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# Over-unity, a Novel attempt.

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posted on Dec, 3 2013 @ 08:13 AM

that the hell ? did you do ANY basic science education ????

Are we not forgetting the velocity of the ball, which means the ball weighs more on its way down than it does on its way up ?

no the mass of the ball remains the same

and unless you somehow change the acceleration due to gravity [ 9.81m/s2 ] - its weight remains the same too

I can only assume that you somehow misunderstand the formula E=M V2 ???? [ energy [ of a falling object ] = mass * velocity squared ] ]

and if so - your premise does not work

while it is true - that the further an object falls - the more it accelerates [ constrained by its terminal velocity [ a function of its cross section / air density ] ]

once it has stopped falling - it has to be returned the further distance

you do realise that - I hope

so - using your " model "

lets assume that a magnet falls - 1m

and ignore air resistance etal - we get a fall time of 0.45 seconds

that's the amount of time that the ball can produce electricity as it passes through the coils

this is the electricity that you expect to pump > 1 litre of air into your " submersible "

PS - neither will mechanical air compression - using the impact of the ball on a piston or other such " solution "

I can see that you don't take well to criticism - but the fact is that your idea is worthless

and you are scientifically illiterate

get over it

posted on Dec, 3 2013 @ 08:18 AM

translation = you cannot handle criticism

any post that disagrees with you is not " negative " or " thoughtless " just because it disagrees with you

I am not going to fact check every post - but mine disagree with you because your science is flawed - but mine is correct

if you cannot handle this - then maybe you should not post

bottom line - do not expect people to agree with a flawed premised - then spit your dummy when they refuse to and point out your errors

posted on Dec, 3 2013 @ 08:28 AM

Yes
That's a neat flashlight tho, may cost you a few bobs more.

posted on Dec, 3 2013 @ 08:40 AM

ignorant_ape

translation = you cannot handle criticism

any post that disagrees with you is not " negative " or " thoughtless " just because it disagrees with you

I am not going to fact check every post - but mine disagree with you because your science is flawed - but mine is correct

if you cannot handle this - then maybe you should not post

bottom line - do not expect people to agree with a flawed premised - then spit your dummy when they refuse to and point out your errors

Firstly, In the opening post I made it clear I made no claims, and implied that I couldn't see the "Fatal flaw" of the design and asked peeps here at ats to study and fully understand the design and explain where that "fatal flaw" lay.

Now go and look at most of the replies.....How were they helpful.

Manfromeurope received 1 star for saying I had a circle of movement but that was not over-unity...though it clearly would have been, had it worked.

Others received stars for posting nothing constructive.

As I have said, maybe posting this here on ats was a mistake, and possibly I should have found a relevant forum for this to have been discussed by people who knew what they were talking about.

posted on Dec, 3 2013 @ 10:31 AM

and you are scientifically illiterate

And you failed to read properly, so what does that make you.

An arrogant as well as ignorant ape....no.

posted on Dec, 3 2013 @ 11:30 AM

StellarX

boncho
You have a flawed premise. Define "over unity" first.

In fact i would suggest that you start and define both perpetual motion ( which i guess presumes no friction and perhaps truly impossible perpetual motion) as well as over unity. If you truly believe that these terms can be used interchangeably perhaps i can help you resolve that state of ignorance/spite...

Stellar

There is a 300 page thread on this already. If you want offer up "Your" version of over unity. Because it appears to be a relative term.

Of course, it has history being used in math/statistics, as "over unity=over 1" meaning that a math problem had a bad equation. "These formulas are wrong I ended up with over 1"

Then you have the perpetual motion crowd, who started saying "over unity" because they realized "perpetual motion" would immediately end any conversation they were having. (Hundreds of years of trying to get a rock to fly does that.)

Then you have the obfuscation of claiming that over unity is something different, new, but really people are just pushing COP, except they simply refuse COP calculations, or the basic principle.

So yes, do enlighten us.

The OP has drummed up a classic perpetual motion machine. And calls it over unity. Great example of the difference.

posted on Dec, 3 2013 @ 11:41 AM

ken10

and you are scientifically illiterate

And you failed to read properly, so what does that make you.

An arrogant as well as ignorant ape....no.

1kg bearing, needs a 1.01kg submersible to bring it back to the top, which requires 1.02 kg of air to lift it.

Your falling bearing is going to produce a small bubble that isn't going to lift the submersible. All you have done by adding in the water is make your contraption a little confusing. That's it. There are no gains from the water being there. Since if you remove the water, you have nearly the same contraption (as the weight distributions still are relative).

posted on Dec, 3 2013 @ 11:45 AM
Well since I have posted this thread I may as well use it to see if I can find the actual "fatal flaw" that kills this design....and please if you don't have anything constructive to say...Don't post.

