I open this topic for discussion on my bench initially, but it should be moved to the Gravity or theoretical discussions section if Peter would like. (I didn't see a way to post a new topic in that heading)
It seems the debate regarding Gravity as an energy source rages on in the physics forums, and some of the explanations for gravity seem counter intuitive and tautological. (at least to me).
I suggest first reading some of the physics forums to get a feel for the arguments.
for a start: https://physics.stackexchange.com/questions/75222/where-does-gravity-get-its-energy-from
and there are others.
I hope we can get a good debate going here. Please stay on topic.
Regards
The simple answer is !YES!
Gravity is stored/potential energy.
To understand this,you have to think way back in time,to when the universe was created.
How do we prove that gravity is a stored/potential energy ?,well that is easy.
First we must abide by the known laws of physics,which state-->energy can be neither created nor destroyed-but only transformed from one state to another.
For this thought experiment,we are going to use two earth sized planets that have the same mass.
Our point of reference will be one of those planets,where it remains motionless in space.
The second planet is heading toward the first planet at a rate of 100 meters a second.
The motionless planet is said to have no kinetic energy,but as the second planet is traveling toward us at 100 meters a second,it will be said to have X amount of kinetic energy.
As the second planet moves closer to the first planet,it's speed increases.
At the same time,the once motionless planet has now started to move toward the second planet,due to the gravitational attraction of masses.
So not only has the kinetic energy value of the second planet increased,the first planet also now has kinetic energy.
As the two planets get closer and closer,the kinetic energy value of the two planets increases.
When the two planets collide,they dissipate many more times the energy than the second planet had at the beginning of the gravitational pull process between the two.
So,if energy cannot be created,and the dissipated energy by the collision of the two planets was far greater than that of the second planets kinetic energy at the start of the process,this can only mean that the extra energy was stored within the two planet's,and that energy was stored as gravity it self.
It should also be noted that this is a non exhaustible/infinite energy source.
What i mean to say is-if a device was invented that produced an output energy,where using only the gravitational stored energy as it's input,and where none of the earths mass was ejected into space,the gravitational force would remain a constant--regardless of the work done by the output of the device.
Wouldn't it be great if we could switch the gravitational force on and off as we wished,or some how invert that force when we needed to do so. :D
Brad
By definition, gravity is a force, not "energy" or a form of energy.
You can give a better description of gravity by adding that it is an "accelerating force".
Here is a good discussion of the units involved:
https://physics.stackexchange.com/questions/297747/units-in-gravitational-constant
The magnitude of the force depends on the distance between the objects, the mass of each object, and the gravitation constant.
So, what is this "gravitational constant"?
https://www.universetoday.com/34838/gravitational-constant/
Hi Brad
I posed a very similar argument that went something like this:
"Consider a vast empty space and two large massive bodies separated by some great distance. With no other forces acting upon our two bodies, the only potential energy at T0 is the feeble gravitational attraction between the bodies. Now we are patient and over time the bodies are pulled to each other. They slowly begin accelerating towards each other and as they do, the gravitational pull goes up according to the square law as the bodies close the distance.We can imagine that at some point the bodies will collide with the release of a huge amount of energy. What was the cause of the huge increase in energy since we started with only a feeble potential energy at T0? We did not do any work on the bodies."
If we consider also just one massive body in an empty space, it apparently has no potential energy until a second body appears. So the gravitational potential energy seems to be created when a second body appears.
Some physics pundits says the potential energy was always there, instilled in the mass at the beginning of creation. That doesn't seem intuitively correct to me as it needs that one "free miracle", i.e. everything created in an instant out of nothing including space and time.
Just naming something e.g. Gravity is caused by Gravitons, doesn't give us anything. The invention of a label does not nourish a hungry mind.
I'm working on some other thought experiments that might be interesting.
Thanks for your reply.
G;
You said gravity is a force, not a source of energy. But it can act on a mass F=MA to produce energy that can be measured in Joules. Without the force, there would be no accumulated energy over time.
Some physics pundits explain it as "to get energy, you had to put work into the object by lifting it "up"against the force of gravity." But in space there is no "up", just a sea of objects that occasionally slam into each other because of gravitational attraction and with the release of a huge amount of energy. Are we back to the "free miracle" that exploded everything and now it is all falling back together releasing all the separation energy? But they tell us space is expanding moving bodies apart against their gravitational attraction?
Regards
Quote from: ion on 2018.06.09, 14:49:24
Hi Brad
I posed a very similar argument that went something like this:
"Consider a vast empty space and two large massive bodies separated by some great distance. With no other forces acting upon our two bodies, the only potential energy at T0 is the feeble gravitational attraction between the bodies. Now we are patient and over time the bodies are pulled to each other. They slowly begin accelerating towards each other and as they do, the gravitational pull goes up according to the square law as the bodies close the distance.We can imagine that at some point the bodies will collide with the release of a huge amount of energy. What was the cause of the huge increase in energy since we started with only a feeble potential energy at T0? We did not do any work on the bodies."
If we consider also just one massive body in an empty space, it apparently has no potential energy until a second body appears. So the gravitational potential energy seems to be created when a second body appears.
Some physics pundits says the potential energy was always there, instilled in the mass at the beginning of creation. That doesn't seem intuitively correct to me as it needs that one "free miracle", i.e. everything created in an instant out of nothing including space and time.
Just naming something e.g. Gravity is caused by Gravitons, doesn't give us anything. The invention of a label does not nourish a hungry mind.
I'm working on some other thought experiments that might be interesting.
Thanks for your reply.
Some times it is hard for some to understand that it is us humans that give names to things we dont understand,and gravity is one of those things.
It's much like those that say math is a universal language.
As far as i know,no one has asked the aliens yet if this is correct :D
No one can deny that it is the gravitational force that increased the kinetic energy value of the two masses in our examples.
Dose it take energy to accelerate a mass ?
I think the answer is quite straight forward.
Brad
Ion you said:-
If we consider also just one massive body in an empty space, it apparently has no potential energy until a second body appears. So the gravitational potential energy seems to be created when a second body appears.
I think if it is only one united body (mass) (one atom with it's revolving electrons), there is no potential energy as the atom and electrons are in peace with one another, there is no possible reaction until another body is placed in the same void. They can be identical atom and electrons making up the two masses, when only one they can't react as there is nothing to react with, place two near one another and things start to react, even if it is only to attract to one another. (near could be millions of miles apart, just takes longer).
The two masses will accelerate the closer they get to one another, and when they finally hit one another, F=MA, as you have stated.
If you increase the acceleration of plutonium (it has a very heavy mass) with an external source (very easy to do) so as it hits another lesser mass, what F can we obtain? usually it is the smaller mass hitting the bigger mass, though it could be two hydrogen atoms, the sun is just that within it's own void, and has a huge gravity. The sun emits EMR which could fry you if it is emitted in the direction of the earth, the ionosphere is our protective cloak, but sometimes some energy gets through and down's electronic equipment, a lot of this has been happening lately.
Now to get back on track, I think gravity is a source of energy if there is a mass for it to react with and has some form of distance be it very small or very large.
This is a topic which is huge, it is not only gravity, waves as in electromagnetic I think will produce energy which is sustainable, one no but two or more!!! ;)
Regards
Mike 8)
Quote from: ion on 2018.06.09, 14:49:24
G;
You said gravity is a force, not a source of energy. But it can act on a mass F=MA to produce energy that can be measured in Joules. Without the force, there would be no accumulated energy over time.
Some physics pundits explain it as "to get energy, you had to put work into the object by lifting it "up"against the force of gravity." But in space there is no "up", just a sea of objects that occasionally slam into each other because of gravitational attraction and with the release of a huge amount of energy. Are we back to the "free miracle" that exploded everything and now it is all falling back together releasing all the separation energy? But they tell us space is expanding moving bodies apart against their gravitational attraction?
Regards
Until the object moves as a results of the gravitational attraction, it only has the "potential" to become energetic (not referring to energy holding it together, just the relationship to gravity). When it start to move, the potential energy become kinetic energy. Without the force, there is no acceleration and no potential to become kinetic energy.
The attraction between two magnets work the same way, but is much stronger, and we know how to generate magnetic force.
In simple terms you can say that energy requires a force for it to be utilized.
Regarding the "free miracle" of creation starting the whole process, I don't believe it happened that way. Solar systems have strikingly similar qualties to homopolar generators.
G said:
QuoteRegarding the "free miracle" of creation starting the whole process, I don't believe it happened that way. Solar systems have strikingly similar qualties to homopolar generators.
I don't believe in the "one free miracle" either, although at times life and the universe seem miraculous.
I was borrowing from Terrence McKenna who was jesting when he coined the phrase regarding current physics stance on the "big bang":
"Give us just one free miracle and we can roll from there"---------------------------------------------------------------------------------------------------------------------------------------------------
OK most physics textbooks say you can't get more out of gravity than you put into it e.g. if you lift a rock, it is like stretching a spring, you will get back exactly the energy you put into it, no more.
What about this:
We launch a body of mass M1 into space towards a massive planet e.g. Jupiter. We expend X amount of energy to get the massive body to just over the gravitational neutral zone between the earth and Jupiter. Now the massive body is attracted to Jupiter which is 317.8 times the volume of earth. Since Jupiter's gravity is larger than the earth's at the surface(2.528x), at some point in time the mass of body M1 strikes Jupiters surface and releases much more energy than it took to get the object to the neutral zone.
Where did the extra energy come from, since we did not put that much potential energy into the object when we lifted it to just slightly over neutral zone?
A good exercise would be to do the math for a 1kG body M1 striking Jupiters surface after accelerating towards it from the neutral zone and compare that to the energy required to get it to the neutral zone from earth.
Regards
Since Jupiter is providing the force, you did not put the energy in. You can only control the mass in this case. You can accelerate with a propulsive force or use a heavier object.
You can use less energy by using the rotation of the earth to help launch the object as well.
What you need is a method to generate an accelerating force like gravity, or a means to alter the mass an object.
Quote from: Grumpy on 2018.06.09, 18:51:38
Since Jupiter is providing the force, you did not put the energy in. You can only control the mass in this case. You can accelerate with a propulsive force or use a heavier object.
You can use less energy by using the rotation of the earth to help launch the object as well.
What you need is a method to generate an accelerating force like gravity, or a means to alter the mass an object.
I agree, you put in energy X and get out many times energy X, bu t not necessarily by returning to the starting point.
My point was that gravity is not like a linear spring otherwise there would be no neutral point and the earth to mass spring would be quite stretched and generating a lot of tension once our mass got to the surface of Jupiter. So the spring analogy only holds for short distances close to the surface of the planetary body.
In the example I gave, it is more like an over center mechanism.
Certainly if you could change the mass of an object it would be game over.
Regards
Each and every action has an equal but opposite reaction, Newtons Third Law
So launch enough large objects at Jupiter yes they accelerate towards it but in theory they would move Jupiter's position also, 2 large heavy balls both attract each other the same amount if they are of equal mass, does that make the gain double from both perspective or half from each balls perspective.
I don't see how you can capture this energy without turning gravity off for a re-gauge cycle.
Quote from: Peterae on 2018.06.09, 21:34:33
Each and every action has an equal but opposite reaction, Newtons Third Law
So launch enough large objects at Jupiter yes they accelerate towards it but in theory they would move Jupiter's position also, 2 large heavy balls both attract each other the same amount if they are of equal mass, does that make the gain double from both perspective or half from each balls perspective.
