Solar Cell Skin

If X ray generated by the Sun from Hydrogen, \(H\) and Helium, \(He\) are captured on solar cells and plant life,

	Z	density (g/m^3)	density/Z (g/m^3)	v boom (m/s)	x ray penetration (m)
Hydrogen, H	1	8.99E-005	8.9900E-005	3.0884E-01	5.8195E-05
Helium, He	2	1.79E-004	8.9250E-005	3.0661E-01	5.7775E-05

where X ray peneration is calculated by taking reference with copper, \(Cu\),

\(S_{p\,H_2}=\cfrac{v_{boom\,H_2}}{v_{boom\,Cu}}*S_{p\,Cu}\)

\(S_{p\,Cu}=0.2\,m\)

\(S_{p\,H}=\cfrac{3.0884E-01}{1061.42}*0.2=5.8195E-05\,m\)

The X ray torus photons collapse on the surface but penetrate only by \(50\,\text{to}\,60\,\mu m\) into the surface; the collection channel that collects the particles cannot be deeper than these depths, with the orientation of the collecting device considered.

If the particles collected are from X rays...

诗赋多词
言当行达
霸溃纹理
钢腐不持
滴水破溅
漫洪没顶

《淹败》

Note: Tidal floods and a broken dam are two different things.

Gravity Sword

A sword that resonate at \(7.489\,Hz\),

given that the speed of sound in steel is \(5000\,ms^{-1}\).

Unfortunately...

\(f.\lambda=v\)

\(l_s=\cfrac{\lambda}{2}=\cfrac{v}{2f}\)

\(l_s=\cfrac{5000}{2*7.489}\)

\(l_s=333.82\,m\)

over three hundred and thirty meters!

If however, we fold the molten smelt repeatedly during the forging process,  such that many half wavelengths spread over the length of the sword we set at,

\(l_s=0.72\,m\)

\(l_s=n.\cfrac{\lambda}{2}=n.\cfrac{v_n}{2f}\)

where \(n\) is the number of folds on the sword, and \(v_n\) is the speed of sound along the sword,

\(n=l_s*\cfrac{2f}{v_n}\)

we assume further that the speed reduced by a factor of \(\cfrac{1}{n^2}\)

\(v_n=\cfrac{v}{n^2}\)

\(n=\cfrac{v_n}{2fl_s}\)

\(n=\cfrac{333.82}{0.72}\)

In practice, the number of folds possible is binary,

\(n_b=\left\lceil\cfrac{log(n)}{log(2)}\right\rceil=\left\lceil 8.856 \right\rceil=9\)

ie,

\(n_p=2^9=512\)

This would be,

\(p=\cfrac{n_p}{n}=\cfrac{512}{333.82}*{0.72}=1.104\)

times the intended \(l_s=0.72\).  So, with nine folds of the smelt, we turn it such that the folds run perpendicularly along the sword and elongate the smelt to a length of,

\(l_b=p*0.72=0.795\)

and file this blade down to \(0.72\).  This way the sword at \(0.72\,m\) will have the intended \(n\) number of folds along its length.

The speed of sound along the blade is an approximated at \(5000\,ms^{-1}\) and we have assumed that the folds on the blade slow the speed of sound along it by a factor of \(\cfrac{1}{n^2}\).

If both these assumptions are true, this sword should be light as a feather.

Tale Spin

\(\nabla.A\ne\nabla.(-A)\),

refers to a point at the periphery of the spinning field, \(P\) and not at its center \(C\).  The flow of \(E\) at the center is reversed when a negative sign is applied.

Strictly speaking,

\(\nabla.A\)

does not represent the effect of spinning \(E\) at point \(P\).  Neither does,

\(-\cfrac{\partial E}{\partial t}\)

nor the more familiar,

\(-\cfrac{\partial B}{\partial t}\)

represents the spinning end of a field \(B\) centered at \(C\).  It is the negative rate of change with time of \(B\) at a fixed arbitrary point \(P\).

Lunch!

Sphere, Cylinder And Egg

Remember this,

if we line them up and send them down a line we have,

\(\psi\) wraps around into a sphere manifest itself as circular waves with centers along the line of travel, forming a cylinder.

At this point, we do not know the direction of this wave (right hand rule or left hand) but we known form symmetry, by rotating the line in any plane containing it by \(180^o\) that this direction reverses.  So, the sense of the wave (clockwise vs anticlockwise) reversed with the follow of the particles along the straight path.

But given one direction of flow, the clockwise or anticlockwise sense remains arbitrary.

Only upon the application of Lenz's Law based on conservation of energy, that the particle is not self propelling like a photon, that the right hand rule for the flow of positive particles is true.

And we are still open as to the nature of this field, cylindrical along the line of travel.

Egg, refers to the deformed shape of the particles as they are propelled in a field.

Have a nice day.

Data Crunch And Rotor Speed

If, the sonic boom at the region on the blade closer to where they connect to the rotor mast of a helicopter, is responsible for its lift,

then, at higher altitudes where the air is thin and water has a lower density value (humility) and a reduced boom velocity, a decrease in rotor speed may just maintain lifts.

Intuitively, the normal reaction to a lost in lift is to increase rotor speed.

