Max Keon wrote:
"Jerry" wrote in message
ups.com...
Max Keon wrote:
-----
-----

My methods are unorthodox, that's why they succeed.

Did you ever fix the gross mechanical problems in your OWLS
anisotropy experiment? Or have you disassembled it, blindly
thinking that you've succeeded?
http://groups.google.com/group/sci.p...31cb900ae33838

Oh, but I have succeeded. Trying to convince the likes of you would
be an exercise in futility. But I can reassemble the device anytime
I like and, as is invariably the case, I will get exactly the same
result.

At the very least, MOUNT THAT RIDICULOUSLY THIN SLAB ON A
THREE POINT SUSPENSION to isolate it from bending stresses.

Jerry

Max Keon wrote:
"Jerry" wrote in message
oups.com...
Max Keon wrote:
This is the resultant graph. The black curve is from experiment, the
red curve is the calculated curve assuming that a gravity anisotropy
exists, while the green curve is the best fit for the calculated
curve which assumes that some mechanical flaw in the device is the
cause. The first character in the full character set is No.0 for
this experiment. http://www.optusnet.com.au/~maxkeon/no-24.jpg

SOMETHING IS CAUSING THE FREE DISC TO ROTATE AS IT DOES, AND THAT
SOMETHING MUST BE IDENTIFIED. IF IT'S NOT A GRAVITY ANISOTROPY, THEN
WHAT IS IT?

You have a rotor being driven by a slippery "clutch" in a 1 atm
environment. There are many questions concerning the nature of the
bearings that form the "clutch" and many questions concerning the
effects of atmospheric drag on the rotor. Even in the absence of
gravitational anomalies, why in the world should you expect the
rotor to rotate at the same rate as the housing?

You have slippage effects, atmospheric drag effects, and drive train
effects to account for. You CLAIM to have eliminated drive train
effects by mounting your bearings in such a manner that, in perfect
conditions, their effects should be equal and opposite. But they can
only cancel due to slippage in the very coupling that is being used
to drive the rotor.

Basically, your entire experiment is a Rube Goldberg setup.

This is the very first paragraph on
the web page describing the experiment:
If the action of gravity is not instantaneous, the forces applied
to the up and down moving sides of a disc rotating on an axis that's
parallel to the earth's surface will not be equal.

[snip]

WHY!?

Jerry wrote:
Max Keon wrote:
"Jerry" wrote in message
ups.com...
Max Keon wrote:
-----
-----

My methods are unorthodox, that's why they succeed.

Did you ever fix the gross mechanical problems in your OWLS
anisotropy experiment? Or have you disassembled it, blindly
thinking that you've succeeded?
http://groups.google.com/group/sci.p...31cb900ae33838

Oh, but I have succeeded. Trying to convince the likes of you would
be an exercise in futility. But I can reassemble the device anytime
I like and, as is invariably the case, I will get exactly the same
result.

At the very least, MOUNT THAT RIDICULOUSLY THIN SLAB ON A
THREE POINT SUSPENSION to isolate it from bending stresses.

Correction. Add the word "floating", i.e.
MOUNT THAT RIDICULOUSLY THIN SLAB ON A THREE POINT FLOATING SUSPENSION

Jerry

"Jerry" wrote in message
oups.com...
Max Keon wrote:
Do you have a theory which predicts that the free disc will fall
behind the rotating housing when the device is positioned as you
require? Or is it just a "Hey let's try it and see what happens"
thing? I was considering setting it up on my garage roof. Then
in my toilet.

Do you have any reason to believe that in the ABSENCE of
gravitational anomalies, that the free disk should rotate at the
same rate as the housing? You have slippage effects, drive train
effects, and air friction effects to account for.

What drive train effects? The rotating housing is rotating at
constant rate. The enclosed air and the free disc are all going
along with it. What forces do you think are in place to DRIVE the
free disc away from its housing and the surrounding air mass? I
know what that force is, but you don't seem to want to know.

