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Try this experiment

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Re: Try this experiment
« Reply #30 on: August 18, 2008, 07:36:16 AM »
Quote from: Robosteve on August 18, 2008, 07:34:09 AM
Quote from: lolz at trollz on August 18, 2008, 07:31:22 AM
no, because that makes sense, it is all relative to the field.  Your light idea relative to every single place some light reflects from for the light that reflected there but not other light that reflected from somewhere else.

I'm afraid I don't follow you at all. The magnitude and direction of the acceleration light experiences as a result of the EA is dependent only on the direction of its velocity vector.

Yes, which is totaly unlike a moving charge in a magnetic field. 
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Parsifal

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Re: Try this experiment
« Reply #31 on: August 18, 2008, 07:42:58 AM »
Quote from: lolz at trollz on August 18, 2008, 07:36:16 AM
Yes, which is totaly unlike a moving charge in a magnetic field. 

The magnitude and direction of the force on a moving charge in a magnetic field depends upon the magnitude and direction of the component of its velocity perpendicular to the direction of the magnetic field. I fail to see how the concept is greatly different in principle.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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Re: Try this experiment
« Reply #32 on: August 18, 2008, 08:23:09 AM »
Quote from: Robosteve on August 18, 2008, 07:42:58 AM
Quote from: lolz at trollz on August 18, 2008, 07:36:16 AM
Yes, which is totaly unlike a moving charge in a magnetic field. 

The magnitude and direction of the force on a moving charge in a magnetic field depends upon the magnitude and direction of the component of its velocity perpendicular to the direction of the magnetic field. I fail to see how the concept is greatly different in principle.

That's because you don't understand the implication of the force you just made up for light. 
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Parsifal

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Re: Try this experiment
« Reply #33 on: August 18, 2008, 08:27:02 AM »
Quote from: lolz at trollz on August 18, 2008, 08:23:09 AM
That's because you don't understand the implication of the force you just made up for light. 

Enlighten me, then.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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Re: Try this experiment
« Reply #34 on: August 18, 2008, 08:47:18 AM »
Quote from: Robosteve on August 18, 2008, 08:27:02 AM
Quote from: lolz at trollz on August 18, 2008, 08:23:09 AM
That's because you don't understand the implication of the force you just made up for light. 

Enlighten me, then.

That force is very simple, it is perpendicular to hte direction of movement, and perpendicular to the field.  Very simple,very sensible.

Your idea is that the force is perpenicular to the direction of movement and in any plane perpendicular to the surface of the planet, in whatever direction is required to turn away from the planet, and then stopping at the point when the direction of travel is perpendicular to the surface of the planet.  Not simple, not sensible. 





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Parsifal

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Re: Try this experiment
« Reply #35 on: August 18, 2008, 09:01:12 AM »
Quote from: lolz at trollz on August 18, 2008, 08:47:18 AM
That force is very simple, it is perpendicular to hte direction of movement, and perpendicular to the field.  Very simple,very sensible.

Your idea is that the force is perpenicular to the direction of movement and in any plane perpendicular to the surface of the planet, in whatever direction is required to turn away from the planet, and then stopping at the point when the direction of travel is perpendicular to the surface of the planet.  Not simple, not sensible.

Oh, it is perfectly simple. It is simple in that the force tries to act in the direction of the dark energy field, but cannot always do so due to the fact that it cannot change the magnitude of the velocity of light, only its direction. So what happens is that it acts so as to minimise the angle between itself and the dark energy field. Allow me to illustrate:



Notice how the magnitude of the force decreases when it acts in a direction that is not parallel to the dark energy field, and reduces to zero when it cannot act at all? Perfectly simple, and perfectly sensible.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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Re: Try this experiment
« Reply #36 on: August 18, 2008, 09:07:11 AM »
So this force is sentient?  It acts not through the minimisation of energy state, but in anticipation of minimisation of energy state? 
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Dark Knight

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Re: Try this experiment
« Reply #37 on: August 18, 2008, 09:07:28 AM »
I think the FE's are the conspirators,  How  much money do you guys make off of this farse?  Who is selling the books and making money off of this?   
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Parsifal

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Re: Try this experiment
« Reply #38 on: August 18, 2008, 09:11:22 AM »
Quote from: lolz at trollz on August 18, 2008, 09:07:11 AM
So this force is sentient?  It acts not through the minimisation of energy state, but in anticipation of minimisation of energy state? 

No. It acts through a mathematical relationship that I am midway through deriving.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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Re: Try this experiment
« Reply #39 on: August 18, 2008, 09:11:40 AM »
oh, then explain this:

Quote from: Rig Navigator on August 17, 2008, 06:58:04 AM
Quote from: Josef on August 17, 2008, 06:34:18 AM
Have we come to the conclusion that the FE explanation to the outcome of this experiment is that light is being bent?

That is their assertion, but it is wrong.

Even a simple set-up, something that will fit on a table, would demonstrate the presence of "bendy" light.



