Author Topic: Woodward's effect  (Read 206278 times)

Offline ppnl

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Re: Woodward's effect
« Reply #960 on: 05/18/2017 03:12 PM »
In MET theory, should you not just substitute the Earth against which you are pushing in your example above, with all the mass in the universe? Then it removes the constraint of having to work harder because you are moving faster in relation to the mass you are pushing against. It also removes the issue of which direction you are travelling in, because there is mass in all directions, if you look far enough.

The problem is constant acceleration. That breaks conservation and physics in general, usually a good sign that the theory is wrong. Why should using the entire mass of the observable universe work differently than using the mass of the Earth? Mach Effect drives would still be an amazing breakthrough even if they didn't provide constant acceleration and operated like any other propulsion system.

I was thinking because the speed relative to the Earth will change, but how do you determine whether the speed relative to the entire Universe is changing?

Well if you are accelerating then your speed relative to everything not accelerating with you is changing. So unless the entire universe is accelerating with you...

Most of the universe is moving at a really high velocity with respect to you. For example our movement through the CMB is pretty high. About 370 k/s. That means you would need gigawatts to get nanogravities. Either that or you could pick a direction that would deliver gigawatts of free energy.


Offline 93143

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Re: Woodward's effect
« Reply #961 on: 05/18/2017 07:56 PM »
According to my calculations, the Doppler effect applied to gravinertial radiation in an expanding universe makes the effective, or interaction-weighted, mean velocity of the matter within the cosmic horizon linear with the velocity of a Mach effect device, at least at low peculiar velocities (and my math is Newtonian and thus probably wrong at relativistic velocities, even if there's any validity to it in the first place).

Obviously, if this effective mean velocity were always equal to the velocity of the device, you'd have conservation of energy without the Lorentz-invariant Mach effect equation having to be wrong.

Unfortunately, due to other factors I lack the time required to delve into the physics and actually establish what would happen in a fully relativistic framework using Hoyle-Narlikar theory (or at least GR) and accounting for any cosmic weirdness an engineer wouldn't automatically think of.  I was hoping Dr. Rodal had done that.

Offline dustinthewind

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Re: Woodward's effect
« Reply #962 on: 05/19/2017 12:34 AM »
The device is switched on and accelerates a ship from 0 to 10,000km/h. Then it is switched off. The ship is now traveling at 10,000km/h relative to its starting position. But the device cannot "remember" this.

When it is switched on again, it will not require more input energy to accelerate from 10,000km/h to 20,000km/h than it required to accelerate to 10,000km/h in the first place.

Hence, it does indeed operate as a local energy generator.

This is a common mistake. Take a conventional BFR (Big F… Rocket). Let's accelerate this spaceship to a decent fraction of the speed of light. The more it approaches the speed of light, the more difficult it is for the ship to accelerate, as the mass of the spaceship increases more than the mass lost as exhaust propellant. Now switch the motor off. You can't say "the ship cannot remember its velocity, so just switch the motor on again to accelerate without the previous relativistic drag." Again, this has to do with the inertial reference frame the ship departed from.

Yes this is true but it is important to remember that there is nothing special about the rocket's frame of reference. Fort example if they were to drop a marker satellite out the door they would find that they could add velocity relative to the marker as easily as ever. It is only from the Earth frame that they seem unable to accelerate. From the rocket's frame it is the Earth that seems trapped in glue. It is the Earth that has relativistic mass.

In Galilean relativity all frames are equally valid. Special relativity extends this equality to a case where the speed of light is constant for all observers. In short you cannot tell how fast you are moving by how hard it is to accelerate. Motion is still only relative and does not exist as an absolute.

How fast would you say the frame of light is near an event horizion?  Just a little further and the frame of reference is moving so fast that light can not escape.  Further away from a source at a set frequency, light moves in opposite directions defining a set frame of reference but detecting two identical sources on each side, while suspended above a gravity well, would lead to a change in the frame of reference for each source.

Since the rate of clocks and the gravitational potential have the same derivative, they are the same up to a constant. ... The changing rates of clocks allowed Einstein to conclude that light waves change frequency as they move, and the frequency/energy relationship for photons allowed him to see that this was best interpreted as the effect of the gravitational field on the mass–energy of the photon. ...

locally the speed of light may appear to be constant but at the expense of change in wavelength.  Consider that in the case of light instead of a loss in local velocity it is perceived a loss in wavelength.  A loss in wavelength is a loss in energy/(energetic mass) and a loss in energy for a particle with rest mass, velocity is lost.  For an all seeing fame if space is moving in a direction, the velocity of light may appear to be slower in one direction as opposed to the other, as opposed to a local observer moving would instead perceive changes in wavelength at constant velocity c. 

Let us consider the frame dragging effect.  For a rotating black hole that pulls space time around it in a vortex the speed of light one way around the black hole is faster than the other. 