So, I have been searching around for real world magnets and pumps to try and do some calculations, BUT I don't think there are any calculations that can be made to find out how much current could be made by a magnetic ball falling through copper coils..... I think this will need to made and tested.

Magnetic Balls = 12mm x 12 grams weight

Air Pump = 800 mlpm 3.3 volt dc

So if we use the 12mm magnetic ball weighing 12grams, that means we need 12mls plus of air to lift the submersible to any height we desire.

And using the air pump described, it would need to run for 1.5 seconds to pump 20mls of air.

am I correct so far ?

If so could the ball dropping through the coils have a chance of producing enough current to do that ???

posted on Dec, 3 2013 @ 11:46 AM

Fatal flaw of the design: Laws of Thermodynamics

posted on Dec, 3 2013 @ 12:36 PM

ken10
Well since I have posted this thread I may as well use it to see if I can find the actual "fatal flaw" that kills this design....and please if you don't have anything constructive to say...Don't post.

So, I have been searching around for real world magnets and pumps to try and do some calculations, BUT I don't think there are any calculations that can be made to find out how much current could be made by a magnetic ball falling through copper coils..... I think this will need to made and tested.

Magnetic Balls = 12mm x 12 grams weight

Air Pump = 800 mlpm 3.3 volt dc

So if we use the 12mm magnetic ball weighing 12grams, that means we need 12mls plus of air to lift the submersible to any height we desire.

And using the air pump described, it would need to run for 1.5 seconds to pump 20mls of air.

am I correct so far ?

If so could the ball dropping through the coils have a chance of producing enough current to do that ???

In the case of the Shake Flashlight It states 10-15 seconds of shaking will give 2-3 minutes of light...So 2.5 seconds shake would provide 30 seconds of light, unable to find specific info though, as to wattage of led etc.

This is pretty much the same principle needed for the design.

edit on 3-12-2013 by ken10 because: (no reason given)

posted on Dec, 3 2013 @ 12:47 PM

ken10

ignorant_ape

making the following assumptions :

1 - mass of ball bearing = 1.0 kg

2 - efficiency of all mechanical systems = 100%

I will give you some figures

the mass of your " submersible " must be > 1.01 kg- to recover the bearing

the mass of water that must be displaced to make the " submersible " ascend the water tube must be >1.01

the air supply must deliver > 1.01 litres of air to displace this water [ the real figure is higher due to water pressure and gas compressibility ]

thus explain how your bearing is going to generate sufficient elictricty by falling - to drive the air pump

Are we not forgetting the velocity of the ball, which means the ball weighs more on its way down than it does on its way up ?

It is this imbalance that I was trying to exploit...and btw I have no intention of building anything (my health doesn't allow it) this is purely an idea that I had and didn't know the answers to.

But thanks for the input

The ball would only appear heavier if it impacted with something BASIC physics !

The potential energy at the top is converted to kinetic energy that would be converted by the coil to electrical energy but you have losses due to heat air resistance etc again all BASIC physics these things will NEVER work.

As for the torch how much energy does the user burn to charge it you have to look at ALL energy losses in
a system not just the ones that take your fancy!!!

posted on Dec, 3 2013 @ 01:17 PM

wmd_2008

ken10

ignorant_ape

making the following assumptions :

1 - mass of ball bearing = 1.0 kg

2 - efficiency of all mechanical systems = 100%

I will give you some figures

the mass of your " submersible " must be > 1.01 kg- to recover the bearing

the mass of water that must be displaced to make the " submersible " ascend the water tube must be >1.01

the air supply must deliver > 1.01 litres of air to displace this water [ the real figure is higher due to water pressure and gas compressibility ]

thus explain how your bearing is going to generate sufficient elictricty by falling - to drive the air pump

Are we not forgetting the velocity of the ball, which means the ball weighs more on its way down than it does on its way up ?

It is this imbalance that I was trying to exploit...and btw I have no intention of building anything (my health doesn't allow it) this is purely an idea that I had and didn't know the answers to.

But thanks for the input

The ball would only appear heavier if it impacted with something BASIC physics !

The potential energy at the top is converted to kinetic energy that would be converted by the coil to electrical energy but you have losses due to heat air resistance etc again all BASIC physics these things will NEVER work.

As for the torch how much energy does the user burn to charge it you have to look at ALL energy losses in
a system not just the ones that take your fancy!!!

If I drop a bouncy ball from 6ft high, it will produce enough energy to lift the ball back up say to half that height....yes.

So what I'm saying is that from an object dropped from 6ft high, I only need to get enough energy to to lift the object 1cm....But I realise it is not as straight forward as that, so I'm trying to see if there is a height at which an object falling can produce enough energy to lift the object 1cm.

