# How to determine Gravity

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## Re: How to determine Gravity

I attempted to factor in equatorial rotation speed of the planets in an effort to see if things would line up. By coincidence(?) only, Jupiter worked right off the bat. I thought "great!". Sadly, the rest of the planets did not give the expected results though.

Here is what I'm working with:

Mercury - 3.76

Venus - 9.32

Mars - 5.22

Jupiter - 107.00

Saturn - 89.64

Uranus - 39.06

Neptune - 37.90

Pluto - 1.83

SUN - 260.68

Probably not of any importance? But here are the rotational speeds at the equator of each planet, in meters per second:

Mercury - 3.03 m/s

Venus - 1.81 m/s

Mars - 241.17 m/s

Jupiter - 12,444.44 m/s

Saturn - 9,690.00 m/s

Uranus - 2,590.00 m/s

Neptune - 2,683.33 m/s

Pluto - 13.11 m/s

SUN - 1.91 m/s

How did you calculate earth? I though earth was the constant on which all the other planets were measured.

Here is what I'm working with:

Mercury - 3.76

Venus - 9.32

Mars - 5.22

Jupiter - 107.00

Saturn - 89.64

Uranus - 39.06

Neptune - 37.90

Pluto - 1.83

SUN - 260.68

Probably not of any importance? But here are the rotational speeds at the equator of each planet, in meters per second:

Mercury - 3.03 m/s

Venus - 1.81 m/s

Mars - 241.17 m/s

Jupiter - 12,444.44 m/s

Saturn - 9,690.00 m/s

Uranus - 2,590.00 m/s

Neptune - 2,683.33 m/s

Pluto - 13.11 m/s

SUN - 1.91 m/s

How did you calculate earth? I though earth was the constant on which all the other planets were measured.

**parkham**- Posts : 21

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## Re: How to determine Gravity

Here are my numbers.

I agree, we are using the earth as the standard.

I would argue that we should use the much better known earth data to predict more accurate planetary densities.

.

The calculations depend on known densities; but the planetary densities are calculated from faulty orbital velocity equations, and Miles knows what else. We expect deviations.Nevyn wrote: Looking at the ratios we can see that the inner bodies are fairly close. Mars starts to jump out a little bit more but the Jovians are really different, especially Saturn. I would tentatively suggest that it is Jupiter and Saturn that are causing this. You see, planets don't exactly orbit the sun, they orbit the sun plus all planets on smaller orbits. This is especially true for the planets outside of Jupiter and Saturn, which sort of act like a second sun. Notice how after Saturn the ratios start to come down again. There isn't enough data to know if that means anything or not but it is suggestive.

I agree, we are using the earth as the standard.

I would argue that we should use the much better known earth data to predict more accurate planetary densities.

.

**LongtimeAirman**- Admin
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## Re: How to determine Gravity

Does that mean that Nevyn's math calculations is correct and reported densities are truly off, sometimes extremely so? If so, then reported gravity is also off as well, yes? It would have to be in order for MM's theory to be true, correct?

It seems the biggest divergence is with the gas / ice giants. I wonder why that is? I thought that it might be something to do with radius, but the Sun's radius is huge and its results aren't horribly off. Also, the Sun is negatively off while everything is positively off (if that makes sense).

I am standing on shaky ground, but for some reason I can't "shake" the hunch that rotational speed * has * to have some part it in all this.

It seems the biggest divergence is with the gas / ice giants. I wonder why that is? I thought that it might be something to do with radius, but the Sun's radius is huge and its results aren't horribly off. Also, the Sun is negatively off while everything is positively off (if that makes sense).

I am standing on shaky ground, but for some reason I can't "shake" the hunch that rotational speed * has * to have some part it in all this.

**parkham**- Posts : 21

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## Re: How to determine Gravity

The e/m field is balanced against gravity - producing the apparent/actual gravity differential. I don't believe variations in the e/m field - such as rotation - are necessary for determining the 'rough' resultant planetary gravity in these examples. The difference between these calculated numbers and the official numbers are much greater than the body's e/m field variations. On the other hand, along with the 'knowns' come mainstream assumptions that we are trying to replace. We do need to consider additional factors such as planetary tilt and orbital inclination; that's why Nevyn gave additional references.

I tend to believe Miles' or Nevyn's work, unless I'm practicing being thrown to the mat.

.

