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The Allais Effect Solved


Bjarne
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Then all I can conclude is that you don't understand what "above" means.

RF will hit "a point" above the ecliptic, - if the moon is above the ecliptic.

But hit a higher point above the eliptic, - by eclipse

PS

we assume that the moon is at the same angle above the earth, let’s just say the moon is 1° above the ecliptic - in all these cases, - relevant for this example

Edited by Bjarne
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RF will hit "a point" above the ecliptic, - if the moon is above the ecliptic.

But hit a higher point above the eliptic, - by eclipse

PS

we assume that the moon is at the same angle above the earth, let’s just say the moon is 1° above the ecliptic - in all these cases, - relevant for this example

 

 

 

 

How does it do this? You said "much higher (above the ecliptic) as for example 4, 8 or 12 hours before (and after) eclipse." The only way for the force to be higher is for the moon to be higher. And you claim it's higher both before and after the eclipse. You also say "much higher". How much?

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How does it do this? You said "much higher (above the ecliptic) as for example 4, 8 or 12 hours before (and after) eclipse." The only way for the force to be higher is for the moon to be higher. And you claim it's higher both before and after the eclipse. You also say "much higher". How much?

10ab.jpg

The triangle connecting the Moon, Earth and Sun can rotate, depending on where the Moon is

Because of that, - the RF will also rotate

Red line = RF (0,0034°) based on 0,6° angle to the Moon

Edited by Bjarne
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10ab.jpg

The triangle connecting the Moon, Earth and Sun can rotate, depending on where the Moon is

Because of that, - the RF will also rotate

Red line = RF (0,0034°) based on 0,6° angle to the Moon

 

 

The answer to how much requires a comparison of one value to another. Giving one value doesn't answer the question. Further, since rotation is not part of the inquiry, as you were talking about how much above the ecliptic the force is, and discussion of rotation does not address the question.

 

I'm asking you to quantify how much the force moves, in the direction perpendicular to the ecliptic. What is its angle at the eclipse, and what is it some time later? You claimed it was much higher. Or better yet, what is the vertical component at these two times.

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The answer to how much requires a comparison of one value to another. Giving one value doesn't answer the question. Further, since rotation is not part of the inquiry, as you were talking about how much above the ecliptic the force is, and discussion of rotation does not address the question.

 

I'm asking you to quantify how much the force moves, in the direction perpendicular to the ecliptic. What is its angle at the eclipse, and what is it some time later? You claimed it was much higher. Or better yet, what is the vertical component at these two times.

First, at all try to understand that 90 % of the rotation (in the case above the ecliptic) happens 12 hours before and after solar eclipse.

 

The closer you (the earth) get to eclipse the more that rotations speed up.

 

So the day of eclipse is where the max upwards pull is a geometric / mathematical fact.

 

To pinpoint that further in decimals can only be after my holidaying the meantime you should think a little serious about what I just wrote

 

 

10ab.jpg

Edited by Bjarne
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First, at all try to understand that 90 % of the rotation (in the case above the ecliptic) happens 12 hours before and after solar eclipse.

I think that 100% of the daily rotation of the earth happens in that 24 hour span (or 50% in a 12 hour span). What I don't see is how rotational motion has any direct effect here.

 

The closer you (the earth) get to eclipse the more that rotations speed up.

 

Rotation speed is pretty much a constant at this level of resolution.

 

So the day of eclipse is where the max upwards pull is a geometric / mathematical fact.

Are we back to the confusion about what maximum means? (You probably mean maximum allowable, but that's not what you keep saying) And the implication that somehow the sun matters to the force the moon exerts? (This configuration will happen with every new and full moon, regardless of an eclipse)

 

 

To pinpoint that further in decimals can only be after my holidaying the meantime you should think a little serious about what I just wrote

I have been serious this whole time. As you have not changed your argument in the face of multiple areas of valid criticism, I think it's you who are not taking others seriously.

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!

Moderator Note

If questions aren't being taken into consideration, and you just keep repeating that you're right, we call that soapboxing, and it's against the rules. This is a science discussion forum, not a blog or a position paper. The preaching must end now or the thread will be closed.

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I think that 100% of the daily rotation of the earth happens in that 24 hour span (or 50% in a 12 hour span). What I don't see is how rotational motion has any direct effect here.

 

Who wrote "rotation of the Earth", except you?

 

The resulting force is rotating, - was that not what I wrote?

 

 

Are we back to the confusion about what maximum means? (You probably mean maximum allowable, but that's not what you keep saying)

 

Right, implied after all what have been written.

