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  #91  
Old 08-22-2010, 01:20 AM
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Harvey Harvey is offline
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Quote:
Originally Posted by Aaron View Post
I have a question for anyone that can answer it - Harvey, anyone?

It takes 1 joule of energy to lift a small apple to 1 meter.
Over how much time? Let's just say 1 second and lets
say it is in a vacuum so air resistance isn't even a factor.

How much does the apple weigh? It would weight about
1/4 pound or 4 ounces - or whatever weight is necessary
to require 1 joule to lift it to a meter in a second but it should be
fairly close to that.

My question(s) is this:

How many joules of energy will it take to lift the exact
apple to 0.75 meters over 0.75 seconds?

How many joules of energy will it take to lift the exact
same apple to 0.50 meters over 0.5 seconds?

How many joules of energy will it take to lift the exact
same apple to 0.25 meters over 0.25 seconds?

To this day, thousands of people have not been able to
answer this simple question.

If it requires more information to calculate the answer,
let me know what more is needed,

Hi Aaron,

The Joule is a measure of energy. Energy is a measure of Power multiplied by Time.

So, if you use 1W of power for 1 Second, then you have used 1J of Energy.

This structure helps us to understand that the energy dimension already has time embedded in it.

For those reasons, when we say we are using 1J for a given period of time it can be a little confusing because you are essentially adding a time factor to something that already has a time factor in it.

For example - if you say that your 4x4 burns 4 Gallons of fuel for every 60 miles you drive, you can equate that to a specific value of energy (1 Gallon = 1.3 x 10^8 J ). So you could transfer back and forth between miles per gallon (15 MPG) or Miles per 1.3 x 10^8 J (same number - 15 M/1.3 x 10^8 J)). But if you want to relate that to time - you must introduce it from your speed, because it is already embedded in that value. @60mph, it will take 1 Hour. At 30mph it will take 2 hours. At 15 mph it will take 4 hours. But the energy is the same in all 3 of those scenarios, the only thing that is different is how long it took you to use it.

But now let's really talk about the root basis of your question: Acceleration

If I am going to move a mass from one location to another then I have a distance. It will take a specific Force and a Specific amount of time to move from Zero velocity to some constant value that I can apply over that distance.

Back to your 4x4 - You take off from a standing stop and Accelerate to 60mph and then you drive for 58 minutes @ 60mph. We already know how much energy you used to cover that 59miles. But we don't know what it took to get that vehicle up to 60mph from a stand still and we don't know how long that took or how much distance it took. Let's suppose it took a full mile (nice easy acceleration). If your acceleration was constant (linear) , what would it need to be to get to that speed in that distance? Acceleration = Delta Velocity divided by Delta Time. Since we know that we are going 60mph, that is our Delta Velocity (we are supposing here that our acceleration is along a straight line). Now, we have put in a very specific value that forces the time to pop out, the distance of one mile. The distance formula: d = 1/2 ( Vf + Vi ) × t will tell us what the time is in hours where Velocity Final (Vf) = 60 miles per hour and Velocity Initial (Vi) = 0.

This shows us: t in hours = 1 mile / 0.5 x 60 = 1/30 Hours or 120 Seconds.

So our Acceleration must give us 1 mile in 120 Seconds. How do we get the instantaneous acceleration? Using the formula a = (Vf - Vi)/t we get 60MPH / 120 or 0.5MPH per second which would look like 0.5 Miles/hour/s which is the same as saying 8.8 inches per second per second or 8.8"/s². But, in real terms, you would watch your speedometer increase by 1/2mph for each second you are accelerating up to 60MPH.

Now it would be good to know the mass of the vehicle to get to the Joules involved during the acceleration. The Joules are the dimension associated with work and work and work is Force times Distance. Therefore, if we know the mass, we can calculate the force from the acceleration (F=ma) and determine the Work in Joules (W=Fd).


To make things easy I am going to switch to SI units here - meters, kilograms, and seconds(mks).

1 mile = 1609m
Let us say that the 4x4 has a mass of 2268kg (Typical F250 4x4). F=ma: This means our Engine provides a Force of (2268kg * (0.5*1609m)/hours/seconds)Nm or (2268 * 0.2235m/s²)= 506.84Nm of force to accelerate to 60MPH during that two minute period. Of course that is just to move the mass, it does not account for any friction or drag or other typical losses in the drive system. But it is interesting that it represents only 1/150 of the energy contained in a single gallon of gasoline. In other words, we could stop and accelerate like this every 2 miles for 300 miles and only use 1 gallon of fuel if we could apply 100% of the fuel energy to moving the mass.

The work in Joules then is W=Fd where F = 506.84 and d = 1609, so W = 815497.515J

Now for your questions:

If you have 1J of energy in moving an apple of some unknown weight one meter in one second then perhaps we can suppose that it was moved under a consistent acceleration and has some value of velocity when it reaches that one meter mark. Since the definition of one Joule is ((kg · m)/s²)m or (kg · m²) / s² then your apple must weigh 1kg because the only way to get 1J in this case is to multiply (1kg * 1m²) / 1s² anything else in the kg slot will not result in 1J. So our apple has a 1kg mass which on Earth equates to about 2.2Lbs in weight (remember weight and mass are two different things)

So if Wikipedia's Practical Examples says it takes 1J to 'lift' a small apple one meter, then does this mean that they think a small apple weighs 2.2Lbs? You will notice that the article does not specify a time limit. It is similar to our 4x4 using 4 gallons of gas to go 60 miles regardless of how long it took to get there. So there is clearly something different between the energy it takes to move 1kg 1m in 1s and the energy it takes to lift a small apple with no time limit.

The difference is in the 'lift'. Here we have a continuous negative acceleration of 9.8m/s² pulling on that mass. While it takes 1J of energy to move 1kg 1m in 1s horizontally, it would take more to move it vertically against gravity - 9.8 Newtons multiplied by 1 meter, or 9.8J to lift it that one meter. So when they say small apple, they are saying 1kg / 9.81 = 102 grams (3.6 oz) so that the lift is only 1J. Notice that this is time independent. It does not matter how long it takes to lift that 102g, it will always take 1J to lift it 1m against gravity. W = Fd and the Force is 9.8N and the distance is 1m.

So where W = Fd, and F = ma, and a = dV/dt so Work in Joules = d(m(dV/dt)) has time embedded in it. So even when the apple is stationary on a table, there is still a negative acceleration on it of 9.8m/s², or a Force of -9.8 Newtons. This implies that there must be a normal force of +9.8 Newtons (or acceleration of +9.8m/s²)in the opposite direction to balance the two forces to stale mate.

