Jump to content

Another theory for gravity


Recommended Posts

Hello,

Its been a couple of years since i came up with this idea. Just want to share with you a video that i made explaining this theory. Basically, it is explained from an electrostatic perspective of a neutral atom that i call the neutrostatic field. It is the parasitic electric field from a dipole atom that we consider 'neutral'. But both charge is not occupying the same space at the same time so the net field cannot be absolute zero at any distance from the atom. Since the positive nucleus and the negative electron is separated by the bohr radius, then their is always a net weak field in any space around it. Currently I'm attempting to do some simple calculation to support this theory and will share the result once finished. Let me know if this theory has some potential, thanks.

 

Link to post
Share on other sites
5 hours ago, Muhammad Faidzul said:

Currently I'm attempting to do some simple calculation to support this theory and will share the result once finished.

It seems highly unlikely that you would be able to explain observations and experimental results regarding gravity using electromagnetic theories. But please feel free to post your attempts. 

5 hours ago, Muhammad Faidzul said:

Let me know if this theory has some potential, thanks.

General Relativity is well established. If you are aiming at replacing that theory with something that has the same predictive power but is based on electromagnetism then my guess is that the probability of success is zero.

Some quick notes:
Gravity can't be shielded.
Gravity is not modelled as a force in General Relativity
Elementary particles with zero charge are affected by gravity
Etc.

Experts in this forum may also explain more fundamentals regarding the mathematics of tensors; source of gravitation is a second-order tensor, the source electromagnetism is a first-order tensor

 

Edited by Ghideon
added note on tensors
Link to post
Share on other sites

Ghideon has essentially said it already (+1) - gravity isn’t actually a force (except as an approximation in the Newtonian domain), so you can’t model it using a vector field approach, such as the one you have presented. It also exhibits different dynamics than electromagnetism, some superficial similarities notwithstanding.

It is, in fact, possible to mathematically show that no vector field theory can accurately model gravity, since such models are intrinsically incapable of encoding the necessary degrees of freedom (for fundamental, albeit technical, reasons). You need at the very least a rank-2 tensor theory, such as GR for example.

Link to post
Share on other sites
9 hours ago, Muhammad Faidzul said:

Just want to share with you a video

I am going to say +1 for your efforts.

Not only has this been one of the few speculations posted in accordance with the rules here (including the way you posted the video) but also it is a respectable train of thought that should not be lightly dismissed.

However there are some points your hypothesis (not theory) has failed to address.

1)Neutrons have no charge whatsoever, yet are subject to gravitational effects.
This is a serious obstacle.

2) Whilst I take you point about the generation of your field, you have taken situations that are electrostatic and combined them in a way that is forbidden by Earnshaw's theorem.
Earnshaws theorem basically states the two or more charges in an isolated universe cannot ever be in equilibrium and therefore cannot ever be stationary or static. It is just not possible to have two charges and nothing else (ie no other forces) in the way you describe. Something else has to hold them in position if they are still.

3) The charge distribution you introduce on your sphere containing both positive and negative charges at about 3.5 minutes is incorrect. Thomson's 'plum pudding' model has long been (experimentally) discredited.

4) What would happen if I placed a long strip of continuous conductor between the positive and negative charges in your early diagrams ?
The conductive material would fully shield each charge from the electric field of the other, yet there would still be a measurable gravitational effect between them.

Edited by studiot
Link to post
Share on other sites
9 hours ago, Muhammad Faidzul said:

Just want to share with you a video that i made explaining this theory.

!

Moderator Note

You also need to share the information in the forum so that people don’t need to watch the video

 
Link to post
Share on other sites
8 hours ago, Ghideon said:

General Relativity is well established. If you are aiming at replacing that theory with something that has the same predictive power but is based on electromagnetism then my guess is that the probability of success is zero.

My apologies. As an infant in physics, i can agree with you that i have zero probability in replacing General Relativity. But as any normal person, I will randomly come up with wild guesses on how stuff works. therefore i am here to just spill it out.

8 hours ago, Ghideon said:

Experts in this forum may also explain more fundamentals regarding the mathematics of tensors; source of gravitation is a second-order tensor, the source electromagnetism is a first-order tensor

I am bounded by my limited knowledge in this field as I am just a mechatronics graduate. Therefore for my calculation, I can only present in high school physics standard. Please correct me where necessary.

