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

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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.

No. But if you imagine for a moment a perfectly elastic reversal after say 4 inches. . Then you have exactly the same in the opposite direction.

I have already established this process is not direction sensitive provided it is on the orbital shell.

How though you might say do you get this perfectly elastic reaction?

Provided the connection to the other mass is along the orbital shell , then the return trajectory will be along the orbital shell and thus in fact be effectively perfectly elastic ( as if there were an elastic brick wall there . )

Mike

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

Then this should be easily testable. Put an oscillator on a sensitive scale and see if the reading decreases as you ramp up the frequency and/or amplitude.

If true, it's a wonder nobody has noticed this before. You should be able easily to measure this at only a few percent of full effect.

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Then this should be easily testable. Put an oscillator on a sensitive scale and see if the reading decreases as you ramp up the frequency and/or amplitude.

If true, it's a wonder nobody has noticed this before. You should be able easily to measure this at only a few percent of full effect.

Yes I have already done a few experiments and results at very low frequency ,using a quartz crystal shock sensor and digital plotter .

Here is the apparatus , I will dig out the results & graphs !

Mike

Edited by Mike Smith Cosmos
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If true, it's a wonder nobody has noticed this before. .....

But they have noticed it . 1000 of people are working on it . The problem is it's going on all the time up there 500 miles out of sight out of mind . We have become so accustomed to it , we think this is the only way possible . All I am saying is the physics is the same down here , using two objects in opposite directions at the same time , with a clever bit of mechanical coupling , at a smaller scale .

It's here, up there , at a bigger scale, in one direction . Noticed ! And moved on .. Forgotten by many as they get on with life as is. .

IF this was solvable , what I am proposing , it would be pretty revolutionary as a means of travel !

Mike

Edited by Mike Smith Cosmos
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All I am saying is the physics is the same down here , using two objects in opposite directions at the same time , with a clever bit of mechanical coupling , at a smaller scale .

THIS is what no one has noticed (because it doesn't exist).

It's not even as if it is an original idea. It comes up on science forums fairly regularly. (There is another thread concurrent with this one suggesting it.)

It wouldn't take long for someone with a bit of mechanical skill to test this. But none of the people who claim it works do this for some reason.

So, please, go ahead: build a working model and then come back and tell us the results.

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But they have noticed it . 1000 of people are working on it . The problem is it's going on all the time up there 500 miles out of sight out of mind .

They are not oscillating back and forth like a tuning fork. It's your conjecture that the situations are equivalent, and requires evidence to support it.

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/Snip/

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.

/Snip/

Well Mike, you are wrong, there is no upward pushing force caused by gravity or generated by horizontal velocity.

It's obvious from your sketch that the object is PULLED DOWN by Earth's gravity and if Earth was flat it would hit the surface.

Basic physics says that a horizontally fired bullet will hit the ground at the same time as a dropped one.

(Neglecting air resistance and surface curvature.)

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Basic physics says that a horizontally fired bullet will hit the ground at the same time as a dropped one.

(Neglecting air resistance and surface curvature.)

I guess this is the thing that people who propose this miss: they seem to think that the object is held up, defying gravity, because of its velocity. It isn't; it is still falling continuously.

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Basic physics says that a horizontally fired bullet will hit the ground at the same time as a dropped one.

(Neglecting air resistance and surface curvature.)

And the special thing about orbits is that the surface curvature matters. The object falls but keeps perpetually missing the earth, owing to the curved surface.

There is no mystical effect and no levitation going on. No lift.

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I guess this is the thing that people who propose this miss: they seem to think that the object is held up, defying gravity, because of its velocity. It isn't; it is still falling continuously.

There appears to be a relentless drive to say : "

That " Father Christmas ...God .. & ....Centrifugal force ... Do not exist "

When I was teaching Physics in school , they had even removed. Centrifugal Force from the text books .