You claim that
you have eliminated drive train effects by mounting the needle
bearings such that, being driven through the inner race on one
end and through the outer race on the other, the differential
effects should cancel each other. This is doubtful.

You use the term "driven" where there is no drive. The free disc,
the enclosed air mass and the housing, rotate as a unit.

Mount two needle bearings symmetrically about the rotor and
drive the rotor IN OPPOSITE DIRECTIONS via the inner race.
If compensation is perfect, then the rotor should exhibit ZERO
steady state rotation.

In contrast, I predict that the rotor, provided that it is well
balanced, will exhibit net steady state rotation that is a
significant fraction of the driving angular velocity. The
compensation of bearing anomalies will not be perfect, because
needle bearings are individual items, despite being mass
produced, and exhibit individual characteristics. (actually,
from your description, I am not certain that what you mean by
"needle bearings" corresponds with the standard use of the
term. No matter. The argument still holds.)

http://www.optusnet.com.au/~maxkeon/needle.jpg is a photograph
of the accessible bearing assembly between the free disc and the
rotating housing. The needle point bearing is exactly that, a
needle point in a point cavity on the flat spring bearing tensioner.
The other end of the free disc axle shaft is the point cavity, while
the housing carries the needle point. The free disc weighs 59 grams
and the load applied on the axle end by the flat spring is 108
grams. So you see, there is no clearance between the mating parts
anyway.

The rotor is being driven through two slippery "clutches", i.e.
the bearings, of differing characteristics. In your original
arrangement, by driving the rotor through the inner race of one
needle bearing, and through the outer race of the needle bearing,
you HOPE that drive train effects will exactly compensate. I say
that they will NEVER precisely compensate.

And it will not always give the same result, regardless of how the
bearings are assembled will it. But the resulting curve **is**
always the same! And you're using the term "driving" again. I don't
think you understand this experiment at all. Try reading it properly.

-----

Max Keon

"Eric Gisse" wrote in message
oups.com...
Max Keon wrote:
[snip]

Do not try to sidestep the issues by answering questions I did not ask
and by ignoring the ones I did.

1) You have no error bars and thus no idea if you are actually
measuring what you think you are measuring. You use "mabey", "could",
"possibly" and variations thereof many times which shows to me you
don't really know what is going on.

But the result is always as predicted.

2) You think you know how your device works, but I don't. Yes - I want
an actual diagram explaining how it works rather than have to piece
together from the disjoint notes. That is the price you pay for wanting
people to take you seriously.

Try reading it properly.
http://www.optusnet.com.au/~maxkeon/gravity.html

3) You still have no theoretical justification for your predicted
effect, nor do you have any for why the effect would exist in the first
place. You say it is due to "the theory", but you fail to say what "the
theory" is, much less even answer my question.

It's a fundamental prediction of the zero origin concept.
But I really don't think you should go there.

4) The acceleration due to gravity in Austrailia is not 9.8m/s^2. The
variance in your local g and 9.8m/s^2 is many orders of magnitude
larger than the effect you seek to measure. If I gussed right and you
are in Victoria, g is 980.1364 cm/s^2.

Very impressive figures.

-----

Max Keon

Max Keon wrote:
"Jerry" wrote in message
oups.com...

Do you have any reason to believe that in the ABSENCE of
gravitational anomalies, that the free disk should rotate at the
same rate as the housing? You have slippage effects, drive train
effects, and air friction effects to account for.

What drive train effects? The rotating housing is rotating at
constant rate. The enclosed air and the free disc are all going
along with it. What forces do you think are in place to DRIVE the
free disc away from its housing and the surrounding air mass? I
know what that force is, but you don't seem to want to know.

You claim that
you have eliminated drive train effects by mounting the needle
bearings such that, being driven through the inner race on one
end and through the outer race on the other, the differential
effects should cancel each other. This is doubtful.

You use the term "driven" where there is no drive. The free disc,
the enclosed air mass and the housing, rotate as a unit.

What do you mean, "there is no drive"? The rotor is frictionally
coupled to the housing through the "needle bearings".

