Free Energy from Earth's Gravity & Water Supply Discovered

[RutinaRottis]

What has potential energy ido with E=MC2? ..... This "ffeijdrug," assumes that basic rules of fysics are not correct....

ps Don't talk the talk if you can't walk the walk

[arjungupta9]

you have to do some work against gravity to lift water to ht of 5m and you will use gravity for 4m you lost 20% there man

find a better solution

[kingfaisal550]

your calculation is wrong i.e first you have to determine how many kg pressure of water is need to move your turbine ok so at 90 degree angle your pipe comes from the uper tank of water if the hieght above is 10 meter so your pressure in kg is 1 kg it is standard so at 55 meter your water pressure coming from hieght by gravity will be 5.5 kg,,,, so you first need to make a huge water tank... this is third class technology

[ffeijdrug]

@kingfaisal550 From Author: Flow rate out of the pipe is 9.8m/s divided by 5m times 4kg = 7.84kg/sec,

[rover5321]

this is retarded, hydroelectric power plants do the same thing, except nature is the pump which replaces the water in the lake WITHOUT loosing energy... there is no need to reinvent the wheel. 

[zoomal1]

It doesn't work

[ffeijdrug]

@zoomal1 It ALREADY WORKS in >nature<

[arjungupta9]

what about the work to be done against gravity for lifting water ??? any bright ideas ???

[TheChristiyanto]

@zoomal1 Agree

[butch14ish]

systems like these do work ,i have developed one myself,don,t forget that tork is very important.

[ffeijdrug]

From the author: Typical electric motor is 85% efficient, pump can be over 90% efficient and generators are typically over 90% as well. When the math is done, there is STILL 'excess' electricity produced.

[superdiza]

Here's a quick solution.

BUILT IT. SHOW IT RUN.

that's it.

your math makes no sense.

[Sonyoooo3]

This idea looks like my theory of using the ocean and lett the wind,wave,solar, moving water under the ocean do its job to pump water like a floating fountain ?

[Sonyoooo3]

Maybe in the future it will be fountain inside or outside the house for making free energy that comes from rain?

[nambinhvu]

Put your thumb over the tip of the hose, you'll get more energy XD

[nambinhvu]

o.O the pump doesn't work like that, the end the water comes out of has to be lower than the end the water is being sucked up from, or the water will be pulled back in.

[butch14ish]

@nambinhvu no, thats what we learned in school

[MIchaelGoguen]

the video of the fan spinning seems to reverse, do some weird things when I watch it... if it is spinning at a consistent speed 'in real life', it may be that the frame rate of recording is making it harder to see the consistent motion... sunlight is probably better than electric light at 60hz for this, too... ... I think upping the frame rate of recording/ quality might help, but I don't know if frame rate is lost when put on youtube, or if the frame rate of the screen viewing would affect too.

[ffeijdrug]

@MIchaelGoguen LOL, it's an optical illusion that it seems to reverse directions. It looks that way to the naked eye too. As the fan speeds up, the eyes can't discern which way it is going. Weird illusion.  It was actually going same direction all the time.

[ThePiones]

The article about siphons on wikipedia should be useful. Don't forget the explanation using the Bernoulli principle.

[ThePiones]

Dude, the velocity of the fluid the same in all of the pipe. Since the flow is constant, and the area of the cross section is constant, the velocity must be constant. Just google volumetric flow rate

[victor15richardson]

@ffeijdrug also if the water is accelerated up at 9.8m/s/s it will not move because the system will be at equilibrium as Force(up)=-Force(down). Therefore acceleration upward>9.81m/s/s for an upward acceleration, and so the net force at the pump will equal the force at the top. The extra force it gains when falling with gravity, will be equal to that lost at the pump overcoming gravity. Thus the efficiency would be 1.00 (ignoring friction and changing the height water drops from 4m to 5m).

[ffeijdrug]

@victor15richardson See the info through the first reference link in info. The unit must be primed, that is, the water accelerated up to a given velocity in the pipe. Once this is done , it only requires constant [and lesser] force to maintain the flow from then on. Yes, it would take more than 196 joules of energy to accelerate the water flow, initally, but then once that is accomplished it would only take 196J to 'maintain' the flow. Constant force on water = to the mass of the water.

[victor15richardson]

@ffeijdrug. Hey mate looks like your put alot of thought into this. I am very interested. However i have absolutely no idea how you found the velocity to be 9.8m/s at the end of the pipe?? please explain.

[ffeijdrug]

It is not in use today, because the science behind it is not in the science books yet. The principle is 'somewhat' in use now, via hydroelectric dams. The water is going to go downhill anyway, so they take advantage of that and tap the flow to make electricity. Same with this device. Water will fall, regardless. Back to resevoir. So you tap it on it's way down. Water through a siphon on a farm also sends water downhill, untapped. It too could be tapped. Any downhill flow could.

[newage4energy]

Hey man, this guy really believes in his system. Like most, someone will ask if hes built one, will add whats missin, pretty good responses for 3 month old post.