Hi Peter
In the thought experiment I proposed, I was talking about e.g. a single 1 kg mass, certainly not enough to modify Jupiters orbit.
I suppose mutual attraction might be similar to mutual inductance in net energy storage or release. 1H + 1H =4H, but we'd have to go over the formulae and my head is tired right now.
This is a topic that is very interesting and is going to get very complicated very quickly..
Let's start here:
Quote from: Grumpy on 2018.06.09, 18:51:38
Since Jupiter is providing the force, you did not put the energy in. You can only control the mass in this case. You can accelerate with a propulsive force or use a heavier object.
You can use less energy by using the rotation of the earth to help launch the object as well.
What you need is a method to generate an accelerating force like gravity, or a means to alter the mass an object.
Forgetting about the mass rotation aspect for a moment and concentrating on the linear relationships we ask the question:
Why does kinetic energy increase quadratically, not linearly, with speed?https://physics.stackexchange.com/questions/535/why-does-kinetic-energy-increase-quadratically-not-linearly-with-speed
G also hit the nail on the head when he said:
"Until the object moves as a results of the gravitational attraction, it only has the "potential" to become energetic (not referring to energy holding it together, just the relationship to gravity)."
Here we are dealing with two frames of reference, internal and external. The energy holding it together is the internal frame of reference, and this can be very great.. think of plutonium as a massive potential energy store. When compressed very quickly with a great force we have a nuclear weapon which as we all know goes into a chain reaction that liberates that internal energy and throws out a massive energy wave into the external frame of reference, the environment. In this example the Mass is simply blown apart and the energy is released to act on other matter. For me this is an excellent validation of the Big Bang model and one of the reasons I support it.. God is not invoked in this, for me. If we think about ion's earlier statement about the gravity being attractive but the majority of the mass in the universe we are told is moving apart, well this makes sense if it began from a highly compressed mass of massive density, a singularity. What we see with localised mass's coalescing is the stabilisation of two mass's relative speed so that they can move closer together due to the attractive force of the mass's overcoming momentum and inertia due to gravitational attraction. This is also the same in reverse when we launch a rocket into space and have to escape the Earth's gravitational attraction.
The DST triangle is important, speed distance and time.
Interestingly hhop gen 3 and gen 4 both have aspects that meet G's two conditions, and in the case of gen 4 it is purely down to definition of the relationships of the fundamental states of matter in the device within a gravitational field which provides the prime mover energy source. The fact that a pressure gradient is created across the input and output of the system of gen 4 means the mass can move and therefore as it is accelerating the energy liberated increases with the square of the velocity before the system returns to equilibrium. Gen 3 changes the mass but relates that change to height change within the gravitational field and therefore the potential energy of the device.
Quote from: Peterae on 2018.06.09, 21:34:33
I don't see how you can capture this energy without turning gravity off for a re-gauge cycle.
You can't turn gravity off ever but you can place the states of matter within it into an imbalanced force equation and exploit their relationship to each other as they seek to return to equilibrium.
https://study.com/academy/lesson/what-is-specific-gravity-definition-formula-calculation-examples.html
"When the specific gravity is greater than one, the object will sink, and when the specific gravity is less than one, then the object will float. If the specific gravity is equal to one', this means that the object will neither sink nor float -- it will hover in the liquid."
Here is a thought to chew on:
Normal beam balance scales work quite well and are accurate in the close range earth gravitational field.
When we increase the size of our beam balance scale, it becomes a device with hysteresis, an over center mechanism.
It will seek one of two possible states, but be nearly impossible to balance.
What size does our beam balance scale have to be to experience the hysteresis effect?
Regards
This has already been largely covered, but..
I see gravity, magnetism, and atomic spin are all 'free' in the sense that they provide an infinite, constant force. The problem we humans have lies in 'gating' or switching these forces.
If I stuck a magnet on a fridge, that magnet is providing a free, constant force against the steel. That force (lets say 1lb) will continue forever with no external input. We just have no way of switching or taking advantage of it. Likewise, gravity provides a similar constant free force (there's nobody sitting at the center of the earth driving a 'gravity generator'). The point being that thy are perpetual forces, regardless of our ability to gate/control them.
Quote from: evolvingape on 2018.06.09, 22:16:53
You can't turn gravity off ever but you can place the states of matter within it into an imbalanced force equation and exploit their relationship to each other as they seek to return to equilibrium.
Let me provide a thought-experiment example to exploit this. Lets first assume the Einstein E=MC^2 is accurate, implying that energy and mass are equivalent.
- Take 2 epic-sized capacitors and place them on opposite ends of a balance-beam. (for the sake of the thought-experiment, lets assume building-sized caps)
- Charge one capacitor.
- E=MC^2 dictates that since one capacitor now contains more energy than the other, it will have more mass and thus the balance beam should fall toward the charged capacitor.
- Next, move the energy to the other capacitor using superconductive cable.
- Now the other capacitor has more mass, and the balance beam should fall in the other direction.
- Rinse and repeat for a 'gravity powered generator'. Powered by 'conventional' physics even ;D ;D.
IMHO this is the type of 'gating' we are looking for. Ways to increase/decrease the effects of a free constant energy source. Creating what Tesla described as an 'energy sink' for one of the above free forces, and attaching a waterwheel to it. ;)
Quote from: Reiyuki on 2018.06.10, 15:54:12
Lets first assume the Einstein E=MC^2 is accurate, implying that energy and mass are equivalent.
Interesting thought experiment.. but it won't work. Energy is equivalent to mass multiplied by the speed of light squared, so it has to be moving..
In the super cap see saw example the energy in the cap is not moving relative to the second cap and so no increase in mass is possible relative to each other.
If we go back to the beginning we can say that a void full of nothing has no space dimensions and therefore no time. No frames of reference..
Now a mass appears and is alone in this void, in the mass internal frame of reference it has height, width and length.. three space dimensions, and because there is a distance between the two boundaries there is time governed by the speed of light. It has internal gravity compressing the parts that make up the mass together.
Now we add a second mass and it has internal frame of reference the same as the first mass.. but.. there is now a relationship between the two of them, that of attraction through gravity.
These two mass's now start moving towards each other and accelerate developing kinetic energy. Kinetic energy increase with the square of the velocity and so this is the mechanism through which energy can be brought into the model.
Each of the mass's has an internal frame of reference that creates 4D space/time and because the two are related in the void through the force exerted on them by their own gravitational attraction the void (space) has 4D space/time.
As everything in the universe is constantly in motion in relation to each other and moving at very high velocity there is the source of energy, deceleration is equal to a loss of energy and acceleration is equal to an increase in energy, but it's all governed by the square of the velocity law..
Where the energy goes and where it comes from I don't know but it is the movement of mass's through 4D space/time that creates the conditions for it to manifest or disappear.
https://en.wikipedia.org/wiki/Electron_rest_mass
The electron rest mass (symbol: me) is the mass of a stationary electron. It is one of the fundamental constants of physics and is also very important in chemistry because of its relation to the Avogadro constant. It has a value of about 9.109×10−31 kilograms or about 5.486×10−4 atomic mass units, equivalent to an energy of about 8.187×10−14 joules or about 0.5110 megaelectronvolts.[1]
Quote from: evolvingape on 2018.06.10, 17:28:03
The electron rest mass (symbol: me) is the mass of a stationary electron.
Stationary relative to what ?
Quote from: evolvingape on 2018.06.10, 17:28:03
Interesting thought experiment.. but it won't work. Energy is equivalent to mass multiplied by the speed of light squared, so it has to be moving..
In the super cap see saw example the energy in the cap is not moving relative to the second cap and so no increase in mass is possible relative to each other.
Good answer, I like that. ;)
Let me then amend the thought experiment and replace the caps with 2 superconducting coils, shuttling energy between each-other. :P
https://www.google.com/url?sa=t&source=web&rct=j&url=https://arxiv.org/pdf/gr-qc/0404005&ved=2ahUKEwjCzrOF3cnbAhWT458KHWrGD9EQFjABegQICRAB&usg=AOvVaw3rmkg-hwDbCiH3Cyn53JuC
Quote from: ion on 2018.06.10, 15:27:39
Here is a thought to chew on:
Normal beam balance scales work quite well and are accurate in the close range earth gravitational field.
When we increase the size of our beam balance scale, it becomes a device with hysteresis, an over center mechanism.
It will seek one of two possible states, but be nearly impossible to balance.
What size does our beam balance scale have to be to experience the hysteresis effect?
Regards
here is a book on balances, 5.6 talks about hysteresis
https://publications.csiro.au/rpr/download?pid=procite:ef45dd58-cfba-4532-8477-34262ae8f703&dsid=DS1
most hysteresis is attributed to friction
Other source indicate magnetic, temperature, and mechanical stresses also affect balances
Where does "size" fit in?
G asked
QuoteWhere does "size" fit in?
This was in reference to this post: http://www.overunityresearch.com/index.php?topic=3638.msg68213#msg68213
Where I said:
QuoteHere is a thought to chew on:
Normal beam balance scales work quite well and are accurate in the close range earth gravitational field.
When we increase the size of our beam balance scale, it becomes a device with hysteresis, an over center mechanism.
It will seek one of two possible states, but be nearly impossible to balance.
What size does our beam balance scale have to be to experience the hysteresis effect?
When the beam balance scale is made large enough, e.g. the arms are large and the pivot point is high in altitude) our weights on either side (or even the empty buckets on either side) will have to experience the square law of gravitation, so as one weight descends close to the surface of the earth it becomes heavier, conversely the weight on the other end of the scale is raised up and becomes lighter.
The beam balance scale will be forced to snap into one or the other position.
So the question was: at what size would this effect be pronounced?
I should have been clearer in my statement.
Well, somewhere around 200 miles up, weightlessness occurs. So, I would guess that the hysteresis would become an issue at a point lower than this, but I don't know if anyone has every bothered to calculate this out.
If you want to create a unidirectional force, just spin a magnetic top. The precession of the rotating magnetic field will create a downward force, opposite gravity. (This is based on Willie Johnson's Gyroscopic Force Theory.)
I think you need a phase change (not electrical phase but material phase) to use gravity to drive a motor. If you split some water at some depth into H and O these could rise due to buoyancy, then be recombined at the surface back into water. That rise could drive a motor. Methinks EA's hhop does something like this. Another phase change is pair production or annihalation where zero rest mass photons change into non-zero rest mass matter particles or vice versa.
I think a more promising line of investigation is some phase change within other conservative fields such as magnetic vector potentials that have no Curl, where there is no magnetic field. But that's another story.
Smudge
Quote from: Grumpy on 2018.06.10, 22:00:39
Well, somewhere around 200 miles up, weightlessness occurs. So, I would guess that the hysteresis would become an issue at a point lower than this, but I don't know if anyone has every bothered to calculate this out.
If you want to create a unidirectional force, just spin a magnetic top. The precession of the rotating magnetic field will create a downward force, opposite gravity. (This is based on Willie Johnson's Gyroscopic Force Theory.)
It is not true that weightlessness occurs somewhere around 200 miles up. There is only around a 10% reduction in gravity at 200 miles up. The weightlessness in the ISS (250 mi altitude) as we see on TV is because the centrifugal force of flying at 17,200 mph offsets the remaining 90% gravitational pull. The tiny leftover component of force resulting in the microgravity of the ISS.
In theory you should be able to detect the hysteresis effect I was referring to with a modest sized scale right at the surface of the earth if you used magnetic bearings on your scale (no friction, no stiction). It becomes easier to demonstrate with a scale where the pivot is much higher in altitude and the arms are much longer.
