Attention to all Gravity and Aether Researchers

[Harvey]

Thank You Stevend,

That is some very interesting stuff and the panel of investigators and advisors is quite impressive - even Buzz Aldrin is there

I would be very interested in whether there is anything documented yet in support of propulsion by HFGW other than a "star undergoing asymmetric octupole collapse". That seems a bit dramatic

I am always interested in learning new things

[Aaron]

joules to lift

So using 0.102kg as our apple weight we now see the time involved does not change the energy involved because the energy has the time embedded in it. So the energy is always just the Force multiplied by the distance and in these questions the force is always 1N (F=ma | F=0.102 * 9.8). Applying any other force, to give an alternate acceleration will always result in a non zero-velocity at your destination altitude.

So, supposing then that the destination is always zero velocity at the end of the move:

.75m * (.102kg * 9.8m/sē) = .75J
.50m * (.102kg * 9.8m/sē) = .50J
.25m * (.102kg * 9.8m/sē) = .25J

Now - if you were looking for the Joules involved where a different acceleration were provided to the end of the movement and you wanted to know the resulting velocity as well as the energy involved to get there, then that is a different set of calculations.

Hi Harvey,

At the peak, for my examples, the apple would be at zero velocity.
It goes up and comes to a stop. When it reaches that height, for a
short moment, the velocity is zero - it is neither going up or down.

Just to paraphrase what I think you're saying... if there is this situation:

If there is an object like a cube of pure lead, a small one that weighed
2 ounces. The joules required to lift it to 1 meter would be:

1.0m * (2 ounces or 0.0567kg * 9.8m/s sq) = 0.55566 joules

Is that right?

What about if I lift it 1/2 that height:
0.5m * (2 oz or 0.0567kg * 9.8m/s sq) = 0.27783 joules

So the same weight to 1/2 the height is exactly 1/2 the joules
required to lift it twice that height.

If I were to then assume these proportions stay, I could then guess
that if I lifted the lead cube to 0.25 meters then I could simply
divide 0.27783 joules in half to get the required joules to lift that
object to 1/2 that height or 1/4 meter:

0.27783 / 2 = 0.138915 joules

If I used the whole formula to double check that, I could use:
0.25 meters (0.0567kg * 9.8) = 0.138915

Please tell me if that is the basic concept. Thanks!

[Aaron]

calculating work

Here is one simple site that also seems to explain it:
Energy and Work

Calculating Work
Work = Mass * Gravity * Heightand is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull.

Work

=

M

*

G

*

H

=

2

*

9.8

*

0.75

=

14.7

Joules

If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.

Work

=

M

*

G

*

H

=

100

*

9.8

*

1

=

980

Joules

While the book sits on the table, no work is being done because no object is moving, even though forces are involved. If a force (like gravity) is applied to an object (like the book) but it does not move, no work has been done.

Work

=

M

*

G

*

H

=

2

*

9.8

*

0

=

0

Joules

Which of the following will result in more work? Running straight up hill, or taking a zigzag path up the hill?
The work will be the same for both paths. The direct path requires more force, but less distance, while the zigzag path requires less force but more distance.

Work = Mass * Gravity * Height

The equation above shows how to calculate the work done. As you can see, your work will be the same – no matter which path you take - because your mass doesn’t change during the trip. Gravity doesn’t change. The final height is the same. Therefore, work is the same.

Imagine you open a door by pushing near the hinges. Now imagine you open the same door by pushing near the handle. Which will result in more work? Again, the work done is the same. As you push the door near the hinges, the force needed to move it is greater, but the distance traveled is less. Pushing near the handle, the force is less, but the distance traveled is greater.

Another Equation for Calculating Work:

One can calculate the work done by an airplane by using the equation: W = F x D
For instance, if a model airplane exerts 0.25 Newtons over a distance of 10 meters, the plane will expend 2.5 Joules.

W

= F x D

= 0.25 * 10

= 2.5 Joules

If you apply a force over a given distance - you have done work. In general, the energy transferred depends on the amount of force and the distance over which that force is exerted. If the object doesn't move, you have not done work, even though you may have expended a great deal of energy! Oh well.