I'm fairly sure the energy could be produced that way, but unsure if that would not create a block somewhere else in the system....water pressure on the pump/submersible, maybe.

posted on Dec, 3 2013 @ 01:23 PM
edit on 3-12-2013 by boncho because: (no reason given)

posted on Dec, 3 2013 @ 01:37 PM

boncho

So if we use the 12mm magnetic ball weighing 12grams, that means we need 12mls plus of air to lift the submersible to any height we desire.

Actually, you need a little more:

At sea level and at 15 °C, air has a density of approximately 1.225 kg/m3 (0.0023769 slugs/ft3, 0.001225 g/cm3)

12G / 0.001225, = you need 9795.91 Mls of air, give or take a ml or two on the positive side so it actually goes UP.

en.wikipedia.org...
edit on 3-12-2013 by boncho because: (no reason given)

Are you sure ?.....I can't see how it could jump from 12 mls to 9795.91 mls.

Or am I misunderstanding you.

1 kilo = 1 liter
1 gram = 1 mls

edit on 3-12-2013 by ken10 because: (no reason given)

edit on 3-12-2013 by ken10 because: (no reason given)

posted on Dec, 3 2013 @ 01:51 PM
Nevermind.
edit on 3-12-2013 by boncho because: (no reason given)

posted on Dec, 3 2013 @ 01:55 PM
Nevermind.
edit on 3-12-2013 by boncho because: (no reason given)

posted on Dec, 3 2013 @ 02:05 PM

boncho

1 kilo = 1 liter
1 gram = 1 mls

Density

1 kilo of bearing weighs 1 kilo.
1 gram of bearing weighs 1 gram.

Water

1 litre of air weighs 1.225 gram. (1.225g per litre)
1 gram of air weighs 0.001225 g/cm3 (0.001225g per mil)

edit on 3-12-2013 by boncho because: (no reason given)

edit on 3-12-2013 by boncho because: (no reason given)

So, if I filled a 1 liter bottle with air, would that not hold a 1 kilo weight afloat in water...or neutral buoyancy.

edit on 3-12-2013 by ken10 because: (no reason given)

edit on 3-12-2013 by ken10 because: (no reason given)

posted on Dec, 3 2013 @ 02:09 PM

ken10

boncho

1 kilo = 1 liter
1 gram = 1 mls

Density

1 kilo of bearing weighs 1 kilo.
1 gram of bearing weighs 1 gram.

Water

1 litre of air weighs 1.225 gram. (1.225g per litre)
1 gram of air weighs 0.001225 g/cm3 (0.001225g per mil)

edit on 3-12-2013 by boncho because: (no reason given)

edit on 3-12-2013 by boncho because: (no reason given)

So, if I filled a 1 liter bottle with air, would that not hold a 1 kilo weight afloat in water...or neutral buoyancy.

edit on 3-12-2013 by ken10 because: (no reason given)

edit on 3-12-2013 by ken10 because: (no reason given)

Ignore me I was thinking about something else.

posted on Dec, 3 2013 @ 02:19 PM

ken10

boncho

1 kilo = 1 liter
1 gram = 1 mls

Density

1 kilo of bearing weighs 1 kilo.
1 gram of bearing weighs 1 gram.

Water

1 litre of air weighs 1.225 gram. (1.225g per litre)
1 gram of air weighs 0.001225 g/cm3 (0.001225g per mil)

edit on 3-12-2013 by boncho because: (no reason given)

edit on 3-12-2013 by boncho because: (no reason given)

So, if I filled a 1 liter bottle with air, would that not hold a 1 kilo weight afloat in water...or neutral buoyancy.

edit on 3-12-2013 by ken10 because: (no reason given)

edit on 3-12-2013 by ken10 because: (no reason given)

Yes, my apologies. I was in a completely different train of thought.

You need only displace the amount of water weight you need to lift. Air is 1.3 times denser under water though so you will need 1.3x whatever you are trying to lift.

posted on Dec, 3 2013 @ 02:22 PM

boncho

ken10

boncho

1 kilo = 1 liter
1 gram = 1 mls

Density

1 kilo of bearing weighs 1 kilo.
1 gram of bearing weighs 1 gram.

Water

1 litre of air weighs 1.225 gram. (1.225g per litre)
1 gram of air weighs 0.001225 g/cm3 (0.001225g per mil)

edit on 3-12-2013 by boncho because: (no reason given)

edit on 3-12-2013 by boncho because: (no reason given)

So, if I filled a 1 liter bottle with air, would that not hold a 1 kilo weight afloat in water...or neutral buoyancy.

edit on 3-12-2013 by ken10 because: (no reason given)

edit on 3-12-2013 by ken10 because: (no reason given)

Ignore me I was thinking about something else.

No problem. At least we moved forward amicably.

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