An interesting experiment in this regard would be to lift a ball 1m and drop it inside a quarter-pipe of 1m radius and see how far it rolls on ice horizontally - maybe 9.8m? Does it matter how fast we raise the ball to the starting point?


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/s²) = .75J
.50m * (.102kg * 9.8m/s²) = .50J
.25m * (.102kg * 9.8m/s²) = .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.

__________________
"Amy Pond, there is something you need to understand, and someday your life may depend on it: I am definitely a madman with a box." ~The Doctor

Last edited by Harvey; 08-22-2010 at 09:42 AM. Reason: Fixed typo was .50m is .25m (Thanx Aaron ;-) )
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  #92  
Old 08-22-2010, 01:44 AM
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Harvey Harvey is offline
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Regarding the Rock and Hole:

The Rock sitting on the ground has some potential value based on it's distance from the center of the Earth. The farther the distance, the greater the potential. Digging a hole next to the rock does not change that potential unless the material removed is some how extricated from the Earth entirely thus reducing the Earth's total mass. And even then, it would be minuscule.

It does not matter if there is a normal force holding the rock, or if we look at a snapshot in time as the rock passes ground level as it hurls into a 7 mile deep hole - the potential energy in the rock is the same (the kinetic energy and momentum will be different however). The potential is always a factor of the force multiplied by the distance and the force is essentially the weight of the rock or the mass times 9.81m/s² and the distance is the altitude from the center of the earth.

So, using the little apple from before at 0.102kg times 9.81m/s² for a Force of 1N we now multiply that by the radius of the Earth (6357000m) and find it has an Energy product of 6,357,000J (Relative of course to the center of the Earth)

So that is the rest potential of the apple sitting on the ground. The farther you move it from the center of the Earth, the greater it's potential. The closer you move it toward the center of the Earth, the lower its potential.

Of course an apple falling out of an airplane with no parachute does not consider that he has more than 6.3MJ of energy because he is more worried about reaching the ground long before he ever reaches the center of the Earth. So he would like to know how much Potential Energy he has relative to the Earth's surface. That value will be the differential between his potential energy at a specific altitude and the potential energy he has at ground level.

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  #93  
Old 08-22-2010, 08:17 AM
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Aaron Aaron is offline
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calculating work

Quote:
Originally Posted by Harvey View Post
Hi Aaron,

The Joule is a measure of energy. Energy is a measure of Power multiplied by Time.

So, if you use 1W of power for 1 Second, then you have used 1J of Energy.

This structure helps us to understand that the energy dimension already has time embedded in it.

For those reasons, when we say we are using 1J for a given period of time it can be a little confusing because you are essentially adding a time factor to something that already has a time factor in it.

For example - if you say that your 4x4 burns 4 Gallons of fuel for every 60 miles you drive, you can equate that to a specific value of energy (1 Gallon = 1.3 x 10^8 J ). So you could transfer back and forth between miles per gallon (15 MPG) or Miles per 1.3 x 10^8 J (same number - 15 M/1.3 x 10^8 J)). But if you want to relate that to time - you must introduce it from your speed, because it is already embedded in that value. @60mph, it will take 1 Hour. At 30mph it will take 2 hours. At 15 mph it will take 4 hours. But the energy is the same in all 3 of those scenarios, the only thing that is different is how long it took you to use it.

But now let's really talk about the root basis of your question: Acceleration

If I am going to move a mass from one location to another then I have a distance. It will take a specific Force and a Specific amount of time to move from Zero velocity to some constant value that I can apply over that distance.

Back to your 4x4 - You take off from a standing stop and Accelerate to 60mph and then you drive for 58 minutes @ 60mph. We already know how much energy you used to cover that 59miles. But we don't know what it took to get that vehicle up to 60mph from a stand still and we don't know how long that took or how much distance it took. Let's suppose it took a full mile (nice easy acceleration). If your acceleration was constant (linear) , what would it need to be to get to that speed in that distance? Acceleration = Delta Velocity divided by Delta Time. Since we know that we are going 60mph, that is our Delta Velocity (we are supposing here that our acceleration is along a straight line). Now, we have put in a very specific value that forces the time to pop out, the distance of one mile. The distance formula: d = 1/2 ( Vf + Vi ) × t will tell us what the time is in hours where Velocity Final (Vf) = 60 miles per hour and Velocity Initial (Vi) = 0.

This shows us: t in hours = 1 mile / 0.5 x 60 = 1/30 Hours or 120 Seconds.

So our Acceleration must give us 1 mile in 120 Seconds. How do we get the instantaneous acceleration? Using the formula a = (Vf - Vi)/t we get 60MPH / 120 or 0.5MPH per second which would look like 0.5 Miles/hour/s which is the same as saying 8.8 inches per second per second or 8.8"/s². But, in real terms, you would watch your speedometer increase by 1/2mph for each second you are accelerating up to 60MPH.

Now it would be good to know the mass of the vehicle to get to the Joules involved during the acceleration. The Joules are the dimension associated with work and work and work is Force times Distance. Therefore, if we know the mass, we can calculate the force from the acceleration (F=ma) and determine the Work in Joules (W=Fd).


To make things easy I am going to switch to SI units here - meters, kilograms, and seconds(mks).

1 mile = 1609m
Let us say that the 4x4 has a mass of 2268kg (Typical F250 4x4). F=ma: This means our Engine provides a Force of (2268kg * (0.5*1609m)/hours/seconds)Nm or (2268 * 0.2235m/s²)= 506.84Nm of force to accelerate to 60MPH during that two minute period. Of course that is just to move the mass, it does not account for any friction or drag or other typical losses in the drive system. But it is interesting that it represents only 1/150 of the energy contained in a single gallon of gasoline. In other words, we could stop and accelerate like this every 2 miles for 300 miles and only use 1 gallon of fuel if we could apply 100% of the fuel energy to moving the mass.

The work in Joules then is W=Fd where F = 506.84 and d = 1609, so W = 815497.515J

Now for your questions:

If you have 1J of energy in moving an apple of some unknown weight one meter in one second then perhaps we can suppose that it was moved under a consistent acceleration and has some value of velocity when it reaches that one meter mark. Since the definition of one Joule is ((kg · m)/s²)m or (kg · m²) / s² then your apple must weigh 1kg because the only way to get 1J in this case is to multiply (1kg * 1m²) / 1s² anything else in the kg slot will not result in 1J. So our apple has a 1kg mass which on Earth equates to about 2.2Lbs in weight (remember weight and mass are two different things)

So if Wikipedia's Practical Examples says it takes 1J to 'lift' a small apple one meter, then does this mean that they think a small apple weighs 2.2Lbs? You will notice that the article does not specify a time limit. It is similar to our 4x4 using 4 gallons of gas to go 60 miles regardless of how long it took to get there. So there is clearly something different between the energy it takes to move 1kg 1m in 1s and the energy it takes to lift a small apple with no time limit.

