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Overunity Machines Forum



What's wrong with this

Started by Floor, December 14, 2014, 12:05:50 PM

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0 Members and 1 Guest are viewing this topic.

tinman

Quote from: MarkE on December 31, 2014, 09:01:05 AM
Of course it gives off the heat of evaporation.  However, its mass:  That stays the same.  (I am ignoring the tiny adjustment for mass/energy equivalence.)
Lets use weight(eg,kg's)Mark,forget mass. So did you see my questions above,and do you agree that every thing that uses electrical energy(eg,an electric motor) runs at unity?.I using electrical energy here,as that is the energy required to convert the liquid to gas-maybe a pressure or vacume pump.

MarkE

Quote from: tinman on December 31, 2014, 06:58:39 AM
Quote: In physics, mass is a property of a physical body which determines resistance to being accelerated by a force and the strength of its mutual gravitational attraction with other bodies. The SI unit of mass is the kilogram (kg).

Mass is not the same thing as weight, even though we commonly calculate an object's mass by measuring its weight. A man standing on the Moon would weigh less than he would on Earth because of the lower gravity, but he would have the same mass (he would have to recalibrate his bathroom scale for lunar gravity).

As we are not standing on the moon Mark,and i clearly stated that we only need to be calculating things that are relative to us here on earth,then the SI unit for mass is the KG. I fail to see why you continually must make things so hard and difficult. You have the habbit of turning the simplest of questions into an extreemly difficult task.
Tinman, I am being picky because misuse of the terms can easily lead to incorrect conclusions in what is an energy problem.  The energy that is required to accelerate a mass between two non-relativistic velocities depends on those velocities and the mass.  The energy that is required to separate a mass from the center of the earth depends on the mass of the earth and the mass of the object, and the object's density with respect to the atmosphere surrounding the object.  If you work only with the latter, your energy balance will in virtually all cases end up wrong.
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So im going to leave you with this question.
lets froget about mass,lets use a measurement term that wont go into some full debate on what something is.
Do you believe it possable to decrease the weight of a vessle where nothing leaves or enters that vessle-->yes Mark,the vessle is air/gas/water tight-->and yes Mark,we are working here on earth and will be testing the device hear on EARTH-->not mar's,not jupiter,and not the moon.
Stipulating that Ge is constant within the space of the problem:  The submerged weight of a vessel depends on:

1) Total mass of the vessel and its contents.
2) Total volume of the vessel.
3) Density of the fluid that the vessel displaces.

At Newtonian velocities, you cannot change the weight of the object without changing at least one of the above three parameters.
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Is it possable to have the same device(vessle)-(once again,air,gas and water tight-nothing enters,and nothing leaves)that dose NOT change shape or physical size,be made to change from being non buoyant to buoyant-->(sink then float,just so as you dont get confused),by changing the state of the fluid inside that vessle.?.
Absolutely not.  See the list above.  You must change one or more of those variables to change either the submerged weight or the buoyancy.
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These are very simple questions Mark,and depending on your answer,will depend on wether or not i continue with this experiment.
Just as 10kg of feathers has the same mass as 10kg of lead shot, 10kg of some material in gas form has the same mass as the same material in liquid or solid states.  If you are thinking hot-air balloon, lower density heated air and higher density cold air exchange  from underneath the bottom opening of the balloon.  When you seal your vessel, no such exchange takes place.  I suggest a very simple experiment for you:

Take a small vessel such as a half liter water bottle and fill it 3/4 or so with crushed ice.  Tightly seal the top.  Thoroughly dry the outside of the bottle.  Hot glue a precision scale to the side of the water bottle.  Get a bucket and fill it half way with room temperature water.  Place the half liter bottle in the bucket.  Snap a picture that captures the where the water surface in the bucket registers on the scale.  Take another picture after the ice has melted.  Guess what?  The scale will read the same value as when the ice was solid.   You can repeat the experiment using a bit of dry ice provided that you seal the cap really well and don't use so much dry ice that the bottle deforms.
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Edits; I should add that i have already done these experiment's,and now it is up to you to decide as to wether or not you can become some one that i (and many others) can work with,or you wish to continually have a high tech dribble contest with other high tech dribbler's,just so as you can see who has the biggest set of technical gonad's.
I am only interested in correct physics.  I admire your experimenting.

MarkE

Quote from: tinman on December 31, 2014, 09:06:52 AM
Lets use weight(eg,kg's)Mark,forget mass. So did you see my questions above,and do you agree that every thing that uses electrical energy(eg,an electric motor) runs at unity?.I using electrical energy here,as that is the energy required to convert the liquid to gas-maybe a pressure or vacume pump.
Tinman the difference in our positions is that you assert that you can change the buoyancy of a fixed volume vessel by changing the state, but not the mass of some of its contents, and I say that unless you change one or more:

1) Total mass of the vessel plus its contents.
2) Total volume of the vessel.
3) Density of the surrounding fluid.

That you cannot affect the weight or buoyancy of the vessel.

tinman

Quote from: MarkE on December 31, 2014, 09:39:48 AM
Tinman the difference in our positions is that you assert that you can change the buoyancy of a fixed volume vessel by changing the state, but not the mass of some of its contents, and I say that unless you change one or more:

1) Total mass of the vessel plus its contents.
2) Total volume of the vessel.
3) Density of the surrounding fluid.

That you cannot affect the weight or buoyancy of the vessel.

Thank you Mark.
That is all i wanted to hear(read).

See you here sunday with an intersting video to show you.
Happy new year to you and all. ;)

P.S-i will be changing the state of the mass contained within the vessle. Hope that still counts as a  no can do on your side of the fence?

MarkE

Quote from: tinman on December 31, 2014, 09:46:06 AM
Thank you Mark.
That is all i wanted to hear(read).

See you here sunday with an intersting video to show you.
Happy new year to you and all. ;)

P.S-i will be changing the state of the mass contained within the vessle. Hope that still counts as a  no can do on your side of the fence?
If you can change the weight or buoyancy without changing at least one of the above three parameters then it will indeed be very interesting.