However, it is not altitude per se that requires a reduced rotor speed, in cold regions a reduction in water density may also require a similar reduction in rotor speed to take flight.

Good night...

Dust And Water Droplets

Fluorescence tubes are rare, but,

With charges flowing towards the left inside the discharge tube, maybe a flow of water around the tube and a ring of light dust particles around the tube in the direction of the \(g\) field can be seen.  A tiny drop of ink might indicate the flow direction of water more clearly.

Water smear and dust tend to stay on the tube when it is on, but these phenomena are attributed to static electricity and not the presence of a \(g\) field proposed here.

Maybe...

Field Party

The fields around a discharge tube may show the way,

Do these fields exist?

Neutral Particles, Neutral Temperature And KaBoom

The flow into a point is the reverse of the flow out of the point by introducing a negative sign only if there is high symmetry like in the case at the center of a sphere where the radius indicates equal vectors.  When the vectors are not equal reversing the flow by introducing a negative sign generates a new flow into the point, not the negative of the flow initially.

If charge is a continuum, when does it end?  How big can a positive charge gets given the right conditions?

What happens when a neutral charge is spinning?

If the manifestation of any one field depended on the relative phase between two, as charge sweeps from negative to zero, to positive, then when the charge is zero... well everything goes zero as if nothing is in spin.

Still there would be three types of neutral particles; electric, gravity and temperature.

Neutral temperature is not at zero temperature as defined on the Kelvin scale.  Neutral particles are not at \(a_{\psi}=0\) but are big at \(a_{\psi\,ne}\).

What if there is a field as a neutral particle spins?  This field is the field at zero phase between the two fields that together manifest the observed field.  It is one field in isolation.  And each type of neutral particle (electric, gravity and temperature) presents one field.

It is possible, that each field instead has three orthogonal components, two of which cancels when two fields are orthogonal, the remaining component presents itself when the particle is in spin.  As charge changes from positive to negative, a phase shift from \(90^o\) to \(-90^o\) occurs (relatively) between the two fields and a different component remains as another set of two components cancel.  In this paradoxical way a neutral particle in spin generates the highest possible fields compared to charged particles.

A neutral particle suggests that all the wave energy is in oscillations.  This is still a wave-particle?  It no longer exert a field in space stationary but it has size \(a_{\psi\,ne}\) and provides the strongest field in spin.  It is the highest negative charge possible.

As long as spins generates a field,

the layered atom model still holds.

Maybe all spinning particles in a nucleus are neutral that, generates the strongest field in spin possible.  And isotopes are of off neutral particles in spin that have a charge, generate a weaker fields in spin and can be bigger (positive) or smaller (negative) than a nucleus of neutral particles.

This makes \(a_{\psi\,c}\) very dangerous as \(a_{\psi\,ne}+a_{\psi\,c}\) makes a neutral charge positive.  A change in charge switches the type of field generated in spin and rearranges the layers inside the nucleus and may cause the disintegration of the atom.

KaBoom!

No Sign Assignment Needed

A hiatus because the model for particles,

and all its variations is not consistent with,

because,

\(t_g\ne -t_c\)

\(t_c\ne -t_T\)

\(t_T\ne -t_g\)

and etc.

Luckily, charge continuum, that charge varies continuously from negative to zero, to positive values  does not implies that the various fields (electric, gravity and temperature) are a entangled mess of opposing fields.  It implies that,

\(\nabla\cdot E=-\nabla\cdot g\)

\(\nabla\cdot T=-\nabla\cdot E\)

and,

\(\nabla\cdot g=-\nabla\cdot T\)

where \(E\), \(g\) and \(T\) are electric, gravity and temperature fields.

If the fields are a mess of opposing fields then, one possible conclusion is that there is only one electric field that produces a magnetic field on spin and gravity is just electric particles interacting at close proximity as waves that attract each other.  Similarly, temperature fields are waves interacting at close distance, and interactions at greater distance manifest as the magnetic fields.

Distance apart being relative to the size of the particle in consideration, on Earth's surface, we are interacting with Earth as waves.

The trouble is, given,

\(\nabla\cdot E=-\nabla\cdot g\)

and,

\(\nabla\cdot T=-\nabla\cdot E\)

we have,

\(\nabla\cdot T=-(-\nabla\cdot g)\)

\(\nabla\cdot T=\nabla\cdot g\)

which contradicts the last statement,

\(\nabla\cdot g=-\nabla\cdot T\)

Taken together, \(\nabla\cdot T=0\) and since all assignments are arbitrary, so is \(\nabla\cdot E=0\) and \(\nabla\cdot g=0\).

Unless,

\(-\nabla\cdot T\ne \nabla\cdot -T \)

and that,

\(\nabla\cdot E=\nabla\cdot (-g)\)

similarly,

\(\nabla\cdot T=\nabla\cdot (-E)\)

and,

\(\nabla\cdot g=\nabla\cdot (-T)\)

that the back (negative) end of a vector spinning is not the negative of the front (positive) end of the same vector spinning.

With this in mind,

where a negative charge particle spins (blue lines) to generate a temperature field, but a positive charge particle spins (red lines) to generate a gravity field.  A reversed spin generates the same field in the reversed direction.

Do we collapse every atom in the universe this way?