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Author Topic: Rosch taking orders on OU Bouyancy device.  (Read 210221 times)

markdansie

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Re: Is the motor and generator atop the Rosch's KPP actually a QMoGen?
« Reply #105 on: April 29, 2015, 11:11:48 AM »

 Is the motor and generator atop the Rosch's KPP actually a QMoGen? Is the buoyancy system unnecessary? - Someone who attended the AuKW demo and is satisfied it is for real, is also convinced that buoyancy is not the heart of the system. Rather, the air compressor motor and the generator are working together like a QMoGen, with the buoyancy system acting like the belt between them. (PESN; April 29, 2015)


Hi Sterling,
unlike your site where most comments are suppressed and posters banned (again) you are always welcome to post your opinions here or at Revolution-Green.com
As you know form many years experience, not one QMoGen you either reported on or promoted with you fee for story journalism approach has ever worked out. Not one has been independently verified or gone into production. So why carry on with this delusional fantasy?


Speaking of fantasy, on a personnel note I was wondering
1. Did Jesus ever come for supper after you invited him ?
2. Did the young ,famous violinist you were stalking ever take you up on a date ?
3. Are you still expecting to be arrested for your self confessed inappropriate sexual behavior ?
4. Do you still believe anyone takes you seriously anymore or wants to be associated with you given the above questions?


I also find it amusing you would post here given the hatchet job and bad mouthing your moderator Stuart is giving Stefan. Stefan is well respected by all and is also entitled to an opinion without emotive and derogatory comments being made.


On a positive note some of your recent stories have been balanced and well written


Kind regards and best wishes
Mark Dansie


Kind Regards
Mark


tinman

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Re: Is the motor and generator atop the Rosch's KPP actually a QMoGen?
« Reply #106 on: April 29, 2015, 11:17:37 AM »

Hi Sterling,
unlike your site where most comments are suppressed and posters banned (again) you are always welcome to post your opinions here or at Revolution-Green.com
As you know form many years experience, not one QMoGen you either reported on or promoted with you fee for story journalism approach has ever worked out. Not one has been independently verified or gone into production. So why carry on with this delusional fantasy?


Speaking of fantasy, on a personnel note I was wondering
1. Did Jesus ever come for supper after you invited him ?
2. Did the young ,famous violinist you were stalking ever take you up on a date ?
3. Are you still expecting to be arrested for your self confessed inappropriate sexual behavior ?
4. Do you still believe anyone takes you seriously anymore or wants to be associated with you given the above questions?


I also find it amusing you would post here given the hatchet job and bad mouthing your moderator Stuart is giving Stefan. Stefan is well respected by all and is also entitled to an opinion without emotive and derogatory comments being made.


On a positive note some of your recent stories have been balanced and well written


Kind regards and best wishes
Mark Dansie


Kind Regards
Mark
Lol-Bang.

LibreEnergia

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #107 on: April 29, 2015, 11:20:37 AM »
if  nRT ln(V2/V1 ) is the work to compress the gas to greater than the pressure at the bottom... what more work is required than that? Oh I see... once the gas does start flowing, then its pressure becomes less and you have to apply more force to maintain the pressure ( or have compressed a larger volume to a greater pressure )...

okay.  So if I compress 5 times the volume of air to twice the required additional pressure it's still less work than the total bouyance force.

and bouyancy is ( mass_displaced - mass_displacing ) * G.  (the air weighs something so should be subtracted from the mass displaced...)

... So, having an understanding of these basic physics principles where is the discontinuity? 


How does one relate the force required to displace X with pressure?  Or.. why does it matter how much water is moved?  The pressure is enough to overcome the additional weight of the water.. and if it was under a solid plunger and moved the whole body of water, why does moving that water imply any more work than it took to presureize the gas?




input... (compress 5x the required air; twice the required exess pressure @ height of 14 buckets)
2.366602913    Work (ft-lb)

(edit: or even 4x the volume and 4x the pressure, so if I take 1 unit of volume out it's still 75% of the increased pressure... something; ya it's some differential)

output...(14 buckets)
36.61705229    lift force (ft-lb/sec^2)

(fixed some annotions, grouping and colored related things)
https://docs.google.com/spreadsheets/d/1YzocJ_dc7pXwHq1Vr9mzXBdS5-nP51xNjlSPsEVPN_M

You seem to have forgotten that your compressed parcel of air needs to be moved down in the water or, if you compressed it on the bottom the water above it is displaced upwards.

