
On Mon, 20 Mar 2006 20:16:53 0000, "George Dishman" <[email protected]>
wrote:
>
>"Paul B. Andersen" <[email protected]> wrote in message
>news:[email protected]
>
><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.
Why don't you ask Paul to set up our light fibre 'coil experiment' as a PhD
exercise.
>
>George
>
HW.
www.users.bigpond.com/hewn/index.htm

