the calculated velocity
curves might be grossly exaggerated.

The end result should be an upper limit on the speed
equalisation distance based on the uncertainty in the
orbital phase and the eccentricity.

I will upload my program to the website George so you can fiddle with it.
It is by no measns complete but is OK for circular orbits.

Click the red button after selecting eccentricity then click either the yellow
one (for my original method) or 'george' for your quick method.
George has the VDoppler correction included...
Increase 'output size' to see the curve at short distances.
If you hold the mouse button down, a vertical line appears on the screen to
compare phases.

http://www.users.bigpond.com/hewn/newvariables.exe

George



"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

Henri Wilson wrote:

The extinction distance is directly proportional to period.
The 0.0007 value is for a period of 0.0042 years.
It becomes 0.007 for 0.042 years, 0.07 for 0.42 years..etc.
...always independent of peripheral velocity.

How can you explain THAT?

As I said 19 and 20 March, the light speed unification
distance is inversely proportional to the rate of pulse
bunching. The more rapidly the pulses bunch, the shorter
the unification distance. What you have found is the
obvious fact that the rate of pulse bunching is inversely
proportional to the period. All else being equal, the
shorter the period, the more rapidly the pulses bunch.
So naturally, the shorter the period, the shorter the
unification distance.

Leonard

On 29 Mar 2007 17:24:58 -0700, "Leonard Kellogg" wrote:


Henri Wilson wrote:

The extinction distance is directly proportional to period.
The 0.0007 value is for a period of 0.0042 years.
It becomes 0.007 for 0.042 years, 0.07 for 0.42 years..etc.
...always independent of peripheral velocity.

How can you explain THAT?

As I said 19 and 20 March, the light speed unification
distance is inversely proportional to the rate of pulse
bunching. The more rapidly the pulses bunch, the shorter
the unification distance. What you have found is the
obvious fact that the rate of pulse bunching is inversely
proportional to the period. All else being equal, the
shorter the period, the more rapidly the pulses bunch.
So naturally, the shorter the period, the shorter the
unification distance.

That's an interesting idea....I'll think about it........but unification - or
classical extinction - should depend only on the properties of the space
through which the light travels...should it not?
Obviously however the speed of a pulse cannot be unified with that of another
that hasn't even been emitted.
I'm somewhat mystified by this.
I don't think unification takes place as rapidly as I originally believed. I no
longer need it to explain why my distances had to always be much shorter than
the Hipparcos ones.






Leonard


"When a true genius appears in the world, you may know
him by this sign, that the dunces are all in confederacy against him."
--Jonathan Swift.

On 30 Mar, 07:07, HW@....(Henri Wilson) wrote:
On 29 Mar 2007 17:24:58 -0700, "Leonard Kellogg" wrote:







Henri Wilson wrote:

The extinction distance is directly proportional to period.
The 0.0007 value is for a period of 0.0042 years.
It becomes 0.007 for 0.042 years, 0.07 for 0.42 years..etc.
...always independent of peripheral velocity.

How can you explain THAT?

As I said 19 and 20 March, the light speed unification
distance is inversely proportional to the rate of pulse
bunching. The more rapidly the pulses bunch, the shorter
the unification distance. What you have found is the
obvious fact that the rate of pulse bunching is inversely
proportional to the period. All else being equal, the
shorter the period, the more rapidly the pulses bunch.
So naturally, the shorter the period, the shorter the
unification distance.

That's an interesting idea....I'll think about it......

I don't think you quite followed what Leonard was saying,
or at least what i think he was saying. This goes back
to the little applet I wrote for you a couple of weeks
ago. Did you never wonder how I was able to do that ?

..but unification - or
classical extinction - should depend only on the properties of the space
through which the light travels...should it not?

Yes it should.

Obviously however the speed of a pulse cannot be unified with that of another
that hasn't even been emitted.
I'm somewhat mystified by this.

