# Possibility for Mass Transport System could take us up a gear.

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Why would that work at the earth's surface? 17,700 mph is the orbital requirement at 500 miles, according to your maths.

Because the physics is no different just adjacent to the earths surface , as it is 500 miles up . The radius difference , from the earth centre to ground level compared to 500 miles up , might need a little adjustment . But the radius of the Earth is 4000 miles so that would now be 4500 miles from earth centre. So the formulae mv (squared )/ r = mg , would need keeping an eye on . This formula when adjusted for meters and seconds instead of miles and hours , should give 17,000 mph . Which is why most space launches of spacecraft go down range to approx 17,700 mph to make earth orbit . But you can do this at ground level ,in your back yard , provided you do not travel too far ( say 4 inches ,would be o.k. ) . But to reach a root mean squared value of 17,700mph , you will need something fairly massive oscillating 4 inch peak at 40,000 times a second ( 40khz ) . It's doable , but do not stand too close. This oscillating mass needs keeping in a vacuum, or it could be like a rock festival gone A.W.O.L

Mike

Edited by Mike Smith Cosmos

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Because the physics is no different just adjacent to the earths surface , as it is 500 miles up . The radius , from the earth centre to ground level to 500 miles up , might need a little adjustment . But the radius of the Earth is 6000 miles so that would now be 6500 miles from earth centre. So the formulae mv (squared )/ r = mg , would need keeping an eye on .

Mike

The physics is no different, but the specific answers surely are. BTW, the radius of the earth is 4000 miles (6400 km).

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The physics is no different, but the specific answers surely are. BTW, the radius of the earth is 4000 miles (6400 km).

Oops! Got my miles and kilometers muddled.

When I last did the sums the surface speed required was not much more that that required 500 miles up.

It is a popular misconception that gravity goes to zero up at 500 miles , seeing all these spacemen drifting about in space stations. They , and everything else that orbits in low earth orbit is travelling at 17,700 mph approx , that's why they appear weightless . The gravitational constant is not a lot different 500 miles up than it is at ground level . What we are battling with is these distortions in space caused by this massive lot of rock , water and metallic elements making up the Earth . We need to re-distort them back to a more neutral condition by doing something less significant but far more local . ( ie . This vibrating mass at 40 kilohertz. )

The other way of course is to put another massive rocky/ metallic object like the moon to re-distort ,which of course is what the moon does , part way between earth and moon . But apart from doing your ' sling shot ' journeys across the solar system , we need another method locally about the low earth orbit, or takeoff position .

Mike

Edited by Mike Smith Cosmos
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It is a popular misconception that gravity goes to zero up at 500 miles , seeing all these spacemen drifting about in space stations. They , and everything else that orbits in low earth orbit is travelling at 17,700 mph approx , that's why they appear weightless .

But undoubtedly not a common misconception here.

The gravitational constant is not a lot different 500 miles up than it is at ground level .

It's exactly the same. It's a constant.

g, the acceleration, is smaller. You change r by 12.5%, and the acceleration depends on r2, so it's not that small of an effect. It's more than 20% smaller.

What we are battling with is these distortions in space caused by this massive lot of rock , water and metallic elements making up the Earth . We need to re-distort them back to a more neutral condition by doing something less significant but far more local . ( ie . This vibrating mass at 40 kilohertz. )

The other way of course is to put another massive rocky/ metallic object like the moon to re-distort ,which of course is what the moon does , part way between earth and moon . But apart from doing your ' sling shot ' journeys across the solar system , we need another method locally about the low earth orbit, or takeoff position .

Mike

What "distortions"? Are you invoking general relativity? If so, why? Newtonian physics works just fine to explain all of this. If not, what is your proposal?

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For gravity to be repulsive you need strange situations like negative energy densities or negative mass; this could be in the form of the cosmological constant. Or you need something like (slow-roll) inflation where you have an inflaton field and a false vacua.

Ajb

Sounds interesting : could you possibly elaborate a bit on:

A) Negative energy densities

B)Negative Mass

C) Form of the cosmological constant

D) Slow roll inflation - inflation field and false vacua

Thanks in anticipation

Mike

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What "distortions"? Are you invoking general relativity? If so, why? Newtonian physics works just fine to explain all of this. If not, what is your proposal?

As I understand it .Field theory was particularly used by Einstein to produce his General theory of relativity and explained Gravity as distortions in space time . The common pictorial representation of this is to show a trampoline like 3 dimensional grid , with what looks like displaced grid lines where mass is strong . ThIs is sufficient for me to say " if we want to mitigate this effect we need to equalise out any distortions that we come across . I see it more as a uniform 3 D field lattice that would be cubic strait lines ( ) , which of course are no more than similar to magnetic field lines that can be followed for direction by a mini compass , if we wish to trace them from North to South about a bar magnet. Or fields of electric charge induced by static electric , which can be followed by even human hair ( ). Now we are dealing with gravitational fields some of equal field strength and others at right angles . Not so easy to see what is going on , as we seem to know less about gravitational forces. Sufficient to say most people pick up a flavour by the sort of trampoline images bandied about.