We should learn more of Willie Johnson's Gyroscopic Force Theory. Maybe start a topic if you haven't already .
Regards
Yeah, I should have looked for a more reliable source regarding gravity and altitude.
Even if a friction-less scale won't balance, and exhibits hysteresis (noise), what does that say about the nature of gravity?
Does gravity have a frequency?
For a mass to be accelerated,work must be done.
For work to be done,energy is required to perform that work.
Two masses accelerating toward each other in space,due to gravitational attraction,means work is being done,and a source of energy is provided to perform that work.
Brad
Quote from: Grumpy on 2018.06.12, 02:40:58
Yeah, I should have looked for a more reliable source regarding gravity and altitude.
Even if a friction-less scale won't balance, and exhibits hysteresis (noise), what does that say about the nature of gravity?
Does gravity have a frequency?
There really is no known state of true weightlessness(except for a short period of time), as gravity from all bodies extends to infinity, so an object in a gravitational "neutral zone" will sooner or later be pulled in the direction of stronger gravity, as there is no perfectly neutral zone.
The hysteresis I am referring to is the deadband or force required to change state of a binary device. e.g. an op-amp with some positive feedback, a Schmitt trigger, an ordinary switch with over center mechanism, a snap disc etc.
https://en.wikipedia.org/wiki/Schmitt_trigger
Nothing has "weight" until it's mass is subjected to a gravitational force, which, like you say, is everywhere to some degree.
Back to the subject at hand, I see no way to harness naturally occurring gravity as a source of energy with the exception of one method.
If you setup a homopolar generator so that the disc precesses (leaning sideways so that the disc is at an angle) then gravity can be used to generate a current in the discs as it rotates (i.e. precesses). The output should be slightly more than the energy required to rotate the disc, but gravity is pretty weak so don't expect too much.
Fields can be rotated much easier than physical discs, and I have a hunch that the early TPU's used gravity in the same way, but the field was much faster and cheaper than rotating a disc. On that note, gravity can be a great free source of energy if you simulate a rotating conductive disc with a rotating electric field and use a conductor to induce a current.
Quote from: Grumpy on 2018.06.12, 17:36:59
Nothing has "weight" until it's mass is subjected to a gravitational force, which, like you say, is everywhere to some degree.
Back to the subject at hand, I see no way to harness naturally occurring gravity as a source of energy with the exception of one method.
If you setup a homopolar generator so that the disc precesses (leaning sideways so that the disc is at an angle) then gravity can be used to generate a current in the discs as it rotates (i.e. precesses). The output should be slightly more than the energy required to rotate the disc, but gravity is pretty weak so don't expect too much.
Fields can be rotated much easier than physical discs, and I have a hunch that the early TPU's used gravity in the same way, but the field was much faster and cheaper than rotating a disc. On that note, gravity can be a great free source of energy if you simulate a rotating conductive disc with a rotating electric field and use a conductor to induce a current.
Grumpy
How will gravity generate a current in the disc ?
Brad
Quote from: TinMan on 2018.06.13, 23:23:47
Grumpy
How will gravity generate a current in the disc ?
Brad
Because the disc is precesing (rotating like a top) and per this theory, you can replace the magnetic field of a conventional homopolar generator with gravity, but only if you precess the disc rather than just rotate it.
One of the main points of this theory is that electric current only occurs if the electrons are forced to precess. Precession usually requires two orthogonal forces, but since the disc is precessing, only one force is required.
It's really hard to make mechanically.
The Gyroscopic Force Theory is a complete unified theory. Unifying gravity, magnetism and other forces.
Quaternions are used in this theory and this makes it difficult for anyone who has never used them. I've never learned to use them, so I just pick interesting pieces out of the theory and ask Willie about them. (Willie is writing a book on electrodynamics now, so he doesn't have a lot of time, but is eager to explain things when he has time.)
I introduced the GFT here back in 2010:
http://www.overunityresearch.com/index.php?topic=419.msg6825#msg6825
This is the old chapter on homopolar motors/generators (before the book was published):
http://www.overunityresearch.com/index.php?topic=79.msg3015#msg3015
Quote from: Grumpy on 2018.06.14, 13:44:04
Because the disc is precesing (rotating like a top) and per this theory, you can replace the magnetic field of a conventional homopolar generator with gravity, but only if you precess the disc rather than just rotate it.
One of the main points of this theory is that electric current only occurs if the electrons are forced to precess. Precession usually requires two orthogonal forces, but since the disc is precessing, only one force is required.
It's really hard to make mechanically.
The Gyroscopic Force Theory is a complete unified theory. Unifying gravity, magnetism and other forces.
Quaternions are used in this theory and this makes it difficult for anyone who has never used them. I've never learned to use them, so I just pick interesting pieces out of the theory and ask Willie about them. (Willie is writing a book on electrodynamics now, so he doesn't have a lot of time, but is eager to explain things when he has time.)
I introduced the GFT here back in 2010:
http://www.overunityresearch.com/index.php?topic=419.msg6825#msg6825
This is the old chapter on homopolar motors/generators (before the book was published):
http://www.overunityresearch.com/index.php?topic=79.msg3015#msg3015
Interesting subject.
Thanks O0
author=Grumpy link=topic=3638.msg68267#msg68267 date=1528825019]
QuoteBack to the subject at hand, I see no way to harness naturally occurring gravity as a source of energy
I am reminded of the !faster down wind than the wind! machine.
So many said it could not be done--but it was.
Only then did we see many say--of course it can work,there is nothing special about it.
But they only said this once it was done.
We also have a massless particle(the photon) that can knock another particle(the electron) through a PN junction,and resulting in a flow of electrical current.
Who would have thought we could turn light into electrical power--no one 100 years ago i recon.
Gravity is a unidirectional force,where it can only pull objects with mass toward other objects with mass-right?-->well no. Gravity can actually push mass away from another body of mass as well-->alternating gravitational force O0
Only once it is done,will we here the--yea,but there's nothing out of the ordinary there.
Brad
In 1887, Heinrich Hertz discovered that electrodes illuminated with ultraviolet light create electric sparks more easily.
In 1905, Albert Einstein published a paper advancing the hypothesis that light energy is carried in discrete quantized packets to explain experimental data from the photoelectric effect.
A lot of people think Tesla's radiant energy collector was an example of the photoelectric effect, but his plates were insulated, and the voltage generated was high voltage that would discharge across a spark gap, so it must have been something else.
How is gravity going to push something away from something else? Are you talking about the sling shot effect?
Quote from: GrumpyIn 1887, Heinrich Hertz discovered that electrodes
illuminated with ultraviolet light create electric
sparks more easily.
In certain military electronics when gaseous voltage regulator
tubes were in common use, a small neon lamp was placed
in near proximity to the Voltage Regulator Tube to assure
reliable ionization under all conditions. The small amount
of light from the neon lamp assisted ionization in the tube.
Quote from: Grumpy on 2018.06.14, 16:33:30
How is gravity going to push something away from something else? Are you talking about the sling shot effect?
No.
You just place one fluid (gas) inside a more dense fluid (water),-->buoyancy
A basket ball is pulled toward the earth,until you place it in water.
It is then pushed away from the earth,due to the very force that pulled it toward earth in the first place.
Brad
https://en.wikipedia.org/wiki/Buoyancy
Buoyancy is a force separate from gravity.
Quote from: Grumpy on 2018.06.15, 00:42:36
https://en.wikipedia.org/wiki/Buoyancy
Buoyancy is a force separate from gravity.
What causes the buoyant force?
Is there a buoyant force without gravity?
Brad
Quote from: TinMan on 2018.06.15, 01:54:05
What causes the buoyant force?
Is there a buoyant force without gravity?
Brad
OK, so buoyancy requires gravity to pull the floating object into the fluid, but it's not a source of energy. There are those wave generators though, that use the ocean wave energy.
Quote from: Grumpy on 2018.06.15, 02:17:33
OK, so buoyancy requires gravity to pull the floating object into the fluid, but it's not a source of energy. There are those wave generators though, that use the ocean wave energy.
My point is this--we can turn a unidirectional force into an alternating force. We now have an alternating force that can be turned into a rotating torque.
The catch being here,is how to get the buoyant object into the bottom of the fluid,without using up the energy gained in the rest of the cycle.
Much the same problem as getting a wind powered vehicle to travel directly into the wind that is the force driving the vehicle.
Oh wait--they have already done that :D
Brad
The problem with the picture from wikipedia is that it is somewhat oversimplified and somewhat misleading as it does not show the force of gravity on the water, rather uses density of the water to illustrate. (no arrows shown pushing down on the water to create what we know as pressure in the water)
A very small body of water(say the size of a swimming pool) floating in zero G would be held together by it's own gravity and the surface tension of the liquid. Our basketball inserted into this body would only have a very weak buoyancy effect.
The same body of water in earths surface gravity would have very a strong buoyancy effect.
Note that the actual mass/unit volume of the water has not changed appreciably in both cases but if we use the old weight /unit volume to determine density it makes density variable depending on which planet you are doing the measuring on unless you've standardized the definition to only consider earth gravity as your reference.
I could be wrong but I would say that mass/volume alone does not directly account for buoyancy e.g. mass/volume of water in zero G as explained above. Rather, it is directly related to gravity plus mass/volume. A higher mass/volume medium will exhibit a stronger buoyancy effect to a lower mass/volume inserted object in a higher gravitational field.
If I am confused , please correct.
author=ion link=topic=3638.msg68339#msg68339 date=1529066135]
QuoteA very small body of water(say the size of a swimming pool) floating in zero G would be held together by it's own gravity and the surface tension of the liquid. Our basketball inserted into this body would only have a very weak buoyancy effect.
How would the basket ball have any buoyancy value in zero G ?.
If the body of water is in a zero G environment (lets say space),to which surface of the water would the ball rise to ?--which is the top of the body of water in space ?
What you have done though ION,is prove what i stated in post 30 .Quote: Gravity can actually push mass away from another body of mass as well.
The basket ball submerged in the body of water in space,has no(very little-if any)buoyant value.
As the body of water gets closer to earth,the buoyant value of the ball increases.
As the gravitational force increases on that body of water,so dose the buoyant value of the basket ball.
Gravity wants to pull the water down,which causes the ball to be pushed up.
Could there be a proof in the above that can prove gravity is an energy source :-\
Let me think about what you said a little ION.
Brad
From Brad:
QuoteHow would the basket ball have any buoyancy value in zero G ?.
If the body of water is in a zero G environment (lets say space),to which surface of the water would the ball rise to ?--which is the top of the body of water in space ?
Only it's own (self) gravity would cause some small buoyancy effect.
Depends on where it was located reference to the water's center of gravity when let go.
QuoteWhat you have done though ION,is prove what i stated in post 30 .Quote: Gravity can actually push mass away from another body of mass as well.
That was my intent. You have well stated that gravity can have a reverse effect.
QuoteThe basket ball submerged in the body of water in space,has no(very little-if any)buoyant value.
As the body of water gets closer to earth,the buoyant value of the ball increases.
As the gravitational force increases on that body of water,so dose the buoyant value of the basket ball.
Agree on all counts. Kinda proves buoyancy directly related to gravity.
QuoteGravity wants to pull the water down,which causes the ball to be pushed up.
True, as you said it is a reverse acting effect, now we need to figure out how to use that reversal in harmony with the original force.