The difference is in the 'lift'. Here we have a continuous negative acceleration of 9.8m/s² pulling on that mass. While it takes 1J of energy to move 1kg 1m in 1s horizontally, it would take more to move it vertically against gravity - 9.8 Newtons multiplied by 1 meter, or 9.8J to lift it that one meter. So when they say small apple, they are saying 1kg / 9.81 = 102 grams (3.6 oz) so that the lift is only 1J. Notice that this is time independent. It does not matter how long it takes to lift that 102g, it will always take 1J to lift it 1m against gravity. W = Fd and the Force is 9.8N and the distance is 1m.

So where W = Fd, and F = ma, and a = dV/dt so Work in Joules = d(m(dV/dt)) has time embedded in it. So even when the apple is stationary on a table, there is still a negative acceleration on it of 9.8m/s², or a Force of -9.8 Newtons. This implies that there must be a normal force of +9.8 Newtons (or acceleration of +9.8m/s²)in the opposite direction to balance the two forces to stale mate.

An interesting experiment in this regard would be to lift a ball 1m and drop it inside a quarter-pipe of 1m radius and see how far it rolls on ice horizontally - maybe 9.8m? Does it matter how fast we raise the ball to the starting point?


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/s²) = .75J
.50m * (.102kg * 9.8m/s²) = .50J
.50m * (.102kg * 9.8m/s²) = .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,

You're a master at breaking things down!

I am aware time is in the "energy" but bring up an example of 1 second
because I got the apple example from the same wiki reference. I saw
they don't indicate time or exact weight so made up a 1 second
example as a ballpark in the moment for an actual small apple.

Besides the real numbers, the most important thing I would love to see
laid out in plain site is conceptually what is happening.

And of course NO, it does not matter how fast we raise something to
the point at which it is dropped in regards to if that would alter how far
it would roll, etc...

But I would add again that all that work input is completely dissipated
the moment it reaches the peak and starts to fall.

Any work that happens to the ball from that point on did not come from
transformed "energy" that we put it, any work from that point on is from
free environmental input from gravitational potential and is a completely
different potential that is not locked to the input potential.

.75m * (.102kg * 9.8m/s²) = .75J
.50m * (.102kg * 9.8m/s²) = .50J
.50m * (.102kg * 9.8m/s²) = .25J

I'm assuming you meant the last to be .25m - its obvious that is what
you meant I think.

I'm going to revisit this concept with your explanation in a video or
photographic experiment.

But for now, just a quick other thing I want to bring up...

If a capacitor has 1000v and 2uf (stored), it is said to have
1 joule of "energy stored"...lol anyway, I won't pick on that for now
anyway, but with 1 joule of "energy"...

Let's say it is in a capacitor and can be dumped into an electromagnetic
coil and that coil repels a magnet into the air (in a tube or something).
With 1 joule of "energy stored" in a capacitor if a magnet was 10
grams, (with or without losses in air resistance, etc...) it should be
able to repel that magnet to a certain height (over a certain time).

So if 1 joule in a cap launched a magnet higher than 1 joule allows,
that would be over 1.0 cop wouldn't it?

Anyway Harvey, I * DEEPLY * appreciate your willingness to spell this
out in your perspective!

Arm is more qualified to do your response more justice but it is an
important point to have these things defined if we're even going to talk
about them.

It should be brought up where the mathematical and conceptual concept
of a joule of energy came from and what experiments were done in order
to determine 1 joule of energy. If anyone is willing and able - dropping
a weight on water to increase it by a certain temp, etc... those definitions
either in a mathematical perspective or conceptual way is very important
in my opinion.

If the bare basics of the very structure we're supposed to be analyzing
aren't spelled out, then there are less frames of reference for all of us to
relate our perceptions to.
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Aaron Murakami

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  #94  
Old 08-22-2010, 09:33 AM
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Aaron Aaron is offline
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potential difference

Quote:
Originally Posted by Harvey View Post
Regarding the Rock and Hole:

The Rock sitting on the ground has some potential value based on it's distance from the center of the Earth. The farther the distance, the greater the potential. Digging a hole next to the rock does not change that potential unless the material removed is some how extricated from the Earth entirely thus reducing the Earth's total mass. And even then, it would be minuscule.

It does not matter if there is a normal force holding the rock, or if we look at a snapshot in time as the rock passes ground level as it hurls into a 7 mile deep hole - the potential energy in the rock is the same (the kinetic energy and momentum will be different however). The potential is always a factor of the force multiplied by the distance and the force is essentially the weight of the rock or the mass times 9.81m/s² and the distance is the altitude from the center of the earth.

So, using the little apple from before at 0.102kg times 9.81m/s² for a Force of 1N we now multiply that by the radius of the Earth (6357000m) and find it has an Energy product of 6,357,000J (Relative of course to the center of the Earth)

So that is the rest potential of the apple sitting on the ground. The farther you move it from the center of the Earth, the greater it's potential. The closer you move it toward the center of the Earth, the lower its potential.

Of course an apple falling out of an airplane with no parachute does not consider that he has more than 6.3MJ of energy because he is more worried about reaching the ground long before he ever reaches the center of the Earth. So he would like to know how much Potential Energy he has relative to the Earth's surface. That value will be the differential between his potential energy at a specific altitude and the potential energy he has at ground level.

Harvey,

I get that the distance from the center of the Earth plays a role.

But I would see that the rock is not what has any potential value.
I don't see it as a matter of semantics either. The "value" of potential from
the center of the earth in my opinion should be assigned to the
gravitational potential itself and not to an object that is subject to that
gravitational potential. Of course the properties of the apple will be taken
into consideration but that in and of itself has no change, reduction,
or increase in any amount of potential being "stored."

Also, the apple in free fall is near motionless relative to the flow of
the gravitational potential
. It achieves an equilibrium of sorts when it
cannot accelerate any faster.

Under acceleration, the aether is pushing it down at a speed with the
apple resisting a certain amount meaning the apple is not moving as fast
as the gravitational potential. If the apple got to a point where it did not
accelerate any faster, it would be at an equilibrium with the flow of the
gravitational potential - just to keep the example simple.