All this misleading information about firing a Canon off a mountain fast enough, and it just keeps falling " said Isaac Newton " I believe , is all very well if that is what you want to do ( fire a cannon ball off a mountain . ) and I am not saying that does not occur but it is misleading people into thinking centrifugal force does not come into being when we tamper , as gravity does , with strait line inertial motion.

But if you want to work at human scales . Inertia in a strait line , acted on by a gravitational force radially , producing a reactive centrifugal force is as real as most things get . Centrifugal force does exist by dint of the speed of the mass giving strait line inertia. Because that's what our dear colleague Isaac Newton said ( for what ever reason , things want to continue to move in a strait line.) Like a rugby player running for the goal mouth . Try to push him off his course , sideways , and you will feel that force in your chest .

This is the centrifugal force we are using to combat gravity . The greater the velocity v The bigger the force ( even squared ) . To say it does not exist is strictly untrue . It might be derived by motion , mass, velocity , inertia , like everything else is derived from something .

But unlike Father Christmas it does exist . ( Centrifugal force exists ... When ...bla..de...bla ) we are all in a ' when ' , nothing is very permanent .

Mike

It all goes on ' here ' :-

Both sides

I could concede you put two cannons back to back in the middle ! Just a bit cumbersome , I was thinking more like some small lightweight powerful electro magnetic pulser !

Edited by Mike Smith Cosmos
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But if you want to work at human scales . Inertia in a strait line , acted on by a gravitational force radially , producing a reactive centrifugal force is as real as most things get . Centrifugal force does exist by dint of the speed of the mass giving strait line inertia. Because that's what our dear colleague Isaac Newton said ( for what ever reason , things want to continue to move in a strait line.)

If this force exists, why do orbiting bodies not move in a straight line? Or are you just claiming that Newton was wrong?

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But if you want to work at human scales . Inertia in a strait line , acted on by a gravitational force radially , producing a reactive centrifugal force is as real as most things get . Centrifugal force does exist by dint of the speed of the mass giving strait line inertia. Because that's what our dear colleague Isaac Newton said ( for what ever reason , things want to continue to move in a strait line.) Like a rugby player running for the goal mouth . Try to push him off his course , sideways , and you will feel that force in your chest .

This is the centrifugal force we are using to combat gravity . The greater the velocity v The bigger the force ( even squared ) . To say it does not exist is strictly untrue . It might be derived by motion , mass, velocity , inertia , like everything else is derived from something .

But unlike Father Christmas it does exist . ( Centrifugal force exists ... When ...bla..de...bla ) we are all in a ' when ' , nothing is very permanent .

Mike

It all goes on ' here ' :- image.jpg

Both sides

I could concede you put two cannons back to back in the middle ! Just a bit cumbersome , I was thinking more like some small lightweight powerful electro magnetic pulser !

Mike, using the diagram as reference, when the mass is at the extreme right hand oscillation position, it will be at a slightly lower point than where it is drawn - in fact it will be on the path of the dotted line.

So at its extreme right hand oscillation position, the mass will be at a lower point to where it is drawn on the diagram.

When the oscillation then returns the movement of the mass to the left, the starting point of the mass is now slightly lower - with respect to the static point that is indicated by the meeting point of the blue line and the green line. This lowering with respect to the static crossing point of the blue line / green line is the reason why the object will fall on each oscillation - and so the oscillating mass will fall under gravity.

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If this force exists, why do orbiting bodies not move in a straight line? Or are you just claiming that Newton was wrong?

No I would not dare to say that Isaac Newton was wrong . I have his Principea it props the bed up.

I am not quite sure what you are saying , that I am saying .

I am saying that when a mass moves in a strait line . It has inertia in the direction of the strait line . When gravity acts ORTHOGANALLY or at right angles to that strait line , so as to move it towards the centre of a sphere like the earth IT will in so doing generate a force .. Centrifugal force acting in opposition . Or away from the centre. But this is physics as I know it. But it is real as is anything else. Which most things are not real anyway , they appear to be the result of something else. ( as here centrifugal force is as a result of gravity and strait line inertia being changed . I don't think this is new . I think it is how we thought in olden days .