[light-travel]

Ideas:

"high frequency gravity waves" matches the concept "propulsion"

"propulsion from hfgw" matches "propulsion from propulsion"

if you grab something, say a outcrop of rock, you can gain momentum of your body forwards or upwards by gripping the rock-outcrop and pushing back with the further-away part of your hand and forwards with the nearer part of your hand. As you vary the pressure of front and near parts of your hand gripping the rock, you can propel your body forwards. A "propulsion backwards" or backwards pressure generated into the rock allows you to transmit force to your body assisting forwards movement.

I think 1 joule is energy to raise 1ml of water at 4 C by 1 degree C

[gravityblock]

Advanced propulsion systems use the atmospheric pressure of a planet to defy gravity. They create an external vacuum around the spacecraft in the direction of travel. The spacecraft experiences no friction or heat due to always moving inside a vacuum or low pressure. How to create an external vacuum? With cathode rays (Lenard Rays). The cathode rays will intersect the anode rays at an angle of 45 degrees. The vacuum creating effect is, however, not strictly due to the intrinsic speed of the ion, but to the atmosphere's ability to absorb ionized particles. While negative ions are absorbed by the atmosphere, the positive ones move towards the negatively charged surface of the saucer, at which point the electrons pass into the vacuum.

In an ordinary cathode ray tube the electric current reaches a saturation point which shows that all the atmospheric particles contained within the tube have been ionized. This is due to the limited amount of electrolyte within the confines of the tube. In the case of the flying saucer the electrolyte is made up of the whole atmospheric envelope of the Earth which never reaches saturation point. The ionized "bubble" surrounding the saucer is attracted and absorbed by the surrounding atmosphere with tremendous force and in its place only a vacuum is left, into which the saucer moves, impelled by the atmospheric pressure of 1.033 kg. per cm2. The whole of the outer edge of the craft acts as a cathode ray emitter. The intensity of the vacuum is proportional to the current used and is controlled by a rheostat.

GB

[gravityblock]

Terrestrial radius (6,378 km) * speed of revolution (106,000 kph) / Speed of rotation of the globe (1,660 kph) = 407,269 km. This gives a value of 407,269 km. for the radius of the total gaseous envelope. If we subtract from this figure the radius of the Earth (6,378 km.) we see that the terrestrial ether extends 400,891 km. beyond the solid surface of the planet. The Moon lies within the fringes of the etheric covering, so that the various phenomena connected with it take place within this covering.

The etheric covering acts as a fulcrum by virtue of which the two opposing forces of attraction and repulsion are able to act upon the Earth. We see, then, that the same force which causes rotation, moves the body through space. In the case of Earth, the rotational force is applied to the solid surface at a distance of 6,378 km. from the axis, but the effect of its movement through space occurs at 407,269 km. from the axis, at which point the surface of the etheric covering attains a speed of 106,000 km. per hour.

Having explained this, we can understand why planets of large volume are situated at a considerable distance from the Sun. By taking note of their distance from the Sun and their volume, we can discover their true density, and this will also give us the magnetic force of its poles. Thus the planet Jupiter is of low density and, having a large diameter, it is more subject to the force of repulsion than that of attraction. If it were true that matter attracted matter in direct proportion to the mass of the bodies, Jupiter, with a volume 1,330 times greater than Earth and 331 times as much mass, should be much closer to the Sun than Earth is.

GB

[Harvey]

Hi Aaron,

Yes, your calculations are correct.

Please keep in mind however that the constant being used (9.81m/sē) is only applicable near sea level. The greater our 'h' factor the lower that number will become and strangely, if we delve below the surface of the Earth it takes on a completely different curve because of lateral mass which is not evenly distributed. To be accurate, any point below sea level (like Death Valley or Indio, California) would need to be calculated as the vector sum of forces from all directions spherically. If you go down a deep mineshaft with the ocean on one side and a mountain on the other it could really get interesting

I find the last two sentences of that quote from the website interesting because it illustrates something with regard to lateral motion of mass. Because we are working with vectors, moving in one direction adds while moving in the opposite direction subtracts. Now, if we are pushing east for 10 meters and then our buddy pushes west for 10 meters, then the two forces are opposite as is the two directions. So we may look at the object and say that it has ended up right back where it started so the net distance is zero. And if we add the distances mathematically, we see this is true (+10) + (-10) = 0. The error commonly made is that persons would like to take this 'net' value and use it in the energy calculation while also summing the forces (+F) + (-F) = zero F. Or they may reason that if they push East they are doing positive work, and when the push West they are doing negative work and therefore the work nets to zero. While this may be true in the sense of accomplishment (nothing is accomplished by moving the object back to the starting place), it is not true in the case of work.

