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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 ?
ignorant_ape
reply to post by ken10
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
and you are scientifically illiterate
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
ken10
reply to post by ignorant_ape
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.
Alfa1 received 5 stars for posting something about magnets "damping"...Which I had already addressed in my opening post.......So was the fatal flaw of the design there ?
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 ???
ken10
ignorant_ape
reply to post by ken10
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
wmd_2008
ken10
ignorant_ape
reply to post by ken10
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!!!
boncho
reply to post by ken10
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)
boncho
reply to post by ken10
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)
ken10
boncho
reply to post by ken10
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)
ken10
boncho
reply to post by ken10
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)
boncho
ken10
boncho
reply to post by ken10
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.