I tend to believe Miles' or Nevyn's work, unless I'm practicing being thrown to the mat.

.

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## Re: How to determine Gravity

Nevyn:

Why do we have to use constants based on the earth in order to determine gravity and e/m field of other planetary bodies?

Why do we have to use constants based on the earth in order to determine gravity and e/m field of other planetary bodies?

**parkham**- Posts : 21

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## Re: How to determine Gravity

parkham wrote:Does that mean that Nevyn's math calculations is correct and reported densities are truly off, sometimes extremely so? If so, then reported gravity is also off as well, yes? It would have to be in order for MM's theory to be true, correct?

It is important to remember that the mainstream numbers are

**calculated**based on

**assumptions**that Newton's equations are perfect and the only force in play. We have never measured the gravity of the sun, we just assume some things and back-calculate. Same with all of the planets except Earth.

However, Miles makes a similar assumption in Newton's gravity equation. He assumes it is correct and then goes further and pulls it apart to show how size and density work individually.

It is also important to realise that we don't live in a gravity only universe and the way charge plays its part is complicated. We are talking about how to calculate the surface gravity of bodies which is very different to determining orbits. The mainstream only has gravity so they stop there and assume everything is fine and dandy. We have charge and gravity and both are required for an orbit to balance. This is why Jupiter and Saturn are so important. They are huge and have large charge fields that affect the rest of the field. This will push all of the planets further out from them further away and possibly the inner planets will be pushed further in, although the Sun's charge gets denser the closer you are to the Sun so it is a reduced effect than that imposed onto the outer planets where the charge from Jupiter and Saturn combine with the Suns charge.

parkham wrote:It seems the biggest divergence is with the gas / ice giants. I wonder why that is? I thought that it might be something to do with radius, but the Sun's radius is huge and its results aren't horribly off. Also, the Sun is negatively off while everything is positively off (if that makes sense).

I am standing on shaky ground, but for some reason I can't "shake" the hunch that rotational speed * has * to have some part it in all this.

The Sun works because it is the center of the solar system. It does not depend on any other bodies. It sets the field and everything else is determined with respect to it. You can almost say the same about Jupiter and to a lesser extent, Saturn. The combined charge fields of Jupiter and Saturn basically reset the field. Saturn and Jupiter can sometimes be thought of as a single entity. They play an intricate game with each other because Saturn is smaller than Jupiter so it wants to go below it. A stable solar system has smaller planets orbiting inside of bigger planets. We don't live in such a system and the Jovians are the main deviation. Everything above Jupiter wants to move below it in order to find its stable position. Saturn tries to do so and has enough size to make some headway but ultimately, Jupiter overrules it and pushes it back out. There may come a day when it doesn't and the whole solar system will be thrown into turmoil for a while.

The rotation rate of the Sun and each planet does have a part to play because it can affect the charge density of that body. In this problem though, they are constants so we can ignore them. We don't have an equation to determine the charge density that includes the rotation rate so we couldn't use it anyway. This will need to be developed at some stage. Prime area to work in if you want to submit a paper to Miles for publishing. I'm not quite sure where to start with that. Probably create a spreadsheet containing the basic properties of each planet such as radius, density, gravity, etc. Now that I start to think about it, I've already started in my post above with the table of gravity values. Add in the known rotation rates and any other properties you think are applicable and see what you can calculate from it all. Use Miles approach of calculating relative values and see how they relate to each other. There may be fame, but probably no fortune, waiting for anyone that wants to put the time into it.

However, given all of that and probably some more I have forgotten, my calculations could still be incorrect. Miles equations may be on the right path but not quite complete yet. Or it all could be just plain wrong, but I don't tend to believe that. The equations I have used mostly came from the 'Moon gives up a secret' paper and that was only concerned with the Earth and Moon system. We are extrapolating up to the whole solar system and so much more comes into play that I think it feasible to forgive some discrepancies as long as there are reasons for it. If you think about it, the equations have worked for those situations that are most like the Earth/Moon system and fail when other entities have affects large enough to change the situation.

**Nevyn**- Admin
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## Re: How to determine Gravity

parkham wrote:Nevyn:

Why do we have to use constants based on the earth in order to determine gravity and e/m field of other planetary bodies?

Because we have measured data for the Earth but no other planets.

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## Re: How to determine Gravity

I can give a slightly better answer than that.