Here is where the mistake is commonly made. The do not properly break down the equation before summing the values. (+F x +D) + (-F x -D) is not the same thing as ((+F) + (-F)) x ((+D) + (-D)). The former is correct, the later is not. This is also where many people introduce errors in DC circuits that have AC in them.

So when calculating the Energy we must tally the products of the vectors before summing those products and this correctly gives us a positive energy value for both directions. (a negative multiplied by a negative produces a positive). This way, when we move the object East we get a positive energy value and when our buddy moves the object West our buddy also gets a positive energy value. Then those two are summed for the total energy involved.

Let's go vertical. Let us make gravity our buddy. We move our object up, our buddy moves our object down. For each movement we get an addition to the energy used. Remember that these motions are vectors. If there is no movement up or down then no energy was used in the vertical direction. While we may be able to put energy in laterally, gravity cannot unless we are subsurface.

So we may look at gravity like force on a sail of a sail boat, not the wind itself, but just the pressure it produces. A boat can tack into the wind, but not into the pressure. While gravity cannot provide a lateral force, if an incline is used a net lateral vector can be accomplished.

Now to blow the lid off of classical 'WORK' and show a way to extract work from gravity repeatedly

Gravity is a spherical force.

That force is equal along a spherical radius.

The radius of the Earth is such, that a level straight line connecting two points (A and B) 300 miles apart will have its midpoint subterranean by a value we will call Y. In this case A and B are exactly the same elevation.

From a gravitational perspective, Y is lower than A and B even though it is level. This is due to the fact that gravity is a spherical force.

Therefore, if we cut a level train track between A and B, and we place a train on that track, it will accelerate toward the midpoint between A and B. Once at the midpoint, gravity will decelerate the train such that when it reaches B it will be stopped. It does not matter how much weight you load your train with at A or B, it will always reach the other side at zero velocity. Oh . . . the track is made of levitation magnets so there is no friction and the train is inside a vacuum tube so there is no drag .

This train can move back and forth 24 hours a day, 7 days a week, 52 weeks a year and never require on Joule of energy to move it and yet, "work" is being done.

Why are we still using fossil fuels to move trains?

Because it is cheaper than digging troughs in the earth.

[gravityblock]

@Harvey,

Is the sun the center of the solar system, or is there a magnetic center within the solar system in which the sun and the planets orbit? I will wait for your response, then I will show how the sun is not the center of the solar system. The earth loses a year every time it completes a cycle, or in other words it loses one day in every seventy years. When the cycle of the spiral movement is completed every 25,784 years, a complete calendar year is lost.

This great spiral that the Earth describes is not only responsible for the 1,223 seconds difference between the tropical and sidereal years, but it also affects every other body in the system, including the Sun. Even the Sun, which is looked upon as the center of the system, itself revolves around a magnetic center, and this center also has a spiral movement of its own which corresponds to the movement of the planets.

The Sun moves around the magnetic center in an orbit of 6,250,000 miles in diameter and completes one revolution in 355 days. Thus it is that the ancient astronomers based their calculations on a year of 355 days. which is the true solar year, and not the movement that the Earth describes around the magnetic center, which takes 365 days.

GB

[Harvey]

@Harvey,

Is the sun the center of the solar system, or is there a magnetic center within the solar system in which the sun and the planets orbit? I will wait for your response, then I will show how the sun is not the center of the solar system. The earth loses a year every time it completes a cycle, or in other words it loses one day in every seventy years. When the cycle of the spiral movement is completed every 25,784 years, a complete calendar year is lost.

This great spiral that the Earth describes is not only responsible for the 1,223 seconds difference between the tropical and sidereal years, but it also affects every other body in the system, including the Sun. Even the Sun, which is looked upon as the center of the system, itself revolves around a magnetic center, and this center also has a spiral movement of its own which corresponds to the movement of the planets.