The thrust of my investigation has been to follow the lead that "CURVATURE " seems to be fundamental to Gravity. Here also often forced motion about a curve produces lines tangential to a curve and at right angles .( ) . Also that ORTHOGONALITY rears its head . ( at 90 degrees )

Mike

Edited by Mike Smith Cosmos
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Inertia seems fundamental to the movement of mass .

So if an object with mass is moving ( by) , as part of an oscillation , from left to right . By Newtonian reasoning and acceptance , it wants to keep moving in that straight line . Unless acted upon by a force .

If a steady force , gravity acts ORTHOGANALLY , at right angles to the movement of the mentioned object , it will attempt to move push the object downward ( could be named centripetal force ) .

However due to the tremendous inertia of the object , this will be resisted , ( could be named centrifugal force ) .

As the speed of our object is 17,700 mph the sums involving these equations namely (mvsquared/r =mg) the object will levitate!

See following sketch.

Mike

Edited by Mike Smith Cosmos
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Yeesh of course inertia is fundamentally movement the two terms are synonymous. Come on your experienced in physics, learn the fundamentals.

Inertia means movement doh.

In all honesty pick up a few textbooks, some of your ideas are way misconstrued. It's not that hard, I taught myself you can too

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Yeesh of course inertia is fundamentally movement the two terms are synonymous. Come on your experienced in physics, learn the fundamentals.

Inertia means movement doh. In all honesty pick up a few textbooks, some of your ideas are way misconstrued. It's not that hard, I taught myself you can too

What is wrong with this :-

Mike

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?

If my understanding is incorrect . A satellite would not stay in orbit ! Surely ?

Mike

Edited by Mike Smith Cosmos
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I've watched your approaches to science. Pick up some textbooks. Enough said.

Impress me show me some peer reviews and mathematics coupled with observations. Then I will listen

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I've watched your approaches to science. Pick up some textbooks. Enough said. Impress me show me some peer reviews and mathematics coupled with observations. Then I will listen

But MORDRED these formulas were used in textbooks . At one stage not too long ago the balance of forces in the atoms , in text books showed electrons in orbit around the nucleus as the balance of electrostatic forces pulling in ( like gravity in my equation ) and centrifugal forces pushing out as the electrons spun in orbit around the nucleus ( like my centrifugal force ) . We have become a little blind as we have moved to energy bands . Although another way of looking at it . Surely the fundamental physics remains hidden underneath. Centrifugal forces are still there . I am sure the astronauts can vouch for that in their centrifuge training for 'G' forces .

Mike

Edited by Mike Smith Cosmos
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Kk, fair enough. However you need to study the relations of each force seperately. For example there is one unique property of gravity compared to the electroweak that I will let you research to discover. It is one of the fundamental reasons why we cannot develop a TUV as opposed to a GUV

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Kk, fair enough. However you need to study the relations of each force seperately. For example there is one unique property of gravity compared to the electroweak that I will let you research to discover. It is one of the fundamental reasons why we cannot develop a TUV as opposed to a GUV

I will have to discuss this later . ( out for the day painting)

I am sure my formulae would work with the moon in orbit around the earth . Have never done it . But I am sure the past masters did .

See you later

Mike

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Sounds good PM me or forum I will help best I can (PS I oil paint lol)

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The thrust of my investigation has been to follow the lead that "CURVATURE " seems to be fundamental to Gravity. Here also often forced motion about a curve produces lines tangential to a curve and at right angles .( ) . Also that ORTHOGONALITY rears its head . ( at 90 degrees )

Yes, gravity is explained by curvature. And curvature appears elsewhere in nature, but that does not imply that it's caused by gravity.

As the speed of our object is 17,700 mph the sums involving these equations namely (mvsquared/r =mg) the object will levitate!

No, it will orbit. Levitation implies there is (or can be) no center-of-mass motion. And there's nothing here that requires any new physics.

But MORDRED these formulas were used in textbooks . At one stage not too long ago the balance of forces in the atoms , in text books showed electrons in orbit around the nucleus as the balance of electrostatic forces pulling in ( like gravity in my equation ) and centrifugal forces pushing out as the electrons spun in orbit around the nucleus ( like my centrifugal force ) . We have become a little blind as we have moved to energy bands . Although another way of looking at it . Surely the fundamental physics remains hidden underneath. Centrifugal forces are still there . I am sure the astronauts can vouch for that in their centrifuge training for 'G' forces .