As you rotated the experiment in the vertical axis, the interference pattern would be changed by the "bend" of the light waves.  Since this is not observed, there is no bend in light waves.
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Re: Try this experiment
« Reply #40 on: August 18, 2008, 09:12:18 AM »
Quote from: Robosteve on August 18, 2008, 09:11:22 AM
Quote from: lolz at trollz on August 18, 2008, 09:07:11 AM
So this force is sentient?  It acts not through the minimisation of energy state, but in anticipation of minimisation of energy state? 

No. It acts through a mathematical relationship that I am midway through deriving.

How come you never finish deriving these mathematical relationships?
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Parsifal

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Re: Try this experiment
« Reply #41 on: August 18, 2008, 09:15:50 AM »
Quote from: lolz at trollz on August 18, 2008, 09:11:40 AM
oh, then explain this:

Quote from: Rig Navigator on August 17, 2008, 06:58:04 AM
Quote from: Josef on August 17, 2008, 06:34:18 AM
Have we come to the conclusion that the FE explanation to the outcome of this experiment is that light is being bent?

That is their assertion, but it is wrong.

Even a simple set-up, something that will fit on a table, would demonstrate the presence of "bendy" light.



As you rotated the experiment in the vertical axis, the interference pattern would be changed by the "bend" of the light waves.  Since this is not observed, there is no bend in light waves.

I do not yet have an explanation for that particular phenomenon.

Quote from: lolz at trollz on August 18, 2008, 09:12:18 AM
How come you never finish deriving these mathematical relationships?

Because I am pondering the best way to solve this for f(t):

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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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Re: Try this experiment
« Reply #42 on: August 18, 2008, 09:18:44 AM »
Yea, tell me how that goes.  Whatever happened to your one for downward force being reduced by gravity from stars? 
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Parsifal

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Re: Try this experiment
« Reply #43 on: August 18, 2008, 09:23:42 AM »
Quote from: lolz at trollz on August 18, 2008, 09:18:44 AM
Yea, tell me how that goes.  Whatever happened to your one for downward force being reduced by gravity from stars? 

I was working on the EA derivation first. It is my highest priority with respect to FES at the moment.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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ghazwozza

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Re: Try this experiment
« Reply #44 on: August 18, 2008, 09:25:46 AM »
Quote from: Robosteve on August 18, 2008, 09:23:42 AM
Quote from: lolz at trollz on August 18, 2008, 09:18:44 AM
Yea, tell me how that goes.  Whatever happened to your one for downward force being reduced by gravity from stars? 

I was working on the EA derivation first. It is my highest priority with respect to FES at the moment.

What problem are you trying to solve, exactly? Maybe I could help.
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Parsifal

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Re: Try this experiment
« Reply #45 on: August 18, 2008, 09:35:17 AM »
Quote from: ghazwozza on August 18, 2008, 09:25:46 AM
What problem are you trying to solve, exactly? Maybe I could help.

I am trying to figure out what f(t) is. It is a function of t as the notation suggests, but I do not know what the function is. I have managed to relate it to its derivative function, however:



Where c and r are constants.

Basically it is one of two parametric equations that describe the parabolic arc in which light is bent with t as the parameter representing time. x = f(t) and y = g(t). Once I know f(t) I can work out g(t), and from there everything will fall into place. Note that it is not the most conventional parametric form of a parabola. I am inclined to suspect that f(t) may be something like an inverse hyperbolic sine curve.

Additional information:

f(t) is an odd function.
f(0) = 0
f'(0) = c
f''(0) = 0
« Last Edit: August 18, 2008, 09:37:27 AM by Robosteve »
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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ghazwozza

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Re: Try this experiment
« Reply #46 on: August 18, 2008, 09:44:20 AM »
OK, I'll grab a piece of paper and start scribbling.
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ghazwozza

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Re: Try this experiment
« Reply #47 on: August 18, 2008, 09:55:15 AM »
How did you arive at that differential equation? Or, what are your initial assumptions for how light must behave in EAT?

I think I have an answer if we assume that the vertical component of light's velocity increases linearly with time (constant acceleration) and that it's velocity must always be c.
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Parsifal

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Re: Try this experiment
« Reply #48 on: August 18, 2008, 10:20:42 AM »
Quote from: ghazwozza on August 18, 2008, 09:55:15 AM
How did you arive at that differential equation? Or, what are your initial assumptions for how light must behave in EAT?

I think I have an answer if we assume that the vertical component of light's velocity increases linearly with time (constant acceleration) and that it's velocity must always be c.

First, I calculated that y = x2 / 2r if it is to approximate the effect of a round Earth at short distances (r being the radius of the RE). This means that dy / dx = x / r and therefore dy / dt = (x / r) * (dx / dt). Substituting x = f(t) and y = g(t), you get g'(t) = f(t) * f'(t) / r.

Next, if the speed is to remain constant, then by Pythagoras' theorem f'(t)2 + g'(t)2 = c2. Squaring the above equation, c2 - f'(t)2 = f(t)2 * f'(t)2 / r2. Rearranging this gives:



What was the answer you got?
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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ghazwozza

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Re: Try this experiment
« Reply #49 on: August 18, 2008, 10:50:14 AM »
Hmm, I did it differently. First, I used x(t) and y(t) instead of f(t) and g(t), just because it is more descriptive.

