It is always dark, Light is an illusion and not a thing!

[theorist-constant12345]

From the video I provided, of the laser experiment, the visual experimental observation within a certainty shows the defining of dark, and energy. We can clearly observe both states at the same time, we are not blind in the dark. We simply have no energy to see.
You can clearly see that the energized area is visual and you can clearly observe, that the laser beam vanishes into the dark once un-propagated and in its weaker state of zero net charge, and maximum velocity.

 

[billvon]

From the video I provided, of the laser experiment, the visual experimental observation within a certainty shows the defining of dark, and energy. We can clearly observe both states at the same time

"Dark" and "energy" are not two states. "Dark" is the absence of visible light.

we are not blind in the dark.

We are blind if we cannot see any light. This is true whether or not that lack of seeing light is caused by darkness, or our eyes being covered, or our eyes malfunctioning.

We simply have no energy to see.

That is quite literally correct. We do not have enough energy, in the form of photons, entering our eyes to be able to see.

You can clearly see that the energized area is visual and you can clearly observe, that the laser beam vanishes into the dark once un-propagated and in its weaker state of zero net charge, and maximum velocity.

The reason you cannot see the laser is that the air is cleaner than the water. If the water was perfectly clean and the air was dusty, you would see the beam in the air but not the water. Thus your hypothesis fails.

 

[origin]

And the reaction of EM radiation to permittivity and permeability properties is?

OK so you do not know what permittivity and permeability is.

permittivity and permeability is the function and work ,
The resulting function of work, being the action of propagation, that is the effect of reduced speed of flow.

You do not know what work is.

Where has force over distance, TFR, explains a prism, where as refraction does not explain a square transparent block, with refractive properties the same has a prism. having no ability to change the function of ''white light'', and only by angular displacement of force, is the work possible, to change the function.

You do not know how a prism works.

Permittivity and permeability have no affect on ''white light'' through an equally same refractive index block, in the dispersion, unless angular displaced.

You apparently do not know english either.

 

[origin]

From the video I provided, of the laser experiment, the visual experimental observation within a certainty shows the defining of dark, and energy. We can clearly observe both states at the same time, we are not blind in the dark. We simply have no energy to see.
You can clearly see that the energized area is visual and you can clearly observe, that the laser beam vanishes into the dark once un-propagated and in its weaker state of zero net charge, and maximum velocity.

Holy crap, you are so confused. We see the laser beam because some of the beam is scattered in water so some photons hit our eyes from the laser.

It is a lucky thing you are not capable of understanding how incredibly ignorant you sound - you would be so embarrased.

 

[theorist-constant12345]

Soon I will find the correct wording.
Dark is an incorporeal impenetrable gloom to sight that is impalpable , that is filled by impalpable Em radiation, that our eyes use to couple to our brains, any matter within the line of sight, by change, in the constant of impalpable spacial volume , making the impenetrable gloom, penetrable for sight, by a temporal energy transition.

 

[Beer w/Straw]

Nice sermon!

 

[origin]

Soon I will find the correct wording.

Well this sure ain't it

Dark is an incorporeal impenetrable gloom to sight that is impalpable , that is filled by impalpable Em radiation, that our eyes use to couple to our brains, any matter within the line of sight, by change, in the constant of impalpable spacial volume , making the impenetrable gloom, penetrable for sight, by a temporal energy transition.

It seems that you are trying to say something about a scenario like this.

If there were no other stars but the sun and you were in space near the sun but not looking at it you would see only blackness even though there were untold billions of photons right in front of your face.

Is it something like this you are trying to say?

 

[theorist-constant12345]

Well this sure ain't it

It seems that you are trying to say something about a scenario like this.

If there were no other stars but the sun and you were in space near the sun but not looking at it you would see only blackness even though there were untold billions of photons right in front of your face.

Is it something like this you are trying to say?

Sort of yes, except with the exception, of that we do not have enough dimension to stop the Em Radiation flooding around us.

 

[origin]

Sort of yes, except with the exception, of that we do not have enough dimension to stop the Em Radiation flooding around us.

Unfortunately that makes absolutely no sense at all.

 

[theorist-constant12345]

Unfortunately that makes absolutely no sense at all.

In simple terms, we are not wide enough to completely block the light and some can bend around us.

I will just answer this thread with a simple question,

If light has to reflect off matter into our eyes, so that we can see that matter,
Why is it that in experiment, a laser has no reflective ray trace in a smoke filled room?. Experiment shows no reflective beam. Experiment shows us that we are seeing a change in a constant, an energy constant sent through a no, or low energy constant, that is greater than the constant.

 

[origin]

In simple terms, we are not wide enough to completely block the light and some can bend around us.

That also makes no sense.

I will just answer this thread with a simple question,

If light has to reflect off matter into our eyes, so that we can see that matter,
Why is it that in experiment, a laser has no reflective ray trace in a smoke filled room?.

It does! When the laser hits smoke some of the photons are reflected off of the smoke particles and that is what you see.

Experiment shows no reflective beam.

It most certainly does. We use lasers at my work and you clearly see the beams when they pass through smoke or dust.