Maybe I misunderstand what you are saying ?

Ps orbiting bodies left to there own devices would prefer to travel in a strait line . It is only the interference by gravity with its ( towards the centre or centripetal force ) force , that converts the strait line trajectory into an orbit . And in so doing generates centrifugal force . The two balance and so appear weightless. There is no resultant force left to be measured

Mike

Edited by Mike Smith Cosmos
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This is the centrifugal force we are using to combat gravity . The greater the velocity v The bigger the force ( even squared ) . To say it does not exist is strictly untrue . It might be derived by motion , mass, velocity , inertia , like everything else is derived from something .

Feel free to come back and tell us when you have a working model.

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Feel free to come back and tell us when you have a working model.

I shall by then be either going skyward or in a lunatic asylum .

Mike

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I am saying that when a mass moves in a strait line . It has inertia in the direction of the strait line . When gravity acts ORTHOGANALLY or at right angles to that strait line , so as to move it towards the centre of a sphere like the earth IT will in so doing generate a force .. Centrifugal force acting in opposition .

How big is that centrifugal force in relation to that of gravity?

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How big is that centrifugal force in relation to that of gravity?

At equilibrium, ie when a mass is in orbit then the force due to gravity . ( F= ma ) where a is the acceleration due to gravity ,namely 32 ft/sec/sec known as g .

So then the centripetal force = centrifugal force

mvsquared/r = mg

m's cancel So v squared/r = g

v squared =rg

Which the last time I worked it out

v = 17,700 mph

Mike

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Mike - if you have a test object travelling at a tangent to the earth's surface and you claim that your inward radial force (centripetal) is equal to your outward radial force (centrifugal) then there is no net force on the the object and it should go in a straight line (ie not follow a curve.) There must be a force on the the object that remains unbalanced - this you run through f=ma and you get the radial acceleration towards the centre which causes a n object to orbit

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There must be a force on the the object that remains unbalanced - this you run through f=ma

Shouldn't that be balanced? An unbalanced force in either direction would result in change in orbit.

Mike there is two aspects conservation of angular momentum and the force of gravity. Both are vector quantities where the sum of the two equals zero determines your stable orbit. Key note Angular momentum not linear in this case. This does not mean one cancels the other, only that the centrifugal force =the centripetal force, both forces remain active

Think of a pitcher throwing a baseball, the arm represents gravity, the radius of the arm swinging the angular momentum. Remove the arm(gravity) the ball flies in a straight Line path.

Edited by Mordred
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Shouldn't that be balanced? An unbalanced force in either direction would result in change in orbit.

No, you need an unbalanced force (and more specifically, a centripetal force) for a circular orbit. If some other force balanced this, you should get straight line motion.

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Yeah your right I had to go over the calcs again wasn't thinking it clearly, lol wife was distracting me..( can't believe I forgot Newtons first law,, facepalm)

Edited by Mordred
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Mike - if you have a test object travelling at a tangent to the earth's surface and you claim that your inward radial force (centripetal) is equal to your outward radial force (centrifugal) then there is no net force on the the object and it should go in a straight line (ie not follow a curve.) There must be a force on the the object that remains unbalanced - this you run through f=ma and you get the radial acceleration towards the centre which causes a n object to orbit

I have always thought of this balance of forces as being an instantaneous , or , integral ,slice of time or slice of distance around the circumference. Instantaneous and the line would be a tangent for that instance only . The next integral moment , the ingoing centripetal force due to gravity would be at a different angle to the last as it again points to the centre of the earth . This time or next instant would have a different angled tangent . The summation of all these tangents ( though individually strait ) ,the summation is a curve . At 17,700 mph this curve should be parallel to the surface . The corresponding instantaneous or integral centripetal force at each instant will be equal to the corresponding gravity induced centrifugal force at each instant.