Miles likes to use relative values in his calculations. Well, I'm sure he would prefer to use concrete values (ie measured), but we often don't have that data available, or at all. He is often looking for relationships between things so relative values are a good way to work. Sometimes the absolute values don't look like they are related, as Miles points out in the 'Moon gives up a secret' paper, but the relative values show a different story and with a little bit of work you can find out why the numbers don't seem to add up to what you want.

Miles has shown, time and time again, how the mainstream gives up too early. They might look at the gravity of the Earth and Moon, say, and compare the values, look at their radii and compare those and find that the relationship doesn't work. But Miles has the charge field and this changes the assumptions and gives you something to fill the gap which he seems to have done quite nicely. Instead of seeing the numbers that don't match and moving on to some other problem, Miles asks why don't they match, what could cause them to be different. He then tries to find answers to those questions, and the math to show the gap. If he does so, then we get a paper on it, but I'm sure there have been cases where he didn't.

Miles likes to use relative values in his calculations. Well, I'm sure he would prefer to use concrete values (ie measured), but we often don't have that data available, or at all. He is often looking for relationships between things so relative values are a good way to work. Sometimes the absolute values don't look like they are related, as Miles points out in the 'Moon gives up a secret' paper, but the relative values show a different story and with a little bit of work you can find out why the numbers don't seem to add up to what you want.

Miles has shown, time and time again, how the mainstream gives up too early. They might look at the gravity of the Earth and Moon, say, and compare the values, look at their radii and compare those and find that the relationship doesn't work. But Miles has the charge field and this changes the assumptions and gives you something to fill the gap which he seems to have done quite nicely. Instead of seeing the numbers that don't match and moving on to some other problem, Miles asks why don't they match, what could cause them to be different. He then tries to find answers to those questions, and the math to show the gap. If he does so, then we get a paper on it, but I'm sure there have been cases where he didn't.

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## Re: How to determine Gravity

Thank you very much. I'm going to re read your responses again before responding.

**parkham**- Posts : 21

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## Re: How to determine Gravity

OK. I've decided to go back to uft.html and really, really, really read it. I thought I understood but, I still don't.

Would someone please clarify H = m(A + a) where we would be dealing with the earth and the moon? Starting at " Let us use the letter H. "

I assume that m = mass of moon 73.50, M = mass of earth 5,973.60, A = acceleration of earth (9.81?), a = acceleration of moon (1.62?)

Would someone please clarify H = m(A + a) where we would be dealing with the earth and the moon? Starting at " Let us use the letter H. "

I assume that m = mass of moon 73.50, M = mass of earth 5,973.60, A = acceleration of earth (9.81?), a = acceleration of moon (1.62?)

**parkham**- Posts : 21

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## Re: How to determine Gravity

.

Airman says. I welcome any opportunity to reread papers and answer questions. I didn’t see your last post till a couple of hours ago. Maybe a halfway reread will do before I quit for the day.

Why are you using numerical examples for H?

F = GMm/r^2

In order to derive

I'll finish reading, but that’s my tentative answer.

.

**parkham said. OK. I've decided to go back to uft.html and really, really, really read it. I thought I understood but, I still don't.**9. The Unified Field Theory. http://milesmathis.com/uft.html Newton's gravitational equation is shown to be a compound equation that includes the E/M field. I mathematically separate the two field equations, do transforms on them, and create simple and useful Unified Field Equations. 24pp.

**Would someone please clarify H = m(A + a) where we would be dealing with the earth and the moon? Starting at " Let us use the letter H. "**Airman says. I welcome any opportunity to reread papers and answer questions. I didn’t see your last post till a couple of hours ago. Maybe a halfway reread will do before I quit for the day.

**I assume that m = mass of moon 73.50, M = mass of earth 5,973.60, A = acceleration of earth (9.81?), a = acceleration of moon (1.62?)**Why are you using numerical examples for H?

**H = m(A + a)**is not a formula for determining field strengths. Miles created a gravitational form intended to be used as a means of expanding Newton’s formula,F = GMm/r^2

In order to derive

the E/M field equation that was buried in Newton’s equationE = (m/R2 )[GM – AR2 – aR2]

I'll finish reading, but that’s my tentative answer.

.

**LongtimeAirman**- Admin
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## Re: How to determine Gravity

.

Airman. One of Miles' defining characteristics, in my opinion, is his ability to continually astonish me. I read, and reread and never see any of his papers the same way twice.