Experiment shows us that we are seeing a change in a constant, an energy constant sent through a no, or low energy constant, that is greater than the constant.

Uneducated gibberish.

 

[theorist-constant12345]

That also makes no sense.


It does! When the laser hits smoke some of the photons are reflected off of the smoke particles and that is what you see.


It most certainly does. We use lasers at my work and you clearly see the beams when they pass through smoke or dust.

Uneducated gibberish.

I own a laser, I can clearly see in a smoke filled room there is only an incident ray, and there is no reflective ray.
If you use a mirror, or a total reflective surface, then yes the beam will have a reflective , redirected beam. I observe no evidence of a reflective beam through smoke off a wall for example from a laser.

 

[theorist-constant12345]

I will add this description from Wiki.

''Night vision is the ability to see in low light conditions. Whether by biological or technological means, night vision is made possible by a combination of two approaches: sufficient spectral range, and sufficient intensity range. Humans have poor night vision compared to many animals, in part because the human eye lacks a tapetum lucidum''.

Consider the effect of sufficient spectral range, and sufficient intensity, being created in the day time, to make up for the lacking in a tapetum lucidum.

At night there is an insufficient magnitude of both intensity and frequency, in the day time, and approaching daytime, or approaching nigh time, twilight is the increasing or decreasing in the magnitudes, comparing to the glooming translucently.

I believe I have offered several observable axioms. Bio logical comparison, lasers etc.


Consider , Humans have poor night vision compared to many animals, so the logical axiom is that we need a greater sufficient amount of magnitudes to see in the dark, having effectively night vision by temporal means.

 

[theorist-constant12345]

Bump - Have I won this argument?

 

[billvon]

Bump - Have I won this argument?

Nope. People just gave up on you. You don't even understand reflection.

 

[theorist-constant12345]

Nope. People just gave up on you. You don't even understand reflection.

I do not understand reflection , hmmm, let me see,

a reflective surface , a none reflective surface, Photons have to reflect of matter , Photons have to be remotely sensed by our eyes, reflecting of the matter into our eyes.
Really, very strange how my laser has no reflective ray heading towards my eyes, any Photons from the incident laser beam, that may or may not disperse into the atmosphere, will have little to no intensity for our sight use.
We are not blind in the dark, there is just no energy to see by or to detect. Dark is a constant , and by interactions of energy in the constant is how we see.

 

[billvon]

I do not understand reflection

No, you do not. From an earlier post:

"I own a laser, I can clearly see in a smoke filled room there is only an incident ray, and there is no reflective ray."

You do not understand that you are seeing light reflected from the smoke particles in the beam.

 

[theorist-constant12345]

No, you do not. From an earlier post:

"I own a laser, I can clearly see in a smoke filled room there is only an incident ray, and there is no reflective ray."

You do not understand that you are seeing light reflected from the smoke particles in the beam.

You are incorrect, if the incident ray through the smoke was fragmenting and particles were dispersing through the smoke from the linearity of the beam, then you would see the fragmentation in the smoke, which you do not, you do not also understand that if there were a small amount of fragmentation it would be to weak to see.
There is no evidence by experimental observation that suggests a beam fragments through the smoke into your eyes, you see the incident ray has a singular linearity in the dark, so stop trying to lie, and defend a complete lie that has no experimental observation to concur the lie.
Light reflects off reflective surfaces, and does not reflect off none reflective surfaces , colour is propagation, or man made frequency manipulation of the constant in a constant.
Your spectrum is nothing more than energy difference.

 

[paddoboy]

Light reflects off reflective surfaces, and does not reflect off none reflective surfaces , colour is propagation, or man made frequency manipulation of the constant in a constant.
Your spectrum is nothing more than energy difference.

You have a lot to learn... and you have a huge ego to manage.
Just as there is no such thing as a perfect reflector, so to is there no such thing as a perfect absorber other than BH's.
Colour of any object, simply in the first instance, depends on the nature of the EMS that falls upon it, and secondly, the reflective property of that object. Certainly not the posturing bullshit you have suggested.
Still, I imagine that is why this is in pseudoscience.

 

[billvon]

You are incorrect, if the incident ray through the smoke was fragmenting and particles were dispersing through the smoke from the linearity of the beam, then you would see the fragmentation in the smoke, which you do not

You do indeed see reflection in the smoke. You do indeed see attenuation due to the reflection of the beam in the smoke.

There is no evidence by experimental observation that suggests a beam fragments through the smoke into your eyes, you see the incident ray has a singular linearity in the dark

A laser shining directly into your eye is an incident ray from the laser. If you see the beam from the side, then it is due to reflected light. That's the definition.

Light reflects off reflective surfaces, and does not reflect off none reflective surfaces

Correct. Smoke particles have reflective surfaces. That's why you can see smoke, both in normal light and in lasers.

colour is propagation

No, color is color and propagation is propagation. They are different.

Your spectrum is nothing more than energy difference.

Close to correct! Different photons at different places on the spectrum (i.e. different colors, or different frequencies) do indeed have different energies. A spectrum is made up of photons of different energies, based on their wavelength.