The central bearing will tend to under-rotate the rotor relative
to the housing. The external bearing will tend to over-rotate the
rotor relative to the housing.

Since most of the weight is on the central bearing, the frictional
coupling between housing and central bearing will dominate, and
the under-rotation that you measure between rotor and housing is
probably due to this mismatch in coupling strengths.

As I stated before, you have a Rube Goldberg setup.

Jerry

Max Keon wrote:
"Eric Gisse" wrote in message
oups.com...
Max Keon wrote:
[snip]

Do not try to sidestep the issues by answering questions I did not ask
and by ignoring the ones I did.

1) You have no error bars and thus no idea if you are actually
measuring what you think you are measuring. You use "mabey", "could",
"possibly" and variations thereof many times which shows to me you
don't really know what is going on.

But the result is always as predicted.

Irrelevant because you can't explain how you predicted it and because
you use so much uncertain language. Hell, you aren't even using the
correct g.

How is a reader supposed to actually believe that you measured what you
expected to measure?


2) You think you know how your device works, but I don't. Yes - I want
an actual diagram explaining how it works rather than have to piece
together from the disjoint notes. That is the price you pay for wanting
people to take you seriously.

Try reading it properly.
http://www.optusnet.com.au/~maxkeon/gravity.html

I guess readability isn't a concern of yours.


3) You still have no theoretical justification for your predicted
effect, nor do you have any for why the effect would exist in the first
place. You say it is due to "the theory", but you fail to say what "the
theory" is, much less even answer my question.

It's a fundamental prediction of the zero origin concept.
But I really don't think you should go there.

Why do you stall when asked to explain the thought process behind your
prediction?


4) The acceleration due to gravity in Austrailia is not 9.8m/s^2. The
variance in your local g and 9.8m/s^2 is many orders of magnitude
larger than the effect you seek to measure. If I gussed right and you
are in Victoria, g is 980.1364 cm/s^2.

Very impressive figures.

So you don't even care that you are using the wrong g?


-----

Max Keon

"Eric Gisse" wrote in message
oups.com...
Max Keon wrote:
"Eric Gisse" wrote in message
oups.com...
Max Keon wrote:
[snip]

Do not try to sidestep the issues by answering questions I did not ask
and by ignoring the ones I did.

1) You have no error bars and thus no idea if you are actually
measuring what you think you are measuring. You use "mabey", "could",
"possibly" and variations thereof many times which shows to me you
don't really know what is going on.

But the result is always as predicted.

Irrelevant because you can't explain how you predicted it and because
you use so much uncertain language. Hell, you aren't even using the
correct g.

How is a reader supposed to actually believe that you measured what you
expected to measure?

2) You think you know how your device works, but I don't. Yes - I want
an actual diagram explaining how it works rather than have to piece
together from the disjoint notes. That is the price you pay for wanting
people to take you seriously.

Try reading it properly.
http://www.optusnet.com.au/~maxkeon/gravity.html

I guess readability isn't a concern of yours.

Well, some degree of intelligence is required.

3) You still have no theoretical justification for your predicted
effect, nor do you have any for why the effect would exist in the first
place. You say it is due to "the theory", but you fail to say what "the
theory" is, much less even answer my question.

It's a fundamental prediction of the zero origin concept.
But I really don't think you should go there.

Why do you stall when asked to explain the thought process behind your
prediction?

That's no stall. From what I see of you here, it would be just a
waste of cyberspace for you to visit that web page because you are
not even remotely capable of comprehending what it has to say.
Here's the address. http://www.optusnet.com.au/~maxkeon/the1-1a.html
I eagerly await your comments. That's no joke.

4) The acceleration due to gravity in Austrailia is not 9.8m/s^2. The
variance in your local g and 9.8m/s^2 is many orders of magnitude
larger than the effect you seek to measure. If I gussed right and you
are in Victoria, g is 980.1364 cm/s^2.

Very impressive figures.

So you don't even care that you are using the wrong g?