The end result is that the energy expended in sinking or creating a buoyant object is exactly the same amount that can be recovered by allowing the object to rise.

d3x0r

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #108 on: April 29, 2015, 11:24:09 AM »
if  nRT ln(V2/V1 ) is the work to compress the gas to greater than the pressure at the bottom... what more work is required than that? Oh I see... once the gas does start flowing, then its pressure becomes less and you have to apply more force to maintain the pressure ( or have compressed a larger volume to a greater pressure )...

okay.  So if I compress 5 times the volume of air to twice the required additional pressure it's still less work than the total bouyance force.

and bouyancy is ( mass_displaced - mass_displacing ) * G.  (the air weighs something so should be subtracted from the mass displaced...)

input... (compress 4x the required air; 1.5 the required total pressure  @ height of 14 buckets)
13.98181442    Work (ft-lb)


output...(14 buckets)
36.61705229    lift force (ft-lb/sec^2)


so if I compress to 1.82 atmospheres 4x the air it's 13.9 ft-lb and the output is still nearly 3x that... (4kw in 12kw out is in the ballpark of product performance)


taking out 1/4 of the volume of it results in 1.36 atm (75% of 1.82atm), and only 1.05 atm is required with current setup of 0.5 inch 1 foot tubes.


so... it's still probable.

d3x0r

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #109 on: April 29, 2015, 11:26:12 AM »
You seem to have forgotten that your compressed parcel of air needs to be moved down in the water or, if you compressed it on the bottom the water above it is displaced upwards.

The end result is that the energy expended in sinking or creating a buoyant object is exactly the same amount that can be recovered by allowing the object to rise.
it's compressed outside of the container, and piped in at the bottom.  does not have to additionally travel through the water.
  I did forget surface tension of the water.




LibreEnergia

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #110 on: April 29, 2015, 11:31:12 AM »
it's compressed outside of the container, and piped in at the bottom.  does not have to additionally travel through the water.
  I did forget surface tension of the water.

What about the column of water above the air that needs to be lifted up to make way for the air? does it just magically float upward?


tinman

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #111 on: April 29, 2015, 11:34:21 AM »
If you'd like to propose a free energy machine that is powered by the steaming pile of crap that equates to this analysis then I'd probably be willing to invest.

Sure, the masses do not change significantly but the distance between them does. The last time I looked the force due to gravity was equal to (G *m1 * m2)/r^2 . If you alter the distance between the two masses then the gravitational potential changes as the inverse square of the distance.

Also mass is NOT proportional to gravity. A 1 kg mass on earth is the same as a 1kg mass on  the moon or floating in space somewhere in the universe.

The quantity you are referring to is weight or mass multiplied by the acceleration due to gravity.
Might i suggest you go and have a peak at exactly what happens to the tide's around the world before you start calling other peoples opinions a steaming pile of crap. Then also have a look into fluid balancing harmonics in a rotating device-such as the earth. Take a flywheel and cut a small piece out of it that would represent our ship,and glue that small piece a little further out toward the outer perimeter of the flywheel. Now spin that flywheel in the vacuum of space and tell me that it will slow down quicker than it would if that piece was placed back to where it came from.

Your fancy numbers(G *m1 * m2)/r^2) mean nothing in this instant,and are only relevant when the forces are confined to earth-->which in this case,they are not,and have no meaning at all in space where there is no gravity-->which is the medium our planet resides in.

So who is talking a steaming pile of crap now.

d3x0r

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #112 on: April 29, 2015, 11:38:15 AM »
What about the column of water above the air that needs to be lifted up to make way for the air? does it just magically float upward?


well see that's why I'm asking how does the pressure relate to that?  I mean... if the pressure is more, it's going to displace... the whole point in needing the excessive pressure over 1atm is the weight of the water... but the weight of the water turns out to be a very small percentage of 1atm additional... so at 1.053atm it can displace 21 inches of water easily.


(increasing the pipe size at this time does improve additional gain... to go 0.5inch which is 2.6x to 3inch is from 4.86x and an 8 inch pipe is 7.5x... increasing the length doesn't help (the height of water is computed based on the pipe diameter) but then requires much more pressure.. and almost 2HP to compress...)


so at 8 inches the height is 10.5 feet, and the pressure required is 1.32atm minimum... (10.5feet is 3.2m and 10m is +1 atm)

MarkE

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #113 on: April 29, 2015, 11:43:53 AM »
if  nRT ln(V2/V1 ) is the work to compress the gas to greater than the pressure at the bottom... what more work is required than that? Oh I see... once the gas does start flowing, then its pressure becomes less and you have to apply more force to maintain the pressure ( or have compressed a larger volume to a greater pressure )...

okay.  So if I compress 5 times the volume of air to twice the required additional pressure it's still less work than the total bouyance force.

and bouyancy is ( mass_displaced - mass_displacing ) * G.  (the air weighs something so should be subtracted from the mass displaced...)

... So, having an understanding of these basic physics principles where is the discontinuity? 


