1 Joule of Energy

[morpher44]

everything NOT the apple

The apple takes up space in our 3d world. If it is moved from point A to point B. Point A, now void of the apple, must respond by moving molecules of air, electrons, dust, etc. into that void as part of the wake caused by the apple moving out of that space. As the apple travels to B, further air is displaced, a wind current is established, electric fields are alerted, there is an inductive/magnetic change on the path. Molecules are spun.
There is a HUGE amount of unaccounted for energy.

[morpher44]

apple on a magnet on bismuth

So the 20cm assumes earth's gravitational pull, I would assume.
Surely lifting a small apple 20cm while weightless in space requires less
than a joule.
If the apple is sitting on a large toroidal magnet suspended above
bismuth, and I place a large magnet above the apple, I could probably
lift the apple expending only the effort to push the magnet out onto a platform 20cm above so that it lines up.
If the magnetic pull is large enough, the apple, on the lower magnet should rise.
Rolling the apple off the magnet and letting it fall its 20cm to give me a joule of work should be easy at this point.

Of course I could only do this one-time ... because now I have the problem of two magnets stuck together.
If only I had a way to disconnect the magnets expending minimal energy.

[Guest]

It's not very hard
Gravity is an equilibrium force
What you put in is the maximum you can get out, just like with magnets etc.

Unless you can find a way around that like shielding

[Aaron]

example

If we put in 1 joule, there is actual work being done to lift it to 20cm. Once it is at 20cm, the work that was done has already been done and cannot be undone - so we have to leave that much work recorded as having been done.

Let's say there is 10% losses, then 0.9 joules of work was done. We can record definitely that we put in 1 joule (we paid for 1 joule worth of work) and got 0.9 joules of work for example.

That 0.9 joules of work that was done will never change - yes unless we travel in time and other possibilities but for simplicity, a very simple linear example will do.

Now 0.9 joules of work has been done and I suppose we can say that after that 10% loss, the apple is now at an energetic potential difference from the ground of 0.9 joules. It has 0.9 joules of potential energy.

We now have in our record log or history book that 1 joule went in and 0.9 joules of work was done. AND, we still have 0.9 joules of potential energy.

With losses, lets say 10%, the apple drops and loses 10% in air resistance, etc... So it had 0.9 joules of potential energy, lost 10% and was able to do 0.81 joules in work until it comes to a dead stop and everything is in equilibrium.

We can now record that AFTER the apple was sitting at 20cm high, from that point until it dropped and came to a dead stop, it definitely did 0.81 joules in actual work that was witnessed, measured and recorded.

After everything is still, there is no more potential left over since it is all equalized now back to zero.

On the way up, 0.9 joules of work was done.
On the way down, 0.81 joules of work was done.

0.9 + 0.81 = 1.71 joules of work total was done.

The apple is sitting on the ground with zero happening.

We give 1 joule to the system - 1 joule that we had to pay for. Gravity and whatever other environmental input contributed to the work, mostly on the way down but nevertheless, nature provided input to the system in actual WORK that is measurable and we never had to input that extra energy ourselves.

So our system shows us that we put in 1 joule and got 1.71 joules of total work measurable out of the system.

If the system has no losses, which is theoretically possible, 1 joule goes in and the apple exhibits 1 joule of work going up and 1 going down. That is 2 joules of work that was demonstrated - 2 joules worth of actual mechanical motion/work over time and we only had to put in 1 joule of energy. Then this example is 100% efficient with a COP of 2.0. Nothing has been violated.

[Peter Lindemann]

Veljko Found the Way

It's not very hard
Gravity is an equilibrium force
What you put in is the maximum you can get out, just like with magnets etc.

Unless you can find a way around that like shielding

PArAdOX,

Did you watch the YouTube clip I posted in Post #9, above? Veljko found the way around the "gravity equilibrium" issue. He lifts the weight on the pendulum ONCE, and it swings and FALLS multiple times, as well as moving the cart multiple times. This shows that he can invest ONE UNIT of energy to raise the "potential" of the weight, then release that "potential energy" by letting the weight swing down, translate the dropping "potential" to "actual momentum" to move the cart, AND retrieve most of the "potential energy" to be used again. COP>1.

