Inertia Machine [Archive]

IFMU

5th Jul 2007, 02:02

Jiff, reconsider using centripetal accel...
I think what Mart is saying here is you can't just look at the accel/decel term. When you accelerate the bobweights to a higher speed, there is a higher centrifugal (centripetal? I'm not smart enough to know the difference) force pulling the thing down due to the m-omega^2 thingy, and less centrifugal force pulling it up due to the lower rotational speed on top. Or whatever m is called for rotational inertia, I forget. I think this is what denies us the free lunch.
But, If I'm wrong, then I think they could program the automatic vibration control on the S92 and make the payload limited only by the structure of the floor.
-- IFMU

Graviman

5th Jul 2007, 11:40

Got it in one, IFMU.

Centripetal acceleration generates centrifugal force (not that your point wasn't perfectly clear).

500e

5th Jul 2007, 20:14

friction:oh:

Paradism

5th Jul 2007, 20:52

It all looks double dutch to me. Is this another one of the impossible machines like the perpetual motion machine? If you are interested in machines that should not work, do a search for "Howard Johnsons Magnetic Motor". He has patents for it and has demonstrated working models. In short, they are motors that have no power input, yet perform work. Scientists say it shouldn't work, but it does, allegedly.

SilsoeSid

5th Jul 2007, 22:46

Having drawn the diagram from the initial post, to try and understand it a bit more, wouldn't a device like this have to be kept in a controlled environment with strict limitations in stability of the operating environment? Surely air temp/press and indeed the equipment temperature itself, would make some significant differences in the operating ability of the machine.

Also, wouldn't any movement produce a gyroscopic force, resulting in more friction/stiction on one axis than the other?

In turn, this would produce an imbalance in the rotational ability of the axes resulting in 'Force A' being vectored at a possibly uncontrollable rate!

IFMU

6th Jul 2007, 02:04

In turn, this would produce an imbalance in the rotational ability of the axes resulting in 'Force A' being vectored at a possibly uncontrollable rate!
So, what I think you are saying is that if it doesn't actually levitate, it might still make a good sex toy.

-- IFMU

NickLappos

6th Jul 2007, 04:03

Without going to great lenghts to refute the specifics, the device described is simply another rehash of a perpetual motion machine.

The device needs only have a free body diagram drawn around it to show that no momentum escapes or enters the device, with lots of fun little masses interacting on each other. I can simply illustrate the device by sitting in a chair and grasping the seat with my hands, then lifting up with all my strength. The net lift is a tad closer to zero than the intrepid inventor would wish.

Jiff

6th Jul 2007, 10:19

Nick,

I disagree with your opinion and I would ask you to take page one to the electrical shop put two servo motors face mounted in a plate ( beam ) then add two reaction weights accelerate them to the maximun rpm acheavable in
.5 of a revolution then at that point decelerate them to a stop 180 deg later ie there start point.

Holding onto a chair with both hands is only practice for holding the pan after a curry or AMS into a G if the teabag gets it wrong.........

Jiff

Shawn Coyle

6th Jul 2007, 16:46

If I remember correctly, one of the demonstrations of the machine done several years ago was to hang it by an electrical cord, and then turn it on. The device would move to the side and slightly up and stay there as long as electrical power was applied, in disagreement with several known laws of physics.

NickLappos

8th Jul 2007, 03:28

As long as tension was on the cord, Shawn, the device could lean to one side (the weights would perhaps have an average CG somewhat displaced when they spun than when they were still.)

The device cannot invent momentum, it must gather it from somewhere and move it somewhere else. Machines like this have been designed since the beginning of time. Mr. Newton would not be amused that 500 years after he explained why things work, there is still a cult that can impress some folks with sketches and such.

Jiff

9th Jul 2007, 19:23

Nick,
The device dose not utilise momentum, during the first half of a rotation, force is created as a result of a torque reaction created by accelerating a given mass. As that mass reaches the half way point, 180 deg, force in the same direction as the first half of a rotation is produced by decelerating that mass and producing a torque reaction in the opposit direction.
Well when Mr Newton got his brains bashed out by an apple it was inertia that made him look to the stars and not momentum, but then again I would love to hear an accurate discription of the difrence between inertia and momentum.
But what you really need is a great big pair of apples to demonstrate this...
Jiff

MarcK

9th Jul 2007, 20:34

Sorry, your invention is anticipated by the "Dean Drive", for which there is considerable discussion in Analog Science Fiction magazine, around 1964, then edited by John Campbell.

forget

9th Jul 2007, 21:43

http://i21.photobucket.com/albums/b270/cumpas/Untitled-2-10.jpg

NickLappos

10th Jul 2007, 05:23

Jiff,
Unfortunately, your mis-understanding of the concepts of force, inertia and momentum need a bit of review if you think you can divorce inertia from the (non) resultant liftingforce or the time rate of change of the system's momentum. Newton would explain (has expalained) that the concepts of a lifting force, the inertia and the momentum are all locked in, in spite of your confusion.

