On 16 Mar 2007 03:19:00 -0700, "George Dishman"
wrote:

On 16 Mar, 09:32, HW@....(Henri Wilson) wrote:
On 16 Mar 2007 00:53:29 -0700, "George Dishman" wrote:
On 16 Mar, 02:33, HW@....(Henri Wilson) wrote:
On Fri, 16 Mar 2007 00:18:00 -0000, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message
.. .
It SHOULD include it but if as you say above your program
does not take this into account, that could explain why it
is missing the VDoppler effect.

I'm working on it. I'll eventually find what's happening.
...
It will be done....but it isn't as simple as one would think.

I can see that your approach might make it tricky. Let
me know when you crack it.

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

This shows how the phase of the TRUE velocity maximum changes with distance
when compared with the maximum bunching. The latter (maximum pulse arrival
rate) is wrongly assumed to indicate the maximum doppler shift.

It is hard to tell but it looks as though the
pulses in the top line are regularly spaced.

That line is the layout immediately after emission (one orbit) All the pulses
are (almost) equally spaced.

If
so it is wrong, they should be bunched closest
at the 90 degree mark because each one travels
less distance to the observer than the previous
pulse (speed is constant at c+v) and most widely
spaced at the 270 degree mark where each travels
farther (at c-v).

George, each line represents a distance further from the source. The diagram is
not wrong.

The maximum bunching occurs at the 0/360 mark. (furthest from observer).
Widest spacing is at the 180 mark. That's how ADoppler works.

The bottom line looks right in
that the acceleration dominates and the maximum
is at the 0 degree mark (which is also where the
Shapiro delay peaks).

All the lines are correct. They show the layout at different distances,
increasing down the screen..

The program merely shows how each pulse moves after emission, given that its
velocity is c+vsin(x/T).


I think this answers your questions George.

No, my first question was to find the row where
the combination of the velocity and acceleration
effects gives a maximum bunching at 45 degrees
and you haven't shown that line.

The acceleration term dominates from the start.
The source velocity is so small that the pulses are virtually evenly spaced
after one orbit. Do you see that? That is represented by the top line....with
the first pulses at the RHS.

As distance increases, the pulses emitted at the 90 mark move towards the
leading ones, causing bunching there.




Specifically though you need that incorporated
into your program so that we can find the distance
at which the measured phase shift occurs and you
may then want to also consider elliptical orbits
which will change both velocity and acceleration
versus phase and yaw due to Kepler's laws. The
diagram has been helpful as it shows you are still
not taking the velocity effect into account and
hopefully will give you a steer on how to do that,
but it is only a step to fixing the error in your
program.

George you are totally confused.
The diagram is correct.

Astronomers unwittingly use the pulse arrival rate as a measure of doppler
shift.
They believe the higher the arrival rate, the faster the radial velocity
towards Earth.

As you can see, this is is completely wrong, both in magnitude and phase wrt
the brightness curve. The true maximum radial velocity occurs at the 90 degree
mark.

Once you do that, you need to be able to say what
extinction distance produces a phase shift of the
order of 10^-5 degrees so you need to be able to
get numbers out of it, but we can estimate it from
the 45 degree figure with a bit of simple trig.

George you don't seem to understand. I suggest you write your own program.
It is quite simple really...for circular orbits.

Set up an array of speeds around the orbit:

For K=0 to 360
lightspeed(K) = 2000 * (Sin(pi/180 * K) * vone)
next

(the starting point is as shown in my diagram.)

then:
using a timer, repeat the loop:

For j = 0 To 90 (gives 90 pulse)
If j = 22 Or j = 45 Or j = 67 Then n = 0 Else n = 255 (red line showing
90,180, 270 points)

Line ((140 * j) + (sec * lightspeed(360 - (4 * j))), 2000)-((140 * j) + (sec *
lightspeed(360 - (4 * j))), 2200), RGB(255, n, 0)
Next

sec= sec+1


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.