Peter

[Guest]

That's a very interesting demonstration. He might really be onto something there. It seems that gravity and inertia play significant roles in his experiment. I would not call that video conclusive evidence, but I definitly think it would be worth the time to investigate his work!

Force modulation provides us with free energy. We can do it with electrostatic force fields [Edwin gray tube, Water spark plug], electromagnetic fields [Tom Bearden's MEG], but gravity... wow

[Aaron]

bouncy ball

Peter,

That is a great vid that clearly demonstrates a more efficient approach.

For example, I just tested a rubber bouncy ball. When I drop it from any height, after it hits the ground, it bounces back up to 1/2 the height, when it falls then, it rebounces up to 1/2 of that height and so on until it stops.

I don't know the actual amount of energy to lift this ball 1 meter. It weighs about the same as a small apple and if 1 joule raises that 20cm, then I'll just ballpark it and say 5 joules to raise this ball 1 meter or 100cm.

5 up (1 meter)
5 down
2.5 up (50 cm)
2.5 down
1.25 up (25 cm)
1.25 down
.75 up (12.5 cm)
.75 down
.375 up (6.25 cm)
.375 down

(it does a few smaller bounces until it stops at this point - but the above is enough to show the point and this ball is very inefficient as it is very hard and one side even has a flat spot.)

---------------
19.75 joules in work done
for 5 joules invested.

COP 3.95 at 50% efficiency in this example.
It lost half the potential on every rebounce.

There were a few bounces not accounted for and
there will be some other losses on the way up and down
so these 2 will cancel each other out to some degree
so essentially COP 3.95 and 50% efficiency is a
pretty good estimate.

The point is that in a ringing oscillation that decays
little by little, there is free energy coming out of
the woodwork.

It has always been known that it is only necessary
to make up for the loss to keep a coil ringing, a spring
bouncing, etc...

So if on the second bounce, I were able to inject
2.5 joules of input to assist the ball back up to
the original 1 meter, it will bounce back down 1
joule and bounce back up 0.5 meter, which is
5 + 2.5 joules = 7.5 joules of work continuous
per cycle for every 2.5 joules of input that I have
to pay for.

2.5 joules input at the peak point on the second
bounce constantly will continue to produce 7.5
joules of work every single time.

This is free energy from nature and I'm glad to see
wikipedia promote the truth for once.

[ABCStore]

On the way up, 0.9 joules of work was done.
On the way down, 0.81 joules of work was done.

On the way up the work was "consumed" for the lack of better word. It's your system input (running battery). On the way down the work was "performed" (charging battery).
If I gave you and $100 bill and you gave it back to me the cash flow (what you call "work") is $200 but your net gain in 0.

Pendulum is a totally different scenario.

ABC

[Aaron]

efficiency versus cop

ABCstore,

I think it might be possible that you are confusing the concepts of efficiency with cop.

Work is a potential moving from one potential to another.

Work is performed on an apple on the way up. Our personal input is partially dissipated in losses and some is stored in the apple as more potential once it is at a resting height.

On the way down, more work is performed as that potential is dissipated in losses and some is transferred to anything else in the way. For example, maybe there is a little rock that the apple hits and this rock is kicked up onto a log. That rock then stores potential, which is able to do more work. Yes, it continues to diminish each time but nevertheless, work is constantly being done which is MORE than WE input. Doesn't this imply there is other input since a system can't have more TOTAL out than in?

Every one of my examples are all 100% efficient or less but COP can still be over 1.0.

COP ONLY takes into account the potential we provide to the system and COP doesn't include free environmental input (gravity, whatever).

When we take our input + free environmental input and see what comes out, that is looking at efficiency and all free energy systems are under 100% efficient. BUT, the output can exceed what we contribute.

Your cash flow analogy is a good one but I think you're comparing apples and oranges.

