Buoyancy - a comparison between Gravity and the Standard Vibration Model

This is the model of buoyancy using Newtonian Mechanics. It is easy to see how this simple model captures the community communal approval.  

Tomorrow I will show how the Standard Vibration Model does a better job of describing what is observed in atmospheric interactions without using gravity or mass. 

Anonymous has left a new comment on your post "The N-Body Problem":

Now, buoyancy.

Buoyancy does not fly in the face of gravity. Gravity, when acting on an object in isolation, holds. The theory of gravity says nothing more than this... that two masses, when considered in isolation, attract one another with some force. There might be other interactions adding additional forces, but somewhere in the big pile of forces acing on a mass is a contribution from gravity, and it conforms to a simple equation.

Buoyancy is a phenomenon that occurs in a large, complex system... A single solution to the simple 2-mass gravity law is not suffucicient in this situation, and nobody backing gravitational theory is saying that it is. But it does contribute in an important way. Allow me to attempt to explain...

If we consider the balloon and earth in complete isolation, it would accelerate downward at G*m-sub-earth/r-sub-earth-balloon^2, as expected.

But this isn't a two body system. There is a third player... air. And in this system, the three bodies are acting on each other. The gravitational force between air and the balloon can be neglected, since the balloon is so small in this instance and the air is so widely distributed (I hope we don't conflict on that assumption... but if we do, I'll happily defend it).

So let's start with the air. Being acted on by gravity, it is all trying to be pulled downward. This is creating pressure at all places within the air, at increasing values at lower altitudes, since there is more weight on top of the air at lower heights. This pressure is acting against gravity. It is the resistance of the air to being compressed into a singularity, primarily driven by kinetic forces of the matter being compressed. Everywhere there is a force pushing down (gravity) and everywhere there is a force pushing up resisting it. Thus we acheive equilibrium and a stable atmosphere.

Enter the balloon. Imagine, if you will, we identify the spherical surface, centered at the earth's origin, who's radius extends just to the bottom of the balloon. Thus, we can talk about the air beneath that altitude (which I'll call lower air), and the air above that altitude (higher air). So the lower air is all the air around the world beneath the altitude of the balloon, and higher air is the rest (which is where the balloon is). Everywhere over earth's surface, the lower air is pushing up at a certain force... the force required to hold the higher air bay... exactly the same force, but opposite in direction, as gravity is generating from the higher air. Without the balloon, it's a uniform equillibrium. But now, in one cylindrical column of higher air, there's a balloon... a balloon that is lighter than the surrounding air.

So now there's a particular place where the higher air is lighter. And so the lower air is no longer in equilibrium at that place. It's pushing up with the same force as all the rest of the lower air, but the air above it in that particular spot is pushing down with less force due to the lighter mass balloon. And so the net force, the combination of the upward pressure and the gravitationally-induced downward weight, is non-zero, and upward acceleration results.

This continues until a new equilibrium between air pressure and gravity is reached, which will occur when the balloon rises to a state of equi-density with it's surrounding.

And all of this was explained nicely with gravity as a key contributor.


End of Anonymous' description.  
------------------
The Standard Vibration Model


It is assumed that the reader has already read the papers 
The Structure of Baryons
Variables Involved in Baryonic Motion


NASA Heavy lift balloons
The interesting thing about these heavy lift balloons is that they are capable of carrying a 1000-kg instrument to approximately 33 km giving little or no day/night altitude variation and ultimately 100-day flights[1].

At launch these balloons are filled with helium to 5% of the volume of the balloon. Any more than that and the balloon would rupture during the ascent. As the balloon rises in the atmosphere the helium expands to fill 100% of the volume of the balloon. The balloons are corrugated allowing for this expansion.

This would require the expansion of the helium atoms during ascent. This would mean the length of the gluon between quarks is a shorter length before launch and a longer length at altitude. The distance between atoms remains the same allowing for a smooth expansion.

I am going to add a drawing here to explain the distance continuity between helium atoms.

Literally the helium expands. There is almost no leakage since the helium atoms are now larger than the holes in the balloon. They have to pop the balloon to return the payload back to Earth.

------------

[1] NASA Heavy Lift Balloon Program

http://sites.wff.nasa.gov/code820/balloonroadmapreport.pdf

The N-Body Problem

Anonymous said...

Mass is supposed to be dimensionless. It has nothing to do with volume. The job of density is to relate mass and volume to one another. It is not the responsibility of mass to do so on it's own.

F=ma is structured as it is because, experimentally, the acceleration of an object is uniquely determined by only the force being applied to it and it's mass, independent of volume.

If I have two spheres of differing density but identical mass, and applied an identical force, F=ma tells us that they'll both experience identical accelerations, and this is exactly what we see in experiment.

Your equation factors in volume. So F=da tells us that if one instead has twice the density of the other, it will experience half the acceleration. So according to F=da, if the two spheres weighed exactly the same, but one was twice as dense, meaning half the size, it would accelerate at half the rate. Or put another way, F=da means I can squish an object, without changing it's weight, and it will accelerate more slowly under equal load. Is this the behavior you're predicting? Have you been able to verify it?
------

So a balloon filled with 1kg of helium and a balloon filled with 1kg of lead will have experience identical accelerations. That does not occur. 

----------

Or do you have another definition of density other than d = m / v? For instance, you say: "Density is the vibration intensity within a volume in relation to the density of the surrounding medium." It's relative? So the density of an object changes when it's surrounding density changes?