 

image.png.77620557eb26ede40d81fe8423ca83b2.png

For the figure above, there are two dipoles. I am interested in finding the force Fn. From my understanding from using superposition principle, I can calculate Fn by adding and subtracting the force of individual charge. Therefore:

image.png.a4ca367dcf154ceb647ffb13cdb48cb8.png

k = coulombs constant = 9x10, q = elementary charge = 1.6x10⁻¹⁹

Let :

d = 4x10⁸ meter (rough distance between earth and the moon)

b = bohr radius 5x10⁻¹¹ meter for M1

c = bohr radius 5x10⁻¹¹ meter for M2

 

Using wolfram calculator:

image.png.e556b75e41f254afcab5e0c4b3a65839.png

Now let’s calculate using our normal gravitational equation. M1 = M2 = Proton mass + Electron mass = 1.67x10⁻²⁷ + 9.1x10⁻³¹ = 1.67x10⁻²⁷

 

image.png.f053a13cff24a43ffc7246da78f6d603.png

G = gravitational constant = 6.7x10⁻¹¹

M1 = M2 = 1.67x10⁻²⁷ kg

Let: d= 4x10⁸ meter (rough distance between earth and the moon)

 

Using wolfram calculator:

image.png.ff5f0fd80e457c7b247a9ceaf71aeebf.png

At d= 4x10⁸ meter, the value of the neutrostatic force Fn (1.35x10⁻⁸² N) isn’t far off from the gravitational force Fg (1.17x10⁻⁸¹ N).

 

But if we reduce distance d, we will notice that Fn will overwhelm Fg by several magnitudes.

Below is a table I calculated using both formula with varying the distance 'd' between the objects.

image.png.620839179be73e8fd56362591ed2c94b.png

So the result doesn't follow the same value as the gravitational equation. But the interesting thing here is, at just 4 meters apart, the neutrostatic attraction is around 1,000,000,000,000,000 time stronger than that of the gravitational attraction. So, if the calculation is correct, the neutrostatic field is something that will certainly stand out more compared to gravity since the vacuum of space will still hold an atom per several meters away. In the formation of stars, the interstellar clouds of hydrogen might use this force to squeeze together into a sphere. When enough matter exist, the center of the sphere would be under high pressure and undergo fusion. The interstellar cloud must be very cold for this to happen. Otherwise, the motion of the dipoles might prevent them from polarizing with each other.

 

4 hours ago, studiot said:

Earnshaws theorem basically states the two or more charges in an isolated universe cannot ever be in equilibrium and therefore cannot ever be stationary or static. It is just not possible to have two charges and nothing else (ie no other forces) in the way you describe. Something else has to hold them in position if they are still.

Perhaps the continuous motion of the dipoles is what reduces the net neutrostatic forces between the two dipoles in my calculation. At the end of the video, I showed the field in motion in the Gravitational heating effect if you're interested. I wonder if there is still net attraction when the fields oscillate.

 

4 hours ago, studiot said:

The charge distribution you introduce on your sphere containing both positive and negative charges at about 3.5 minutes is incorrect. Thomson's 'plum pudding' model has long been (experimentally) discredited.

The sphere is not a plum pudding model of an atom. It is a visualization of charge distribution on a conductive sphere.

4 hours ago, studiot said:

1)Neutrons have no charge whatsoever, yet are subject to gravitational effects.
This is a serious obstacle.

I am aware of this since the neutrostatic field mainly rely on the bohr radius, this hypothesis (not theory) does not follow the equivalent principle when the bohr radius is significanly over or below the 5e-11meter distance. Therefore in plasma or in neutron state (I imagine neutrons have its electron very close to its proton), it couldn't hold up.

 

4 hours ago, studiot said:

4) What would happen if I placed a long strip of continuous conductor between the positive and negative charges in your early diagrams ?
The conductive material would fully shield each charge from the electric field of the other, yet there would still be a measurable gravitational effect between them.

If you mean by placing a conductor in the middle of the dipole, then the space permittivity will increase and reduce the neutrostatic field. Similarly if we can just simply reduce the bohr radius of the atom it would have the same effect. A neutron might fit the description. And for this, I will acknowledge that it is an obstacle with mainstream science.