You need to step back a little and look at the problem
a different way. The VDoppler as you said produces a
relatively small brightening effect so for high values
we can assume ADoppler is dominant. The equation for
ADoppler without speed equalisation is is 1/(c^2-da)
where d is the distance from the source to the observer
and a is the instantaneous acceleration towards the
observer at the time of emission. The value c^2/a is
then the "critical distance". Obviously that depends on
the acceleration which in turn depends on the period.
Note also though that the component of the acceleration
towards the observer also depends on the pitch.

What that means is that for a high brightness, the
speed equalisation distance has to be an exact fraction
of the "critical distance" which means the properties
of the space the light passes through depend on the
inclination of the orbit.

Basically you have to invent this "speed equalistion"
factor and set it to an orbit dependent value to avoid
de Sitter's argument. You can set a low value but then
you get no brightening and Doppler effects are no
different to conventional values, but to get any of
the effects you have been claiming over the years, you
have to have the "properties of space" being entirely
dependent on the source acceleration and the inclination
of the orbit.

Inclination is particularly telling. It means if we see
a star with high variability, the speed equalisation
distance must be very close to the critical distance,
and that means another observer looking at the same
star form an inclination a few degrees less would see
multiple images. However there is nothing special about
us so we should see some stars showing multiple images
if this model was correct. As you know, we don't.

The solution is that speed equalisation must happen
over a relatively short distance and there aren't any
significant brightening or ADoppler effects.

I don't think unification takes place as rapidly as I originally believed. I no
longer need it to explain why my distances had to always be much shorter than
the Hipparcos ones.

As you can see, the requirement is actually that it
is a lot shorter than you thought.

With a very rough estimate based on your figure of
0.0007 light years for 45 degrees and a phase
uncertainty based on the time spread of 74ns on
a PRF of 2.295ms, I get a speed equalistion distance
of 54 light seconds. That should be typical of the
"property of space" for all stars.

George

On 30 Mar 2007 03:25:40 -0700, "George Dishman"
wrote:

On 30 Mar, 07:07, HW@....(Henri Wilson) wrote:
On 29 Mar 2007 17:24:58 -0700, "Leonard Kellogg" wrote:


Henri Wilson wrote:

The extinction distance is directly proportional to period.
The 0.0007 value is for a period of 0.0042 years.
It becomes 0.007 for 0.042 years, 0.07 for 0.42 years..etc.
...always independent of peripheral velocity.

How can you explain THAT?

As I said 19 and 20 March, the light speed unification
distance is inversely proportional to the rate of pulse
bunching. The more rapidly the pulses bunch, the shorter
the unification distance. What you have found is the
obvious fact that the rate of pulse bunching is inversely
proportional to the period. All else being equal, the
shorter the period, the more rapidly the pulses bunch.
So naturally, the shorter the period, the shorter the
unification distance.

That's an interesting idea....I'll think about it......

I don't think you quite followed what Leonard was saying,
or at least what i think he was saying. This goes back
to the little applet I wrote for you a couple of weeks
ago. Did you never wonder how I was able to do that ?

..but unification - or
classical extinction - should depend only on the properties of the space
through which the light travels...should it not?

Yes it should.

Obviously however the speed of a pulse cannot be unified with that of another
that hasn't even been emitted.
I'm somewhat mystified by this.

You need to step back a little and look at the problem
a different way. The VDoppler as you said produces a
relatively small brightening effect so for high values
we can assume ADoppler is dominant. The equation for
ADoppler without speed equalisation is is 1/(c^2-da)
where d is the distance from the source to the observer
and a is the instantaneous acceleration towards the
observer at the time of emission.

I get c^2/(c^2-da) ....no worries...

The value c^2/a is
then the "critical distance". Obviously that depends on
the acceleration which in turn depends on the period.
Note also though that the component of the acceleration
towards the observer also depends on the pitch.