----------

I do have an idea of what the complex density equation would look like but I lack the resources needed to provide the necessary data to substantiate the equation. I have created a general equation of baryonic density in the paper 'Variables involved in Baryonic Density'
----------
Anonymous said...
Regarding constants, there is no problem there. G or c are perfectly valid constructs, and don't point to any sort of illegitimacy of the theory.

If anything, they're simply the product of an unfortunate choice of units. We arbitrarily chose grams and seconds and miles and furlongs and hogsheads and bushels and meters, and those arbitrary decisions are reflected in the constants.

If our units aligned better with the universe, the silly constants would drop out of the equations.

For instance, everyone's favorite e = m c^2, under a particular choice of units for mass, would simply become e = m, which is really much more to the point, isn't it? Energy is mass is exactly what the equation says. c^2 was just a constant, and has no effect on the behavior of the equation.
-----------

Constants are like perfect solids. They don't exist in the universe. It is not possible to have a natural sphere. I explain this in many postings throughout the blog. 

--------------------------------

And the law of gravity, under a particular choice of units for mass, would become F = m1 m2 / r^2, which states exactly the idea intended, that each mass increases the effect linearly, and the distance decreases it quadratically.

And you might argue that I'm just hiding the constant in the unit, and yes! I am! But there's nothing wrong with that. The unit itself was arbitrary to begin with. The constant just scales the equation to fit the units of the universe, but the structure of the equation--it's behavior over the domain--stays the same.

If the equation is fundamentally wrong, there could be no constant that would result in correct behavior. The constant can't make an invalid equation valid, it can only scale the resultant units.

Anonymous said...
Regarding multi-body... Gravity works just fine for N bodies. It's simply:

F-sub-i = G * m-sub-i * Sigma-sub-j [ ( m-sub-j / r-sub-ij ^ 2 ) * r-hat-sub-ij ]

where i and j range over 1..N, and r-hat-sub-ij is the normal vector pointing from i to j (I sincerely apologize for the phonetic syntax... not sure if your blog supports math markup). Or in other words, the force on each body is the superposition of the law of gravitation applied to each other body. There is no trickery here, and interestingly, to solve for a body, all I have to do is sum the mass/distance-scaled vectors to each body, then apply my mass.

--------------
This is the best solution to the N-Body Problem I have seen. 

The hat is an iterative process. It is a mathematical way of describing the mean after all the summations are completed iteratively. I agree that there is no trickery here. You would have one solution to a many iteration problem. This will help me in describing multiple bosons interacting in one ruleset. Thank you for that. 


But that is not the real problem with gravity. Gravity does not allow for the interactions of magnetism, electricity, temperature or density. Each of these forces does affect the motion of an object. Each of these forces are (in this model) associated with a specific boson. This model does not need, and works well without; gravity or the higgs boson. In my opinion the higgs is an attempt to protect gravity by validating mass. Gravity describes a force that has not been found. Gravity is also 1 dimensional, it cannot be dimensionless. It is a scalar, it has a number. 


We have found the planets are ordered by density.


Thanks for reading my work. 
Aaron

Equality is highly overrated part 2

Anonymous said...

Calculus seems analog, not digital. It is the mathematics of limits. The entire foundation of calculus is the study of what happens when we take something discrete and push the intervals to the infinitesimal.

DNA seems digital, not analog. There are a finite number of bases arranged in a finite number of pairs. In what way is that analog?

And as to equality, you offered some examples of how everyday things aren't equal, but can you offer any concrete examples of where physics goes wrong in assuming equality when it shouldn't?

----------------

I will again start from the bottom and work to the top. Concrete examples are easy to find. 

The most simple example of everyday things are not equal is Einstein's work. 

Energy (a field) cannot equal a scalar * a vector squared. When you square a vector your result will be a vector. Multiplying a vector by a scalar again leaves a vector. Thus E does not equal mc^2, ever.

--------------------------  

Newton's Gravity cannot produce results for n bodies. The reason this occurs is because mass is a one dimensional object. One dimensional objects cannot represent any 3 dimensional object. There is just not enough information in mass to describe how gravity and magnetism interact. 

-------------------------

Fourier's Law of heat conduction.

Fourier's Law states that heat flows always travel from hot to cold. If you take a steel bar and bend it into a square u shape, then place one end into liquid nitrogen and the torch the other end to the exact temperature change in the positive direction, you will see that cold flows to hot and hot flows to cold. You will see the frost on the bar exceed the midpoint of the bar. This problem has more to do with capacitance and conductivity of the baryons.

------------------------

Lets look at that DNA problem. The data in DNA is digital. The motion of that data from molecule to molecule is analog. Analog to Digital and Digital to Analog conversion is required for motion of data. Until that last statement is fundamentally understood it will be impossible to describe the structure, motion and interactions of Z bosons. This is even true for computers. For a processor to send data to RAM it needs to send the data through a series of conversions to get the data to the memory and reverse the conversion to return the data back to the processor.

I am not prepared to describe Z boson analog motion any more than I already have in this blog. I am tired of being paid to be a disabled epileptic. $13500 per year is not enough to live on. I cannot afford to seek the treatment I need. I can barely run one computer let alone my whole system. I would like to produce experiments that prove what I am saying is true/false, but I lack the resources to build these simple systems or to be of benefit to society. 

---------------------------

If we understood fractals and non-linear equations we would see that calculus does not describe interactions. Calculus requires a universal time clock, this is not found in the universe. dt binds the equation to a standard time. This is not to say Calculus is not useful, it is just not useful in describing interactions of analog or non-related data. I really should have never brought up the issue with calculus, this discussion will only bring up disdain in readers.

Thanks for reading my work.

Aaron Guerami