The model with electrons orbiting like planets is wrong. It makes predictions that are wrong, even though it gets the energy right. We didn't move to energy bands for the model of the atom. That's in solid-state physics, for large collections of atoms. The QM solution to the hydrogen atom still has well-defined energies for the states. It also gets the angular momentum states right and accounts for Zeeman splitting of the levels, too, which are missing from the Bohr model.

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The model with electrons orbiting like planets is wrong. It makes predictions that are wrong, even though it gets the energy right. We didn't move to energy bands for the model of the atom. That's in solid-state physics, for large collections of atoms. The QM solution to the hydrogen atom still has well-defined energies for the states. It also gets the angular momentum states right and accounts for Zeeman splitting of the levels, too, which are missing from the Bohr model.

Yes, I am aware of all that you say regarding the atomic analysis of electron models. I don't think I could compete with all the quantum mechanical analysis of atomic theory at this time. I threw it in a bit of the 20th century history development of ideas about electron orbits.

However I am more interested at this ,moment , for human scale vibrations , in partial arc motion , at world size radii.

Here then , when we plug in a figure for radius of 4000 odd miles to get to say bench height in a lab , we can set up ( with a humungous amount of mechanical engineering ) a transducer with a horizontal displacement of 4 inches ( both ends 4 inches , each in anti phase with the other end ). Power it up with a drive of 40 kHz . Stand back and watch the said transducer vibrate in a horizontal orientation " In Orbit " but here , near the Earth surface. ( if I have done my sums correctly . )

Mike

Edited by Mike Smith Cosmos
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Here then , when we plug in a figure for radius of 4000 odd miles to get to say bench height in a lab , we can set up ( with a humungous amount of mechanical engineering ) a transducer with a horizontal displacement of 4 inches ( both ends 4 inches , each in anti phase with the other end ). Power it up with a drive of 40 kHz . Stand back and watch the said transducer vibrate in a horizontal orientation " In Orbit " but here , near the Earth surface. ( if I have done my sums correctly . )

A device pushing and pulling two weights apart and together will not have a moving center of mass and thus no momentum.

It will neither orbit nor levitate, regardless of orientation, frequency, stroke length or height above Earth surface.

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Would this only take effect at that frequency and amplitude? i.e. No effect at 39.9 kHz, or no effect at 3.9" of amplitude? Or is it, as the equations imply, simply a matter of their speed, so the right combination would work, and the weight would be seen to decrease as you approach the values for levitation?

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A device pushing and pulling two weights apart and together will not have a moving center of mass and thus no momentum.It will neither orbit nor levitate, regardless of orientation, frequency, stroke length or height above Earth surface.

.

The transducer elements (2) are the main mass of the device . Any containing ( possibly vacuum casing slightly curved to the contour of the earths surface ) housing would be comparatively light compared with the two moving masses of the transducer device. This could be viewed symbolically as two masses moving in opposite directions along the ( close to earth ) orbit. They are however linked elastically( not literal) such that the masses change direction sinusoidal fashion . However because of the V squared term in the equation of motion , the centrifugal force equalising the centripetal force (gravity ) is always positive .

Thus is achieved rectification of an analogous sine wave electrically . Hence although the peak speed of mass is likely to be approx 22,000 mph , and the change direction low ' 0 ' , the RMS ( root mean squared ) speed should be approx 17,700 mph . Giving a mean force of zero ( however in orbit , at near to ground level ) . Presumably to raise or lower , would hopefully work, by more or less energy , input into the transducer system.

Would this only take effect at that frequency and amplitude? i.e. No effect at 39.9 kHz, or no effect at 3.9" of amplitude? Or is it, as the equations imply, simply a matter of their speed, so the right combination would work, and the weight would be seen to decrease as you approach the values for levitation?

.

I would believe so , as the cross explanation , just given to 'Spyman ' ( your second --"or "-- comment/ question )

The Grey tube illustrated below ( non massive yet strong ) is symbolically the container for , the device for achieving and maintaining " Near to Earth Orbit ". The key component/s and by far the dominant Mass is/are , the two masses illustrated Brown . The two grey lines joining the two masses are symbolic of the mechanism to ( a ) accelerate the opposing masses( zero to peak ) and (b) decelerate the opposing masses( peak to zero ) . This could be named transducer. It is important to note that these lines ,as with the container follow the curvature required eg, ( earth surface curvature) , in this way forces between large masses run along a tangential line . Also the orthogonal lines ( 90 degrees ) to the masses , follow the same direction as the earth radius ( gravity field lines core to far outer space ) .