Hold a circle in front of you at any angle. (or an ellipse)
Rotate you head until you find an axis in the plane of the circle that
horizontal to the line between your eyes and is also perpendicular to the LOS.
(one always exists)
ALL the radial velocities and the accelerations around the orbit are then
multiplied by the same factor, cos(pitch), where the pitch angle refers to the
rotation around the above axis.

What that means is that for a high brightness, the
speed equalisation distance has to be an exact fraction
of the "critical distance" which means the properties
of the space the light passes through depend on the
inclination of the orbit.

That's OK. Cos(Pitch) is included in the velocity figure.

Basically you have to invent this "speed equalistion"
factor and set it to an orbit dependent value to avoid
de Sitter's argument. You can set a low value but then
you get no brightening and Doppler effects are no
different to conventional values, but to get any of
the effects you have been claiming over the years, you
have to have the "properties of space" being entirely
dependent on the source acceleration and the inclination
of the orbit.

George, frankly I cannot see where you got the idea that the ratio of VDoppler
to ADoppler is in any way connected to the 'extinction distance'.
The '45 degree' point is just a result of the minute difference in travel time
due to the distace being modified by Rsin(xt). It is just a second order
trigonometrical fact, quite negligible at normal star distances.

Extinction is a property of the space through which the light has to travel.

Inclination is particularly telling. It means if we see
a star with high variability, the speed equalisation
distance must be very close to the critical distance,
and that means another observer looking at the same
star form an inclination a few degrees less would see
multiple images. However there is nothing special about
us so we should see some stars showing multiple images
if this model was correct. As you know, we don't.

Sorry George, I think you have gone off the rails here.

The solution is that speed equalisation must happen
over a relatively short distance and there aren't any
significant brightening or ADoppler effects.

It doesn't have to happen over a very short distance at all.
You are using the wrong values for your radial velocities. In reality they are
much lower.
I suspect that DeSitter based his calculations on similarly wrong radial
velocities. I'll look it up.

I don't think unification takes place as rapidly as I originally believed. I no
longer need it to explain why my distances had to always be much shorter than
the Hipparcos ones.

As you can see, the requirement is actually that it
is a lot shorter than you thought.

With a very rough estimate based on your figure of
0.0007 light years for 45 degrees and a phase
uncertainty based on the time spread of 74ns on
a PRF of 2.295ms, I get a speed equalistion distance
of 54 light seconds. That should be typical of the
"property of space" for all stars.

I don't know what you are talking about....and I don't think you do either
George.
Your pulsar's true radial velocity (orbit speed x cos(pitch)) is only a few
metres per second.


George


Einstein's Relativity - the greatest HOAX since jesus christ's mother.

Henri Wilson wrote:

[grammatical errors corrected to improve readability]

Hold a circle (or an ellipse) in front of you at any angle.
Rotate your head until you find an axis in the plane of the
circle that is horizontal to the line between your eyes,
and is also perpendicular to the LOS. (one always exists)
ALL the radial velocities and the accelerations around the
orbit are then multiplied by the same factor, cos(pitch),
where the pitch angle refers to the rotation around the
above axis.

Rotating one's head is irrelevant. The rotation that you
describe (A "roll" of either the head or the projected
ellipse) simply puts the long axis of the projected ellipse
on the viewer's X axis. That is convienient but has no
effect on the process of multiplying radial velocities and
accelerations around the orbit by a factor of cos(pitch).

You said this previously and I do not understand why George
did not point out its irrelevancy at that time.

Do I understand your terminology correctly as saying that
the "pitch" of an orbit is zero when seen edge-on and 90
degrees when seen face-on?