 

Anyway, all of this is just my two cent :)

 

Edited by Muhammad Faidzul
Link to post
Share on other sites
36 minutes ago, Muhammad Faidzul said:
4 hours ago, studiot said:

1)Neutrons have no charge whatsoever, yet are subject to gravitational effects.
This is a serious obstacle.

I am aware of this since the neutrostatic field mainly rely on the bohr radius, this hypothesis (not theory) does not follow the equivalent principle when the bohr radius is significanly over or below the 5e-11meter distance. Therefore in plasma or in neutron state (I imagine neutrons have its electron very close to its proton), it couldn't hold up.

 

A Neutron has no electron.
So it is a particle with mass but no charge.
So why does it respond to your proposed gravitational neutrostatic force ?

 

37 minutes ago, Muhammad Faidzul said:
4 hours ago, studiot said:

The charge distribution you introduce on your sphere containing both positive and negative charges at about 3.5 minutes is incorrect. Thomson's 'plum pudding' model has long been (experimentally) discredited.

The sphere is not a plum pudding model of an atom. It is a visualization of charge distribution on a conductive sphere.

 

I understand that, so what is this ?

 

ng2.jpg.2cb5e13865cb4b2efb01db56398dc5a9.jpg

 

I was actually referring to this diagram

ng3.jpg.8f3f7f7455961fb1ac026ae67dc3d680.jpg

 

In both cases you have mixed up (arranged) the charges in an impossible way.

 

ng1.jpg.ff8053a3aad3708f6aab41aa71112dcb.jpg

 

Here is a clip of your dipole.

A problem with this is that you have shown forces acting as though the positive and negative have equal masses.
An electron and a proton have equal  charge but very unequal masses.
The mass of a proton is nearly 2000 times that of an electron.
The mass of say a Uranium ion and an electron are tens of thousands of times different.

 

48 minutes ago, Muhammad Faidzul said:
5 hours ago, studiot said:

4) What would happen if I placed a long strip of continuous conductor between the positive and negative charges in your early diagrams ?
The conductive material would fully shield each charge from the electric field of the other, yet there would still be a measurable gravitational effect between them.

If you mean by placing a conductor in the middle of the dipole, then the space permittivity will increase and reduce the neutrostatic field. Similarly if we can just simply reduce the bohr radius of the atom it would have the same effect. A neutron might fit the description. And for this, I will acknowledge that it is an obstacle with mainstream science.

Yes I mean placing a conductive shield between the positive and negative poles of a dipole.
But size has nothing to do with this.
You could use the original device of Cavendish to do this.
Charge one sphere positively and the other negatively and fit the shield between them.
You would still find the same gravitational attraction as measured on the torsion balance.

 

These are meant to be constructive criticisms of your proposal, not "I know more Physics than you so shut up little boy"  type answers.
You do not need advanced maths or physics to test your ideas and see that they are wrong because they contravene measurable reality.

 

Link to post
Share on other sites
8 hours ago, studiot said:

So why does it respond to your proposed gravitational neutrostatic force ?

I didn't and i couldn't claim it does. That's why I said it does not hold up to the equivalent principle.

 

8 hours ago, studiot said:

I understand that, so what is this ?

In the previous image, i showed a dipole and a double dipole. therefore for this diagram, it was meant to visualize a body with multiple dipole. You can pair every positive charge with a negative charge in that diagram. Even if it isn't perfectly symmetrical as shown, the weak field lines will still show up radially outwards. Given enough charge pairs that can make a big sphere, the weak neutrostatic field will likely be tangent perpendicular to the sphere.

 

8 hours ago, studiot said:

A problem with this is that you have shown forces acting as though the positive and negative have equal masses.

Up until now i have only shown that positive and negative have equal charge. For the gravitational calculation above, i neglected the negative charge mass because it is negligible.

 

8 hours ago, studiot said:

Charge one sphere positively and the other negatively and fit the shield between them.
You would still find the same gravitational attraction as measured on the torsion balance.