What happens during the other parts of the cycle ,where the instantaneous velocity goes to 22,000 mph and down to zero , I can only predict a mean value hence RMS predicted 0.707 of peak which is approx 17,700 mph . I think because this is happening 40,000 times a second it should smooth out ,or shake the apparatus to destruction!

In the electrical analogue, one would have a smoothing capacitor to maintain the potential at a constant DC voltage having been rectified from a continuous sine wave AC .IN THIS CASE I AM OBTAINING ( or hoping to obtain ) a rectified v squared value of centrifugal force in a similar to full wave electrical rectification . What the capacitor equivalent is I am not sure ( inertia , momentum , angular momentum ? )

Mike

Centrifugal force waveform

Ps the centripetal force waveform would be ?..? Must be the same , but how can it be ?

Edited by Mike Smith Cosmos
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I have always thought of this balance of forces as being an instantaneous , or , integral ,slice of time or slice of distance around the circumference. Instantaneous and the line would be a tangent for that instance only . The next integral moment , the ingoing centripetal force due to gravity would be at a different angle to the last as it again points to the centre of the earth . This time or next instant would have a different angled tangent . The summation of all these tangents ( though individually strait ) ,the summation is a curve . At 17,700 mph this curve should be parallel to the surface . The corresponding instantaneous or integral centripetal force at each instant will be equal to the corresponding gravity induced centrifugal force at each instant.

So it's a circle, not straight, meaning that there is a net force. You even tacitly agree to this, when you set GMm/r2 = mv2/r That's saying the net force is a centripetal force, and the net force is gravitational. There is no other force in the problem.

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So it's a circle, not straight, meaning that there is a net force. You even tacitly agree to this, when you set GMm/r2 = mv2/r That's saying the net force is a centripetal force, and the net force is gravitational. There is no other force in the problem.

.But hang about , you have introduced another equation , ( I am not saying you should not introduce another equation) , namely the force between two masses , namely a single mass m , and the earth mass M , As well as G the universal gravitational constant , as opposed to g the acceleration due to gravity. This is just stating there is a force of attraction between two masses.

But by dint of the fact we are thundering off down range at 17,700 mph , all be it , for 1/4 of 1/40,000 of a second we invoke the other equation of motion that mg= mvsquared/r , then the centripetal force will invoke the reactive centrifugal force , thus neutralising the attraction between the earth and our little mass. m ( or at least one of them ) . The other one linked to it loosely is hurtling down range in the opposite direction similarly at 17,000 mph and invokes the exact same response ( but in the opposite direction of travel ) . If this were not the case, we or NASA would not have to send rockets down range at 17,700 mph to attain orbital velocity!

I think !

Note both forces are present , in opposition so as to neutralise , so there is no net force . Device is effectively in an oscillating bit of orbit , just above the earth surface , not 500 miles up.

So the theory is saying the device should hover in near earth orbit at bench hight. ( or perhaps 100 feet up to be on the safe side , see diagram below ) Neither going very far as a whole ,but internally the two masses are ,on average travelling at approx 17,700 mph which is escape velocity. ( That is if one could build the darn thing ! )

Mike See picture below .

Start building :

.

.

Wow !

.

Phew! Just look at it go !

Mike

Edited by Mike Smith Cosmos
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..

Note both forces are present , in opposition so as to neutralise , so there is no net force . Device is effectively in an oscillating bit of orbit , just above the earth surface , not 500 miles up.

Mike

2. "No net force" means the following: because F=ma we get 0=ma, m cannot equal zero so therefore a must equal zero. If a equals zero then the object continues in a straight line ie v is constant. This is pretty much the first lesson in a mechanics course

3. A change in direction is a change in v - ie a non-zero a , and thus a non-zero net F.

The above is pretty much axiomatic to most mechanics problems. Please confirm you are not arguing with it. Only then can we continue to explain where you are wrong about gravity

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