The relevant physics assumptions here are : - the laws of physics as regards these masses is ' that they are subject to the same centrifugal and gravitational forces , in principle , here near the surface of Earth , as is 500 miles etc , up at conventional orbitals. Also that the forces are the same of each of both masses , irrelevant as which direction they are pointed opposed but as long as they are across the sphere of equal radius . The overarching idea is that the two masses are in some way connected and coordinated in an anti phase arrangement , oscillating at resonance ,about a short distance on the surface of the sphere ,previously mentioned.

Below with values of balancing force as they move from zero to max and zero to max ( no negative )

Mike

Separate though relevant

.--------------- appendix (A) ----------------

Appendix (a) It has also been posited here that a similar arrangement MAY exist is some way , with electron pairs . I appreciate this is another subject all together. But I mention it here , as 'one ' does need to get 'ones ' ideas from somewhere . And although the main thrust of the idea came from the surprising operation of the way I uncovered how tuning forks and resonance works , very soon in the development of the idea , electron spin and pairs came blasting into the arena . So it is difficult to totally separate them .

Edited by Mike Smith Cosmos
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Also , my reasoning on satellite motion is that the satellite does NOT need to complete its orbit for the physics to be true. It is true for a part of its orbit . Similarly in the opposite direction it would be true. Combine the two , and you have the mechanism described previously with two opposing partial arcs .

Mike

Hi Mike, good to see you posting again.

With the two masses oscillating (at orbital speed), the problem is that this is not the same as if the two masses are in orbit around the earth !?

When gravity acts on a mass in orbit, there is a vector force (gravity) towards the earth. The bit that is different for your oscillating masses to a mass in orbit, is that the oscillating masses change direction (back and forth) - and when they do, they require the base to push up. If the base did not 'push up' when the mass changes direction, then its downward direction it gains due to gravity would still be present when it reverses direction, and the mass would not be moving flat to the earth's surface, but slightly towards it. (An orbiting mass doesn't change its orbital direction - other than the action of gravity - and so it doesn't have this problem.)

A simple way to see the problem, is to have the two masses connected by a rod and to spin them in a circle (at orbital speed) - rather than oscillate. The spinning mass will not levitate.

Edited by robinpike
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Hi Mike, good to see you posting again.

When gravity acts on a mass in orbit, there is a vector force (gravity) towards the earth. The bit that is different for your oscillating masses to a mass in orbit, is that the oscillating masses change direction (back and forth) - .

But surely during the section of time 'say' when one of the masses is moving at 17,700 mph during a 'bit ' of partial arc .

Who knows where it came from ? How it got there? Where it is going in the immediate future?

Surely for that integral part of time , if the conditions are met , by the physics , :-

The mass is travelling along a tangential orbital line, it's speed is 17,700 mph , the gravity (vector force as you put it ) is centripetally acting on the mass , from a strong inertial point of view the mass would soon keep going in a strait line. Thus it will respond , re actively with a centrifugal force acting against the gravitational force . This in such a way as to ( for that integral of time at least ) , by the physics and maths calculation , it will maintain orbit , even though this is ' near ' the surface of the Earth'

Surely if we were to look at all the integral steps ( some at greater speeds , up to 22,000 mph , and down to 0 mph ) we would come up with this RMS mean of 17,700 mph that I have explained in the previous paragraph.

The pushing up surely is the reactive force ( centrifugal force ) caused by gravity attempting to push the inertial mass in toward the earths' centre. Some times ( at say 19,000 mph ) the pushing by gravity is insufficient , other times it over sufficient say 12,000 mph . Other times just right at 17,700mph. Possibly needless to say the summation of all these time integrals is what I have proposed , namely an average or mean ( root mean squared value ) always positive by dint of mvsquared /r Direction INSENSITIVE on the surface of the orbital sphere. So a change in direction is IMMATERIAL , surely .

Mike

It is the strait line inertia that causes the (centrifugal ) reaction to gravity ( centripetal ) . The compromise or ongoing calculation is an orbital curve of some sort .

Less than 17,700 mph ...downward (lower orbit ) ......Equal to 17,700 orbital ......more than 17,700 Upward ( higher orbit )

Edited by Mike Smith Cosmos
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If I am understanding your set up correctly, the problem is the oscillation.

If the mass was not oscillating, but continuing on, then if moving at the orbital speed for its height, it would (in principle) orbit the earth - since the downward direction caused by gravity matches the fall off of the surface of the earth due to the curvature of the earth.

But once you change to the oscillation set up, the small downward movement incurred due to gravity as the mass moves in one direction, on oscillating into the reverse direction, that downward movement is now 'into' the curvature of the earth - not matching it.

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