When the apple hits the ground, there is resistance the apple encounters
and the aether pushes on the mass of the apple with a certain amount
of real force - force in the literal sense of mass being acted upon.

The resistance of the apple to the flow of the aether from free fall is the
potential difference and that resistance is work. There was a bit of
work with air resistance but not as much as a direct impact.

If we have a 6 inch pipe of water pressurized at 100psi and put the apple
in the flow it will accelerate until it is at the speed of the water or
at least close enough depending on how much the apple resists the
movement - lets just call this free fall.

If that pipe is 10 feet or 100,000 miles long, the potential in that apple
isn't changing - assuming the water is the same along the entire
length, lets just compare that to 9.81. If it traveled 100 feet or 1 million
feet, there is no change in the potential in that apple.

The apple isn't feeling a push on it when "in the flow" so for all practical
purposes, there is no potential difference at all - (no matter how far
or close to the grate it is - there is no difference in potential).

If the water goes through a grate and the apple can't get through,
it is then and only then that the water exerts a real push on the apple
and that push is from the resistance of the apple resisting the
push from the water.

So essentially, none of the potential from the water was usable by the
apple UNTIL it was able to resist it. Then and only then was there a
potential difference and then work was manifest.

So an apple in free fall is in near equilibrium with the aether. It is not using
any of that potential and there is no potential difference relative to the
falling apple. When the apple hits the ground, then and only then does it
resist the downward push of the aether on the mass of the apple.
Then and only then does that free gravitational potential supply the
source potential to cause actual work in measurable joules of energy.

Again, I don't see it as semantics. When lifting an object, that doesn't
create the potential difference and it doesn't address what the actual
source of potential is.

The higher it goes, it has more potential in the abstract form but not
in the actual.

If the rock is further from the surface, I say it has access to less
potential, and I am correct in saying that because that does not
include any variable of the rock having more distance to fall. In that
time and place that an object is sitting STILL at a higher altitude,
there is less local potential it has access to.

Any reference to the concept of something "having" more potential
implies a local phenomena of more potential gathering, having, storing,
increasing, etc... when in fact, the local nature of the potential the
object has access to in it's locale is actually decreasing at a rate
at whatever proportion to the distance from the earth.

That picks at the "storing potential" concept more than anything but
does not take away from the fact that you are correct that there is
more "potential" the further from the surface. Again, that is why I state
that the higher an object is, the less potential is there at the object,
which is completely opposite from the idea that as an object is raised,
it "stores" more potential.

But the "more" potential the higher up part I think in aetheric terms is
to state that the higher an object is, the longer it can be in the
flow of the dynamic river of gravitational potential in order to build up speed so there is a greater "potential" difference upon impact meaning more
work is done - resisting air, impacting and damaging the object, whatever...

I'm not trying to
correct you Harvey, I'm just paraphrasing in the perspective of an
aetheric gravity model - mine at least - everyone can state their own
based on their model.

You make a good point about the dirt being removed from Earth and
thereby reducing its mass overall. But of course for all practical purposes,
it would be insignificant - or minuscule as you mention.

The apples has quite a bit of "potential" that far from the center of the
Earth - assuming there is even a gravitational pull there. It could get
lighter and lighter down there for all we know if we get close enough
to the center.

Again, using the surface as the reference point or "ground", the higher
up the apple goes, the longer it would have access to gravitational
potential if it fell and there is no intrinsic increase in potential in the apple itself,
meaning there absolutely is no greater potential "stored" in the apple
the higher it goes.

It is AFTER the apple falls that more of anything is realized but not before.

The text book definition of this is seeing it backwards and is not a matter
of semantics. It is like saying the universe revolves around the Earth.
All the calculations to show what is doing what will still work out to
determine proportions and relationships when but seeing it for what it
is - it is the Earth that revolves around the sun, which revolves around
the center of the galaxy, etc...
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Aaron Murakami

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  #95  
Old 08-22-2010, 10:09 AM
Cloxxki Cloxxki is offline
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Epic post Harvey. If you don't teach for a job, perhaps you should.

"An interesting experiment in this regard would be to lift a ball 1m and drop it inside a quarter-pipe of 1m radius and see how far it rolls on ice horizontally - maybe 9.8m? Does it matter how fast we raise the ball to the starting point?"


Sorry for highlighting the one thing that doesn't seem perfect in your post. If a drop an objec from ANY height, onto a quarterpipe and level exit of ice (low drag) surface, the resulting horizontal displacement until the roll is over, is, In my understanding indefinately. No specific value it to be expected, in any multiple or fraction of a gravity constant, in any unit of measure.
Furthermore, the radius of the quarterpipe is irrelevant unless friction is greatly pressure dependant.
Or, I misunderstood your explanation, in which case I apologize.
More likely, you were just messing with our minds with your example, in which case I didn't fall for it.

My simple understanding of raising an object vertically has always been tightly connected with velocity change. If yo first accelerate the apple upwards, say 0.5m at 9.8m/ss, and then let go of it, another 0.5m later, another same amount of time later, it has reached 1m, with Vend=Vstart.
If you put in the joules for the first 0.5m of constant accelleration, you get a bit of extra height "for free". In case of matching gravity, inversed, in my little logic, you get the same speed and distance, making it easy even for me to comprehend.
Would you accelerate at only 1m/ss, you'll need to raise to much closer to 1m high, in order to get to that 1m with a small free end lift, where vertical velocity is exchanged for height until it stalls.
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Old 08-22-2010, 10:42 AM
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Hi Aaron,

There certainly is a particular issue associated with the local perspective of gravitational potential energy. The issue is akin to a conversation I once had with a fellow on You-Tube named Desertphile with regards to magnetic potential energy and my first You-Tube videos entitled "Magnets Doing Work".

The issue has to do with creation and transference and it really underscores what you are saying about how things are perceived.

In my conversation with this fellow I posed a question to him regarding magnetic potential energy - I asked him, if I create a new magnet, and the magnetic field is said to be infinite by classical terms, then does this create potential energy with respects to every magnetic material in the universe? He refused to answer but instead turned the conversation to gravity.

So, in trying to help him realize that we live in an 'open system' we call the universe, I asked him what the potential energy was for an incoming meteor and asked when exactly was that potential given to the meteor. For some reason he treated me as some sort of nemesis.

But that is the reality of it. Every object in the universe contains a potential relative to every other object in the universe and that potential begins at the time that energy is converted to matter (or matter is transmuted).

