[jerrygg38 (OP)]

They were called British Units in my physics books. Oh well.

[Bored chemist]

They were called British Units in my physics books. Oh well.

It may have escaped your notice, but the British Empire used them. Some bits- notably the USA still do.

[jerrygg38 (OP)]

It may have escaped your notice, but the British Empire used them. Some bits- notably the USA still do.
GG: It was so long ago I hardly remember anything from the past. Luckily I can still find my way to the bathroom.

[Kryptid]

The left side is easy enough to convert. The answer is in newtons and the imperial unit of force is the pound-force. The conversion factor is 1 lb_f = 4.448222 N, so 3.4136859 x 10^-44 N equals 7.67427 x 10^-45 lb_f.

Here’s the conversion for the right side:

μ₀ = 2.8250387 x 10^-7 (slug•ft²/s²•A²)/ft
Q = 1.602176634 x 10^-19 C (the same as in metric)
c = 983,571,056.43 feet per second
Vb* = 3.9916996028 x 10^-28 feet per second
r = 2.4278216 x 10^-10

And here is the math for the right side:

=((2)(μ₀)((Qc)/137.036))(4)(pi)(Q)(Vb*)(cos 30^o))/r²
=((2)(2.8250387 x 10^-7)(((1.602176634 x 10^-19)(983,571,056.43))/137.036))(4)(1.602176634 x 10^-19)(3.9916996028 x 10^-28)(cos 30^o))/((2.4278216 x 10^-10)²)
=((5.6500774 x 10^-7)(((1.602176634 x 10^-19)(983,571,056.43))/137.036))(4)(1.602176634 x 10^-19)(3.9916996028 x 10^-28)(cos 30^o))/((2.4278216 x 10^-10)²)
=((5.6500774 x 10^-7)((1.575854564 x 10^-10)/137.036))(4)(1.602176634 x 10^-19)(3.9916996028 x 10^-28)(cos 30^o))/((2.4278216 x 10^-10)²)
=((5.6500774 x 10^-7)(1.14995663 x 10^-12)(4)(1.602176634 x 10^-19)(3.9916996028 x 10^-28)(cos 30^o))/((2.4278216 x 10^-10)²)
=((6.497343966 x 10^-19)(4)(1.602176634 x 10^-19)(3.9916996028 x 10^-28)(cos 30^o))/((2.4278216 x 10^-10)²)
=((2.598937586 x 10^-18)(1.602176634 x 10^-19)(3.9916996028 x 10^-28)(cos 30^o))/((2.4278216 x 10^-10)²)
=((4.16395707 x 10^-37)(3.9916996028 x 10^-28)(cos 30^o))/((2.4278216 x 10^-10)²)
=((1.66212658 x 10^-64)(cos 30^o))/((2.4278216 x 10^-10)²)
=((1.66212658 x 10^-64)(0.8660254))/((2.4278216 x 10^-10)²)
=(1.4394438362 x 10^-64)/((2.4278216 x 10^-10)²)
=(1.4394438362 x 10^-64)/(5.8943177214 x 10^-20)
=2.4420873 x 10^-45

I'm surprised that they are on the same order of magnitude, but the first value is still more than three times larger than this one (assuming I didn't make any errors). So still no match.

Oh, by the way, the metric unit for angular measurement is the radian, not the degree. In this particular case, it doesn't matter because the cosine of either is the same number. Under other circumstances, it would indeed be important to get it right.

[jerrygg38 (OP)]

To Kryptid
The top equation is 7.67427E-45 according to your calculations
The bottom equation is 2.44207873E-45 according to your calculations
The ratio is 3.142515   now pi = 3.14159
The pi shows in the equation but is missing in your calculation
The second number should be  7.67204E-45
This is an error of 0.029 percent from the first number.  I am glad that you were able to supply all the numbers. Unfortunately I threw out my 50 year old college physics book a little over a year ago. It had the slugs and feet conversion factors. I went to Barnes and Noble looking for a physics book but  they did not have a good book like I had long ago. The internet confuses me. So you have provided the necessary conversions. I am also having some trouble doing all the multiplications you did. I used to proof read my Professor/ Sperry Boss's books. I always found the errors. But I am not as good as I used to be.