You cannot shield the electric field of a dipole this way. if you plan of putting a flat piece of shield in between the charge spheres, the charged spheres will attract to the shield instead due to the polarization or charge migration within the shield. If a hollow spherical shield if placed instead, only the interior of the hollow sphere is shielded. But still, the weak neutrostatic field still penetrates the sphere because the resolution of the alternating field lines is at an atomic level and due to the weak fields, only polarization of individual atoms of the sphere will occur. Hope this ugly drawing could help visualize.

image.thumb.jpeg.1f67ae13c21abec5be39264df17aeeb9.jpeg

For the normal shielding, the charge migration causes an imbalance to the spheres electric field. The superposition of the exterior field and the interior field due to charge migration cancels each other out within the sphere. Therefore the total field is 0.

image.thumb.jpeg.d18c2f998c5fc41d3878e1d8da42043f.jpeg

As for the neutrostatic field, not only it does not cancel, but it slightly increases the field strength within the sphere itself. Even if the field strength is strong enough to cause charge migration, the small resolution of the alternating field lines does not allow the charge to migrate between left and right of the sphere. therefore the field will just penetrate as above.

Edited by Muhammad Faidzul
Link to post
Share on other sites
10 hours ago, Muhammad Faidzul said:

I didn't and i couldn't claim it does. That's why I said it does not hold up to the equivalent principle.

 

In the previous image, i showed a dipole and a double dipole. therefore for this diagram, it was meant to visualize a body with multiple dipole. You can pair every positive charge with a negative charge in that diagram. Even if it isn't perfectly symmetrical as shown, the weak field lines will still show up radially outwards. Given enough charge pairs that can make a big sphere, the weak neutrostatic field will likely be tangent perpendicular to the sphere.

 

Up until now i have only shown that positive and negative have equal charge. For the gravitational calculation above, i neglected the negative charge mass because it is negligible.

 

You cannot shield the electric field of a dipole this way. if you plan of putting a flat piece of shield in between the charge spheres, the charged spheres will attract to the shield instead due to the polarization or charge migration within the shield. If a hollow spherical shield if placed instead, only the interior of the hollow sphere is shielded. But still, the weak neutrostatic field still penetrates the sphere because the resolution of the alternating field lines is at an atomic level and due to the weak fields, only polarization of individual atoms of the sphere will occur. Hope this ugly drawing could help visualize.

 

For the normal shielding, the charge migration causes an imbalance to the spheres electric field. The superposition of the exterior field and the interior field due to charge migration cancels each other out within the sphere. Therefore the total field is 0.

 

As for the neutrostatic field, not only it does not cancel, but it slightly increases the field strength within the sphere itself. Even if the field strength is strong enough to cause charge migration, the small resolution of the alternating field lines does not allow the charge to migrate between left and right of the sphere. therefore the field will just penetrate as above.

 

Perhaps if you paid as much attention to what I said as I did to what you presented you would not make such inappropriate answers.

I did try to help you rather than just dismissing your stuff out of hand as others did.

Clearly there is no point continuing.

Link to post
Share on other sites
20 hours ago, Muhammad Faidzul said:

Therefore for my calculation, I can only present in high school physics standard. Please correct me where necessary.

How come the ratio Coulomb force divided by Gravitational force depends on distance in your calculations?  I have not gone through the details but please note that when adding numbers of very different magnitudes (b, d and c in your case) you can run into issues with rounding. 

I also would like to add some more general comments:
-What is the connection to gravity in your idea? You have shown Coulomb forces and force from gravitation but no clarification why this is "Another theory for gravity" as the topic is named.
-What happens when you add a third body to your example and compare electromagnetic and gravity?
-What happens when you scale up your idea? If* your claims were true wouldn't we observe the following? Just as there are tidal effects on earth water there would be massive electrical phenomena related to rotations and orbits? Wouldn't large scale rearrangement of charge in celestial bodies have observable effects; for instance on earth magnetic field, weather and the operation of satellites?

The above is intended to inspire some thoughts about how you may compare your idea and accepted mainstream theories and models. You may be able to see topics you wish to study further or what questions to ask to be able to spot the issues.

 

*) Disclaimer: very big if, to allow for discussion, not as support for the idea proposed.

 

Edited by Ghideon
grammar
Link to post
Share on other sites
13 hours ago, studiot said:

 

Perhaps if you paid as much attention to what I said as I did to what you presented you would not make such inappropriate answers.

I did try to help you rather than just dismissing your stuff out of hand as others did.

Clearly there is no point continuing.

Sorry for the misunderstanding. I only tried to explain on my opinion on the given question. Sorry for failing to understand your question. Your help is much appreciated. Thank you.