So we can say that lifting a spaceship off the surface of Earth imparts potential energy to the spaceship but it is entirely subjective. Because the potential is only measured against some reference. When that exact same spaceship reaches Jupiter do we assume that some magical transference has occurred and now the spaceship embodies some special energy field that increases it's potential by 318 times? No. And there is the other problem - transferring our reference. The simple act of moving our reference from Earth to Jupiter somehow causes all this potential energy to manifest itself.

Similar things happen in electronics. You can have a split rail power supply and measure both positive and negative voltages relative to the power ground reference. But as soon as you move your reference from ground to the negative supply, now you get the sum of the two supplies added together. So while it may seem like we just doubled our supply voltage, the reality is that the potential was always there and all that needed to be done was move the reference. It may surprise us to discover that all of the voltage on both supplies is a negative voltage when we move our reference to Earth ground.

So while we may perceive things a certain way, it does not mean they are as we perceive them. We may lay on the ground looking 'up' at the stars when in fact we may be hanging under the Earth looking 'down' . Try that sometime and imagine you are hanging under the Earth We say that energy can be stored as potential energy because it is how we perceive it. In reality the potential has always been there and we are simply moving to the other potentials relative to a specific reference. Fortunately in electronics we can determine what a true zero reference is because it can be demonstrated chemically to show a zero charge. But how do we do that gravitationally? What is our reference or touchstone for true zero potential?

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Old 08-22-2010, 11:49 AM
light-travel light-travel is offline
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gravity = natural teleportation?

First post here:

thoughts:

to move an object, requires pushing, or pulling.

With gravity, objects move without pushing or pulling.

To teleport an object, you need to give it "imaginary momentum', so that it "switches off gravity" (by creating its own space-time continuum or "gravity"?)(----------), and "imaginary displacement" (so it switches gravity on only relatively- it will then "jump" to another location where a complimentary pattern of vibration has been created to the pattern generated in the object at its initial location.....???).

How and why a ball bounces is related to how to levitate something- if instead of "bounce" you get "a jarring action", this is one-step below that needed to teleport.

Partial teleportation (i.e. where part of the movement of an object is via a teleport-factor), and partial levitation (where part of an object's state on the Earth may be levitation (so it presses less) may be quite common.

(Shockwaves in rocks in multi-collisions could "factorise their mass (at least partly) causing the rock to develop "a rare-earth charge" (or gravito-electric/magnetic 'charge')(gravity = imaginary electro-magnetic effect?)(electro-magnetism fits apparently the idea "juxta-position" (abc -> acb); imaginary juxtaposition fits the idea "square is the distance") so press less on the Earth?)(A spring-weight measurer does a similar trick so cannot be directly used to detect if this effect is occuring?)

"Natural (partial) teleportation" may occur in some rocks in multi-rock slides, but only be detectable by looking at the relative positions of many rocks vis-a-vis the boundary of the area they end up in. It would show up as "extra displacement" not able to be accounted for by traditional physics, but measuring the field of rocks via conventional physics would probably discharge the effect so proving it would require a different way...?
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Old 08-22-2010, 11:58 AM
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Quote:
Originally Posted by Harvey View Post


So, in trying to help him realize that we live in an 'open system' we call the universe, I asked him what the potential energy was for an incoming meteor and asked when exactly was that potential given to the meteor. For some reason he treated me as some sort of nemesis.

But that is the reality of it. Every object in the universe contains a potential relative to every other object in the universe and that potential begins at the time that energy is converted to matter (or matter is transmuted).

But how do we do that gravitationally? What is our reference or touchstone for true zero potential?


Hi Harvey

When you defend opinions against what others believe be true, there is a tendency to antagonize or ignore you, which in my view is even worse, than consider you an archenemy.
That is the “knowledge potential referential frame” that we choose, and if we do not understand that it is easy to move to a different frame, we just stay on it and there is no gains.
I believe more in a transmutation by association of matter with kinetic energy, (the matter reduces velocity and associates with other matter) than in the energy-matter equivalence.
Thanks to you, I understood that space-time is just an effect with time as reference (?) and can in fact be the “Aether”.
If the Aether does not act on each other but in the transmuted matter only, then, will we find the Gravity zero potential when we found the God´s particle?
Imagine if we can move ourselves to that reference, then we can be everywhere?
It is what the spirit does when leaves the body?

Thank you
David
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Old 08-22-2010, 08:53 PM
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Hi GB,
Sorry for my late answers to your questions below. I knew that the Wave Structure of Matter (WSM) model that I use the most had some answers but I had to go learn what they were, draw some diagrams and then check it with the folks who make the model (not me) and then wait for confimation. Note that when I use the terms "space" or "vacuum" you can just substitute the work "aether". They all just refer to a medium in which stuff happens. I'll use aether here since it's the one you seem to prefer.

Quote:
Originally Posted by gravityblock View Post
Thanks Aromaz,
Does anyone have anymore "What if's" to add to Aromaz's theoritical gravitational model? Please post your own personal model, so we may all learn together. One of the models or combination of models may be correct. It wouldn't be a surprise if gravity is due to a variety of effects and causes interacting with each other.
GB
The simplified, complete answer to what is gravity along with pretty diagrams can be found here:
http://wsminfo.org/faq.htm#WHAT_IS_GRAVITY
That's my own website, even though the model isn't mine.

But since not everyone looks at webpages, in short: gravity is a movement of particles from lower density aether to higher density aether. In the WSM model, the aether is filled with waves what are constantly being exchanged back and forth between particles, which are themselves composed only of waves riding on the medium, aether. A wave is emitted from a particle with a certain starting amplitude but that amplitude drops the farther it gets from the particle. Waves arriving at a particle arrive with increasing amplitude. So basically, the farther from the particles, the lower the wave amplitude. And conversely, the closer to the particles, the higher the wave amplitude.

The Aether density, really the Aether energy, at any one location is calculated as:
density = mc^2 = hf = c x sum_for_waves_from_all_particles(amplitude^2 / distance_from_particle^2)
So basically the density is proportional to the square of the amplitude of all waves from the universe at that location.

If you have a large mass of particles like the Earth then the amplitudes of the waves contributed by those particles will be relatively high and so the density, and hence gravity, will be high. If you are far from any mass then the wave amplitudes will all be smaller since the amplitude is lessor the farther they are from the originating particles. So out in the middle of space the density, and hence gravity, will be smaller.