Rotational frame-dragging (the Lense–Thirring effect) appears in the general principle of relativity and similar theories in the vicinity of rotating massive objects. Under the Lense–Thirring effect, the frame of reference in which a clock ticks the fastest is one which is revolving around the object as viewed by a distant observer. This also means that light traveling in the direction of rotation of the object will move past the massive object faster than light moving against the rotation, as seen by a distant observer.

if there is any acceleration going on it should be with respect to the local frame of light.

Sorry dusty I have no idea what point you are trying to make. You appear to say some things about general relativity mixed with some word salad. Since I was talking about special relativity in a flat space it seems at best to be a change of subject.

I can't make sense of "how fast is the frame of light". But if you are asking what the speed of light is near a black hole the answer is complicated. Locally the speed of light is always the same. Measuring the speed of light at a distance can give different values because of the warped coordinate system. Also because of this warped coordinate system you can't have a globally valid inertial frame at all.

But over a sufficiently short distance in a sufficiently flat space general relativity reduces to special relativity. Since there are no black holes around I think we can ignore general relativity for the moment.

My last sentence sums it up.  The rest is elaboration about various points that support that last statement. 

Offline tdperk

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Re: Woodward's effect
« Reply #963 on: 05/19/2017 09:41 AM »


Just to clarify, I do believe that the MET will result in greater local energy gain than the electrical power input into the device. Consider the following example, used many times before in this discussion.

The device is switched on and accelerates a ship from 0 to 10,000km/h. Then it is switched off. The ship is now traveling at 10,000km/h relative to its starting position. But the device cannot "remember" this.

When it is switched on again, it will not require more input energy to accelerate from 10,000km/h to 20,000km/h than it required to accelerate to 10,000km/h in the first place.

Hence, it does indeed operate as a local energy generator.

Yes exactly. There simply isn't any way around it. You could use the drive to move asteroids and then use the energy to melt down the asteroids to extract the metals needed for massive structures.

But you also have to believe a simple tabletop device is a method for hacking the universe to draining an infinite amount of power from the universe for our use. The experimental evidence is so close to noise that God would need glasses. And the theory for it is... lets just say "inconsistent" so I don't get  nasty messages.

So... good luck with that.

" The experimental evidence is so close to noise that God would need glasses. And the theory for it is... lets just say "inconsistent" so I don't get  nasty messages. "

The theory behind the MET follows directly from the same observations by Mach which prompted Einstein to develop SR and GR.

The experimental evidence is significant to 6 sigma.

Offline ppnl

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Re: Woodward's effect
« Reply #964 on: 05/20/2017 08:07 AM »


Just to clarify, I do believe that the MET will result in greater local energy gain than the electrical power input into the device. Consider the following example, used many times before in this discussion.

The device is switched on and accelerates a ship from 0 to 10,000km/h. Then it is switched off. The ship is now traveling at 10,000km/h relative to its starting position. But the device cannot "remember" this.

When it is switched on again, it will not require more input energy to accelerate from 10,000km/h to 20,000km/h than it required to accelerate to 10,000km/h in the first place.

Hence, it does indeed operate as a local energy generator.

Yes exactly. There simply isn't any way around it. You could use the drive to move asteroids and then use the energy to melt down the asteroids to extract the metals needed for massive structures.

But you also have to believe a simple tabletop device is a method for hacking the universe to draining an infinite amount of power from the universe for our use. The experimental evidence is so close to noise that God would need glasses. And the theory for it is... lets just say "inconsistent" so I don't get  nasty messages.

So... good luck with that.

" The experimental evidence is so close to noise that God would need glasses. And the theory for it is... lets just say "inconsistent" so I don't get  nasty messages. "

The theory behind the MET follows directly from the same observations by Mach which prompted Einstein to develop SR and GR.

The experimental evidence is significant to 6 sigma.

Meh, you should be careful idolizing Mach. He seems to have been as much a philosopher as a scientist. That may be why Mach's principle was never a quantitative testable idea. It was not driven by experimental science so much as the demands of his philosophy. The same philosophy caused him to deny the existence of atoms. So...

It's kinda ironic that Einstein's work on Brownian motion pretty much sealed the deal for the existence of atoms. And Mach apparently totally rejected Einstein's theory of relativity.

As for the six sigma thing - it kinda misses the point. Nobody is claiming that they got positive results due to a confluence of random errors. Rather it is the small size of the results with the noisy environment that make it difficult to eliminate systemic errors. Statistical significance calculations cannot account for unknown systemic errors.

Pons and Fleischmann initially reported excess heat 20% higher than could be accounted for by any chemical reaction. Eventually they claimed to be able to generate one kilowatt per cubic centimeter of palladium for 50 days straight. What is the statistical significance of that? 2000 sigma? That was almost 30 years ago.

"The first principle is that you must not fool yourself and you are the easiest person to fool." - Richard Feynman


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