[Kryptid]

To Kryptid
The top equation is 7.67427E-45 according to your calculations
The bottom equation is 2.44207873E-45 according to your calculations
The ratio is 3.142515   now pi = 3.14159
The pi shows in the equation but is missing in your calculation
The second number should be  7.67204E-45
This is an error of 0.029 percent from the first number. 

You're right, I did make a mistake. Interesting how it matched up so well.

[jerrygg38 (OP)]

Kryptid said
You're right, I did make a mistake. Interesting how it matched up so well.
GG: Thanks for your effort. You did a good job matching the MKS and British systems. The equation is a standard type force equation so it should work in any system although I never thought of such things.

[Bored chemist]

Mass M = Charge Q x Light Speed x constant

By observation, no; it is not.
The charge of a proton, an antimuon, a +ve pion, and a positron are all the same.
The masses are all different.

[jerrygg38 (OP)]

Bored Chemist says:
By observation, no; it is not.
The charge of a proton, an antimuon, a +ve pion, and a positron are all the same.
The masses are all different.
GG: Yes they have different masses. It is necessary to calculate how many positive, negative and bipolar dot-waves are in each particle. At the dot-wave level the charge and the mass are related by a constant which depends upon the system of units you are in. For the MKS system a dot-wave has a mass of 1.96867E-71 kilograms and a charge of 3.47119E-60 Coulombs

[Kryptid]

GG: Thanks for your effort. You did a good job matching the MKS and British systems. The equation is a standard type force equation so it should work in any system although I never thought of such things.

The units still don't match up, though. The left side is measured in newtons (which are kilogram-meters per second squared). The right side (after much simplification) is measured in kilograms per second squared.

[Bored chemist]

Bored Chemist says:
By observation, no; it is not.
The charge of a proton, an antimuon, a +ve pion, and a positron are all the same.
The masses are all different.
GG: Yes they have different masses. It is necessary to calculate how many positive, negative and bipolar dot-waves are in each particle. At the dot-wave level the charge and the mass are related by a constant which depends upon the system of units you are in. For the MKS system a dot-wave has a mass of 1.96867E-71 kilograms and a charge of 3.47119E-60 Coulombs

Since it will always be possible to make up numbers of "angels dancing on pins" or "dot waves" to get the numbers to tally, it is impossible to falsify the claim.
So the claim isn't science.
This is a science web site.
Please come back with evidence, or not at all.

[jerrygg38 (OP)]

To Kryptid
Naked 2.21.20 5 pm
The left side of the equation is kilograms meters per second squared. This is mass time acceleration. F= mass x acceleration =kilograms meters/ second squared
 The right side of the equation for units is
F =   Uo Q Co Q Vb*/ RR  where Co is the speed of light and Vb* is the velocity of expansion of the hydrogen atom Bohr radius and R is the distance between the two atoms
So we have Uo coulombs meters per second x coulombs meters per second / meters squared.
 Breaking the R R into two parts we get
F = Uo [coulomb x meters/(second x meters)]  x [coulomb meters/(second x meters)]
We can then get rid of the meters
F = Uo [(Coulomb/second) (Coulomb/second)
F = Uo I I
 The force per unit length between two current carrying wires on the internet is
F/L = Uo Ia  Ib/ 2 pi R
Since L and R are both meters, the force units are
F = Uo Ia Ib = electrical permeability times  the current in wire a time the current in wire b
   In any event the right side of the equation is a standard form of electrical force equation which is mass time the acceleration.

[Kryptid]

To Kryptid
Naked 2.21.20 5 pm
The left side of the equation is kilograms meters per second squared. This is mass time acceleration. F= mass x acceleration =kilograms meters/ second squared
 The right side of the equation for units is
F =   Uo Q Co Q Vb*/ RR  where Co is the speed of light and Vb* is the velocity of expansion of the hydrogen atom Bohr radius and R is the distance between the two atoms
So we have Uo coulombs meters per second x coulombs meters per second / meters squared.
 Breaking the R R into two parts we get
F = Uo [coulomb x meters/(second x meters)]  x [coulomb meters/(second x meters)]
We can then get rid of the meters
F = Uo [(Coulomb/second) (Coulomb/second)
F = Uo I I
 The force per unit length between two current carrying wires on the internet is
F/L = Uo Ia  Ib/ 2 pi R
Since L and R are both meters, the force units are
F = Uo Ia Ib = electrical permeability times  the current in wire a time the current in wire b
   In any event the right side of the equation is a standard form of electrical force equation which is mass time the acceleration.