The reason particles move from lower density aether to higher density aether is because, as I said above, particles are really just made up of the waves that arrive at and leave the particles. Waves moving through the higher density region move more slowly than waves moving through the lower density region. So the waves arriving at a particle from the side of the particle closer to the Earth will be moving slower than the waves arriving on the other side of the particle, farther from the Earth. Since particles are really just made up of these waves, by definition the particle is now closer to the Earth. I'll try an attach the last diagram on the above webpage to this post; it shows it well.

Quote:
Originally Posted by gravityblock View Post
Thanks Aaron and Harvey.
I think this thread is moving forward and in the right direction. Thanks to All. Keep the information and ideas flowing. Can gravity be controlled, manipulated, reversed, negated, or turned on/off to our benefit? If yes, then a theory for how?
GB
How can gravity be manipulated? As I said above, the Aether density is really Aether energy. If you look at the above formula again you'll see that it's even equal to mc^2. The amount of average energy at any location is equal to the mass of the universe (m) x c^2. And it's really the energy in all those waves. If you tap into that energy, then you decrease the amount of energy i.e. the density. Since gravity is a movement of particles from a lower density aether to a higher density aether, and you've just reduced the density by stealing some of that energy, you've just reduced local gravity!

Where'd the energy come from? The waves. Those waves were headed to some particles and now those waves are diminished in energy. So whatever energy exchange that was going on between the particle that emitted that wave and the receiving particles will be diminished. So if that exchange was making the partiticles vibrate, i.e. heat energy, you've just cooled them. If that exchange was attraction of a negative charged particle to a positive charge particle then you've just stolen some of the electrical power (current, voltage) in whatever circuit was doing something.

When John Searle allegedly fired up his SEG and connected a load to it, two things happened. One was the machine got colder and the other is it flew. So he may have been tapping Aether energy. David Hamel's allegedly created a plasma and then took off (more on my particle creation idea below.) Floyd Sweet, at the prompting of Tom Bearden, also measured a weight loss when his device was running. And then there are all those reports of car's stalling close to UFOs.

So how to tap the energy so you can get the side effect of manipulating gravity? My idea is to convert that energy to particles, i.e. particle creation. My webpages describing it are at:
http://rimstar.org/sdenergy/vacuum_energy
But as a result of doing this research so I could answer your questions, I realized the calculations on those webpages are wrong. I should have been using the above formula for density instead of just summing the amplitudes. So I have to go redo the calcs. However, with the way I was calculating it before, I couldn't see where the net energy gain was coming from. In terms of the math there was no net gain, all the waves from the universe just cancelled each other out since I was just summing positive and negative amplitudes. But now since I'll be squaring the amplitudes the gain is very clearly from all the waves from more distant particles not involved in the device. So I seriously owe you a big thanks for asking these questions and causing me to look more closely again at WSM! Now, hopefully the new calcs will show that it's doable.
-Steve
http://rimstar.org http://wsminfo.org
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Old 08-23-2010, 01:14 AM
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Quote:
Originally Posted by light-travel View Post
First post here:

thoughts:

to move an object, requires pushing, or pulling.

With gravity, objects move without pushing or pulling.

To teleport an object, you need to give it "imaginary momentum', so that it "switches off gravity" (by creating its own space-time continuum or "gravity"?)(----------), and "imaginary displacement" (so it switches gravity on only relatively- it will then "jump" to another location where a complimentary pattern of vibration has been created to the pattern generated in the object at its initial location.....???).

How and why a ball bounces is related to how to levitate something- if instead of "bounce" you get "a jarring action", this is one-step below that needed to teleport.

Partial teleportation (i.e. where part of the movement of an object is via a teleport-factor), and partial levitation (where part of an object's state on the Earth may be levitation (so it presses less) may be quite common.

(Shockwaves in rocks in multi-collisions could "factorise their mass (at least partly) causing the rock to develop "a rare-earth charge" (or gravito-electric/magnetic 'charge')(gravity = imaginary electro-magnetic effect?)(electro-magnetism fits apparently the idea "juxta-position" (abc -> acb); imaginary juxtaposition fits the idea "square is the distance") so press less on the Earth?)(A spring-weight measurer does a similar trick so cannot be directly used to detect if this effect is occuring?)

"Natural (partial) teleportation" may occur in some rocks in multi-rock slides, but only be detectable by looking at the relative positions of many rocks vis-a-vis the boundary of the area they end up in. It would show up as "extra displacement" not able to be accounted for by traditional physics, but measuring the field of rocks via conventional physics would probably discharge the effect so proving it would require a different way...?
It is interesting that we can move objects with electric fields, magnetic fields and gravitational fields in very similar ways.

Electric fields can be unipolar - but to push or pull it needs to have a same or opposite pole and the force between the two can be measured and cause action at a distance.

Magnetic fields are always (so far) dipoles and like the electric fields they must have a same or opposite polarity magnetic field to interact with in order to push or pull. These to can be measured and cause action at a distance.

Gravitational fields are always (so far) unipolar. But unlike Electric and Magnetic fields - the Gravitational field will act on any mass regardless of the Electromagnetic state and it does not need an opposite or same polarity to cause the pull. It too can be measured and cause action at a distance.

Young children learn early how to cause an action at a distance when they blow out a flame on a candle. So intuitively, they learn that action at a distance can be caused by moving air locally and having that move to the object and cause the action at a distance from the source.

We know that water flows in the turbines at a hydroelectric plant and yet by using wires that can cause an action at a distance where the energy is applied.

We know that if we turn on our water hose, we can move leaves with the stream. Or we can use a blower or compressed air etc.

So intuitively, we understand the concept of using a medium to transfer energy from a source to a load. So we rationalize that if a magnetic field is like wind or water that we can put energy into and get an action at a distance. And likewise, we can view Gravity in similar terms as a fluid like field that we can put energy into and get that energy to move to a distance and cause an action. The moon puts energy in and moves our tides from a distance. And interestingly, the Sun has a greater gravitational force than the moon at our location here on the Earth. But the gradient for the moon is steeper than the gradient for the Sun, so a larger potential exist across the Earth and this cause a greater differential to exist for the water to align with. Interesting stuff


There is one thing you mention that is quite worthy of evaluation: "a jarring action"

There is a way to look at this that is not easy to explain in simple terms, but I will try.

I will say that there are five states of position relative to time and each higher state is a rate of change of the lower state. In Calculus, each state is called the derivative of the prior state and is given a notation which we call a "Prime". The symbol for Prime is the single closing quote. So if we were to say P-Prime, we would write it as P'.

Zero state: Position. P
This is a single motionless point in a coordinate system relative to a reference. There is no lower state, thus this is not a derivative.