I'm sorry. No disrespect intended. What's your question?

.

**parkham said. OK. I've decided to go back to uft.html and really, really, really read it. I thought I understood but, I still don't.**Airman. One of Miles' defining characteristics, in my opinion, is his ability to continually astonish me. I read, and reread and never see any of his papers the same way twice.

**The Unified Field Theory**is a particularly rich paper. I'm far from understanding it. For example, even with Miles simplifying things, relativity still throws me.**Starting at " Let us use the letter H. "**is a bit confusing. At first it appears you're asking for an interpretation, but following up with numbers makes me think you want a verification. Your number assumptions look good to me, especially considering our previous discussion. Miles uses them in an example on the next page and elsewhere in the paper.I'm sorry. No disrespect intended. What's your question?

.

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## Re: How to determine Gravity

Miles Mathis, uft.html wrote:

gE - EE = 9.8 m/s2

gM - EM = 1.62 m/s2

RE/RM = gE / gM = 3.672

gM = .2723 gE

EE /EM = 1/3.6724 = .0055

EM = 181.81 EE

But that last equation is assuming that the Earth and Moon have the same density. So I must now correct for density. Notice we are correcting the E/M field for density, not the gravitational field.

DE /DM = 5.52/3.344 = 1.6507 = 1/.6057

EM = 110.12 EE

How did he get: gM = .2723 gE & EM = 181.81 EE

**parkham**- Posts : 21

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## Re: How to determine Gravity

.

The Unified Field Theory references another paper.

8.

The two numbers are based on a relative comparison of the earth and moon radii.

parkham said. How did he get: gM = .2723 gE & EM = 181.81 EE

The Unified Field Theory references another paper.

8.

**The Moon Gives up a Secret.**http://milesmathis.com/moon.html Numbers from the Moon are analyzed to show that "gravity" is a compound field. 4pp.

For this reason, a large spherical E/M field will vary as 1/r^4, if measured from a distance...

Gravity varies ONLY as the radius of the object, and no longer as the distance of separation.

Given these two postulates we can proceed directly to the math. Let us first make a prediction, using the postulates above. I am claiming that that I can show that the gravitational fields of the Moon and the Earth are directly proportional to their radii. Let us do the math to show what the Moon’s gravitational field would have to be if that were true.

gE / gM = 3.672

9.8 m/s2 / gM = 3.672

gM = 2.669 m/s2

But the current number for gM is 1.62 m/s2. That seems like a huge amount of acceleration to make up, and I can understand your doubts. When I first did the math I thought there was little chance the numbers would work, to be honest. I was just following an idea. But watch closely:

We know that the total field of the Earth at its surface creates an acceleration of 9.8 m/s2 and we hypothesize that this is the gravitational field minus the E/M field [the gravitational field is an attractive field and the E/M field is a repulsive field]. And we know the same for the Moon.

gE - EE = 9.8 m/s2

gM - EM = 1.62 m/s2

I have also postulated that the gravitational part of this acceleration should be proportional to the radii.

gE / gM = 3.672gM = .2723 gE

And I have just postulated that the E/M field is proportional to 1/r^4.

EE /EM = 1/(3.672^4) = .0055EM = 181.81 EE

But that last equation is assuming that the Earth and Moon have the same density. So I must now correct for density.

DE /DM = 5.52/3.344 = 1.6507 = 1/.6057

EM = 110.12 EE

So, we just substitute:

.2723 gE - 110.12 EE = 1.62 m/s2

gE - EE = 9.8 m/s2

.2723gE - .2723EE = 2.6685 m/s2 [subtract the two equations]

-109.85EE = -1.0485 m/s2

EE = .009545 m/s2

EM = 1.051 m/s2

gM - EM = 1.62 m/s2

gM = 2.671 m/s2

You can see that the math bore out my prediction exactly. Once we correct for the presence of the E/M field, the Earth and the Moon have gravitational fields that are exactly proportional to their radii.

We did not get an exact match in the third decimal place only because we used 9.8 m/s2 for gE in the first equation. We must now add .009545 to that, and if we do we get 2.671 m/s2 in the first equation as well.

The two numbers are based on a relative comparison of the earth and moon radii.