I think I see what you did now. You took a known gravitational force, compared it to the known electromagnetic force in the same system, and then came up with a number to relate those two values. Unfortunately, that doesn't tell us anything profound. I can make the exact same kind of equation comparing the Earth's gravity with Mars' gravity, for example. It certainly isn't evidence for the validity of your model.

By the way, how does your model account for the existence of the strong nuclear and weak nuclear forces? Why do some particles interact with the strong nuclear force but not others? Why can dot-waves come together to form certain types of particles but not others? Surely it would be possible to arrange dot-waves to produce negatively-charged particles with only one-tenth or one-one-hundredth of the electron's mass. So why don't those particles seem to exist? How does it account for conserved properties such as lepton number or baryon number?

[jerrygg38 (OP)]

Naked 2.23.20 8 AM
Kryptoid asks:
By the way, how does your model account for the existence of the strong nuclear and weak nuclear forces? Why do some particles interact with the strong nuclear force but not others? Why can dot-waves come together to form certain types of particles but not others? Surely it would be possible to arrange dot-waves to produce negatively-charged particles with only one-tenth or one-one-hundredth of the electron's mass. So why don't those particles seem to exist? How does it account for conserved properties such as lepton number or baryon number?
GG: The concern of my model is the underlying structure of the universe. It is composed of a multiplicity of two things. As the dot-waves are compressed they form photons, sub-particle, and particles. The complexity increases greatly. Atomic physics describes the interactions of particles and photons composed of huge numbers of dot-waves. I have no answers for your questions. That is for the physicists who specialize in such things.
  We can never measure a single dot-wave. We measure things related to our electrons which has a charge of Q. As far as an electron with only one tenth the mass, that would require it to have one tenth the charge as well. Someday five hundred billion years from now, the electron will radiate enough dot waves to do that. Then it will explode and the universe will be gone.

[jerrygg38 (OP)]

Kryptid said:
I think I see what you did now. You took a known gravitational force, compared it to the known electromagnetic force in the same system, and then came up with a number to relate those two values. Unfortunately, that doesn't tell us anything profound. I can make the exact same kind of equation comparing the Earth's gravity with Mars' gravity, for example. It certainly isn't evidence for the validity of your model.
GG: The whole purpose of the calculation was to produce a time of the universe since big bang equivalent to what the astronomers got in 1981-3. At that time it was around 15 billion years according to the limited data I had. This was before I had a home computer and the internet. All I had is very limited information.
  A few years ago I had to add the cosine of 30 degrees to get the 13.78 billion years which is closer to the astronomers today.  The whole purpose was to get the radius of the universe from the big bang to what is observed by the Hubble effect. This enables me to find the longest wavelength so that I can find the mass of a dot-wave based upon the radius of the universe.
   Since I needed the cosine 30 degrees it tells me that the expanding field of the hydrogen atom is an AC type field rather than a DC type field.
   By the way, on another group I pay a yearly fee to get rid of the ads. Is there some way to do that here?
   

[Kryptid]

The whole purpose of the calculation was to produce a time of the universe since big bang equivalent to what the astronomers got in 1981-3.

Is that so? So when you made that thread about how you could calculate the age of the Universe, you actually weren't doing that at all. All you were doing was working backwards from the known age.

A few years ago I had to add the cosine of 30 degrees to get the 13.78 billion years

You can't arbitrarily add numbers to an equation in order to make it work. If the numbers were not implied by the model in the first place, then there is no reason for them to be added after the fact. One of the factors that has made relativity so successful is the fact that its equations have matched observational data without any fudge factors added in.

The concern of my model is the underlying structure of the universe

The strong and weak nuclear forces are a part of that underlying structure. They are just as important as electromagnetism and gravity. If you have no problem dealing with those two forces, then why are the strong and weak nuclear forces a problem?