You are really grasping at straws. If I choose to use 9.8 or
9.801364 for the gravity rate, this is how the results compare.

g = 9.8
Revs per second = 10.51875
Gravity rate up = ((c+v)^2/c^2)*g = 9.800000747013589 m/sec.
Gravity rate down = ((c-v)^2/c^2)*g = 9.79999925298644 m/sec.
The gravity anisotropy is 1.494027149107069D-06 m/sec.

g = 9.801364
Revs per second = 10.51875
Gravity rate up = ((c+v)^2/c^2)*g = 9.80136474711756 m/sec.
Gravity rate down = ((c-v)^2/c^2)*g = 9.801363252882467 m/sec.
The gravity anisotropy is 1.494235092991403D-06 m/sec.

I'm nowhere near that level of precision yet. All that is proven
so far is that the speed at which the action of gravity is applied
is nothing like infinite and is somewhat greater than zero.
So what are you on about?

-----

Max Keon

"Jerry" wrote in message
ups.com...
Max Keon wrote:
"Jerry" wrote in message
oups.com...

Do you have any reason to believe that in the ABSENCE of
gravitational anomalies, that the free disk should rotate at the
same rate as the housing? You have slippage effects, drive train
effects, and air friction effects to account for.

What drive train effects? The rotating housing is rotating at
constant rate. The enclosed air and the free disc are all going
along with it. What forces do you think are in place to DRIVE the
free disc away from its housing and the surrounding air mass? I
know what that force is, but you don't seem to want to know.

You claim that
you have eliminated drive train effects by mounting the needle
bearings such that, being driven through the inner race on one
end and through the outer race on the other, the differential
effects should cancel each other. This is doubtful.

You use the term "driven" where there is no drive. The free disc,
the enclosed air mass and the housing, rotate as a unit.

What do you mean, "there is no drive"? The rotor is frictionally
coupled to the housing through the "needle bearings".

Needle *point* bearings, which are indeed coupled to the housing
through friction. It takes a certain rotation rate to break the
*constant* friction which is applied to the axle end by the spring
loaded cavity. That friction maintains the fixed disc-housing
relationship until the rotation rate reaches a point where the
anisotropy in the gravity force is enough to break that bond.

Your dilemma is that the weight of the free disc on the bearing and
its clearance can only apply a specific torque with which to drive
the disc. Unless your gravity force changes, that torque won't
increase as the rotation rate increases. So, in your universe, how
is the bearing friction overcome?

The central bearing will tend to under-rotate the rotor relative
to the housing. The external bearing will tend to over-rotate the
rotor relative to the housing.

Since most of the weight is on the central bearing, the frictional
coupling between housing and central bearing will dominate, and
the under-rotation that you measure between rotor and housing is
probably due to this mismatch in coupling strengths.

Good. We're making progress at last. We all agree that the free disc
will now rotate at a lesser rate than the housing. Now replace that
axle with one which is pointed on both ends, while the inner housing
now carries a cavity. I assume we all agree that the free disc will
now rotate in advance of the housing?

But that just doesn't happen.

As I stated before, you have a Rube Goldberg setup.

You are really not understanding this experiment are you, whatever
your agenda may be.

-----

Max Keon

Max Keon wrote:

------- ----------------
R l ----------- l R
O l l l l O
T l l FREE l l T
A l l DISC l l A
T l l l l T
I l l-------\ l l--------\ I
N l----------- \--l--------- \ N
G l l \ G
l l l
H l l / H
O l----------- /--l--------- / O
U l l-------/ l l--------/ U
S l l l l S
I l l FREE l l I
N l l DISC l l N
G l l l l G
l ----------- l
------- ----------------

Picture the above assembly rotating as a unit, with gravity forcing
the free disc bearing ends onto the rotating housing bearing
components. It shouldn't be too difficult to see that one end of the
free disc bearing is going to roll in advance of the housing while
the other will be retarded by the same amount.

You diagram shows that the bearing loads are unequal.

Since most of the weight is on the central bearing, the frictional
coupling between housing and central bearing will dominate, and
the free disk will lag behind the housing due to this mechanical
artifact.

Jerry