QuoteCould there be a proof in the above that can prove gravity is an energy source :-\
Well, we are all still working on that, aren't we ;)
I often put forth my own erroneous ideas to be corrected by those on the forum that have greater training and insight (e.g Smudge). In this way I learn, however it is not without the risk of me also having to wear the dunce hat from time to time.
regards
author=Grumpy link=topic=3638.msg68334#msg68334 date=1529029053]
QuoteOK, so buoyancy requires gravity to pull the floating object into the fluid, but it's not a source of energy.
I was referring to a ball rising to the top of a column of water,but anyway O0
Stay with me here Grumpy,and i'll show you gravity being an energy source--and your going to help me confirm it O0
Lets say we have a tube and some other components, that weighs say 10,000KGs in total.
That tube is 1 meter in diameter,and 10 meters high.
On top of this tube is an ideal generator,which has an ideal propellor attached to it.
Inside we have an ideal battery pack,which is drained of all it's energy<-- the other components.
Now,we are going to raise this device up into space until we reach a point where the gravitational pull from the earth is only 1% of what it is here on the earths surface.
We are stopping at this 1% position,so as the device will fall back to earth by it self.
On the way back down to earth,the !ideal! propellor is turning the !ideal generator,which in turn is charging the !ideal! battery bank.
The energy required to lift the device to the 1% gravity point is X amount.
What will be the energy recaptured in our !ideal! battery bank on the devices descent back down to earth,where we assume no other losses in the descent ?.
Brad.
Quote from: ion on 2018.06.15, 12:35:35
The problem with the picture from wikipedia is that it is somewhat oversimplified and somewhat misleading as it does not show the force of gravity on the water, rather uses density of the water to illustrate. (no arrows shown pushing down on the water to create what we know as pressure in the water)
A very small body of water(say the size of a swimming pool) floating in zero G would be held together by it's own gravity and the surface tension of the liquid. Our basketball inserted into this body would only have a very weak buoyancy effect.
The same body of water in earths surface gravity would have very a strong buoyancy effect.
Note that the actual mass/unit volume of the water has not changed appreciably in both cases but if we use the old weight /unit volume to determine density it makes density variable depending on which planet you are doing the measuring on unless you've standardized the definition to only consider earth gravity as your reference.
I could be wrong but I would say that mass/volume alone does not directly account for buoyancy e.g. mass/volume of water in zero G as explained above. Rather, it is directly related to gravity plus mass/volume. A higher mass/volume medium will exhibit a stronger buoyancy effect to a lower mass/volume inserted object in a higher gravitational field.
If I am confused , please correct.
OK,using IONs above post,i have found that physics it self confirms gravity is an energy source.
It also confirms a statement i made elsewhere,where when work is done by the gravitational field,this work has no effect on the volume of energy stored in that gravitational field.
Gravity is an inexhaustible source of energy.
To go even further-->gravity can create energy,and physics backs this up O0
Brad
I think everyone is better off utilizing "other" modes of induction to create useful electricity than harnessing a single-direction accelerating force.
http://www.overunityresearch.com/index.php?topic=2561.msg40498#msg40498
You take that wonderful electricity, more out than you are using to make it, and generate a gravity-like accelerating force with it.
Sha-Zam!
Go back through history and take a look at the butcher job they did on electrodynamics. They threw out half of it! Yet most scientist keep pushing forward with the half they left us with, which is incomplete.
Quote from: Grumpy on 2018.06.17, 15:25:59
I think everyone is better off utilizing "other" modes of induction to create useful electricity than harnessing a single-direction accelerating force.
http://www.overunityresearch.com/index.php?topic=2561.msg40498#msg40498
You take that wonderful electricity, more out than you are using to make it, and generate a gravity-like accelerating force with it.
Sha-Zam!
Go back through history and take a look at the butcher job they did on electrodynamics. They threw out half of it! Yet most scientist keep pushing forward with the half they left us with, which is incomplete.
Is this not !OverUnityResearch! ?
Is there no interest here about researching an !overunity! device,where the sea of gravity can provide endless clean energy ?. This can all be done on paper,using the current laws of physics.
If some one can show just one instance of gravity creating energy,would that not open the doors for other methods to be obtained ?.
I mean,the universe is full of energy,and that had to come from some where-did it not?.
Oh well,if there is no interest in the subject at hand,i'll just go on with something else.
Brad
Brad, you said:
QuoteIf some one can show just one instance of gravity creating energy,would that not open the doors for other methods to be obtained ?.
I mean,the universe is full of energy,and that had to come from some where-did it not?.
Oh well,if there is no interest in the subject at hand,i'll just go on with something else.
Brad
Well, Brad, you know I am interested in the subject and the purpose of this thread was to arouse interest and arguments.
In post #7 I think I created an experiment that shows you can get more out of what you put into lifting an object, the trick being that you send the weight to the neutral zone then let a different gravitational body of more mass attract the object.
There are probably some good refutations of this argument that I have not seen posted.
I also posted some other arguments. Maybe these arguments are sophomoric, but I am just learning.
Regards
Quote from: ion on 2018.06.18, 02:18:19
Brad, you said:
Well, Brad, you know I am interested in the subject and the purpose of this thread was to arouse interest and arguments.
In post #7 I think I created an experiment that shows you can get more out of what you put into lifting an object, the trick being that you send the weight to the neutral zone then let a different gravitational body of more mass attract the object.
There are probably some good refutations of this argument that I have not seen posted.
I also posted some other arguments. Maybe these arguments are sophomoric, but I am just learning.
Regards
My mathematical experiment is simply raising a mass to a given hight,and then lowering that mass back to it's starting position,where the value of that mass never changes.
To perform these calculations,we stick to all the known laws of physics.
We will need two people to carry out the math,and present there findings here.
I believe we can show those pesky laws (thermodynamics and the conservation of energy)do not always apply,and gravity can be a source of infinite energy.
I am hoping Pm and Smudge will take up the small mathematical challenge-or anyone here that wishes to give it a go.
It just seems that now everything has to comply with the known laws of physics,and it has become an automatic response now.
What are the ramifications of showing that energy can be created ?
Brad
I have previously suggested a reason for the Bessler wheel to work where the falling mass gains spin energy rather than the usual linear velocity energy, then uses force precession to transfer that angular momentum to the big wheel. That was shot down by TK who pointed out that forced precession requires energy. However the recent mention of the Aspden effect on another thread has suggested another approach.
Aspden discovered that he could spin-up a flywheel and the input energy needed to do this was equivalent to the subsequent stored energy in the flywheel. However if he quickly stopped the flywheel and then spun it up again it needed far less energy to spin it up a second time. Somehow the aether had achieved some vortex property that allowed that second spin-up to require less input enegy. His experiments showed that the aether held this property for several minutes.
Now go back to my Bessler wheel where the falling mass has small pinions that engage with an inclined or near vertical rack so as to gain angular energy of almost m*g*h. It is caught by the inside of the big wheel where it is trapped and can't independently rotate thus imparting that m*g*h energy to the big wheel. It has left behind in the aether that hidden Aspden vortex that current physics doesn't recognize. That mass gets carried back to the top and on its second fall requires less energy to spin up so it achieves a greater spin rate than previously. Here is the possibility of obtaining perpetual motion. Maybe Bessler's wheel was not the scam that most people think.
Smudge
Quote from: Smudge on 2018.06.18, 10:03:25
I have previously suggested a reason for the Bessler wheel to work where the falling mass gains spin energy rather than the usual linear velocity energy, then uses force precession to transfer that angular momentum to the big wheel. That was shot down by TK who pointed out that forced precession requires energy. However the recent mention of the Aspden effect on another thread has suggested another approach.
Aspden discovered that he could spin-up a flywheel and the input energy needed to do this was equivalent to the subsequent stored energy in the flywheel. However if he quickly stopped the flywheel and then spun it up again it needed far less energy to spin it up a second time. Somehow the aether had achieved some vortex property that allowed that second spin-up to require less input enegy. His experiments showed that the aether held this property for several minutes.
Now go back to my Bessler wheel where the falling mass has small pinions that engage with an inclined or near vertical rack so as to gain angular energy of almost m*g*h. It is caught by the inside of the big wheel where it is trapped and can't independently rotate thus imparting that m*g*h energy to the big wheel. It has left behind in the aether that hidden Aspden vortex that current physics doesn't recognize. That mass gets carried back to the top and on its second fall requires less energy to spin up so it achieves a greater spin rate than previously. Here is the possibility of obtaining perpetual motion. Maybe Bessler's wheel was not the scam that most people think.
Smudge
I spent a couple of days researching this !Aspden! effect some years back--I think Grum did as well,if memory serves me correct.
My measurement method was very accurate,but the results showed only a very small reduction in needed energy and time to get the flywheel back up to speed.
My findings for the motor with bearings were that it was nothing more than the bearings in the motor getting up to running temperature,where the grease in the bearings became warmer and thinner during the first 5 minutes of running,thus taking less time and energy to get up to speed in the next run.
To confirm this,i let the motor (which had the flywheel directly attached to its shaft)sit for 2 hours,so as it was cold again. This time,before starting my first run,i heated up the motor a bit with the heat gun,so as the motor and bearings would be at the temperature they would be after a 5 minute run.
The result was that the flywheel got up to the same speed,in the same time as it did in the !Aspden effect! run.
My second set of tests used a motor with no bearings,but brass bushes instead.
I used a very light machine oil as lubricant on the bushes.
in 5 run's,there was no difference in time or required power to get the flywheel up to a set RPM.
In both tests,i used my PSU,with a set voltage,and current limited to a predetermined value.
Brad
QuoteI guess if we had some kind of gravity shield we could put it above an object and see if the object floats which would maybe prove the force is pushing down on the object. Or put the gravity shield below the object and if it floats would that prove gravity is an attraction force?
Respectfully,
Carroll
Hi Carroll
You don't need to float objects, just put your "gravity shield" under one side of a wheel with horizontal axle, it will spin when you shield one side. Just note the direction of spin to determine push or pull. This could be carried out with a lightweight wheel in a vacuum with magnetic bearings. Now where is that "Gravity Shield"........ I have a few ideas.
Brad
The Gravity Toy proposed on my bench might work by accumulating excess angular momentum while staying strictly aligned with the vertical force of Gravity. A normal weight rolling down a ramp does not do this. I have some good ideas for improving this device by adding variable pitch to the descending spinning weight to present a better variable "impedance match" to gravity force.
I would be very interested in seeing your idea for gaining excess energy by returning a weight to it's starting point.
Regards
Hello All.
Yes, I did look into the " Aspen effect " my results were inconclusive but the second run up showed a marked decrease in time. I heated everything up above ambient temperature so the rig was actually cooling down.
Cheers Graham.
As a post script, I was wondering if device orientation might play a part? Mine was East West at the time of testing.
I was told that some very big and well funded labs could find no gain in the Aspen effect ?
I am aware of no specifics regarding the testing ,except that it was claimed to be "exhaustive"
for what its worth ?
Quote from: ion on 2018.06.18, 02:18:19
Brad, you said:
Well, Brad, you know I am interested in the subject and the purpose of this thread was to arouse interest and arguments.
In post #7 I think I created an experiment that shows you can get more out of what you put into lifting an object, the trick being that you send the weight to the neutral zone then let a different gravitational body of more mass attract the object.
There are probably some good refutations of this argument that I have not seen posted.
I also posted some other arguments. Maybe these arguments are sophomoric, but I am just learning.
Regards
I think some are that stuck in there way's,that they have let schooling get in the way of there education.
Lets have a close look at one of those laws-->energy cannot be created ???