As far as an electron with only one tenth the mass, that would require it to have one tenth the charge as well. Someday five hundred billion years from now, the electron will radiate enough dot waves to do that. Then it will explode and the universe will be gone.

That violates conservation of charge.

[jerrygg38 (OP)]

Naked 2.23.20 10 pm
Discussion with Kryptid:
GG: The whole purpose of the calculation was to produce a time of the universe since big bang equivalent to what the astronomers got in 1981-3.
Kryptid: Is that so? So when you made that thread about how you could calculate the age of the Universe, you actually weren't doing that at all. All you were doing was working backwards from the known age.
GG: You are probably right. I do not always find the right words to express what I am doing. That is why at Sperry I had technical writers explaining how my designs worked to the Government. (As best as they could).
   I had concluded that the equivalent electrical force between two hydrogen atoms related to the current of expansion of the Bohr Orbit of one atom and the spinning electron’s current in the second atom and vice versa. This enabled two parts of the equation in which a very tiny current of expansion is multiplied by the much larger Bohr orbit current of the electron as it revolves around the proton.  So most of the equation is readily written from standard electrical theory.  This produces an equation with a factor of less than ten to the Astronomer’s answer.
   Assuming that the astronomers are correct, and then I can find the correction factor using the 4 pi term and the Cosine term.  4 pi is common to many electrical equations so it looked like a good fit. The Cosine was added years later when better numbers became available to me from astronomical data. A better mathematician no doubt would be able to produce a better explanation for the equation.
   To me the equation is an electrical representation of the gravitational force between to hydrogen atoms. The Quantum physicists no doubt have better answers for their own purposes. My purpose is to produce an electrical model equivalent.  It is my belief that the universe is governed by simple algebraic equations like Newton’s equations.
   As I see it, the actual time since big bang in clock ticks is nearly infinite. The tiny hydrogen atom of long ago had much more charged and much more mass. Over time it radiated away and the equations would still be valid in terms of a variable clock if everything followed an e^-x function. If not then things get even more complicated.
  If the equation had no real value then we could not take a very tiny expansion velocity and the high speed electron in the Bohr orbit and get the expansion rate of the universe within a ball park factor of ten or less.
   Your example of using a common equation relating an electrical force to a mechanical force is meaningless. The Bohr Orbit equation is
K Q Q/RR  =  M V V/ R
  This equation solves for the speed of rotation of the electron and the corresponding radius of orbit as per Bohr work. It is a centripetal force type equation and not a gravitational equation.
  My equation is quite different. When I wrote it down in 1981 it amazed me and my fellow engineers and physicists who I discussed it with. It appears to me now that the equation is not the cause of the gravitational force but it is an effect of the gravitational force. Thus it is a valid equation but it is not the driving force.
   Einstein’s answers seem pretty good except the force curving his space-time comes from the fifth dimension. The oscillation of particles in the fourth dimension at our light speed pushes against the higher light speed dimension. It is this dimension that pushes back and curves Einstein’s space time.
   Anyway someday scientists will come to realize that. However it may take another one hundred years.
Quote from: jerrygg38 on Today at 14:52:27
GG: A few years ago I had to add the cosine of 30 degrees to get the 13.78 billion years
Kryptid: You can't arbitrarily add numbers to an equation in order to make it work. If the numbers were not implied by the model in the first place, then there is no reason for them to be added after the fact. One of the factors that have made relativity so successful is the fact that its equations have matched observational data without any fudge factors added in.
GG: I have no complaints with relativity. It is an excellent model for a four dimensional universe. It does not explain what causes gravity. Long ago Einstein had to use a constant in his work which he was not happy with at the time. Years ago I doubted Einstein’s work but over time I have come to realize that his answers are the best possible describing function for a lot of the universe. He did not understand that black holes radiate away into the fifth dimension. Yet in time people will understand that.
   As far as me matching the equations to fit, I have no ability to produce three dimensional equations to describe the interaction between the hydrogen atom currents. All I can do is to write the approximate answer and match it to a known answer.
  In any event that is the way my mind works. On tests I write down the answer and then work backwards to the question.  Yet I had to bring my mind to a highly excited state to do that. That was a long time ago. Right now I am just a slow minded elderly person who is going to get his right eye cut open Friday and I am hopeful that I will have two working eyes afterwards.
Quote from: jerrygg38 on Today at 13:30:49
GG: The concern of my model is the underlying structure of the universe
Kryptid: The strong and weak nuclear forces are a part of that underlying structure. They are just as important as electromagnetism and gravity. If you have no problem dealing with those two forces, then why are the strong and weak nuclear forces a problem?
GG: I had some answers in my Doppler Space time of 2000 for these forces but I have stopped working on these. The big problem is that I have no lab experience with such things. For gravity, I can do some simple experiments at home to study it. I can take a simple gyroscope toy and experiment with it. I can push an object and get an idea of what is happening.
   At work I could build many electrical things. At home I could do the same. Then I could design things by inspection and test them and see them work.  I have to be with things and touch things. I am not a scientist but an engineer. And I would not be happy being a scientist.
   So I build universes in my mind. I study variable light speeds, variable constants, etc. Then I pick what is most likely possible. Mostly my work is intuitive. Something seems right to me. Then my unconscious mind brings me a solution in my dreams or sometimes in audio/visual communication. Then I study the possibility.
   Over time things change. I study the principles of string theory and see that they look for other dimensions. Some physicists who work on pure theory and seek new ideas often ask for a copy of my books when I email them.  They look for new ideas out of the box so they say.
    Anyway I like to paint houses and fix things.  It takes me a lot of time to do a small job now. So on one hand I am building universes in my mind and fixing a toilet with my hands.
Quote from: jerrygg38 on Today at 13:30:49
GG: As far as an electron with only one tenth the mass, that would require it to have one tenth the charge as well. Someday five hundred billion years from now, the electron will radiate enough dot waves to do that. Then it will explode and the universe will be gone.
Kryptid: That violates conservation of charge.
   GG: The conservation of charge works well for a closed system universe.  Once you have a fifth dimension the charge Q will fall and be radiated into the fifth dimension. The conservation of energy fails as well as the energy exits the fourth dimension (our light speed Co) and enters the fifth dimension. The total energy is constant when you add both dimensions. Mass is not conserved since the mass of a dot-wave in the Cs dimension is very small The same is true of the charge of a dot-wave when you enter the fifth dimension.
   The fifth dimension permits the universe to oscillate perpetually. The big bang inversion will occur over and over again forever. Cold death evaporates away. Time is meaningless. At the beginning is meaningless. It is a different universe that I see. I build practical universes in my mind. Are they correct? Time will tell.