How does one relate the force required to displace X with pressure?  Or.. why does it matter how much water is moved?  The pressure is enough to overcome the additional weight of the water.. and if it was under a solid plunger and moved the whole body of water, why does moving that water imply any more work than it took to presureize the gas?




input... (compress 5x the required air; twice the required exess pressure @ height of 14 buckets)
2.366602913    Work (ft-lb)

(edit: or even 4x the volume and 4x the pressure, so if I take 1 unit of volume out it's still 75% of the increased pressure... something; ya it's some differential
 edit2: hmmm ya that needs to be looked at it's probably the missing work)

output...(14 buckets)
36.61705229    lift force (ft-lb/sec^2)

(fixed some annotions, grouping and colored related things)
https://docs.google.com/spreadsheets/d/1YzocJ_dc7pXwHq1Vr9mzXBdS5-nP51xNjlSPsEVPN_M
Your premise fails right when you start conflating force and energy.

LibreEnergia

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #114 on: April 29, 2015, 11:44:53 AM »
Might i suggest you go and have a peak at exactly what happens to the tide's around the world before you start calling other peoples opinions a steaming pile of crap. Then also have a look into fluid balancing harmonics in a rotating device-such as the earth. Take a flywheel and cut a small piece out of it that would represent our ship,and glue that small piece a little further out toward the outer perimeter of the flywheel. Now spin that flywheel in the vacuum of space and tell me that it will slow down quicker than it would if that piece was placed back to where it came from.

Your fancy numbers(G *m1 * m2)/r^2) mean nothing in this instant,and are only relevant when the forces are confined to earth-->which in this case,they are not,and have no meaning at all in space where there is no gravity-->which is the medium our planet resides in.

So who is talking a steaming pile of crap now.

Astounding!  this is "free energy" at it's best. Not content to produce power from steaming bullshit you multiply the effect by digging an even bigger hole for the theory to reside in.

My advice is to stop digging before the fall becomes painful.


d3x0r

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #115 on: April 29, 2015, 11:46:06 AM »
Your premise fails right when you start conflating force and energy.
well... they are both expressed as answers to
1) how much work is done by bouyancy
and
2) how much work is required to compress air


I realize that the result is in different units; but which equation is wrong? 

MarkE

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #116 on: April 29, 2015, 11:47:03 AM »
You seem to have forgotten that your compressed parcel of air needs to be moved down in the water or, if you compressed it on the bottom the water above it is displaced upwards.

The end result is that the energy expended in sinking or creating a buoyant object is exactly the same amount that can be recovered by allowing the object to rise.
And let's not forget, you can only reclaim all of the energy if:  There are no viscous losses and you take infinite time to reclaim the energy.

MarkE

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #117 on: April 29, 2015, 11:49:00 AM »

so if I compress to 1.82 atmospheres 4x the air it's 13.9 ft-lb and the output is still nearly 3x that... (4kw in 12kw out is in the ballpark of product performance)


taking out 1/4 of the volume of it results in 1.36 atm (75% of 1.82atm), and only 1.05 atm is required with current setup of 0.5 inch 1 foot tubes.


so... it's still probable.
It is no more probable then compressing a spring returns free energy, or hoisting a weight attached to a counter weight via a rope over a pulley.

MarkE

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #118 on: April 29, 2015, 11:53:50 AM »
well... they are both expressed as answers to
1) how much work is done by bouyancy
and
2) how much work is required to compress air


I realize that the result is in different units; but which equation is wrong?
Force and energy are expressed in different units because they are unique from each other.  They do not equate with one another.

1) Zero, nada, de minimus, squat. 
2) A lot more than anyone has ever been able to recover letting it expand again.

So from 1) you get:  The machine can't perform any net useful work. And, from 2) you get:  The machine cannot transfer work between a source and a load efficiently.  It is a very expensive room ornament / heater.

LibreEnergia

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #119 on: April 29, 2015, 11:53:52 AM »

well see that's why I'm asking how does the pressure relate to that?  I mean... if the pressure is more, it's going to displace... the whole point in needing the excessive pressure over 1atm is the weight of the water... but the weight of the water turns out to be a very small percentage of 1atm additional... so at 1.053atm it can displace 21 inches of water easily.


(increasing the pipe size at this time does improve additional gain... to go 0.5inch which is 2.6x to 3inch is from 4.86x and an 8 inch pipe is 7.5x... increasing the length doesn't help (the height of water is computed based on the pipe diameter) but then requires much more pressure.. and almost 2HP to compress...)


so at 8 inches the height is 10.5 feet, and the pressure required is 1.32atm minimum... (10.5feet is 3.2m and 10m is +1 atm)

The amount of energy you will need to impart to the air will be (at minimum) the amount of energy required to lift a mass of water represented by the column of water with the horizontal component of the surface area of the float times the depth that it is submerged to. Fairly easy to calculate.

Using air will require more than that due to the losses to heat when compressing the air, and depends on the efficiency of the compressor.