 

 

 

And the implication that somehow the sun matters to the force the moon exerts? (This configuration will happen with every new and full moon, regardless of an eclipse)

 

 

What do you mean ?

 

 

 

I have been serious this whole time. As you have not changed your argument in the face of multiple areas of valid criticism, I think it's you who are not taking others seriously.

 

I am also learning my friend, nothing wrong with that, - or ?

Now I am changing my mind again.

This is how science is, and always should be. Never get stucked in the same old mud hole.

 

I checked the data tonight, and can see that the Moon only will be 0,4° above ecliptic by solar eclipse 21 of August (USA) this will only exposed about 22μGal of the DFA, but enough to measure the anomaly with pendulum as well as with the 2 gravimeter experiment near artic.

 

However by lunar eclipse the 7 of August this year the moon will be 0,8° above the ecliptic that day and moment, measurement with pendulum will of course also be an option, the best result by using a pendulum will be between the 40° and 50 ° latitude (much higher than by the solar eclipse)

 

In both cases these anomalies can be measured by the gravimeter experiment, which under all circumstances is a better , more precise and trustworthy option.

However the lunar eclipse will reveal a much larger exposed DFA - Certainly near 40μGal.

The only problem is that the gravimeter cannot be at the best position all the time, due to the tilt and rotation of earth.

However this small deviation from expectation is really peanuts compared to the significant anomaly that we certianly will see.

 

Next time meassurement is possible will first be January 2019

 

And finnally, the moon is changing position almost 1° degree per 24 hours, not 0,34° like you wrote before, - this too is very important. The moon can reach + or - 5,15° within 7 days

Edited by Bjarne
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Who wrote "rotation of the Earth", except you?

 

No, you just wrote rotation, and the only thing of consequence rotating here is the earth. The moon's movement is revolution, not rotation. (The moon's rotation takes an entire orbit)

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No, you just wrote rotation, and the only thing of consequence rotating here is the earth. The moon's movement is revolution, not rotation. (The moon's rotation takes an entire orbit)

 

 

 

10ab.jpg

The triangle connecting the Moon, Earth and Sun can rotate, depending on where the Moon is

Because of that, - the RF will also rotate

Red line = RF (0,0034°) based on 0,6° angle to the Moon

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That triangle has nothing to do with the vertical component of the force. If you would learn some basic physics you would understand this.

I agree that if , - let us say the position of the Earth, Moon and sun and therefore the triangle is frozen, - the vertical component and the resulting force is always the same angle and magnitude.

 

10ab.jpg

 

Now rotate the triangle (fig 10a) 180°

Do we both agree that after this 180° rotation of the triangle (fig 10a) - (which mean only the moon have change position, and now is below the ecliptic instead of above) - the vertical component / force - is now pointing south, Not north ?

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I agree that if , - let us say the position of the Earth, Moon and sun and therefore the triangle is frozen, - the vertical component and the resulting force is always the same angle and magnitude.

 

10ab.jpg

 

Now rotate the triangle (fig 10a) 180°

Do we both agree that after this 180° rotation of the triangle (fig 10a) - (which mean only the moon have change position, and now is below the ecliptic instead of above) - the vertical component / force - is now pointing south, Not north ?

 

 

You said the force was moving higher. Much higher. This would be an example of going lower. And there is no 180º in that diagram, so what is rotating 180º? You never explain these details. You know them, but people reading the page cannot read your mind. It's especially confusing when you start making new stuff up instead of continuing the current discussion.

 

You never answered my question about how much higher it moves. How much does the moon move, vertically, and how long does that take?

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You said the force was moving higher. Much higher.

To keep it simple, I am saying that when you rotate the triangle, you will also rotate the point where the resulting force is pointing

 

This would be an example of going lower.

 

Thats right and such triangle "rotation" takes about 2 weeks

 

And there is no 180º in that diagram, so what is rotating 180º? Y

 

Just to figure out whether we could agree that this happens in the real world also.

 

You never explain these details.

 

Well I tried too.

 

You know them, but people reading the page cannot read your mind.

 

True, it is naturally, - space is very spacy, can be difficult to explain what I mean, and English is not my first language. Furthermore, I was very lazy in school.

 

It's especially confusing when you start making new stuff up instead of continuing the current discussion.

 

Remember I lean all about resulting force etc. here at the forum. I was force suddenly to implement that in the theory.

 

You never answered my question about how much higher it moves. How much does the moon move, vertically, and how long does that take?