Last edited by LongtimeAirman on Tue Jun 28, 2016 12:35 am; edited 2 times in total (Reason for editing : added a couple of exponential ^ marks because the current font is not superscripted)

**LongtimeAirman**- Admin
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## Re: How to determine Gravity

I figured out gM = .2723 gE

rM = radius of moon = 1,737,100.00 m

rE = radius of earth = 6,371,000.00 m

rM / rE = 0.2727 gE - All he had to do was put that in his steps, but he didn't. I guess he just assumed we all could follow along?

1,737,100.00 / 6,371,000.00 = 0.2727

Still not sure how he got EM = 181.81 EE

rM = radius of moon = 1,737,100.00 m

rE = radius of earth = 6,371,000.00 m

rM / rE = 0.2727 gE - All he had to do was put that in his steps, but he didn't. I guess he just assumed we all could follow along?

1,737,100.00 / 6,371,000.00 = 0.2727

Still not sure how he got EM = 181.81 EE

**parkham**- Posts : 21

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## Re: How to determine Gravity

.

parkham, Look at my previous post. You showed it too, again based on the radius ratio 3.672.

parkham, Look at my previous post. You showed it too, again based on the radius ratio 3.672.

.And I have just postulated that the E/M field is proportional to 1/r^4.

EE /EM = 1/(3.672^4) = .0055

EM = 181.81 EE

**LongtimeAirman**- Admin
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## Re: How to determine Gravity

eM / eE = 1 / (.2723 ^4) = 181.81 eE

OR

gE / gM = 3.672

EE /EM = 1/(3.672^4) = .0055

EM = 3.672^4 = 181.81 eE

For adjustment relevant to density:

DE /DM = 5.52/3.344 = 1.6507 = 1/.6057

181.81 * .6057

EM = 110.12 EE

I'm attempting to reason through this now, it all seems so

edit: been looking at this, and I think I'm missing the obvious again. Would someone help me along with explanations of how he's getting these numbers?

.2723 gE - 110.12 EE = 1.62 m/s2 = -109.85 EE I can see that.

.2723gE - .2723EE = 2.6685 m/s2 hunh? 9.81 / 2.6685 = 3.672 which is rE / rM how does that fit in here?

OR

gE / gM = 3.672

EE /EM = 1/(3.672^4) = .0055

EM = 3.672^4 = 181.81 eE

For adjustment relevant to density:

DE /DM = 5.52/3.344 = 1.6507 = 1/.6057

181.81 * .6057

EM = 110.12 EE

I'm attempting to reason through this now, it all seems so

*circular*to me though:edit: been looking at this, and I think I'm missing the obvious again. Would someone help me along with explanations of how he's getting these numbers?

.2723 gE - 110.12 EE = 1.62 m/s2 = -109.85 EE I can see that.

.2723gE - .2723EE = 2.6685 m/s2 hunh? 9.81 / 2.6685 = 3.672 which is rE / rM how does that fit in here?

So, we just substitute:

.2723 gE - 110.12 EE = 1.62 m/s2

gE - EE = 9.8 m/s2

.2723gE - .2723EE = 2.6685 m/s2 [subtract the two equations]

-109.85 EE = -1.0485 m/s2

EE = .009545 m/s2

EM = 1.051 m/s2

gM - EM = 1.62 m/s2

gM = 2.671 m/s2

**parkham**- Posts : 21

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## Re: How to determine Gravity

Hi Parkham,

Mathis also steps through this in this paper:

http://milesmathis.com/third7.html

A Redefinition of Gravity

Part VII

Also this paper gives some background:

by Miles Mathis

http://milesmathis.com/weight.html

Mathis also steps through this in this paper:

http://milesmathis.com/third7.html

*The Third Wave - Part VII*A Redefinition of Gravity

Part VII

Mathis wrote:We know that the total field of the Earth at its surface creates an acceleration of 9.8 m/s^{2}and we hypothesize that this is the gravitational field minus the E/M field. And we know the same for the Moon.

gE-EE= 9.8 m/s^{2}

gM-EM= 1.62 m/s^{2}

I have also postulated that the gravitational part of this acceleration should be proportional to the radii.

gE/ gM= 3.67^{2}

gM= .2723 gE

And I have just postulated that the E/M field is proportional to 1/r4.

EE/EM= 1/3.672^{4}= .0055

EM= 181.81 EE

But that last equation is assuming that the Earth and Moon have the same density. So I must now correct for density.