[MrFudgeisgood]

I would like to point out that if a "gravity powered generator" was this simple wouldn't it be in use today

[saintr17]

I was just thinking if you needed speeds at 13-14m/s at the turbine on the same concept what about using a restrictor valve, decrease from 2 inch pipe forced through a nosel will increase the velocity of the water. less volume at higer velocity should increase productivity

[saintr17]

I did. the gravity calculator included in the links say that in 1second the water will fall 4.9 m/s so the time to fall 4m is .9035 seconds. then if you use any of the physics calculators online to find vertical acceleration due to gravity. vert displacement @ 4m, time .9 seconds(not .4 sec ), inital vert velocity 9.8m/s, acceleration is 9.8m/s/s gives you 11.9m/s. i love this concept which is why i did the numbers not to debunk but so you can recalculate to get truer results.

[ffeijdrug]

@saintr17 The way to figure the time of freefall is to divide: 4/9.8 So freefall time is slightly less that .4 seconds. g is 9.8m/s/s so a free fall from standstill of 1 second would give 9.8m/s velocity. Remember that the water is 'already' going downward at 9.8m/s, so freefall time is shortened to point four seconds or less. The total energy output cannot exceed unity. So it cannot exceed 2 times 196joules no matter how low to the resevoir the turbine is. Thanks for ur input!

[saintr17]

I like the concept but some of your math is wrong. with 9.8m/s as the initial vertical velocity add acceleration due to gravity at 9.8m/s/s , with a displacement of 4m. Time: 0.9035079029052512s giving you total vertical velocity of 11.901234567901 meter/second^2

[ffeijdrug]

@saintr17 What AshenSalvation had said , before removing his comment, was that

end velocity is 9.8m/s plus 9.8m/s/s(elapsed time of freefall). Elapsed time of

freefall is aproximately point four seconds to turbine. Use the gravity calculator

linked in info, as stated by me below to calculate.

[KirkMcLoren]

sorry - no cigar

[ffeijdrug]

@KirkMcLoren And your field of expertise is??? WHY no cigar? Explanation please.

[ffeijdrug]

Water exits the pipe at 9.8m/s. It then falls 4m to turbine. Time of freefall is aprox. 4/9.8 seconds. Using the SECOND link in info, insert .4 into the question: How fast is an object going after falling for t seconds? Then add this to 9.8m/s to arrive at the 13.8m/s stated. AshenSalvation commented on this, but now seems has remvoved comment. I don't think he realized he was exactly correct, but proving the end vel. , not disproving it.

[DFWITDEPARTMENT]

nature uses solar power(residual energy) to convert water into a water vapor that rises into clouds and falls.

break this down more in your head and u should be able to figure out that lifting 1 pound of water is not going to give you 2 pounds worth of energy

and the directing the end of the pipe, you're only redirecting the excess energy produced by the pump.

it takes equal amount of energy to lift the water as u could potentially generate from the water's fall. and there is always a loss.

[ffeijdrug]

The water only becomes Potential energy if it comes to a stop at the top. In this design the water does not stop, but comes out of the pipe, downwards at 9.8m/s. Thus it may be consedered Kinetic energy of 196J, not Potential.

[RICHLES01]

The system does work to the laws of energy conservation, Potential energy Pgh, mass x gravity height. It takes a very powerful pump to lift to the height you want. These pumps will use far more energy than they can create via a generator. You need to build and you will find that it will not work. I tested and built many devices similar and at the end they dont work. Thanks for the work though, free energy will be cracked one day.

[Khyber68]

@ffeijdrug how did you calculate the 13.8 m/s for the water falling down?

[ffeijdrug]

@Khyber68

See link in description. Urls of references are listed there.

[Khyber68]

@ffeijdrug ya, I did that and the velocity after 4 m should only be 8.86 m/s, so where did you get the 13 m/s?

[ffeijdrug]

@Khyber68

It would be coming out of the upward curved pipe at 9.8 m/s to begin with, then the free fall of 4m.

[overunitydotcom]

Did you calculate in the hydrostatic paradoxon which also

needs more energy to pump up

small pipes diameter dimensions waterflow ?

The hydrostatic paradoxo says, that it only depends on the height of the water,

not on its pipe diameter...so I guess your principle will not work ?

[ffeijdrug]

@overunitydotcom

Right, dependent on height. Weight of water in pipe is 4kg. Since E=2mc^2 full unity would be 200% efficiency, which is impossible to attain. The actual efficiency of this model would be less than the 193% stated. Would be 150% or so, because no pump or turbine or generator is 100% efficient. This works because gravity is not subject to the law of conservation of energy. It is an exception.

[ffeijdrug]

@overunitydotcom

see parameter based relativity, on my channel, for info on 2mc^2

[ffeijdrug]

@beac310

See reference urls in description, to see how the calculations were made.

[ffeijdrug]

Yes, hm, seems too good to be true, I know, but see the 2 sites I used to develop it. Just sent them to ya via email.

[worldspacenews]

hm?