 

I would be happy to get deeper into specific details, however it is necessary first at all to be sure whether you understand what I really mean.

An important point here is that the "triangle rotation” I speak about" - happens in the real world.

It is very important to notice that the larges "rotation” of that triangle happens the day on eclipse.

Do you agree to that?

Edited by Bjarne
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Could someone not write a program to simulate all the forces on a pendulum and see what it kicks out during a simulated eclipse and at other times?

Edited by DrP
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To keep it simple, I am saying that when you rotate the triangle, you will also rotate the point where the resulting force is pointing

 

Yes. And the relevant angle is the one between the ecliptic and the moon. That's the vertical force.

 

But you have said in the past that this vertical force points in directions here it is impossible to point. And won't answer when asked questions about it. You just keep repeating the same nonsense.

 

Thats right and such triangle "rotation" takes about 2 weeks

 

Tha Allais effect takes a few hours, so what relevance does this have?

 

Just to figure out whether we could agree that this happens in the real world also.

 

Something that has no relevance to what you're discussing, for multiple reasons, does happen.

 

 

Well I tried too.

 

 

True, it is naturally, - space is very spacy, can be difficult to explain what I mean, and English is not my first language. Furthermore, I was very lazy in school.

 

Remember I lean all about resulting force etc. here at the forum. I was force suddenly to implement that in the theory.

 

To conclude that you have explained some effect while not understanding the mechanisms involved is very, very backwards.

 

I would be happy to get deeper into specific details, however it is necessary first at all to be sure whether you understand what I really mean.[/size]

An important point here is that the "triangle rotation” I speak about" - happens in the real world.

It is very important to notice that the larges "rotation” of that triangle happens the day on eclipse.

Do you agree to that?

 

The triangle rotation does not measure the vertical force, which is what you are interested. If you think it does, you need to conclusively demonstrate that. i.e. use it in a calculation that gives you a value that's useful in the discussion.

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To keep it simple, I am saying that when you rotate the triangle, you will also rotate the point where the resulting force is pointing

 

Yes.

It is important where exactly the resulting force is pointing, at least if we want to know whether the earth is accelerated upwards, downwards or horizontal.

10ab.jpg

The more the resulting force points above the ecliptic, the more the moon will exert an upwards pull of the earth.

The day of eclipse (and always when the moon is crossing the ecliptic) - the change of where the resulting force is pointing happens very rapidly.

This is one of two reason to why the duration of the Allais Effect is as it is.

 

Tha Allais effect takes a few hours, so what relevance does this have?

 

As discussion before, the 2 hours duration is a local phenomenon.

The duration can be tracked almost 24 hours (if lucky) (by the gravimeter experiment) without you need to move location.

However when measuring with a pendulum, it is more complicated because the rotation of the Earth can bring you too much above or below the perfect place of measurement , as well you will change the perfect interaction axis, that we discussed before. To solve this you can have pendulum measurement taken on the path of the moon, around the earth.

Responsible for the duration seen from an overall perspective, is partly the "rotation of where the resulting force is pointing" and the change of the location of the Moon (0, 4º per 12 hours)

For example let us say that the moon is 0, 8º above the ecliptic. 12 hours later it is only 0, 4º above. - Now you have 2 factor both weakening the anomaly duration length

Therefore, you have to look at each specific episode to be able to predict the impact of these to factors.

 

The triangle rotation does not measure the vertical force, which is what you are interested. If you think it does, you need to conclusively demonstrate that. i.e. use it in a calculation that gives you a value that's useful in the discussion.

I already calculated that, and have shown it here, left is to agree how rapidly the rotation of the resulting force really happens by eclipse.

Edited by Bjarne
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It is important where exactly the [/size]resulting force is pointing, at least if we want to know whether the earth is accelerated upwards, downwards or horizontal. [/size]

You have asserted a number of times that it is only the vertical acceleration that matters.

 

[/size]

10ab.jpg

The more the resulting force points above the ecliptic, the more the moon will exert an upwards pull of the earth.[/size]

The day of eclipse (and always when the moon is crossing the ecliptic) - the change of where the resulting force is pointing happens very rapidly. [/size]

This is one of two reason to why the duration of the Allais Effect is as it is.[/size]

 

The upward component can only depend on the height of the moon above the ecliptic.

 

The moon crosses the ecliptic twice per orbit, regardless of whether there is an eclipse.

 

 

As discussion before, the 2 hours duration is a local phenomenon.

