Rosch taking orders on OU Bouyancy device.

[sterlinga]

(http://peswiki.com/images/d/d5/Motor-generator_Ronny-Korsberg_sq_95x95.gif)

Featured: Buoyancy > Rosch > GAIA > Demo >
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)

[allcanadian]

@TK

Look. The whole contraption is based on one side being heavier than the other
side, right? So just get rid of the water altogether, it just creates drag you
don't need. Drop a series of heavy balls into the buckets on the descending side
and have them roll out on the bottom. Use an Archimedes Screw to elevate the
balls back up to the top so they can be dropped back into the descending
buckets. Power the Archimedes Screw with a simple pulley-belt linkage off the
top sprocket of the bucket-chain. Have the bottom sprocket drive your generator.
There will be so much mechanical advantage from the Screw-Pulley system that
you'll have to install a brake mechanism to keep it from speeding up to
self-destruction.

I believe you are making the same flawed argument as most here and we start with a fairly complex buoyancy machine and then you say why not metal balls as it's the same thing with weight on one side and if we use this same flawed logic we could take the next step. Why not just say it is the same as bouncing a ball and expecting it to bounce higher?. The problem here is that obviously they are not the same thing and to presume they are is absurd. I understand this is the same false premise most critics like to use however this old same/same argument game doesn't imply a great deal of intelligence in my opinion.

Obviously the most intelligent person would be the one smart enough to have figured out how to get the air into the bottom of the tank without all the losses normally involved. There is one question that matters here and that is how can we get the air across the pressurized boundary condition into the water without the normal losses however you wouldn't touch that one with a ten foot pole would you?. Because you have literally no idea do you and in fact you have no idea where to even start.
So how about we have an intelligent conversation for a change of pace?, tell me how would you get the air into the tank without all the losses normally involved?... any idea's?.

AC

[MarkE]

@TKI believe you are making the same flawed argument as most here and we start with a fairly complex buoyancy machine and then you say why not metal balls as it's the same thing with weight on one side and if we use this same flawed logic we could take the next step. Why not just say it is the same as bouncing a ball and expecting it to bounce higher?. The problem here is that obviously they are not the same thing and to presume they are is absurd. I understand this is the same false premise most critics like to use however this old same/same argument game doesn't imply a great deal of intelligence in my opinion.

You assert a distinction that you fail to establish.

Obviously the most intelligent person would be the one smart enough to have figured out how to get the air into the bottom of the tank without all the losses normally involved. There is one question that matters here and that is how can we get the air across the pressurized boundary condition into the water without the normal losses however you wouldn't touch that one with a ten foot pole would you?. Because you have literally no idea do you and in fact you have no idea where to even start.
So how about we have an intelligent conversation for a change of pace?, tell me how would you get the air into the tank without all the losses normally involved?... any idea's?.

AC

Unless you wish to rely on magical thinking, the minimum work to force the air into the buckets is identically the increase in mgh of the water that the air displaces.  That places the whole affair at zero sum gain in the best case.  This is all very black letter.  If you are going to make the ridiculous request that someone propose to you a means of imparting free energy to air, why not cut to the chase and just ask for a free source of electricity to run your appliances?

[d3x0r]

You assert a distinction that you fail to establish.Unless you wish to rely on magical thinking, the minimum work to force the air into the buckets is identically the increase in mgh of the water that the air displaces.  That places the whole affair at zero sum gain in the best case.  This is all very black letter.  If you are going to make the ridiculous request that someone propose to you a means of imparting free energy to air, why not cut to the chase and just ask for a free source of electricity to run your appliances?

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

[LibreEnergia]

You slow down nothing,because the mass of the earth as a whole has not increased. Regardless of wether that ship was there or not,the earths rotational rate would decrease by the same amount,as the same mass was moved in the rise of the tide. What you say MH is like saying a ton of grass is heavier than a ton of rock. So what will slow down the earths rotation more-->50 000 tons of water being raised 14 meters,or 50 000 tons of ship being raised 14 meters?. Mass is directly proportional to gravity,and as we have the same mass amount,then nothing has changed as far as gravitational pull go's,and thus the earths rotation would slow no more than it would if the ship wasnt there. The earth has a set amount of mass,and only things like meteorites can increase the mass of the earth. So as our mass remains constant,then so dose the decrease in rotational speed of the earth due to gravitational drag.

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.