Pooling Water From Thin Air

My water post from metallic diaphragm speaker is missing...  But basically, as I remember it,

From the post "A Shield" and "A \(\Psi\) Gun" both dated 27 May 2016,

\(f_{res}=0.061\cfrac{c}{a_{\psi}}\)

were a particle of size \(a_{\psi}\) is resonated at \(f_{res}\).

In the case of hydrogen, its calculated size is \(53\,pm\), with this

\(f_{res}=0.061\cfrac{299792458}{53\times 10^{-12}}=345.0\,PHz\)

where \(PHz=10^{15}\,Hz\)

Since frequencies at integer division of \(f_{res}\), such as \(345.0\,Hz\) (\(\div\,10^{15}\)) or \(34.50\,kHz\) (\(\div\,10^{13}\)) will also resonate hydrogen.  Hydrogen gas spilts,

\(H_2\rightarrow2H\)

and reacts with oxygen

\(4H+O_2\rightarrow2H_2O\)

to give water.

With a speaker at the audio frequency of \(345\,Hz\) will become wet at the diaphragm, and a piezo attached to a glass or metallic surface vibrating at \(34.50\,kHz\) will pool water on its surface.

Now that I have this bit of information, what was it needed for?!

The Familiar Squeeze

I must have this before,

If  the velocity of the particle is \(c\) then,

\(v=c.sin(\theta)\) 

And the distortion to the particle and/or its path, due to the force acting in the direction opposite to \(v\) is  zero when \(\theta=0\) because \(v=0\) according to the above.  The force that distorts the particle is along in the plane of the particle's circular motion.  As is the case when \(\theta=90^o\).

The mid point between these two equivalent extremes is at \(\theta=45^o\).

Is this the same squeeze as before?  Maybe...

Distorted Particle

This is wrong,

max deformation is not at when  \(\theta=90^o\)  because  \(F=0\)  as \(cos(90^o)=0\).  In fact distortion at \(\theta=90^o\) is zero.

When \(\theta\approx 0\),

\(F\) is around its maximum value is the distortion most pronounced.

What happen when \(\theta=0\)?  \(F\) as the result of \(\omega\times v\) is still perpendicular to both \(\omega\) and \(v\), but symmetry dictates that \(F\) rotates in the plane containing the spin, perpendicular to the direction of travel.  Since in all orientations \(F\) acts against the centripetal force, the particle spin is reduced.  The particle spins slower but is not distorted.

Blowing Plastics, Hot

If burning plastics is exothermic, with hot plastic and a intake of relatively cold air (oxygen).  Maybe to balance temperature charges, we need cold plastic and hot intake of air.  This is consistent with a landfill of plastics where the lower layers of the fill, at great depth, become brittle and decompose.  This layer is cold and the air drawn in through ventilation pipes is relatively hot.

So, freeze plastics in the open and subject it a hot stream of air.  Is the decomposition obvious?  Is methane detectable?  Maybe, dry air, without moisture would reduce methane emission?

Goodnight...

A Multicolored Big Ass,

This is possible,

where the particle spins is not in the plane perpendicular to the direction of travel along \(v\).   In this case, the direction of \(F\) given by \(\omega\times v\) is in the plane of the spin, as \(\omega\) is perpendicular to this plane,

\(F\) is perpendicular to \(v\) and does not retard \(v\) but it is in the direction opposite to the centripetal force that keeps the particle in spin, on the side indicated by the right hand rule (\(\omega\times v\)).  The particle move further from the center on this side,

and on the opposite side moves closer to the center, assuming that \(v\) is constant.  The particle bulge on the side to which it turned.  The opposite side, contracted.

When \(\theta=90^o\), \(F=0\) because \(cos(\theta)=0\) and if \(theta\) is indicative of the deformation on the particle,

in this direction of spin, the particle has maximum distortion at \(\theta=90^o\), \(F=0\).

This might be the interpretation for \(\theta\) in the post "A Pump!" and "Polarization And Invisibility" both dated 25 Jul 2015, etc.

Energy is stored in the deformation and this energy is released when the deformation on the particle relents.

This however is a non-quantized release of energy...unless \(\theta\) is restricted in some way to discrete values.

In the case of light released over a range of energy values; light photons within a ranged bandwidth; multicolored.

What Is \(\theta\)?

The following formula is for the force, \(F\) on a sphere moving through a fluid of density, \(\rho\) at a velocity of \(v\) and a spin about a axis, \(\omega\), 

\(F=\pi^2r^3\rho\omega\times v=\pi^2r^3\rho\omega v cos(\theta)\)

from the book "The Math Behind..." by Colin Beveridge, Cassell,  Octopus Publishing Group Ltd.

For \(\theta=0\), the spiral flattens out and the particle is just in circular motion.  For a spin axis through the center of the particle, \(r=a_{\psi}\); \( r\) is the size of the particle.

Since,  \(cos(\theta=0)=1\),  \(F=F_{max}=\pi^2r^3\rho\omega v\)

If \(v=c\) and \(v=r\omega=c\)

\(F_{max}=\pi^2r^3.\rho.\cfrac{c^2}{r}=\pi^2r^2.c^2.\rho\)

And if this force is the drag force, which is proportional to velocity squared,

\(F_{drag}=F_{max}=k.c^2=\pi^2r^2.c^2.\rho\)

\(k=\pi^2r^2.\rho\)

And in general, \(\theta\ne0\), the spin axis is not through the center of the particle, and the spiral stretches out,

\(k=\pi^2r^2.\rho.cos(\theta)\)

where \(\pi r^2.\rho\) is the mass per unit length in direction perpendicular to the area \(\pi r^2\).  This value multiplied by \(v\), \(\pi r^2.\rho.|v|\) is the mass of the fluid displaced by the particle of radius \(r\) with velocity \(v\) in one unit second.
 
\(k=\pi.cos(\theta).m_{fluid,\,v}\)

\(m_{fluid,\,v}=\pi r^2.\rho.|v|\)

What is \(\theta\)?

Leafy Greens, Carrots And Big Potatoes

This was missing from the blog,

Chlorophyll is obtained from vegetable juice (eg.  spinach) and freeze dried to a powder and mixed with a clear resins.  It is possible to mix in the green juice without drying first.  Copper charge collectors are inserted into the plastic and it is used as a single ended device; a current source.  A ground return can be connected to the plastic.

Later, carrot juice was used instead.  Practically, carrots were dried and grounded to produce a dark orange powder that was used in place of dried leafy greens.  It was expected that heat particles are drained from the resulting device.

 But instead,

a voltage is applied to a suitable collector separation of \(d\) that produces a discharge and the device is used as a carrot light to grow potatoes.  Big potatoes.

Maybe...