 

12 hours ago, Ghideon said:

How come the ratio Coulomb force divided by Gravitational force depends on distance in your calculations?  I have not gone through the details but please note that when adding numbers of very different magnitudes (b, d and c in your case) you can run into issues with rounding. 

In the calculation, I attempted to show the similarities yet the difference between the two equations.

Similarities: Both follows an inverse square law (which depends on distance, hence the distance variable in my calculation)

Difference: Gravity: decreasing distance by 1 magnitude follows an increase in Force by 2 magnitudes

                     Neutrostatic: decreasing distance by 1 magnitude follows an increase in force by 4 magnitudes

 

For the high precision calculator, I used 3 calculators for double checks.

1. Wolfram Alpha (https://www.wolframalpha.com/)

2. Wolfram Mathematica (Computer Software)

3. Full precision calculator (https://www.mathsisfun.com/calculator-precision.html)

 

The reason I showed the calculation is to seek opinion if the equation is flawed. Maybe the 3 calculator that gave me the same answer can be wrong too.

 

13 hours ago, Ghideon said:

What is the connection to gravity in your idea? You have shown Coulomb forces and force from gravitation but no clarification why this is "Another theory for gravity" as the topic is named.

As studiot pointed out, this isn't actually a theory. it is just a hypothesis. I stand corrected on that one.

Initially my objective was to compare the similarities and differences between the forces of those two fields and form a discussion. It wasn't an attempt to replace GR as a whole. I apologize. Maybe I can change the topic title if possible.

I was also hoping for a discussion whether the Neutrostatic field can have any effects on celestial bodies. And if so, what are the magnitude.  I gave the example of a simple dipole is because the hydrogen atom is a dipole and is the most abundant matter in the universe.

 

13 hours ago, Ghideon said:

-What happens when you add a third body to your example and compare electromagnetic and gravity?

 

image.png.eba37bb8810a2783d14e24a132f80e56.png

 

The equation gets a little more complicated as follow:

image.thumb.png.653e45b9b43844b1157502e1e3a95a36.png

 

Now let:

d = 4x10⁸ meter, a = 5x10⁻¹¹ meter , b = 5x10⁻¹¹ meter, c = 5x10⁻¹¹ meter

image.png.a28699c601b89eacd1f97b4698f99883.png

 

Without M3: -1.35x10⁻⁸² N

With M3: -2.7x10⁻⁸² N

 

As comparison to gravitational attraction for body M4=M1+M3=(1.67x10⁻²⁷kg)+(1.67x10⁻²⁷kg)=3.34x10⁻²⁷kg vs M2(1.67x10⁻²⁷kg)

image.png.ef1e51017fd2a0ebe83f52a8710a56d4.png

image.png.c09c3ff8cbf8ea9cc6ff3cf5ea5281db.png

Without M3: 1.17x10⁻⁸¹ N

With M3: 2.34x10⁻⁸¹ N

 

So the conclusion is that both fields will double in force if we double the dipole and both will still be very weak at 400,000,000 meters away.

 

I understand that up until now i only calculated in 1-dimensional plane. To simulate a sphere, we will need to use at least a 2-dimensional plane which involves calculating coulombs law in vector form. Since calculating just 3 bodies in non-vector form above have made the calculation complicated, I am at my limit to calculate coulombs law in the vector form with enough dipole samples to form a moderate sphere. This will also need to include the knowledge on how the dipoles will be arranged naturally (imagine an interstellar hydrogen cloud).

For this matter I came here to find answers and interest if this is even a path worth trying.

Even if we don't find it interesting enough, hopefully a few years ahead someone with interest can stumble upon this topic.

 

15 hours ago, Ghideon said:

What happens when you scale up your idea? If* your claims were true wouldn't we observe the following? Just as there are tidal effects on earth water there would be massive electrical phenomena related to rotations and orbits? Wouldn't large scale rearrangement of charge in celestial bodies have observable effects; for instance on earth magnetic field, weather and the operation of satellites?

I guess that the fields are too weak and the alternating field lines are too close together to cause massive charge migration that is needed to interfere with electronics. Since the 'dipoles' are always in motion, the fields are in constant oscillation within a certain static position outside the sphere. The best it can do is vibrate an atom in that static position if the field is strong enough. An atom will give off light when vibrated. Therefore, I speculate that the corona layer of the sun has something to do with it.

image.png.0a995f9aa2d24e7513df7a0923b3855c.png

Image: https://indianapublicmedia.org/amomentofscience/the-corona.php

 

Again, I would like to apologize if this topic didn't follow the title.