[Kryptid]

Does your model predict that red photons and blue photons contain different numbers of dot-waves?

[jerrygg38 (OP)]

Kryptid asks:
Does your model predict that red photons and blue photons contain different numbers of dot-waves?
GG: The amount of dot-waves within a photon is the equivalent rest mass of the photon divided by 1.96867E-71 kilograms which is the mass of a dot-wave. No two red photons have the same amount of dot-waves. As I see it, photons are soft energy devices and they breathe. As the gravitational field increases they absorb more dot-waves. As the gravitational field decreases they lose dot-waves. They leave the sun shifted more to the blue. As they enter the balance point between the sun's field and the Earths field, they move toward the red. Then they slowly gain dot-waves as they reach toward the Earth. Thus photons continually change energy levels. In general particles such as electrons will maintain their charge Q but they will absorb bipolar dot-waves depending upon their speed and the gravitational density. Therefore no two particles or photons have exactly the same amount of dot-waves.

[Kryptid]

GG: The amount of dot-waves within a photon is the equivalent rest mass of the photon divided by 1.96867E-71 kilograms which is the mass of a dot-wave.

So the number of dot-waves depends on the photon energy, just as I thought. I see a potential problem here: a photon's energy is not absolute but rather varies from one reference frame to another. A spaceship moving towards the Sun will see a particular photon as having more energy than it would if it was moving away from the Sun instead (blueshift vs. redshift). This implies a contradictory number of dot-waves in that photon depending on your reference frame. In one it is made of more and in the other it is made of less. Relativity does allow for certain quantities to differ between reference frames (passage of time, relativity of simultaneity, mass/energy) but I don't think it allows for the total number of objects in a system to differ. If I'm wrong about that, I hope someone more knowledgeable about relativity will correct me on that.