On 23 Sep, 22:09, HW@....(Henri Wilson) wrote:
On Sun, 23 Sep 2007 12:33:14 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in messagenews:kigbf317ahrqpdeknk5lrk0rsv0t7oc9f3@4a x.com...
On Fri, 21 Sep 2007 01:08:23 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message

restoring the subject from which Henry is trying
to run away

On Wed, 19 Sep 2007 21:36:10 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
m...
On Mon, 17 Sep 2007 19:23:54 +0100, "George Dishman" wrote:

Multiple images start at what I called the critical
distance given by d = c^2/a where a is the peak radial
acceleration. Variation of 1.5 mags means a ratio of
4:1 in luminosity which means the speed unification
distance is about 75% of the critical distance. Other
values are

Percentage of
Mag 1:n critical
1 2.5 60.19%
1.5 4.0 74.88%
2 6.3 84.15%
3 15.8 93.69%
4 39.8 97.49%
5 100.0 99.00%
6 251.2 99.60%
7 631.0 99.84%
8 1584.9 99.94%
9 3981.1 99.97%

Isn't it magical how we see variations of up to 9 mags
in some stars yet NEVER see multiple images.

It is true some stars are reported to vary by 7-9 mags. This cannot be
explained solely by c+v effects.

Of course it can Henry, that's the point. Suppose
we take an arbitrary figure of 10 light years
for the speed equalisation distance in the space
surrounding some star. To get 1.5 mag variation
you need a critical distance of 13.35 light years.
To get 9 mag that needs to be 10.0025 light years,
just a 25% increase in the peak orbital
acceleration.

What the hell are you talking about?

Check the numbers, see for yourself. For a given
value of peak acceleration, draw a graph of peak
variation as a function of distance.

What do you think my program does?

If you think it can tell you the above results,
use it. Tell me what distance as a percentage
of (c^2/a) you get a variation peak variation
of 8 magnitudes (peak-to-peak will be about
0.75 magnitudes more).

Tell me whether you would get multiple images if
the distance were 1% greater.

If that is at an inclination of 45 degrees, would
we see multiple images if it were 1 degree closer
to edge-on?

It never gets to that point.

Stop ducking the issue Henry, you claimed my numbers
were wrong so prove it, put up or shut up.

Your numbers are completely wrong...

Then use your program and post your alternatives.

You can use it. ...

Why, don't you know how? I think you are so hopeless
that even this trivial arithmetic is beyond you, but
you claimed my numbers were wrong so it is up to you
to back that up, I'm not going to do it for you. Oh
and if your program gives something other than those
values, you will still need to prove it is your code
that is right and not my numbers, IMO your code is
highly suspect. Bear in mind the numbers are peak,
not peak-to-peak which I think is what your program
usually produces as the summary.

George