DE/DM= 5.52/3.344 = 1.6507 = 1/.6057

EM= 110.12 EE

So, we just substitute:

.2723 gE-110.12 EE= 1.62 m/s^{2}

gE-EE= 9.8 m/s^{2}

.2723gE-.2723EE= 2.6685 m/s^{2}[subtract the two equations]

-109.85EE= -1.0485 m/s^{2}

EE= .009545 m/s^{2}

EM= 1.051 m/s^{2}

gM-EM= 1.62 m/s^{2}

gM= 2.671 m/s^{2}

Once we correct for the presence of the E/M field, the Earth and the Moon have gravitational fields that are exactly proportional to their radii. We did not get an exact match in the third decimal place only because we used 9.8 m/s^{2}for gE in the first equation. We must now add .009545 to that, and if we do we get 2.671 m/s^{2}in the first equation as well.

Also this paper gives some background:

**An Update on Weight in which I falsify F=GMm/R**^{2}by Miles Mathis

Mathis wrote:To find what it must be, we will solve as above. My gravity pseudo-field obeys the inverse square law, but only due to Relativity. So the apparent attraction at one Earth-radius will be 2.68 times 1/3.67^{2}, or .198 m/s^{2}. The E/M field drops off with 1/d^{4}, however, so if the distance increases by a factor of 3.67, the field will decrease by 3.67^{4}or 181.4 times. Since I have shown in another paper that this field has an acceleration of 1.051 at the surface, at one Earth radius it will have an acceleration of .0058. We subtract that from the expansion pseudo-field to get .1924 m/s^{2}. That is 1/50 that of the Earth, which is roughly the volume ratio, not the mass ratio.

Which is to say that the standard model believes the number will be about .12 in that situation. 9.8/81 = .12. I predict it will be almost .2.

http://milesmathis.com/weight.html

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## Re: How to determine Gravity

There is a relatively simple experiment to show charge, and there are also some videos on youtube.

You should do this (if I remember correctly), use an axis that is free to rotate at the center, and put 2 heavy objects at the ends. Then place 2 very dense objects on the ground big enough to be taller than the objects on the axis, one at the left and one at the right of the objects on the axis.

Wait and you will see the axis starting to rotate towards the dense objects on the ground.

This phenomenon is not explained by gravity between objects, because it is so much stronger than it should be. It is due to the dense objects shielding part of the surrounding charge wind. This shielding allows the other side wind to be the only part to operate on the heavy objects and on the axis that is free to rotate, and so it rotates until blocked.

This needs to be done with all the due measurements and calculations (and predictions) to be called an experiment of course.

I can't post links but one of the youtube videos has this code: https://www.youtube.com/watch?v=euvWU-4_B5Y

- title: Gravity Experiment - Author: Nars Guzman

EDIT: and this is probably the same as the Cavendish experiment - but watch this video https://www.youtube.com/watch?v=6Lzd86ZYf_o where an other experiment is described wich is against mainstream gravitational theory (put 3 blocks near each axis ends, 1 at one sid, 2 at the other with a gap between them to let charge pass. Gravity expects the axis to move towards the 2 dense objects but it still prevers the direction of the single dense objects)

You should do this (if I remember correctly), use an axis that is free to rotate at the center, and put 2 heavy objects at the ends. Then place 2 very dense objects on the ground big enough to be taller than the objects on the axis, one at the left and one at the right of the objects on the axis.

Wait and you will see the axis starting to rotate towards the dense objects on the ground.

This phenomenon is not explained by gravity between objects, because it is so much stronger than it should be. It is due to the dense objects shielding part of the surrounding charge wind. This shielding allows the other side wind to be the only part to operate on the heavy objects and on the axis that is free to rotate, and so it rotates until blocked.

This needs to be done with all the due measurements and calculations (and predictions) to be called an experiment of course.

I can't post links but one of the youtube videos has this code: https://www.youtube.com/watch?v=euvWU-4_B5Y

- title: Gravity Experiment - Author: Nars Guzman

EDIT: and this is probably the same as the Cavendish experiment - but watch this video https://www.youtube.com/watch?v=6Lzd86ZYf_o where an other experiment is described wich is against mainstream gravitational theory (put 3 blocks near each axis ends, 1 at one sid, 2 at the other with a gap between them to let charge pass. Gravity expects the axis to move towards the 2 dense objects but it still prevers the direction of the single dense objects)

Last edited by LongtimeAirman on Sun Sep 11, 2016 8:26 am; edited 1 time in total (Reason for editing : Activated links.)

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