For this,we're going to the moon O0. I mean,we have been there before-right?,so no problem there C.C
Anyway,we're on the moon,and we launch a mass which has a weight of 1000KGs on the moon,toward earth.
That same mass will impact the earth with far greater energy than it took to launch it from the moon.
So the question remain's,if gravity is not a source of energy,where did that extra energy come from?
Now,some here will say that-->well,you just lost some of the moons mass,and that loss is where the energy came from.
This of course is wrong,as we lost no mass at all--it was just transfer'd
What the moon lost,the earth has now gained.
The moons mass just decreased,and so it's gravitational value must also decrease.
But,at the same time,the earths mass just increased,and so it's gravitational value also increased-->by the very same amount that the moons decreased. The earths to moons gravitational attraction value has not changed
This means that,if we carried out the same experiment again,it would take the same amount of energy to launch that 1000KG mass from the moon,but the mass would also be slightly larger for that 1000KG weight on the moon,and it would impact the earth with even greater energy,due to the increase in mass,and the increase in earth gravitational value.
But we can stay within earths gravity field,and carry out some mathematical experiments,and still show how gravity can provide energy--or create it,as no gravity will be consumed :D ,if you can say it like that ?.
What i mean is,energy can only be transformed from one state to another.
How exactly would gravity be transformed or consumed, if the mass remains the same ???
Anyway,lets get on with the mathematical experiment.
Brad
Ok,so lets get to our mathematical experiment.
From now on,all masses raised will be in a geostationary position,or orbit,as we will be going pretty high,so as to make the math easier,and show larger results-->much like trying to measure microwatts of power to that of watts of power.
The conservation of energy states that the potential energy (PE)of a mass (M),is equal to the required energy to lift/raise that mass (M) to height (H),and when that mass is lowered to it's starting position,the same amount of energy to that of the PE will be dissipated in one form or another.
!!There will be no energy gain,and no energy loss,but only a transformation from one state to another!!.
Am i right so far?
Example.
If we raise a mass which has a weight of 1000KGs,to a height of 1000 meters,it will take 9800000 Joules of energy to do so,and our PE will be 9800000 Joules.
If we lower that same mass back to it's starting position,we will produce 9800000 Joules of energy in one form or another-->energy is conserved O0
The same rules apply in any gravitational field of any strength.
Example--we will calculate the same mass in a gravity field that is 1/10th that of earths.
It will take 980000 Joules of energy to raise that same mass to a height of 1000 meters. The PE will be 980000 Joules,and 980000 Joules of energy will be produced in one form or another when that mass is returned to it's starting position.
Even passing through lowering gravitational values as we gain height above the earths surface,the energy returned as the mass returns to it's starting point,will be the same amount of energy needed to raise it to that height in the first place-->energy is conserved.
Smudge
Please check and confirm the above-->incase i missed something here.
Thanks
Brad
Quote from: TinMan on 2018.06.18, 13:17:23
Ok,so lets get to our mathematical experiment.
From now on,all masses raised will be in a geostationary position,or orbit,as we will be going pretty high,so as to make the math easier,and show larger results-->much like trying to measure microwatts of power to that of watts of power.
The conservation of energy states that the potential energy (PE)of a mass (M),is equal to the required energy to lift/raise that mass (M) to height (H),and when that mass is lowered to it's starting position,the same amount of energy to that of the PE will be dissipated in one form or another.
!!There will be no energy gain,and no energy loss,but only a transformation from one state to another!!.
Am i right so far?
Example.
If we raise a mass which has a weight of 1000KGs,to a height of 1000 meters,it will take 9800000 Joules of energy to do so,and our PE will be 9800000 Joules.
If we lower that same mass back to it's starting position,we will produce 9800000 Joules of energy in one form or another-->energy is conserved O0
The same rules apply in any gravitational field of any strength.
Example--we will calculate the same mass in a gravity field that is 1/10th that of earths.
It will take 980000 Joules of energy to raise that same mass to a height of 1000 meters. The PE will be 980000 Joules,and 980000 Joules of energy will be produced in one form or another when that mass is returned to it's starting position.
Even passing through lowering gravitational values as we gain height above the earths surface,the energy returned as the mass returns to it's starting point,will be the same amount of energy needed to raise it to that height in the first place-->energy is conserved.
Smudge
Please check and confirm the above-->incase i missed something here.
Thanks
Brad
Right so far, Brad.
Physics note:
PE = MgH
measured in standard units as Joules = Newton*m
where 1 Newton = N = kg*m/s^2.
Potential Energy = PE = latent or stored energy, based on an objects POSITION in a field (such as a gravitational or electric field).
You are OK so far Brad. However you keep expressing a given mass as a weight (like 1000Kg) but weight is a force. A better statement is to refer to the mass itself as a 1000Kg mass. Weight is a function of gravity. It will weigh 1000Kg on the earth but in your 1/10th gravity field it will weigh 100Kg, there will be only a 100Kg force on it. Of course you know this, so just ignore me.
Smudge
Quote from: Smudge on 2018.06.18, 18:35:53
You are OK so far Brad. However you keep expressing a given mass as a weight (like 1000Kg) but weight is a force. A better statement is to refer to the mass itself as a 1000Kg mass. Weight is a function of gravity. It will weigh 1000Kg on the earth but in your 1/10th gravity field it will weigh 100Kg, there will be only a 100Kg force on it. Of course you know this, so just ignore me.
Smudge
Yes,as the gravitational pull decreases with height,the weight of the mass will decrease,but the mass will remain the same.
This is where the clever math is needed.
If we raise a given mass say only 100 meter's ,where that mass remains in the 9.8m/sec* area,then we use that masses weight to calculate our PE when at rest. If we raise that same mass to a height where the acceleration of G is only .98m/sec,then we have to slowly decrease the value of G as the mass is raised.
So a question would be,do we use an average value of 4.41m/sec to make our calculations ?
Not that it matters much,as the end result is-energy is conserved,and the energy dissipated on the descent back to earth,will be the same as the energy required to raise the mass to that height in the first place-->correct?.
This being the case,there will be no need for calculating needed energy to raise the mass to a given height,no need to calculate the PE,as it will be the same as the energy required to raise the mass to that height,and no need to calculate the dissipated energy as our mass falls back to,and makes contact with earth,as that will be the same value as our PE was.
So,your math skills in this experiment is needed to calculate the energy required to make physical changes within the mass once it reaches it's maximum height,and the stored energy within the mass,once it returns back to the earths surface,and is in a rest state ;)
So,if what is in blue is correct,then we will move on.
Brad
Quote from: PhysicsProf on 2018.06.18, 14:06:18
Right so far, Brad.
Physics note:
PE = MgH
measured in standard units as Joules = Newton*m
where 1 Newton = N = kg*m/s^2.
Potential Energy = PE = latent or stored energy, based on an objects POSITION in a field (such as a gravitational or electric field).
Thanks Steve.
Glad you made this point-->Potential Energy = PE = latent or stored energy, based on an objects POSITION in a field (such as a gravitational or electric field).
Our mass will remain within the earths gravitational field O0
Brad
Quote from: TinMan on 2018.06.18, 21:06:01
So a question would be,do we use an average value of 4.41m/sec to make our calculations ?
No! Averaging works OK when the change of force with distance is a linear relationship, but gravity is not linear. Like the Coulomb force gravity follows an inverse square law. It is then necessary to resort to integration techniques to get the total energy requirements.
QuoteNot that it matters much,as the end result is-energy is conserved,and the energy dissipated on the descent back to earth,will be the same as the energy required to raise the mass to that height in the first place-->correct?.
This being the case,there will be no need for calculating needed energy to raise the mass to a given height,no need to calculate the PE,as it will be the same as the energy required to raise the mass to that height,and no need to calculate the dissipated energy as our mass falls back to,and makes contact with earth,as that will be the same value as our PE was.
That is correct
QuoteSo,your math skills in this experiment is needed to calculate the energy required to make physical changes within the mass once it reaches it's maximum height,and the stored energy within the mass,once it returns back to the earths surface,and is in a rest state ;)
What physical changes do you have in mind?
Smudge
Not that it matters much,as the end result is-energy is conserved,and the energy dissipated on the descent back to earth,will be the same as the energy required to raise the mass to that height in the first place-->correct?.
This being the case,there will be no need for calculating needed energy to raise the mass to a given height,no need to calculate the PE,as it will be the same as the energy required to raise the mass to that height,and no need to calculate the dissipated energy as our mass falls back to,and makes contact with earth,as that will be the same value as our PE was.
QuoteThat is correct
QuoteWhat physical changes do you have in mind?
Well.lets get started O0
Brad
The mathematical experiment go's like this-->
We have a sealed cylinder full of water,and with a non compressible hollow ball also inside the cylinder.
The ball is made from aluminum,and has a weight of 10KG,and is filled with air.
The total weight of the mass(our cylinder) is 7500KGs.
1st pic shows the DUT sitting on the earths surface.
We now raise our mass,in a geostationary fashion,until we reach a height where the gravitational value is 1/10th of that on the earths surface-->.98m/sec*
Once this height is reached,the ball is pulled down to the bottom of the tank,and hooked onto the hook-->see pic 2
Once the ball is hooked onto the hook,our mass is returned to earth,where the kinetic energy is dissipated on the way back to earth by way of a parachute to slow it's descent,and heat from atmospheric friction.The remaining kinetic energy is dissipated once the mass impacts earth-->which is only a small impact,so as not to destroy our DUT.
So in other words--a controlled descent O0.
As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved.
Now,all we have to do is calculate the energy used to pull the ball to the bottom of the tank,when our DUT was in that 1/10th strength gravity field,and calculate the !now! stored energy of the submerged ball back here on earth.
My calculations.
Pulling the ball down to the bottom of the tank ,when the acceleration of G is .98m/sec*
Ball displacement = 449L
Weight of displaced water=44.9KGs
Height=9 meters
Energy required= 396 joules.
Stored energy when DUT is back on earths surface
Ball displacement=449L
Weight of displaced water=449KGs
Height=9 meters
Stored energy=39601 joules.
Energy that came from somewhere=39205 joules.
Is this correct?
Brad
P.S--oops
I forgot to factor in the weight of the ball in both calculations.
Smudge
Could you calculate the right values for both
Cheers
Quote from: TinMan on 2018.06.19, 11:25:32
Smudge
Could you calculate the right values for both
You almost got it right. 1 Kg expressed as a force is a force unit that doesn't change with gravity. It is the force on a 1Kg mass here on earth and has a value of 9.81 Newtons. In outer space 9.81 Newtons is still a 1Kg force. So when you expressed your 449L of water as a 449Kg mass you were correct in saying that its weight here on earth is 449Kg force and in your 1/10th gravity is 44.9Kg force. Those two values have to be changed to Newtons to get the energy in Joules and in both cases you must multiply by 9.81. So my calculations are:-
Pulling the ball down to the bottom of the tank ,when the acceleration of G is .981m/sec*
Ball displacement = 449L
Weight of displaced water=44.9KGs force =440.47N
Height=9 meters
Energy required= 3964.2 joules.
Stored energy when DUT is back on earths surface
Ball displacement=449L
Weight of displaced water=449KGs force = 4404.7N
Height=9 meters
Stored energy=39642 joules.
Energy that came from somewhere=35678 joules.
Still thinking about where the energy comes from.
