
"Paul B. Andersen" <paul.b.andersen@xxxxxxxxxxxxxxxx> wrote in message
news:dvmi7g$25p$1@xxxxxxxxxxxxxxxxxxx
<snip detailed analysis>
> Note also that:
> m*(c+v)^2/r = m*c^2/r + 2*m*c*v/r + m*v^2/r
My backofenvelope went like this.
Henry can include an arbitrary constant
for the 'coefficient of friction' and
time in contact (for multiple discrete
reflections) so omitting m and r for the
moment:
(c+v)^2 = c^2 + 2*c*v + v^2
For v << c that simplifies to c(c+2v)
(though the validity of that might be
questionable when the total speed
approaches zero).
Including an arbitrary constant k for
the 'coefficient of friction' and
factors ommitted, if initially the
speed is c+v then
delta_v = k*(c+2v)
> Of course the centripetal forces are different.
> But not much.
> Let us assume that there is a friction slowing
> you down, and that the slowing during one revolution
> is proportional to the centripetal force, we can write:
> Slowing when going with the rotation:
> delta_vf = k*1000.20001 N
> Slowing when going in the opposite direction:
> delta_vb = k*999.80001 N
> Now we know that to explain the Sagnac in an
> IFOG, the difference between the two speeds
> must be in the order of 2v.
Paul, if the mean speed of a beam is c and
it starts at c+v it must obviously finish
at cv. Doesn't that mean the delta needs
to be 2v for each beam and 4v for the pair?
If so, the factor k above obviously needs to
be 1. The beam that starts at c+v finishes
at (c+v)(c+2v) = v while the other ends at
(cv)(c2v) = +v.
> thus, in our analogy:
> delta_vf  delta_vb = k*0.4 N = 2v = 0.002 m/s
> k = 0.005 m/Ns
> delta_vf = 5.001 m/s
> delta_vb = 4.999 m/s
>
> See?
> For the difference to be big enough, you would
> have to slow down to half the speed.
>
> In an IFOG with a much smaller v/c ratio
> it would be even worse.
>
> The very idea is idiotic beyond belief.
I thought he was pulling my leg at
first, it turned out he was serious.
>>>>>>I have shown that to be wrong.
>>>>>
>>>>>You have shown nothing whatsoever. Post the results
>>>>>of your experiment. If you do manage to prove it wrong
>>>>>then you have shown ballistic theory to be wrong.
>>>>
>>>>The light beams DO NOT experience the same 'centrifugal' forces in both
>>>>directions. Right or wrong, George?
>
> Wrong.
> The centrifugal forces do not depend on velocity
> (neither speed nor direction), and it is tiny.
> It is however correct that the centripetal forces
> are very slightly different.
>
> To explain the Sagnac, the light would have to slow
> down to a fraction of it original speed in an IFOG.
> It doesn't.
>
> Face it, Henri.
> The Sagnac falsifies the ballistic theory.
> No way out.
I did warn him several times but he never
even attempted the calculation. There's a
big difference between his handwaving and
a scientific explanation.
George