First State: Velocity. V = P' (spoken "P prime")
Velocity is the rate of change of Position and is the first derivative of position.

Second State: Acceleration. a = V' = P'' (spoken "P double prime")
Acceleration is the rate of change of Velocity and is the second derivative of position. It is also the first derivative of velocity.

Third State: Jerk (or jolt). j = a' = V'' = P''' (spoken "P triple prime")
Jerk is the rate of change of Acceleration and is the third derivative of position, the second derivative of velocity, and the first derivative of acceleration.

Fourth State: Jounce (or snap). s = j' = a'' = V''' = P'''' (spoken "P quad prime")
Snap is the rate of change of Jerk and is the fourth derivative of position, the third derivative of velocity, the second derivative of acceleration and the first derivative of Jerk.

There are two other facetiously added progressions to this pattern that are called 'crackle' and 'pop' but I see no reason to extend this out that far.


Let us evaluate a real world situation where all of these things can occur:

The Drag Race - The car sits on the line waiting for the tree to turn all green. This is Position, P and velocity is zero.

The light turns green and the car begins to move. For a few short inches we have a change in position, P' aka Velocity, V.

But very quickly that Velocity begins changing, and the car begins to accelerate, a. This means that there is a compound change in position, P'' and a change in velocity, V'.

In a couple of seconds the engine tach's out and the driver must shift gears. The momentum of the car keeps the car in acceleration during the shift, but the power curve gives the driver a sudden burst of energy after the gears have meshed and now we get a noticeable Jerk, j as there is a rate of change in acceleration. Right in the middle of that Jerk, the tires lose a minor about of traction and you may even hear a chirp, but when the grab you get . . .

The Snap, s. While the acceleration was changing somewhat consistently during the tire chirp, the Jerk itself has a rate of change as the tires transit from slipping to grabbing. And thus we get the final state.

If each of these rates are known for the precise time they become active, you can accurately identify precisely where the position of the Dragster is at every instant in time.

So what is "a jarring action"? Well, it is most likely a Jerk, or even a snap. Ever watch a child pull a wagon over a tree root and the wheel gets stuck? What does the child do, intuitively? Since the tire stopped with a constant velocity, the child automatically initiates acceleration by rolling the wagon back a bit and accelerates it. The wagon pops up in the air and falls back trapped. So what does the child do now? Rolling the wagon back and then accelerating it, just as the root and the wheels meet they give the handle a little jerk. Presto, problem solved.

This jarring action is a very good application of energy and can have some unexpected results. It is a way to concentrate a lot of energy in a very short period of time.

Case in point:
Recently I wanted a single slice of individually wrapped sandwich cheese. After consuming my snack I had the plastic wrapper in my hand and I thought I would snap it and play with my cat a bit. So holding it between my fingers and thumbs, I crossed my hands and then spread them out quickly expecting to hear a loud audible pop. Although I was holding the material in just a small 1" diameter area, to my surprise I found myself holding two pieces of wrapper with a clean break that looked like a scissor cut perpendicular to the direction of the pull. "Must have been a weak spot in the plastic" I thought. I tried it again - same thing, clean precision break straight across the entire width of the plastic. A third, fourth and fifth time all resulted in the same type of tear. So I pondered, I wonder if I gradually increase the pressure over time if there is a breaking point that will result in this behavior. I pulled the plastic with gradual increasing pressure and soon it began to deform. It stretched and stretched until I had huge bubbles surrounding my thumbs, but no tears or breaks.

While this behavior is specifically related to the ability of the plastic to change shape under pressure in a certain amount of time (I forget what we call that ) the experiment does illustrate how effective the 'Jerk' or 'Jarring Action' can be. We apply this in our daily lives, opening jars, sliding doors, pushing a grocery cart over a clump of gum on the floor, loosening a nut with a wrench etc. And it is one of the reasons for multistage rocket engines that needed to reach escape velocity.
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Old 08-23-2010, 02:35 AM
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Potential and Sound

Potential:
I think we have a description problem here. Similar to that between Photon and Light. Potential as word is suppose describe the maximum potential. A 9mm bullet has the potential to kill a 100Kg human over distance of 30 meters; but while lying on the table it cannot even kill a milligram of mosquito.

Physics description is:
“The energy stored in a body (or system) due to its position in a force field” Now this is a problem, regardless all mathematical equations. Theoretical the maximum energy is the theoretical Einstein E=MC^2. Thus a base ball has the potential energy of its weight times the speed of light. However it is limited to the speed of the bowler’s hand or bat velocity bouncing it away – it just did not reach its full potential.

The potential of the ball next to a big hole is that of mass x speed of light; however it’s real potential is zero; until it gets in motion. Here is another thorn, should such ball for instance be on a perfect smooth level surface, and roll at 1m/s; its potential is not only the direct energy with which it moves forward, but also the drag behind, above and below it.

Sound waves:
Why is it that sounds travels faster in cold temperatures; and more clearly? Well, most will say it is due to higher density of the medium (air) through which it travels. Maybe, it is true because when we look at misty environment the same is also true. But when in smoke from some distant forest fire – the sound is dampened, yet the air is denser than pure clean cold air.

As if this is not enough, sound travels better in upwards direction (away from core of earth) than the opposite. Yet, going down increases density of air – and moisture. In a deep mineshaft of say 100 meters, the person at the bottom can speak normal and be heard on surface; but if the topside person shouts full out, he can barely be heard at the bottom – even if there is a lid on the shaft top.

Just pondering these two issues, Aaron stimulated my lazy brain and recent smog in Thailand added to the aspect of sound waves.
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Old 08-23-2010, 05:13 AM
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Quote:
Originally Posted by Matos de Matos View Post
Hi Harvey

When you defend opinions against what others believe be true, there is a tendency to antagonize or ignore you, which in my view is even worse, than consider you an archenemy.
That is the “knowledge potential referential frame” that we choose, and if we do not understand that it is easy to move to a different frame, we just stay on it and there is no gains.
I believe more in a transmutation by association of matter with kinetic energy, (the matter reduces velocity and associates with other matter) than in the energy-matter equivalence.
Thanks to you, I understood that space-time is just an effect with time as reference (?) and can in fact be the “Aether”.
If the Aether does not act on each other but in the transmuted matter only, then, will we find the Gravity zero potential when we found the God´s particle?
Imagine if we can move ourselves to that reference, then we can be everywhere?
It is what the spirit does when leaves the body?