Smudge
I think it comes down to centre of mass. Initially the mass centre is below the geometric centre then when it returns to earth the mass centre is above the geometric centre. If you go into outer space it requires zero energy to move the hollow ball, and the same would apply to moving a solid mass within a hollow cylinder. We can get similar things with metal magnets where some of the permanent magnetism comes from the spins of conduction electrons. Those itinerant electrons can be dragged towards one end of the magnet thus altering the magnetization density along the magnet. This evidences itself when you measure attractive force v. repulsive force, for a given separation distance they are not the same.
Smudge
It's that change of centre of mass that means you don't bring the system back to its starting point. Thus Brad's statement "As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved." is untrue. Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start. And that difference accounts for the apparent gain.
Smudge
Quote from: Smudge on 2018.06.19, 18:11:10
It's that change of centre of mass that means you don't bring the system back to its starting point. Thus Brad's statement "As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved." is untrue. Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start. And that difference accounts for the apparent gain.
Smudge
Smudge
That difference is only the diameter of the ball-->950mm
The energy lost in this case would only be the energy required to raise 449KGs up 950mm,which is 4180.19 joules.
The second thing to remember is this.
When the ball was pulled to the bottom of the tank in the 1/10th gravity region,the center of mass was raised by the very same amount it was lost here on earth-->950mm
Even if we discount the later,and we subtract that 4180.19 joules from our apparent gain of 35678 joules,we still have 31497.81 joules of energy unaccounted for.
Looking at my second point,then i stand by my statement-->As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved
My second point shows that the center of mass is raised and lowered the very same distance
I do hope that this is given the time and calculations it deserves,instead of just falling victim to that !must conform to the laws of physics! syndrome.
This is OverUnityResearch.
We hare here to find and investigate the exceptions to the rules/laws.
Lets not just throw our hands in the air,and say--that must be where the extra energy is coming from.
Lets carry out the math,and see if there is truly an energy gain.
Your math showed a gain,but your accountability for that gain just dose not add up.
449KGs at a height of 950mm dose not net an energy requirement of 35678 joules.
Brad
Quote from: TinMan on 2018.06.19, 22:38:15
Smudge
That difference is only the diameter of the ball-->950mm
Well not exactly but I agree the difference is quite small.
Quote
The energy lost in this case would only be the energy required to raise 449KGs up 950mm,which is 4180.19 joules.
There I must disagree. We are talking about the centre of mass of the whole thing, so it is the energy required to raise that whole mass that small distance.
Quote
I do hope that this is given the time and calculations it deserves,instead of just falling victim to that !must conform to the laws of physics! syndrome.
I have spent the last 21 years searching for things that do not conform to the accepted laws of physics. When I see an obvious flaw in someone's reasoning I feel obliged to point this out. I used the centre of mass argument to demonstrate that the thing returns to earth in a different state to that when it leaves the earth, hence the energy lost in going to that outer space region is not equivalent to the energy gained when it is returned to earth. You do not have to work out the actual centre of mass to do the calculation. You just concentrate on the two spherical volumes at the top and bottom of the cylinder (all the rest of the system undergo the same raise and fall hence can be ignored. On the way up the bottom spherical volume is filled with water while the top one isn't, the energy needed for that amount of water to reach outer space is a given value. On the way down the the top spherical volume is filled with water but the bottom one isn't. Thus that spherical quantity of water doesn't return to its starting point, it is 9m higher. The difference in energy between that needed to reach outer space and that gained on return is thus the weight of that volume of water multiplied by 9m.
Smudge
Quote from: Smudge on 2018.06.20, 09:03:31
I have spent the last 21 years searching for things that do not conform to the accepted laws of physics. When I see an obvious flaw in someone's reasoning I feel obliged to point this out. I used the centre of mass argument to demonstrate that the thing returns to earth in a different state to that when it leaves the earth, hence the energy lost in going to that outer space region is not equivalent to the energy gained when it is returned to earth. You do not have to work out the actual centre of mass to do the calculation. You just concentrate on the two spherical volumes at the top and bottom of the cylinder (all the rest of the system undergo the same raise and fall hence can be ignored. On the way up the bottom spherical volume is filled with water while the top one isn't, the energy needed for that amount of water to reach outer space is a given value. On the way down the the top spherical volume is filled with water but the bottom one isn't. Thus that spherical quantity of water doesn't return to its starting point, it is 9m higher. The difference in energy between that needed to reach outer space and that gained on return is thus the weight of that volume of water multiplied by 9m.
Smudge
QuoteThere I must disagree. We are talking about the centre of mass of the whole thing, so it is the energy required to raise that whole mass that small distance.
Well,i think we both disagree here,and i think your own math disagree's with you as well.
For this we go back to your other posts.
Your calculations in reply 61
Energy that came from somewhere=35678 joules.
Reply 63-->Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start. And that difference accounts for the apparent gain.
And when you say-->so it is the energy required to raise that whole mass that small distance,im guessing you mean the water only,as the tank it self returns back to the starting point(the ground).
The volume of water in the tank is 7323 liters,and so it's weight is 7323KGs.
To raise a weight of 7323KGs 950mm, takes 68246.7 joules of energy.
The energy you calculated (the apparent gain) was only 35678 joules.
So,something went wrong here ???
Also,would the cycle not be complete when the ball is released,and the water fell to the bottom of the tank,while at the same time the ball rises back to it's starting position ?.
Brad
Quote from: TinMan on 2018.06.20, 10:52:49
Well,i think we both disagree here,and i think your own math disagree's with you as well.
For this we go back to your other posts.
Your calculations in reply 61
Energy that came from somewhere=35678 joules.
Yes that is the buoyancy energy gained minus the energy needed to pull down the float in your 1/10th gravity. That 1/10th gravity is a bit of a red herring, you might as well go far away to zero gravity then there is no energy needed to move mass or floats.
Quote
Reply 63-->Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start. And that difference accounts for the apparent gain.
And when you say-->so it is the energy required to raise that whole mass that small distance,im guessing you mean the water only,as the tank it self returns back to the starting point(the ground).
No I mean the total device.
Quote
The volume of water in the tank is 7323 liters,and so it's weight is 7323KGs.
To raise a weight of 7323KGs 950mm, takes 68246.7 joules of energy.
The energy you calculated (the apparent gain) was only 35678 joules.
You are using that 950mm there and that is not the distance between geometric centre and mass centre. That is just your assumption. As I said, it is not necessary to use that mass centre. The tank and the water outside of those two spherical volumes can be raised to any height you want, then back down again, and energy is conserved. So the mass of that water and of the tank and of the sphere shells is immaterial, we can forget them and just concentrate on those two spherical volumes inside those shells. Imagine two lightweight spheres joined by a 9m long vertical lightweight rod. The bottom sphere contains water. Alongside we have another pair with the top sphere containing water. I think you will agree that if we raise them both to a distant region where gravity is virtually zero the water in the first one travels 9m more than the water in the second one to get to that zero gravity. So there is a difference in energy associated with that 9m difference. If that doesn't convince you I will work out the actual change in mass centre using the details you have provided and do the math for water mass, cylinder mass and spherical shell mass.
Quote
Also,would the cycle not be complete when the ball is released,and the water fell to the bottom of the tank,while at the same time the ball rises back to it's starting position ?.
Yes, that completes the cycle.
Smudge
I know Steve had mentioned he was driving cross country through the Midwest USA
and would be crunching numbers here too as internet access and time permited
he was hoping to be close to a internet port by the weekend.
Quote from: Smudge on 2018.06.20, 09:03:31
...I have spent the last 21 years searching for things that do not conform to the accepted laws of physics...
Smudge
Go back in history and read about the how vectors were chosen over quaternions. This was after Maxwell released his treatise in 1873. Gibbs and Heavyside tossed half of Maxwell's work. What we have now is 100+ years of development based on the half that they kept. What would 100+ years of development of that half give us?
EDIT:
This paper has a nice collections of references to works with quaternions.
https://arxiv.org/pdf/1504.06182.pdf
It's interesting that many of them are related to gravity.
Quote from: Grumpy on 2018.06.20, 15:33:36
Go back in history and read about the how vectors were chosen over quaternions. This was after Maxwell released his treatise in 1873. Gibbs and Heavyside tossed half of Maxwell's work. What we have now is 100+ years of development based on the half that they kept. What would 100+ years of development of that half give us?
EDIT:
This paper has a nice collections of references to works with quaternions.
https://arxiv.org/pdf/1504.06182.pdf
It's interesting that many of them are related to gravity.
Through the years,i have found many ways to show the flaws in the conservation of energy,and producing energy at no cost. I even managed to stump the late MarkE once,with the introduction of a venturi in a tank to tank transfer of compressed air.
There are many ways to show work being done without the loss of energy.
Maybe we should open a thread that looks at all this stuff Grumpy ?.
Brad
Quote from: Smudge on 2018.06.20, 13:31:52
Yes that is the buoyancy energy gained minus the energy needed to pull down the float in your 1/10th gravity. That 1/10th gravity is a bit of a red herring, you might as well go far away to zero gravity then there is no energy needed to move mass or floats. No I mean the total device.You are using that 950mm there and that is not the distance between geometric centre and mass centre. That is just your assumption. As I said, it is not necessary to use that mass centre. The tank and the water outside of those two spherical volumes can be raised to any height you want, then back down again, and energy is conserved. So the mass of that water and of the tank and of the sphere shells is immaterial, we can forget them and just concentrate on those two spherical volumes inside those shells. Imagine two lightweight spheres joined by a 9m long vertical lightweight rod. The bottom sphere contains water. Alongside we have another pair with the top sphere containing water. I think you will agree that if we raise them both to a distant region where gravity is virtually zero the water in the first one travels 9m more than the water in the second one to get to that zero gravity. So there is a difference in energy associated with that 9m difference. If that doesn't convince you I will work out the actual change in mass centre using the details you have provided and do the math for water mass, cylinder mass and spherical shell mass.Yes, that completes the cycle.
Smudge
Smudge
Regardless of whether or not that ball has a resistance placed upon it as it is released and rises,the water will fall back to it's starting position,dissipating the remaining energy to balance out what was needed to raise it in the first place.
It dose not matter if a mass falls from a set height fast over a short period of time,or slow over a long period of time--the end result is the same.
Brad
Quote from: TinMan on 2018.06.20, 23:13:57
Maybe we should open a thread that looks at all this stuff Grumpy ?.
Brad
I did years ago and it didn't go anywhere. Everyone has their own ideas and little interest in others.
Quote from: Grumpy on 2018.06.21, 14:11:56
I did years ago and it didn't go anywhere. Everyone has their own ideas and little interest in others.
Dear G
Could you provide the link to that thread or name of it?
regards
OK, I have looked more carefully into this system and I believe I have the answer. If you look at the tube plus ball before it is launched and compare it with how it looks after it returns from outer space (before gaining the buoyancy energy) the two pictures are different. The hollow ball is in a different position. Now we assume that these two conditions weigh the same, but that assumption assumes that gravity is uniform in our laboratories. But gravity isn't uniform, it reduces with distance from the earth following an inverse square law. The difference in weight between those two pictures is miniscule, but it is there. Before launch the thing weighs very slightly more than it does after return. So it takes slightly more energy to get it to outer space than is gained on its return. But that slightly more applies to the enormous energy consumed over that vast distance, and in fact it is quite significant. It exactly accounts for the gained buoyancy energy. Below is my write up proving this by math. If I have got it wrong please tell me.