If you give me $100 that has $100 in potential buying power and is turned into "work" when it actually buys something. It triggered an actual transaction. Let's say you do give me $100. If you have a dvd player that you are willing to sell me for the $100, that much work was performed and you now have the $100 with a full $100 in potential buying power. You can take that $100 and spend it elsewhere and go buy a muffler for a car. That is more work performed for that $100. The muffler shop owner takes that same $100 and goes and buys a fish tank. More work performed with the same $100.

Yes it takes money to build those things to begin with and they didn't pop out of nowhere (fishtank, muffler, dvd player) but the point is, in our little economic system, we are utilizing all that environmental potential energy input (merchandise) that we didn't have to build.

There is still one single $100 bill, which is able to create - theoretically - an infinite amount of transactions because of all the potential transactions stored in merchandise, etc... that we can utilize and we never had to input anything other than the $100 bill.

Still one bill but many units of work was triggered by it.

[ABCStore]

I see where you're going with this. Even though the apple example is not very good. Any you're talking about an open system.
Now, if you go back and search for the thread I've started months ago you'll see that it is possible to have energy extracted from a completely closed system, harnessing magnetic repulsive force.

ABC

[Shamus]

Thanks for sharing this example--I love it! Same goes for the bouncing ball example. Examples like these really point the way towards open systems and environmental input. Just brilliant.

[quantumuppercut]

I try not to jump in, but I can't help it...

If you can ask gravity one question, what would it be?
I would ask " Gravity, how much work did you do on the apple?"
What if gravity said, "twice the amount of work you've done to push it against me" .
We've been jacked by gravity forever... well, except for Bessler.

mg=2ma

The centrifugal force is mg. The force require to put it in that velocity is ma.

[Guest]

I've always had a feeling that inertia was a type of "capacitor" that could store an excess of gravity's work. Could that be how some of Veljko's devices work?

[Jbignes5]

Hmm...

I think quantumuppercut is right.. What happened to the speed equation? Speed is the factor you are not taking into account here since gravity works on a free falling form the increase in speed of the apple and the amount of joules it takes to increase that speed hasn't been entered into the equation. If I lift the apple at 10cm per second I bet the speed of the apple is more then that falling. Hence it could be that gravity takes alot more to act on an object to increase that speed of the apple.
The classical newtonian law is F=mg
•F is the force of attraction between two objects
•G is the universal gravitational constant; G = 6.67*10-11 N-m²/kg². The units of G can be stated as Newton meter-squared per kilogram-squared or Newton square meter per square kilogram.
•M and m are the masses of the two objects.
I suspect the fall has little input of energy since gravity is a constant the fall might be in the sub joule range and what happens later is actually the energy imparted to get the apple up to that drop point. Making the net 1=1. Even with the result looking like it does more energy it doesn't.
Example given you move an object in space it takes 1 joule of energy to impart the movement and when you want to stop it it takes 1 joule of energy to stop. Adding gravity increases the joules imparted to it but decreases the energy used to fall with the potential energy used at the end when it stops to do "work". The fall is not work performed by the apple but by gravity the energy stored from the lifting of said apple is done at the end minus air resistance that is where you loose stored energy from the mass equation. Thats why a ball falling only goes halfway back up. It takes more to fight gravity to get back up and air resistance plays as well. Losses from the ball impact are another thing but then if there is sand on the surface it shows the energy of the impact as well. All of these add up to 1=1. Gravity is a constand to an extent and cannot be used in such a manner.
A better way to see it is move a ball in space in between two fixed walls and the ball will eventually stop impacting from the loss of the impacts against the ball.
I would think that all together 1=1 with losses or else balls on earth would bounce forever.

[Jbignes5]

Sorry

Sorry about the spelling mistakes but the forum will not let me edit it atm. Try this take a ball on a track in the shape of a U like but wider and drop it from point A being the left side and you will see gravity takes 2 times the amount of the mass of the ball to climb up. Gravity then works against the ball when before we imparted the energy the first drop. Whats left is what it recovered from the stored energy we imparted to it and and proceeds to drop from the new position on the right under it's own eneregy to climb against gravity which should be half as high as we started.
I hope this is some what understandable since I am not a professor but it is within 1=1-loss.