We know (if I understand EAT correctly):
y'(t)2 + x'(t)2 = c2                                       (speed of light is constant)
y''(t) = a                                                    (light accelerates upwards constantly with acceleration a)
y(0) = 0,  x(0) = 0,  y'(0) = 0                        (initial conditions)

y''(t) = a  =>  y'(t) = at  =>  y(t) = at2/2  This is assuming the EA is constant and directed upwards.

So, by pythagoras: (at)2 + x'(t)2 = c2
Re-arrange for x'(t) to get a difficult integral, type into the online integrator to get:

x(t) = (c2/2a) arctan (at/{c2-(at)2}0.5) + (t/2)(c2-(at)2)0.5

which is ugly, but there you go.

I think your initial assumption that light follows a parabola is incorrect.
« Last Edit: August 18, 2008, 10:52:43 AM by ghazwozza »
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Parsifal

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Re: Try this experiment
« Reply #50 on: August 18, 2008, 10:57:50 AM »
Quote from: ghazwozza on August 18, 2008, 10:50:14 AM
y''(t) = a                                                    (light accelerates upwards constantly with acceleration a)

This is an incorrect postulate. There cannot be both constant vertical acceleration and constant speed. The acceleration must be perpendicular to the velocity, such that:

x''(t)2 + y''(t)2 = a2

Though a is not constant; it decreases as |dy/dx| increases.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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ghazwozza

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Re: Try this experiment
« Reply #51 on: August 18, 2008, 11:02:19 AM »
Quote from: Robosteve on August 18, 2008, 10:57:50 AM
Quote from: ghazwozza on August 18, 2008, 10:50:14 AM
y''(t) = a                                                    (light accelerates upwards constantly with acceleration a)

This is an incorrect postulate. There cannot be both constant vertical acceleration and constant speed. The acceleration must be perpendicular to the velocity, such that:

x''(t)2 + y''(t)2 = a2

Though a is not constant; it decreases as |dy/dx| increases.

I was going by your original electromagnetic accelerator post. Maybe I got the wrong end of the stick, maybe the thery has changed a bit.

So what rules govern the path light takes? What direction is a (other than perpendicular to v)? What is its magnitude?

PS: I'm going out now. Hopefully I'll have a look in a few hours.
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Parsifal

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Re: Try this experiment
« Reply #52 on: August 18, 2008, 11:08:07 AM »
Quote from: ghazwozza on August 18, 2008, 11:02:19 AM
So what rules govern the path light takes? What direction is a (other than perpendicular to v)? What is its magnitude?

That is what I am trying to work out. Also, yes I did have a different idea in the beginning. I only later came to realise that the only way to keep the speed constant was to have the acceleration be perpendicular to the velocity.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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Rig Navigator

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Re: Try this experiment
« Reply #53 on: August 18, 2008, 11:12:11 AM »
Quote from: Robosteve on August 18, 2008, 09:15:50 AM
I do not yet have an explanation for that particular phenomenon.

...Because I am pondering the best way to solve this for f(t):

It would save you time working out differential equations if you accepted that the experimental data means that the effect you are trying to model doesn't exist.

Of course, that would probably defeat the purpose of you arguing in the first place.
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Parsifal

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Re: Try this experiment
« Reply #54 on: August 18, 2008, 11:14:06 AM »
Quote from: Rig Navigator on August 18, 2008, 11:12:11 AM
It would save you time working out differential equations if you accepted that the experimental data means that the effect you are trying to model doesn't exist.

Of course, that would probably defeat the purpose of you arguing in the first place.

Perhaps when RE'ers unanimously accept that gravity does not exist, I will consider accepting that the EA does not exist.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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Re: Try this experiment
« Reply #55 on: August 18, 2008, 11:31:38 AM »
let him work it all out.  cos then comes the experiment. 
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Dark Knight

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Re: Try this experiment
« Reply #56 on: August 18, 2008, 11:37:20 AM »
Where do comets come from?  Don't they orbit towards and away from the sun?   That can't happen since everything is going the same speed in the same direction.  Also where do meteors come from?  They to should be going in the same direction,  gravity shouldn't be pulling these things in since it doesn't exist?
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Parsifal

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Re: Try this experiment
« Reply #57 on: August 18, 2008, 12:06:39 PM »
Quote from: Dark Knight on August 18, 2008, 11:37:20 AM
Where do comets come from?  Don't they orbit towards and away from the sun?   That can't happen since everything is going the same speed in the same direction.  Also where do meteors come from?  They to should be going in the same direction,  gravity shouldn't be pulling these things in since it doesn't exist?

All of these questions have been asked and answered.
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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Dark Knight

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Re: Try this experiment
« Reply #58 on: August 18, 2008, 12:35:26 PM »
Ok so answer them.


Oh, I forgot,  its a conspiracy..
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Parsifal

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Re: Try this experiment
« Reply #59 on: August 18, 2008, 12:38:24 PM »
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Quote from: Ævan on August 11, 2013, 01:18:20 PM
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