Smudge
Quote from: ion on 2018.06.21, 16:22:43
Dear G
Could you provide the link to that thread or name of it?
regards
http://www.overunityresearch.com/index.php?topic=2561.msg40498#msg40498
http://www.overunityresearch.com/index.php?topic=2541.msg44301#msg44301
http://www.overunityresearch.com/index.php?topic=2185.msg40305#msg40305
Posts by GFT (Willie Johnson, author of the GFT):
http://www.overunityresearch.com/index.php?action=profile;area=showposts;u=644
(Off the top of my head):
The Gyroscopic force Theory (GFT) is a complete unified field theory that unites all known forces in physics. The GFT was developed through dimensional analysis and the use of quaternions rather than the currently accepted vector methods.
The Forgotten Quaternions
https://pdfs.semanticscholar.org/8058/b8ecc4c2932843fd306f067884e701538903.pdf
The vector algebra war: a historical perspective
https://arxiv.org/pdf/1509.00501.pdf
More on all of this math up to Clifford Geometric Algebra:
https://slehar.wordpress.com/2014/03/18/clifford-algebra-a-visual-introduction/
I gather from quick glances that the GFT could be expressed with Geometric algebra, without the complex numbers, but this is just a guess.
Quote from: Smudge on 2018.06.21, 17:08:02
OK, I have looked more carefully into this system and I believe I have the answer. If you look at the tube plus ball before it is launched and compare it with how it looks after it returns from outer space (before gaining the buoyancy energy) the two pictures are different. The hollow ball is in a different position. Now we assume that these two conditions weigh the same, but that assumption assumes that gravity is uniform in our laboratories. But gravity isn't uniform, it reduces with distance from the earth following an inverse square law. The difference in weight between those two pictures is miniscule, but it is there. Before launch the thing weighs very slightly more than it does after return. So it takes slightly more energy to get it to outer space than is gained on its return. But that slightly more applies to the enormous energy consumed over that vast distance, and in fact it is quite significant. It exactly accounts for the gained buoyancy energy. Below is my write up proving this by math. If I have got it wrong please tell me.
Smudge
Hi Smudge
Thanks you for putting together that PDF on the math for our mathematical experiment.
From what i see,everything is accounted for,except one thing.
To complete the cycle,the ball would have to rise to the top of the tube.
On the way to the top,we can use that buoyant energy to do work--perhaps by way of a cable attached to the ball,that passes through a water tight seal at the bottom of the tank,and onto the drum of a generator--so as the generator turns as the ball rises.
Now,i see you have accounted for the buoyant energy(the ball rising),but i cannot see in the math the added energy of the water falling back to it's starting position as the ball rises-->every action has an equal and opposite reaction.
So,if the balls displacement is 449 liters,and the ball is falling up 9 meters,then 449 liters of water has to fall down the same 9 meters.
Brad
Quote from: TinMan on 2018.06.21, 22:56:51
Now,i see you have accounted for the buoyant energy(the ball rising),but i cannot see in the math the added energy of the water falling back to it's starting position as the ball rises-->every action has an equal and opposite reaction.
So,if the balls displacement is 449 liters,and the ball is falling up 9 meters,then 449 liters of water has to fall down the same 9 meters.
I see what you are getting at and I can't readily account for that. Also I can't see how you could use that fall since it is not a spherical chunk of water falling down but is a distributed flow through the water medium. No water actually falls that 9m. 449 liters get displaced at the top to flow into the reservoir and 449 liters flow from the reservoir into the bottom void. The cumulative effect of all those small water movements is that 9 meter drop. I think maybe there is a change of potential energy as the ball rises and that could account for it. I'll look into that.
Smudge
The simple gravity model for the Earth modeled as a sphere of uniform mass density where outside the sphere we get the inverse square law for the total mass modeled as a point mass at the center of the Earth. That doesn't apply to the gravity inside the sphere where gravity rises linearly from the center. The tiny difference in weight between the two states (7051.00063 Kg and 7050.99937 Kg in my paper) represents a potential energy difference of 39642.21 Joules. (That is the potential energy for falling down a mine shaft that reaches the center). That is exactly the energy for the 449 liters of water falling 9 meters. That can't be a coincidence, so maybe that answers Tinman's question.
For anybody interested in falling down a mine shaft, gravity follows the law K*r/R3 where the K is that in my paper (i.e. outside the earth sphere the gravity law is K/r2), r is the distance from earth center and R is earth radius.
Smudge
Smudge quote:
QuoteFor anybody interested in falling down a mine shaft, gravity follows the law K*r/R3 where the K is that in my paper (i.e. outside the earth sphere the gravity law is K/r2), r is the distance from earth center and R is earth radius.
Smudge, would this work? Assuming of course no molten core, no friction and a uniform gravitational body>
Thought experiment #4
You have drilled a hole through the earth and out the other side. You started in Madrid spain and wound up in Wellington, New Zealand, it's antipode.
Now you get into your protective capsule at the surface in Madrid and fall into the tunnel hole accelerating toward the center of gravity. Like a pendulum,you overshoot the center of mass and begin slowing down until you come to rest at the surface in Wellington N.Z.where you step out of your capsule and trade places with someone who wants to go to Madrid.
You traveled a great distance, almost 8000 miles. Outside of the cost in energy to dig the tunnel, what future energy is expended to transport passengers between the two cities? Of course no work was done according to the standard definition so no energy was expended, yet it seems to solve a practical problem.
Regards
p.s. this is not meant to derail the current experiment, I just brought it up since falling down holes was mentioned.
Quote from: Smudge on 2018.06.22, 14:36:17
The simple gravity model for the Earth modeled as a sphere of uniform mass density where outside the sphere we get the inverse square law for the total mass modeled as a point mass at the center of the Earth. That doesn't apply to the gravity inside the sphere where gravity rises linearly from the center. The tiny difference in weight between the two states (7051.00063 Kg and 7050.99937 Kg in my paper) represents a potential energy difference of 39642.21 Joules. (That is the potential energy for falling down a mine shaft that reaches the center). That is exactly the energy for the 449 liters of water falling 9 meters. That can't be a coincidence, so maybe that answers Tinman's question.
For anybody interested in falling down a mine shaft, gravity follows the law K*r/R3 where the K is that in my paper (i.e. outside the earth sphere the gravity law is K/r2), r is the distance from earth center and R is earth radius.
Smudge
Smudge
I use the word !falls!,as that is basically what the water is doing.
The end result is the same as taking that 449 liters of water from the top,and dropping it 9 meters to the bottom-when the ball rises to the top.
So your calculated energy dissipation of that event(the 449 liters of water dropping 9 meters) is 39642.21 Joules-correct?
This was also not added into your calculations in the PDF,as your calculations were based around the ball still being at the bottom of the tank when made-correct?
So this 39642.21 Joules of energy would need to be added to the dissipated energy value once the ball was released,and had risen to the top of the tank-correct?.
This value is the very same as the buoyant energy value that would also be dissipated as the ball rises to the top,and has the ability to do work while doing so.
Your math in the PDF took this buoyant energy into account,and the math seem to show that everything balanced out--total energy gained was 0.
Quote: Add to that
loss the one tenth needed to pull the ball down when at that outer space region and we find
that all the energies are accounted for.
This is without that 39642.21 Joules of energy that would be dissipated as the 449 liters of water falls back to earth--that 449KGs of weight falling 9 meters where g=9.81
Your above statement dose not apply to the ball in our DUT,and the reason is-Quote: The simple gravity model for the Earth modeled as a sphere of uniform mass density
Where our ball has no mass-it is hollow.
If you are going to say-but the air is mass,then i would reply-->we could pull the inside of that ball down to an absolute vacuum,and it's buoyant value would still be the same.
So it seems to me,that this 39642.21 Joules of energy has not been accounted for in your math yet.
Brad
Quote from: ion on 2018.06.22, 15:23:59
Smudge quote:
Smudge, would this work? Assuming of course no molten core, no friction and a uniform gravitational body>
Thought experiment #4
You have drilled a hole through the earth and out the other side. You started in Madrid spain and wound up in Wellington, New Zealand, it's antipode.
Now you get into your protective capsule at the surface in Madrid and fall into the tunnel hole accelerating toward the center of gravity. Like a pendulum,you overshoot the center of mass and begin slowing down until you come to rest at the surface in Wellington N.Z.where you step out of your capsule and trade places with someone who wants to go to Madrid.
You traveled a great distance, almost 8000 miles. Outside of the cost in energy to dig the tunnel, what future energy is expended to transport passengers between the two cities? Of course no work was done according to the standard definition so no energy was expended, yet it seems to solve a practical problem.
Regards
p.s. this is not meant to derail the current experiment, I just brought it up since falling down holes was mentioned.
A great thought experiment ION O0
So,if we had a steel sphere the size of earth,with a hole through the center,and residing in space where there is no resistance or heat at play--we could move a mass from one side of that sphere to the other without using any energy O0
But,physics tells us that energy is required to accelerate a mass ???
Of course,that energy is returned once you pass through the center,and start to slow down--also forgetting that it takes energy to slow a moving mass. :D
Brad
Quote from: TinMan on 2018.06.23, 00:44:31
So it seems to me,that this 39642.21 Joules of energy has not been accounted for in your math yet.
Let's go back to square one where you offered the thought experiment. You sent the thing into outer space and got it back again at no energy cost (or so you thought). While in outer space at 1/10 gravity you pulled the ball down at an energy cots of 1/10 that above figure (we won't argue about the exact figure since you originally used 9.8 and I used 9.81 for gravity). You then got the above energy out when you let the ball float up. So there appeared to be an energy gain of 9/10 the above figure. I then showed that there was an energy cost in going to outer space, you spent more energy getting there than you could recoup on the way back. That energy loss is exactly 9/10 the above figure. So energy spent and energy recouped exactly matched. You then pointed out that the falling water during that last buoyancy act represented another energy that was unaccounted for. I then showed that during that buoyancy act the potential energy of the system changed by that exact amount, thus accounting for that falling water energy. So why do you still say there is energy unaccounted for?
I am expecting you to come back with the question, "what about the rising water when the ball is pulled down in 1/10 gravity? There is 1/10 the above figure still unaccounted." My answer will be again that the potential energy changes when you perform that act.
Smudge
Quote from: ion on 2018.06.22, 15:23:59
Smudge quote:
Smudge, would this work? Assuming of course no molten core, no friction and a uniform gravitational body>
Thought experiment #4
You have drilled a hole through the earth and out the other side. You started in Madrid spain and wound up in Wellington, New Zealand, it's antipode.
Now you get into your protective capsule at the surface in Madrid and fall into the tunnel hole accelerating toward the center of gravity. Like a pendulum,you overshoot the center of mass and begin slowing down until you come to rest at the surface in Wellington N.Z.where you step out of your capsule and trade places with someone who wants to go to Madrid.
You traveled a great distance, almost 8000 miles. Outside of the cost in energy to dig the tunnel, what future energy is expended to transport passengers between the two cities? Of course no work was done according to the standard definition so no energy was expended, yet it seems to solve a practical problem.
Regards
p.s. this is not meant to derail the current experiment, I just brought it up since falling down holes was mentioned.
You are correct. In that hypothetical situation no energy is required to transport passengers. But if there were no friction and we could always recoup during deceleration the energy spent in acceleration we could move passengers across the surface at no energy cost. We are getting towards that with electric vehicles that use dynamic braking to recharge batteries, regaining some of the otherwise lost kinetic energy.