Thank you
David

Thank you David for the kind words and encouragement

Whether or not the Aether can act on itself has been the subject matter of discussions I have had recently off forum. It must be able to, I think. Otherwise, I don't think a wave could propagate using it as the medium.

I watched a cat in a zero gravity environment where it was flowing at the same speed as the Aether in the zero gravity craft and I noticed that there is a clear connection between matter and space time because as the cat attempted to land on its feet through inertial maneuvering, it actually began to accelerate in the spin. What was the cat pushing against when it was spinning? So there must be a binding between matter and space-time that we interrupt when we accelerate matter.

What would a disturbance in space-time be like? Perhaps a vortex:YouTube - underwater tornado in Aruba

Or rushing wave:YouTube - Pearlies


I imagine it would be a longitudinal spherical wave. Of course so did Oliver Heaviside :The State When the Speed of Light is Exceeded
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Old 08-23-2010, 11:02 PM
light-travel light-travel is offline
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fifth dimension...

Thanks Harvey;

I have limited computer access- will need to print your reply-

some quick comments:

an experiment was successfully carried out on the T.V. program called Top Gear". A snowmobile drove at high speed (70 mph I think) and continued from land on to deep water, and travelled across an expanse of water without sinking. At rest, the snowmobile would have sunk. (There is also a lizard that can "run on water").

I think the "Hutchinson effect" is like "catching": to catch something "it mustn't "jar" (i.e. mustn't 2-d bounce?) - ther shock waves must disperse back-and forth in a relatively "random" way (or star pattern).

A square-shaped magnetic field could cause an object to "jar "externally" (part-levitate?).....

The 'Searle effect" I think is like "bouncing": to "bounce" an object must (in some degree) "square up" (or refuse to jar- i.e. not line-up its internal and external structure-support). A star-shaped magnetic field I think could assist an object to "bounce "externally" ("jar" wholy internally- so creating a strong push from its surface(s) on surrounds- again would look like a levitation tendency...

I think the inverse of electro-magnetism (of "slip" or "area differentiation")
is "magno-electricity" (or "grip" , "volume differentiation": a constant "jarring" (bounce))
(e-m as "a jarring constant" ).

A "grip field" (ability to change between different grips) = a (virtual) gravitational field.

The inverse electro-magnetic state (the magno-electric state or "grip-state") of an object could interfere with a local gravity source (or "the internal structure of an object" could resist gravity (in other words, "an archway")....?

An archway doesn't "fall down", it "falls up" ....???

RE: your very interesting analysis:

my thoughts are (I left velocity as you wrote it, the others I have depicted somewhat differently (perhaps from a 2-d math perspective (Imaginary category, or "jump" perspective (Partial differentiation perspective?)(?) .....)



ero state: Position. P

..or non-position ...?

what is the difference?

Need a triangle (triangulation) to decide...?



First State: Velocity. V = P' (spoken "P prime")
Velocity is the rate of change of Position and is the first derivative of position.

Second State and third states: Acceleration.

Second, Third, fourth State: Jerk (or jolt).

+- Fourth State: Jounce (or snap).

5th (------------) crackle

6th (------------ ---------) pop

Sorry about the gaps- potential value (Re: new technology I'm thinking about) saying that (not states as such)

(Gravity, "5-d", pre-math Einstein General Relativity, are all the same pattern, is my impression)(= "+ - fourth; "state" ")........

........................

........................

............
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Old 08-23-2010, 11:34 PM
light-travel light-travel is offline
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a cat in space

thought:

the cat "pushes against itself"- but cannot do so effectively without rotating (about one (or more than one ?) fixed axis

A spinning object (try walking towards (or away) the center of a merry-go-round) stacks to-away from the center of spinning; by spinning the cat can slingshot its paws (get extra resistance to propel its legs outwards?)

(conserving additional angular momentum so do not have to conserve (all) of it- so can retain some leverage ...?)

Thoughts: to set up an axel jump in ice-skating, one makes "an imaginary step" (raises one foot up as if to walk a step on the ice) - on putting that foot down the other side of one's body tends to lurch forwards- but as one lurches anyway deliberately to make the spin, the other lurch-effect dilutes the lurch-for-the-spin, allowing a measure of control to be achieved re: the spin (It becomes a mass-integrated spin or moment levitation effect- add jumping to this and you get "a controlled landing"- add "a controlled landing" to this and you get a slight imbalance- add putting one leg out on landing the axel--jump and you get "Neutral density bias' (or a gravitational 'transistor')

To make a double axel, the same pattern but with more force although less momentum (create a slight sideways-frontways uncertainty leading into the double-axel ?)
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Old 08-24-2010, 01:03 AM
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stevend stevend is offline
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Hi Harvey, light-travel,
If I recall correctly, jerk is what propulion using and detection of High Frequency Gravity Waves (HFGW) are supposed to use. I think the main researcher is:
http://www.gravwave.com
Funny enough, I did a google search on High Frequency Gravity Waves to find some links for you and the first that came up was a US government paid for report saying the whole HFGW idea is fundamentally wrong. Of course that didn't stop me from keeping looking to find the above link and wouldn't you know it, their first document is one refuting the US government one saying that they were looking at the wrong effect:
http://www.gravwave.com/docs/Q%20&%20A.pdf
So if you see the US government one, ignore it.
-Steve
http://rimstar.org http://wsminfo.org
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Old 08-24-2010, 02:28 AM
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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
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Old 08-24-2010, 09:12 AM
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joules to lift

Quote:
Originally Posted by Harvey View Post
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/s²) = .75J
.50m * (.102kg * 9.8m/s²) = .50J
.25m * (.102kg * 9.8m/s²) = .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!
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Old 08-24-2010, 09:25 AM
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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.
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Old 08-24-2010, 12:16 PM
light-travel light-travel is offline
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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
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Old 08-24-2010, 08:25 PM
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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
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Old 08-24-2010, 08:55 PM
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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
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Old 08-24-2010, 09:04 PM
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Hi Aaron,

Yes, your calculations are correct.

Please keep in mind however that the constant being used (9.81m/s²) 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.
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Old 08-24-2010, 10:21 PM
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@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
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  #114  
Old 08-25-2010, 01:25 AM
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Quote:
Originally Posted by gravityblock View Post
@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
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Old 08-25-2010, 08:32 AM
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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?
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Old 08-26-2010, 06:48 AM
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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.
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  #117  
Old 08-27-2010, 08:09 AM
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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.
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Old 05-03-2013, 05:40 PM
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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.
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Old 05-04-2013, 06:00 PM
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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

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Old 05-04-2013, 09:47 PM
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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.
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