The Sun moves around the magnetic center in an orbit of 6,250,000 miles in diameter and completes one revolution in 355 days. Thus it is that the ancient astronomers based their calculations on a year of 355 days. which is the true solar year, and not the movement that the Earth describes around the magnetic center, which takes 365 days.

GB

Am I being setup?

Ok, I like a bit of fun just like the next guy so I'll bite

Do to the compound motion of celestial bodies there is no absolute center of any orbit over any given period of longevity as there is no stationary reference with regards to any celestial framework.

This combined with the fact that the orbits are elliptical in our solar system forces us to acknowledge that the true center of orbit is different for each body and is a function of the foci used to map the elliptical orbits.

But I still want to see all the very interesting data you have regarding the solar system cycles as it passes through the galactic plane - it sounds quite informative

[Aaron]

2 joules from one

Harvey,

I'm with you so far...

With the apple example, what do you think about Wiki's second part...
that the one joule of energy is released when that same apple falls to
the ground. And it was 20cm, not a meter, but that doesn't matter for
the conceptual examples.

-------------------------

Practical examples

One joule in everyday life is approximately:

• the energy required to lift a small apple 20 cm straight up.
• the energy released when that same apple falls 20cm to the ground.

-------------------------

When lifting an apple with 1 joule, we USE 1 joule to get it to 20cm.
If 1 joule is released when it hits the ground, then 2 joules of work total
were done. 1 joule worth of WORK to lift it and 1 joule of work in air
resistance and impact.

That is a cooperation with nature.

What is your take on it?

[Harvey]

Hi Aaron,

There should be air resistance lifting and falling so some of the energy must be applied to that even during the lift. Likewise, we are moving the mass a given distance, and even if we were in deep space where gravitational effects were negligible, we would still expend some energy to move the mass, and therefore that too is part of the energy making up that 1J. The rest of the energy making up the 1J is that portion that is used to counteract gravity - so we can view it as pushing the apple against the Aether flow upstream if we like.

The same energy must be applied to overcome drag and move the mass the given distance during the fall as was required during the lift. The kinetic energy in the fallen Apple can be calculated then by subtracting those two parts from the 1J, and we can consider it to have come from the Aether flow pushing the apple back Earth if we like.

There are very specific characteristics with regards to gravitational force the prevent us from accurately applying other forms of conservative force. For example, a spring acts with a conservative force - we push it and it pushes back. But the tension curve is much different than gravity so it just isn't a good analogy. A magnetic field acts very similar, with opposite poles attracting. But a true normalized magnetic field falls off with the inverse cube of the distance rather than the inverse square and gravity as we know it is a unipolar force. So using magnetism has its drawbacks as far as analogy goes. The Aether flow into the Earth's center is a good one because it solves those analogy problems and can be easily demonstrated with any object placed 'upstream' even if it enters from outside our atmosphere from some other source of creation

The reality that energy is released is just another way of saying that the stream accelerated it and put energy into it. And the reality that energy is viewed as being stored, is simply a way of saying that we placed the object into the stream so it could be acted on by that stream.

Now what would be cool, is if we could walk along the stream, throw our raft in and ride it back home without having to swim upstream For us, the stream is all around us. But for a rafter, the stream is confined in channels. So for us, with the Aether, we must move upstream, if possible, in channels where the Aether does not flow - so it is a bit inverse of how we are used to seeing water flow. It is more like we are in a constant downpour and looking for a dry spot, a hole in the clouds, so to speak.

[boguslaw]

Energy is disturbance in time flow.That means you have two way to increase energy : by making bigger disturbance or by manipulating time.
Our current energy conservation laws eliminated time dependence.
As Bearden said , we should rather speak about mass-time then mass.
Do something faster and you have energy gain.

[ewizard]

Rebooting this old thread with a fascinating device that says it runs and generates power with only gravity once it has been started.
RAR Energia Ltda

I have to believe that anyone putting that much money into a huge construction like that has down their homework and have built smaller working prototypes. They say they will building one in Illinois this year also.

[artoj]

Big Machine

Thanks for finding this one Ewizard, I can see they have not finished the unit, but here is some sequential pics anyway. Regards Arto

[Ruphus]

An interesting build for sure! Also a fantastic thread. Aaron and Harvey's thoughts on gravity and aether are great. I really like Harvey's thoughts on pressures and gradients. A pleasure to read it all.