If so, your term "pitch" means the same as "inclination",
which is the term everyone else uses in astronomy. Though
it is often measured as angular deviation from face-on
rather than from edge-on. That is how it is used in arXiv
astro-ph/0507420.pdf (Table 1, "Orbital inclination, i")

To double-check that we are talking about the same thing,
see the illustration of "yaw", "pitch", and "roll" near the
top of this page:

http://mtp.jpl.nasa.gov/notes/pointing/pointing.html

Leonard

"Leonard Kellogg" wrote in message oups.com...

http://mtp.jpl.nasa.gov/notes/pointing/pointing.html

Leonard

"Positive roll is right wing down, positive pitch is nose up, and positive yaw is east when heading north."


Positive roll is right wing down = clockwise seen from tail.
Positive pitch is nose up = clockwise seen from port wing.
Positive yaw is east when heading north = clockwise seen from above.

Mathematical angle is positive counterclockwise so you'll never
be sure you are talking about the same thing.
http://www.androcles01.pwp.blueyonde.../Androcube.gif

This has all been explained to Wilson before, his standard response
is "No", which he learns from Bielawski, Draper, Dishman and Poe,
rendering him ineducable.

Bielawski understands Poles are the butt of American jokes but
does not know how far it is from A to A.

"The answer was zero." - Androcles
"No, the answer is not zero.
Distance travelled by photon from A to A is not A-A. End of story." --Bielawski.
End of story.

On 30 Mar 2007 23:12:44 -0700, "Leonard Kellogg" wrote:


Henri Wilson wrote:

[grammatical errors corrected to improve readability]

Hold a circle (or an ellipse) in front of you at any angle.
Rotate your head until you find an axis in the plane of the
circle that is horizontal to the line between your eyes,
and is also perpendicular to the LOS. (one always exists)
ALL the radial velocities and the accelerations around the
orbit are then multiplied by the same factor, cos(pitch),
where the pitch angle refers to the rotation around the
above axis.

Rotating one's head is irrelevant. The rotation that you
describe (A "roll" of either the head or the projected
ellipse) simply puts the long axis of the projected ellipse
on the viewer's X axis. That is convienient but has no
effect on the process of multiplying radial velocities and
accelerations around the orbit by a factor of cos(pitch).

You said this previously and I do not understand why George
did not point out its irrelevancy at that time.

Do I understand your terminology correctly as saying that
the "pitch" of an orbit is zero when seen edge-on and 90
degrees when seen face-on?

Yes...but the rotation is about an axis in the edge-on position....that axis
lying perpendicular to the LOS and in the plane of the orbit.

It is ALWAYS POSSIBLE TO FIND SUCH AN AXIS, no matter what the orbit
configuration wrt Earth.
..

If so, your term "pitch" means the same as "inclination",
which is the term everyone else uses in astronomy. Though
it is often measured as angular deviation from face-on
rather than from edge-on. That is how it is used in arXiv
astro-ph/0507420.pdf (Table 1, "Orbital inclination, i")

To double-check that we are talking about the same thing,
see the illustration of "yaw", "pitch", and "roll" near the
top of this page:

http://mtp.jpl.nasa.gov/notes/pointing/pointing.html

Leonard

I have tried to explain before that I have redefined pitch and yaw to make the
programming of this stuff possible. My method is 100% correct and effective.
For the purpose of brightness variation and measurement, one angle can be
eliminated by simply 'rotating the horizontal', ie., one's head.

Every orbit, eliptical or circular can be described in this way. ...an edge on
orbit multiplied by cos(pitch)...or 'inclination' as you call it.

To verify what I am saying, I suggest you make a paper cutout of an ellipse,
stick it at some odd angle onto the end of a rod and hold it up in front of
you. If you rotate the rod (representing the LOS) you will see that at one
particular angle there will be an axis in the orbit plane that lies
perpendicular to the LOS and parallel to the line between your eyes (the new
horizontal).

In that position, the orbit can be rotated around THAT AXIS through an angle
(my 'pitch') into an edge on position. I define YAW as the angle between the
major axis of the ellipse and the LOS when the orbit is in that edge on
position. My 'zero yaw angle' is also defined differently ...for programming
reasons.