 

Link to post
Share on other sites
43 minutes ago, Muhammad Faidzul said:

Difference: Gravity: decreasing distance by 1 magnitude follows an increase in Force by 2 magnitudes

                     Neutrostatic: decreasing distance by 1 magnitude follows an increase in force by 4 magnitudes

What is the difference between the superposition of electrostatic force and your "Neutrostatic"? You seem to use the formula for electrostatic force; Coulomb law ?

Please show the derivation of the formula you use to get your result. Use algebra instead of a numeric example. 

 

Edited by Ghideon
Clarification electrostatic force / Coulomb law
Link to post
Share on other sites
1 hour ago, Ghideon said:

You seem to use the formula for electrostatic force; Coulomb law ?

Yes

 

1 hour ago, Ghideon said:

What is the difference between the superposition of electrostatic force and your "Neutrostatic"?

There is no difference. It's just a term I named to specify the weak electric field of a neutral atom. Hence I defined the term in my first post above:

On 1/28/2021 at 10:50 AM, Muhammad Faidzul said:

Basically, it is explained from an electrostatic perspective of a neutral atom that i call the neutrostatic field. It is the parasitic electric field from a dipole atom that we consider 'neutral'.

 

The reason I don't use normal "electrostatic force" is because, normally, electrostatic can be any positive or negative standalone charge which has very strong force fields. Whereas the neutrostatic field has to contain both within a neutral atom to define the fields weakness. I'm in no way claiming it as a new force. I'm just defining it for simplicity and attempt to avoid confusion.

 

1 hour ago, Ghideon said:

Please show the derivation of the formula you use to get your result. Use algebra instead of a numeric example. 

Sorry but this is out of my league. Therefore, I cannot satisfy you on this.

Link to post
Share on other sites
1 hour ago, Muhammad Faidzul said:

Sorry but this is out of my league. Therefore, I cannot satisfy you on this.

Ok, no problem, but I had hoped for some evidence for the following claim:

4 hours ago, Muhammad Faidzul said:

Neutrostatic: decreasing distance by 1 magnitude follows an increase in force by 4 magnitudes

Because the above seems to be incompatible with Coulombs law.

You say there is no difference between the superposition of electrostatic force and "Neutrostatic".
Coulombs law is [math] |F|=k_{e} \frac{| q_{1} q_{2} |}{ d^{2} }  [/math]. 
Half the distance d means the force F increase by a factor of four. You seem to claim it increases much more. How?

 

1 hour ago, Muhammad Faidzul said:

I'm in no way claiming it as a new force. I'm just defining it for simplicity and attempt to avoid confusion.

Sorry, I think renaming known concepts from physics is very confusing. I have some basic knowledge about the mainstream physics and find it much easier to base discussions on common definitions rather than personal definitions. 

 

4 hours ago, Muhammad Faidzul said:

I was also hoping for a discussion whether the Neutrostatic field can have any effects on celestial bodies. And if so, what are the magnitude.

Electromagnetic fields in general can have an effect on celestial bodies. Why not open a thread and as questions in the mainstream sections of this forum? 

Edited by Ghideon
Link to post
Share on other sites
1 hour ago, Muhammad Faidzul said:

There is no difference. It's just a term I named to specify the weak electric field of a neutral atom. Hence I defined the term in my first post above:

 

The reason I don't use normal "electrostatic force" is because, normally, electrostatic can be any positive or negative standalone charge which has very strong force fields. Whereas the neutrostatic field has to contain both within a neutral atom to define the fields weakness. I'm in no way claiming it as a new force. I'm just defining it for simplicity and attempt to avoid confusion.

 

It's a pity you won't listen to others as your idea has the germ of truth in it.

But not as you appear to expect.

Strong and weak are relative terms so that neither of the above statements are absolutely true.

But your idea that the force experienced by other charges varies with aspect for atoms (and molecules) is one of the most vitally important subjects in chemistry as it controls both chemical bonding and chemical reactions.

Perhaps you should ask about this as the level of maths required is probably within your grasp.

Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.