The duration can be tracked almost 24 hours (if lucky) (by the gravimeter experiment) without you need to move location.

Those two statements contradict each other. If you can track it for 24 hours then it does not have a 2 hour duration.

 

However when measuring with a pendulum, it is more complicated because the rotation of the Earth can bring you too much above or below the perfect place of measurement , as well you will change the perfect interaction axis, that we discussed before. To solve this you can have pendulum measurement taken on the path of the moon, around the earth.

Responsible for the duration seen from an overall perspective, is partly the "rotation of where the resulting force is pointing" and the change of the location of the Moon (0, 4º per 12 hours)

For example let us say that the moon is 0, 8º above the ecliptic. 12 hours later it is only 0, 4º above. - Now you have 2 factor both weakening the anomaly duration length

Therefore, you have to look at each specific episode to be able to predict the impact of these to factors.

 

If the force takes 12 hours to drop in half, why would the duration of the effect be just 2 hours for a pendulum? Why does the graph of the original Allais results have such a sharp increase and decrease?

 

I already calculated that, and have shown it here, left is to agree how rapidly the rotation of the resulting force really happens by eclipse. [/size]

You've presented an equation or a graph of the predicted pendulum frequency as a function of time? Or even the upward force component as a function of time? Using your triangle? In which post(s) do these show up?

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You have asserted a number of times that it is only the vertical acceleration that matters.

One of the requirement to measure Allais Effects is that the Earth must accelerate upwards, it really doesn’t matter how much, what matter is that the upwards accelerating force not affect the testing body, - and that, - the second requirement, - depend on high the moon is on the sky.

 

The upward component can only depend on the height of the moon above the ecliptic.

 

 

The ecliptic is not decisive for what is up or down in the universe.

The upwards component is an integrated vector of a 2D triangle.

You can rotate that 2D triangle in a 3D universe, and you can always claims that the vertical component always is vertical component.

However, we already rotated that triangle 180º today, - and suddenly the upwards component had change to a downwards component.

Upwards and downwards component are therefore suddenly the same and not the same confused concept.

You can also rotate the triangle 90º, now the vertical component is suddenly "a horizontal component", pointing parallel, - the same direction as the horizontal ecliptic axis.

Therefore, the concept “vertical component” is a relative 2D concept, in a 3D universe.

10ab.jpg

It seems that you are missing one factor; - the earth, - as part of the 2D triangle , so soon you will freeze the Sun and the Moon, you will see that the motion of the earth is causing the 2D triangle and therefore also what you say is a vertical component, - to rotate.

Not only can the opposite position of the moon cause the triangle to “rotate”.

Also change of position of the Earth can cause the triangle to rotate, simply because of Earth is a integrated part of the 2D triangle.

If the earth is changing position, the way it does by eclipse, - the 2D triangle is simply rotating in a 3D universe, not easy when concepts can have 2 different meaning.

maanebane.gif

The Moon orbits the sun,"more than orbiting the earth"

 

Those two statements contradict each other. If you can track it for 24 hours then it does not have a 2 hour duration.

If the force takes 12 hours to drop in half, why would the duration of the effect be just 2 hours for a pendulum? Why does the graph of the original Allais results have such a sharp increase and decrease?

 

 

When lucky 24 hours, ( gravimeter experiments)

Even from an overall perspective the anomaly can be much shorter.

Some Allais effects is measured; few hours anomaly, some 4 hours , some 7 hours.

 

There are several reasons:

1.) The testing device is brought to a position too high or too low relative to the Moon due to the rotaion of Earth

2.) Change of the DFA interacting axis due to the rotaion of Earth.

3.) A very low moon (0,2º) at the moment of max eclipse + declining moon.

4.) A very high moon (1º) at eclipse + inclining moon

5.) If the operator of the pendulum don’t know and choose a wrong swing direction this too can weaken the ability to meassure the anomaly, or having the pendulum to change anomaly-direction.

 

Best swing direction is east west, never completely east west and never north south.

Better read the article, it’s all explained and illustrated at page 4. here

Edited by Bjarne
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I think it is the OP's own website - and yes, I agree with your characterisation

Yes, it is my webpage.

This theory was discussed here at the forum and the thread was closed, because nothing is so far was found wrong with the theory of relativity

Well fine, - no problem with that.

So fare I know the theory of relativity is right now tested on board ISS.

I made a prediction of the outcome, - until then it’s OK to call me whatever you want.

But carefully even a blind chicken can find a grain of gold.

Edited by Bjarne
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