Thus, both acceleration and velocity can be simply multiplied by cos(pitch) to
reduce their component in the direction of the observer. The effect is to
simply reduce the height of my predicted brightness curves but not their
**shapes**, which are determined solely by eccentricity and yaw angle.

Note: It is not possible to resolve the pitch angle from a point source of
light and I know of no method that can determine the pitch component involved
in a measured velocity. So my radial velocity figures automatically represent
(orbital velocity x cos(pitch).


Einstein's Relativity - the greatest HOAX since jesus christ's mother.

"Leonard Kellogg" wrote in message
oups.com...

Henri Wilson wrote:

[grammatical errors corrected to improve readability]

Hold a circle (or an ellipse) in front of you at any angle.
Rotate your head until you find an axis in the plane of the
circle that is horizontal to the line between your eyes,
and is also perpendicular to the LOS. (one always exists)
ALL the radial velocities and the accelerations around the
orbit are then multiplied by the same factor, cos(pitch),
where the pitch angle refers to the rotation around the
above axis.

Rotating one's head is irrelevant. The rotation that you
describe (A "roll" of either the head or the projected
ellipse) simply puts the long axis of the projected ellipse
on the viewer's X axis. That is convienient but has no
effect on the process of multiplying radial velocities and
accelerations around the orbit by a factor of cos(pitch).

You said this previously and I do not understand why George
did not point out its irrelevancy at that time.

Do I understand your terminology correctly as saying that
the "pitch" of an orbit is zero when seen edge-on and 90
degrees when seen face-on?

If so, your term "pitch" means the same as "inclination",
which is the term everyone else uses in astronomy. Though
it is often measured as angular deviation from face-on
rather than from edge-on. That is how it is used in arXiv
astro-ph/0507420.pdf (Table 1, "Orbital inclination, i")

To double-check that we are talking about the same thing,
see the illustration of "yaw", "pitch", and "roll" near the
top of this page:

Leonard, I think Henry has just swapped some definitions
for convenience. His cos(pitch) is the same as the usual
sin(inclination). I'm less clear about his yaw but I'm
fairly sure it is directly related to the longitude of
the ascending node.

George

I think I missed this one, much of it has been
covered elsewhere.

"Henri Wilson" HW@.... wrote in message
...
On 30 Mar 2007 03:25:40 -0700, "George Dishman"
wrote:
On 30 Mar, 07:07, HW@....(Henri Wilson) wrote:
On 29 Mar 2007 17:24:58 -0700, "Leonard Kellogg"
wrote:
Henri Wilson wrote:
....
Obviously however the speed of a pulse cannot be unified with that of
another
that hasn't even been emitted.
I'm somewhat mystified by this.

You need to step back a little and look at the problem
a different way. The VDoppler as you said produces a
relatively small brightening effect so for high values
we can assume ADoppler is dominant. The equation for
ADoppler without speed equalisation is is 1/(c^2-da)
where d is the distance from the source to the observer
and a is the instantaneous acceleration towards the
observer at the time of emission.

I get c^2/(c^2-da) ....no worries...

True, the full TDoppler combines that with the VDoppler
formula of (1+v/c) so the whole is c(c+v)/(c^2-da) and
I tend to think of (c^2-da) as the ADoppler term within
that. The infinity occurs when da=c^2 either way.

The value c^2/a is
then the "critical distance". Obviously that depends on
the acceleration which in turn depends on the period.
Note also though that the component of the acceleration
towards the observer also depends on the pitch.

Hold a circle in front of you at any angle. (or an ellipse)
Rotate you head until you find an axis in the plane of the circle that
horizontal to the line between your eyes and is also perpendicular to the
LOS.
(one always exists)
ALL the radial velocities and the accelerations around the orbit are then
multiplied by the same factor, cos(pitch), where the pitch angle refers to
the
rotation around the above axis.

