
On Mon, 20 Mar 2006 16:37:19 +0100, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxxx> wrote:
>Henri Wilson wrote:
>> On Fri, 17 Mar 2006 14:50:00 0000, "George Dishman"
>> <george@xxxxxxxxxxxxxxxxx>
>> wrote:
>>>>I have pointed out your mistake.
>>>>Walking around a carousel in opposite directions will soon tell you that
>>>>I'm
>>>>right.
>
>Right about what?
>Let's have a closer look at your carousel.
>
>You are obviously very confused about the role of
>the centrifugal 'force', so let us be concrete and
>calculate the forces.
>
>I would advice you not to make a fool of yourself
>and claim that my analysis is wrong.
>Think before writing, and don't repeat stupidities
>like:
>"It is an imaginary force, 'centrifugal', in the rotating frame.
> Its magnitude is the same as the centripetal force in
> the nonrotating frame."
It is not surprising that the definition of centrifugal force is rather vague.
Each author seems to have his own opinion.
What is YOUR definition of centrifugal force, Paul? I think your 'coriolis
force' is my centrifugal force, and my 'coriolis force' is something you appear
to know nothing about. Maybe you live to close to the N pole.
Twirl an object around by hand, on a string.
In the nonrotating frame, both the object and your hand rotate around the
barycentre. A centripetal force is required to accelerate both your hand and
the object towards the barycentre. These two CENTRIPETAL forces are opposite in
direction and show up as a tension in the string = mv^2/r or MV^2/R.
Centrifugal force does not stricty exist in this frame, although in most cases
R is so small that even the best scientists and engineers (including myself)
tend to call the object's 'pull' on the string a 'centrifugal force'. This is
also convetient in the case of a balanced wheel where the barycentre is also
the centre of rotation.
In the rotating frame, The object does not move but the same tension remains in
the string. How can this happen? It is due to the imaginary 'centrifugal
forces', of course.
Coriolis force goes like this.
In the above example, if the string is shortened, the object's rotation rate
will increase due to conservation of momentum.
In the rotating frame however, shortening the string is no big deal and should
do nothing except bring the object closer to the centre. However in practice,
an imaginary force pushes the object sideways during this process. That is the
imaginary CORIOLIS force. It explains why heavy air moving towards the earth's
poles forms a couple with light air moving towards the equator and we get the
familiar rotations around low and high pressure systems.
I hope the students of Norway will benefit from your 'lesson'.
>or:
>" 'centrifugal' often confused with 'centripetal'....
> and they have the same values anyway."
>
>(In the example below the centripetal force is
> ten million times greater than the centrifugal force.)
>
>To be a valid analogy, you will have to run at a vastly
>higher speed than the peripheral velocity of the carousel.
>A 1 m radius IFOG can detect the rotation of the Earth,
>that's a peripheral speed of 2.3*10^13 c.
>But let us be generous, let the peripheral speed of
>the carousel be as high as a 10^4 part of your speed.
>Let the radius of the carousel be r = 10m.
>Let the speed with which you run be c = 10 m/s.
>Let the peripheral velocity of the carousel be v = 0.001 m/s.
>Let your mass be m = 100kg.
>
>The centripetal forces the carousel exerts on
>your feet are now:
>Running with the rotation:
> The centripetal acceleration is (c+v)^2/r
> The force is Ff = m*(c+v)^2/r = 1000.20001 N
>Running in opposite direction:
> The centripetal acceleration is (cv)^2/r
> The force is Fb = m*(cv)^2/r = 999.80001 N
>
>Note that these are the actual centripetal forces
>acting on your feet. These forces could be measured
>and are obviously independent of which frame
>you use to calculate them in.
>
>So what about the 'centrifugal' force?
>Let us calculate the forces in the rotating frame.
>The centripetal acceleration is c^2/r and
>the component of the centripetal force causing
>it is thus Fca = m*c^2/r = 1000 N
>The centrifugal force is m*v^2/r = 0.00001N,
>same in both directions.
That's not how you calculate centrifugal force.. It has the same magnitude as
the centripetal foreces.
>Since the centrifugal
>force is acting outwards, there must be a component
>of the centripetal force counteracting this.
>Fcf = = 0.00001N
>So far, the two components of the centripetal
>force are equal for both directions and amounts
>to 1000.00001N.
>So what's wrong? Why are the forces equal?
>Because there is a third component of
>the centripetal force, namely the one counteracting
>the Coriolis pseudo force.
>This force is 2m*(w X c) where w is the angular
>velocity vector (spin vector) and c is the velocity
>(vector) of the object. In our case the absolute
>value of this force is m*v*c/r, and its direction
>is radially outwards when you are running with
>the rotation and radially inwards when you are
>running in the opposite direction.
>So the third component of the centripetal force
>counteracting the Coriolis force is:
>With the rotation: Fcof = m*v*c/r = 0.2 N
>Opposite direction: Fcob =  m*v*c/r =  0.2 N
>
>So the centripetal forces will be:
>With the rotation: Ff = Fca + Fcf + Fcof = 1000.20001
>Opposite direction: Fb = Fca + Fcf + Fcob = 999.80001
If c=v then Fb = zero.
>
>The centripetal forces are the forces exerted on your
>feet by the carousel. They are obviously the same
>whether you calculate them in the stationary or in
>the rotating frame.
>
>Note also that:
>m*(c+v)^2/r = m*c^2/r + 2*m*c*v/r + m*v^2/r
>
>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.
>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.
I don't like your method or your maths.
You should burn all the books in Norway, they are obviously wrong.
>In an IFOG with a much smaller v/c ratio
>it would be even worse.
>
>The very idea is idiotic beyond belief.
The fact is, I have proved my point. There are two separate effects. 1. The
beam's 'energy centre' is thrown slightly off the 'fibre centre', thus
increasing path length by different amounts for the two beams.
2) Each beam will experience a different amount of 'wall drag' due to slightly
different amounts of centrifugal force pushing them towards the outside of the
fibre.
>
>>>>
>>>>
>>>>>>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.
That proves my point and I win the argument.
I never claimed that htis was the major cause of the sagnac effect.
The main reason it occurs is basically due to the fact that photons have AXES
that don't like turning corners. The interfere strangely with photons whose
axes point in different directions.
>
>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 suppose it proves LET too, eh Paul?...because that's what YOU use to explain
the effect.
>
>Paul
HW.
www.users.bigpond.com/hewn/index.htm