Smudge
Quote from: Smudge on 2018.06.23, 08:44:31
Let's go back to square one where you offered the thought experiment. You sent the thing into outer space and got it back again at no energy cost (or so you thought). While in outer space at 1/10 gravity you pulled the ball down at an energy cots of 1/10 that above figure (we won't argue about the exact figure since you originally used 9.8 and I used 9.81 for gravity). You then got the above energy out when you let the ball float up. So there appeared to be an energy gain of 9/10 the above figure. I then showed that there was an energy cost in going to outer space, you spent more energy getting there than you could recoup on the way back. That energy loss is exactly 9/10 the above figure. So energy spent and energy recouped exactly matched. You then pointed out that the falling water during that last buoyancy act represented another energy that was unaccounted for. I then showed that during that buoyancy act the potential energy of the system changed by that exact amount, thus accounting for that falling water energy. So why do you still say there is energy unaccounted for?
I am expecting you to come back with the question, "what about the rising water when the ball is pulled down in 1/10 gravity? There is 1/10 the above figure still unaccounted." My answer will be again that the potential energy changes when you perform that act.
Smudge
Smudge
When back on earth,and the ball is released,there are two equal and opposite forces taking place.
1-the water pushing the ball up
2-gravity pulling the water down.
There is also acceleration associated with both forces,and both the water and ball have mass.
So we have 2 lots of applied forces accelerating masses.
Your calculations came back even,but that was before the cycle was complete.
Your calculations are based on the fact that the ball is still at the bottom of the tank,and so the center of mass is higher when the DUT lands back on earth.
Now,you accounted for the dissipated energy of the rising ball,but you never accounted for the water displaced by the ball, falling back down to it's very starting position,where the center of mass position is then the very same that we started with.
I have reviewed your math in the PDF 4 times now,and can confirm that the dissipated energy of the displaced water as the ball rises was not included in those calculations--and it must be.
Brad
Quote from: TinMan on 2018.06.23, 13:25:48
Smudge
When back on earth,and the ball is released,there are two equal and opposite forces taking place.
1-the water pushing the ball up
2-gravity pulling the water down.
There is also acceleration associated with both forces,and both the water and ball have mass.
So we have 2 lots of applied forces accelerating masses.
Your calculations came back even,but that was before the cycle was complete.
My calculations did not come back even (before the cycle was complete). My calculations showed that you had to expend energy in order to get to outer space and back again. Then when the cycle was complete you regained that energy and you are now evens.
QuoteNow,you accounted for the dissipated energy of the rising ball,but you never accounted for the water displaced by the ball, falling back down to it's very starting position,where the center of mass position is then the very same that we started with.
Not in the pdf but I did account for it in my later post, it is accounted for by the change in potential energy.
Quote
I have reviewed your math in the PDF 4 times now,and can confirm that the dissipated energy of the displaced water as the ball rises was not included in those calculations--and it must be.
OK I will add to my pdf to include the change in potential energy caused by the ball rising, and it is accounted for as I have said three times now.
Smudge
Quote from: Smudge on 2018.06.23, 13:58:35
My calculations did not come back even (before the cycle was complete). My calculations showed that you had to expend energy in order to get to outer space and back again. Then when the cycle was complete you regained that energy and you are now evens.
Not in the pdf but I did account for it in my later post, it is accounted for by the change in potential energy.OK I will add to my pdf to include the change in potential energy caused by the ball rising, and it is accounted for as I have said three times now.
Smudge
I think we need to go over this again.
Post 61
Ball displacement = 449L
Weight of displaced water=44.9KGs force =440.47N
Height=9 meters
Energy required= 3964.2 joules.
Stored energy when DUT is back on earths surface
Ball displacement=449L
Weight of displaced water=449KGs force = 4404.7N
Height=9 meters
Stored energy=39642 joules.
Energy that came from somewhere=35678 joules.
Still thinking about where the energy comes from.
Post 62
I think it comes down to centre of mass. Initially the mass centre is below the geometric centre then when it returns to earth the mass centre is above the geometric centre.
Post 63
It's that change of centre of mass that means you don't bring the system back to its starting point. Thus Brad's statement "As our mass never changed,then all energy used to raise the mass,was returned during the descent-->energy was conserved." is untrue. Less energy is returned during the descent because the centre of mass is not returned to the starting point, it is slightly higher than it was at the start. And that difference accounts for the apparent gain.
Now,here is where i disagree.
When the ball is allowed to rise to the top of the water,and the buoyant PE is transformed into dissipated energy,example-turning a generator,and driving a lightbulb,once the ball has reached the top,then that volume of water that occupied that area at the top,must now fill the void left by the ball at the bottom.So,no mater how you look at it,449 liters of water fell(equivalent to) a distance of 9 meters.
So,as i stated at the start,we had an energy gain in the buoyant PE,as when one full cycle is completed,the center of mass is the same as it was at the start,and not different like you based your math around.
QuoteSo it seems to me,that this 39642.21 Joules of energy has not been accounted for in your math yet.
Let's go back to square one where you offered the thought experiment. You sent the thing into outer space and got it back again at no energy cost (or so you thought). While in outer space at 1/10 gravity you pulled the ball down at an energy cots of 1/10 that above figure (we won't argue about the exact figure since you originally used 9.8 and I used 9.81 for gravity). You then got the above energy out when you let the ball float up. So there appeared to be an energy gain of 9/10 the above figure. I then showed that there was an energy cost in going to outer space, you spent more energy getting there than you could recoup on the way back.
Now,the highlighted in red.
Yes,and that cost was based around the fact that the center of mass was higher when it returned to earth.
At this very point,all energies added up to a net gain of 0.
But this is wrong.
It is wrong because the cycle is not complete.
There is the PE stored in the ball-->this you took into account,and at that very point,all energies in and out were equal.
What you did not do,is account for the PE stored in the raised water it self.
You stated your self-->the center of mass is higher when it returns back to earth.
So,when the ball rises to the top,the mass(water) that is above the ball,must fall back to the original starting position.
Only then is 1 full cycle complete
Only then can you do your math,where the center of mass will be the very same at the start and end of the cycle.
Once you have done this,you will then see that my original calculations where correct,where there is an excess of 39642 joules of energy delivered by the system.
So,i hope we are on the same page now,as the above is the very events that took place.
I can see no where that you have accounted for any energy after your PDF post,that was not already accounted for in your PDF,as your PDF accounted for all energies(except the water falling back to it's original starting position),and came back as all energies in and out added up to being equal.
So i am lost as to how you can now account for that stored energy in the water,that is above that of the starting level,when you have already given a result of E/in and E/out being equal before i mentioned this stored energy that was not taken into account ???
So,can we try the math again,but where 1 full cycle is complete.
This means that you can eliminate that difference in the center of mass between the start and end of the cycle,as the water level will be exactly the same at the end of the cycle,as it is at the start.
Brad
It is clear to me that the confusion comes from my paper not making things clear enough and you have not understood exactly what I have said. Here is version 2 of my paper that makes it clear what the math has shown and at what part of the cycle. As I said the math showed that energy expended in (a) getting to outer space and back again and (b) pulling that pesky ball down while up there in space was exactly accounted for by the energy gained during the final buoyancy rise. That is after the full cycle. The supply and retrieval of mechanical energy is fully accounted for over the full cycle.
The additional account of energy for the falling water is now included where I show it to be exactly the change of potential energy that accompanies that fall. The rising buoy does two things, (a) it supplies energy that you can retrieve and (b) it causes water to fall. That first energy (a) is accounted for, you lost it in going into space and back again. That second energy (b) is not lost, the system has a changed potential energy of exactly that value. So I think I have covered all bases.
Smudge
Quote from: Smudge on 2018.06.23, 18:40:56
It is clear to me that the confusion comes from my paper not making things clear enough and you have not understood exactly what I have said. Here is version 2 of my paper that makes it clear what the math has shown and at what part of the cycle. As I said the math showed that energy expended in (a) getting to outer space and back again and (b) pulling that pesky ball down while up there in space was exactly accounted for by the energy gained during the final buoyancy rise. That is after the full cycle. The supply and retrieval of mechanical energy is fully accounted for over the full cycle.
The additional account of energy for the falling water is now included where I show it to be exactly the change of potential energy that accompanies that fall. The rising buoy does two things, (a) it supplies energy that you can retrieve and (b) it causes water to fall. That first energy (a) is accounted for, you lost it in going into space and back again. That second energy (b) is not lost, the system has a changed potential energy of exactly that value. So I think I have covered all bases.
Smudge
Here is the problem
Tinman has since pointed out that when the hollow sphere rises under buoyancy an equivalent
amount of water falls the same amount, and the energy associated with that rise has not been
accounted for. However it can be shown that when that happens there is a change of potential
energy of the system, and that change of potential energy accounts for the fall of that water.
This change in potential energy is rarely considered because it cannot be put to practical use,
and the energy associated with that rise has not been
accounted for
No,the energy associated with the fall of the 449KGs of water over 9 meters has not been accounted for.
However it can be shown that when that happens there is a change of potential
energy of the system
Yes,the potential energy change associated with that 449KGs of water falling,is returned to the system,and now the center of mass is exactly as it was at the beginning of the cycle.
This change in potential energy is rarely considered because it cannot be put to practical
use
How can this be true when it is a stored PE,and must be taken into account in order to balance out the energy values :o
That volume of water that is above the starting center of mass line,is stored energy--potential energy,and simply dose not just disappear .
Here is how it is,in a simple version.
Lets say that it takes 100,000 joules of energy to raise our DUT to the 1/10 gravity region-->a small amount to simplify the math.
So our DUT leaves the ground,and 100,000 joules of energy was used to get it to the required height.
Energy required to pull the ball down is
m=44.9KGs
g=9.81
h=9 meters
J=3964.221
Total used at half cycle point=103964.221 joules
DUT returns back to ground,ball not yet released
Energy returned =
100,000 --minus
m=449KGs
g=9.81
h= 9 meters
= 39642.21
Total returned is only 60357.79 joules,due to the !now higher! center of mass
At this point in time,i should also state that the center of mass was raised by this amount when the ball was pulled down in the 1/10th gravity region. The energy to raise that mass in the 1/10 gravity region is 3964.221 joules. To perform the same task here on earth would require 39642.21 joules of energy.
!!!! Take note-->The difference being-->35678 joules!!!! C.C
Now ball is released
Energy returned from falling water is
m=449KGs
g=9.81
h= 9 meters
J=39642.21
Our water level is now exactly the same as it was at the start of the cycle,and all energy returned is equal to that used to raise the DUT,minus the energy required to pull the ball to the bottom of the tank in the 1/10th gravity region..
60357.79 joules + 39642.21 joules - 3964.221 joules.
Total returned is 96035.809 joules
We now calculate the energy gained from the ball when it rose
displacement value=449KGs
g=9.81
h=9 meters
Energy gained is 39642.21 joules.
Total energy returned is
96035.809 joules + 39642.21 joules =135678.019 joules.
Gain = 35678.019 joules.
Now,if we look at your original calculations Smudge,where you calculated the complete cycle,where there is no difference in the center of mass between start and finished (because there just isn't when 1 full cycle is complete,as i have done above) ,look what you got--
Pulling the ball down to the bottom of the tank ,when the acceleration of G is .981m/sec*
Ball displacement = 449L
Weight of displaced water=44.9KGs force =440.47N
Height=9 meters
Energy required= 3964.2 joules.
Stored energy when DUT is back on earths surface
Ball displacement=449L
Weight of displaced water=449KGs force = 4404.7N
Height=9 meters
Stored energy=39642 joules.
Energy that came from somewhere=35678 joules.
So,you calculated a gain of 35678 joules for a complete cycle
I calculated a gain of 35678.019 joules for a complete cycle
Whats the chances of that ;)
Brad