Or sin(inclination) in standard terminology, yes.

What that means is that for a high brightness, the
speed equalisation distance has to be an exact fraction
of the "critical distance" which means the properties
of the space the light passes through depend on the
inclination of the orbit.

That's OK. Cos(Pitch) is included in the velocity figure.

That misses the point. For a given luminosity variation,
you have to be effectively "seeing" the source from a
distance which is a precise fraction of c^2/a. The
"seeing" distance is the speed equalisation distance
for much larger real distances hence the connection.

You said above "I'm somewhat mystified by this." and if
you step back and think about what I said you should
understand the link.

Basically you have to invent this "speed equalistion"
factor and set it to an orbit dependent value to avoid
de Sitter's argument. You can set a low value but then
you get no brightening and Doppler effects are no
different to conventional values, but to get any of
the effects you have been claiming over the years, you
have to have the "properties of space" being entirely
dependent on the source acceleration and the inclination
of the orbit.

George, frankly I cannot see where you got the idea that the ratio of
VDoppler
to ADoppler is in any way connected to the 'extinction distance'.

We covered that elsewhere, the known relationship at
that phase (ADoppler = VDoppler) allows you to set an
upper limit on the distance, it is a measurement
technique, not a physical relationship.

The '45 degree' point is just a result of the minute difference in travel
time
due to the distace being modified by Rsin(xt). It is just a second order
trigonometrical fact, quite negligible at normal star distances.

Extinction is a property of the space through which the light has to
travel.

Yes, but we can measure one and not the other. Using
the phase tells us where the effects are equal.

Inclination is particularly telling. It means if we see
a star with high variability, the speed equalisation
distance must be very close to the critical distance,
and that means another observer looking at the same
star form an inclination a few degrees less would see
multiple images. However there is nothing special about
us so we should see some stars showing multiple images
if this model was correct. As you know, we don't.

Sorry George, I think you have gone off the rails here.

It's a more complex point, you probably need to learn
more about ballistic theory before you follow it.

The solution is that speed equalisation must happen
over a relatively short distance and there aren't any
significant brightening or ADoppler effects.

It doesn't have to happen over a very short distance at all.

It does to get the phase right.

You are using the wrong values for your radial velocities. In reality they
are
much lower.
I suspect that DeSitter based his calculations on similarly wrong radial
velocities. I'll look it up.

Over all known binaries, there should be a statistical
spread of inclination (pitch) angles and the orbital
speeds are constrained by Kepler so probably he used
typical values rather than specifics.

I don't think unification takes place as rapidly as I originally
believed. I no
longer need it to explain why my distances had to always be much shorter
than
the Hipparcos ones.

As you can see, the requirement is actually that it
is a lot shorter than you thought.

With a very rough estimate based on your figure of
0.0007 light years for 45 degrees and a phase
uncertainty based on the time spread of 74ns on
a PRF of 2.295ms, I get a speed equalistion distance
of 54 light seconds. That should be typical of the
"property of space" for all stars.

I don't know what you are talking about..

Well punch the numbers into your program and see what
it tells you. I'm working these out mostly using mental
arithmetic with the occassional calculator number so
there's a big risk but they should be in the ball park.

..and I don't think you do either
George.
Your pulsar's true radial velocity (orbit speed x cos(pitch)) is only a
few
metres per second.

0.0013 m/s you said before IIRC. Sorry Henry, that's not
possible. The source would need to be a supermassive
black hole and nearby stars would have their velocities
grossly changed. The whole galaxy would be reshaped in
fact.

The simpler interpretation is that it is nearly edge on
and the speed equalisation distance is much smaller than
you though, in fact in line with the numbers from the
page we discussed before

http://www.datasync.com/~rsf1/binarie4.htm

where the author gets 0.0045 light years. Of course that
also removes any problems with understanding the Shapiro
delay and apparent eclipsing behaviour of various pulsars.

George