Your diagrams can, however, be used. The reverse side will record many cribbage scores......

PS he Dean drive simply converts rotary to linear motion, it produces no force nor any lift. It is a sophisticated (impractical?) reciprocating crank equivalent.

forget

10th Jul 2007, 07:26

The primary characteristic of the Dean system is that it is a self-contained propulsion system not requiring the loss of mass.

In part from the Popular Science Magazine, Richard Dempewoiff, 1958, "Engine With Built In Wings". "Dean discovered he could develop tremendous one-directional thrust." Doing it sequentially would provide steady propulsion like no other. There would be no reaction against the earth or water or air. It would be a completely sustainable force in any direction it was pointed. It has earthly applications that were imponderable. However after receiving a mathematical report form a major university and its celebrated mathematician, it now had serious space applications also.

In part; the problem was what would happen to a 3000 thousand Ib. Load if it were attached to a 23hp. Dean Drive Unit developing 2,900 rev/min.; assuming no energy losses to friction? It would put that 30001b vehicle into zero gravity.

All other values remaining the same but increasing the hp to 150, and rpm to 4,600 we would have the minimum instantaneous vertical acceleration of that 3000 Ib vehicle to 49ft/sec. (It is a self contained propulsion generator).

Yeah, Right. :D:D:D

http://www.deanspacedrive.org/dean_drive.html

Jiff

10th Jul 2007, 14:11

Forget,

"The primary characteristic of the Dean system is that it is a self-contained propulsion system not requiring the loss of mass."

Well this would also apply to a propellor or a rotor system.

The dean system while haveing a constant speed input will convert rotating mechanical horse power into a sinusoidal linear movement with equal total thrust produced in each direction. It would appear that there is a damper above the moving beam so the peak thrust in each direction may not be the same.

Nick,

Hmmmmm cribbidge hey, looking forward to poker or a ride in a minamalistic G

Jiff

IFMU

11th Jul 2007, 01:52

Well this would also apply to a propellor or a rotor system.
Does it? I would think there is pretty good mass flow going on around a propeller or rotor system. Air goes in, air is accelerated and goes out.

Jiff,

Have you ever written down the equations of motion of the device? Or would they be hampered by their Newtonian limitations?

-- IFMU

Deemar

11th Jul 2007, 12:47

Jiff,
regardless of whether you analyse it from the momentum perspective (as Nick Lappos has done correctly) or from the consideration of forces, the net reaction averaged over an entire cycle will be 0.

There are two forces that you need to look at:

1) The reaction force on the base plate of the machine generated by the acceleration and decceleration of the weights. As you have correctly analysed, this will provide a net average force in one direction over an entire cycle.

2) The centripetal force required to keep the weights moving in a circle (this is the force of constraint provided by the arms the weights are mounted on). As the weights are travelling at a faster speed on one side of the cycle, and slower on the other side of the cycle, the average centripetal force exerted by the baseplate on the rotating masses is not zero.

These two forces will end up cancelling each other out over an entire cycle.

Daniel

212man

11th Jul 2007, 13:15

"....inertia that made him look to the stars and not momentum..."

At least try and use technical terms in their correct sense, if you want to have an informed debate!:ugh: Otherwise kindly poke off!:ok:

NickLappos

11th Jul 2007, 14:17

The torque reactors on the Space Telescope are similar to the device Jiff sketches. They use the conservation of angular momentum to simply rotate the telescope by rotating a spinning mass. Why? Because the exhaust of a reaction rocket motor would create a cloud of gas around the scope that would harm the image, and perhaps deposit on the optics. Of course, the "inertia" engine produces no thrust, it just re-points the direction of the total angular momentum of the scope system. Scroll down to "Pointing System" at this site:

http://www.gsfc.nasa.gov/gsfc/service/gallery/fact_sheets/spacesci/hst3-01/hubble_space_telescope_systems.htm

Jiff

11th Jul 2007, 20:09

212 man,

So which technical term was in the wrong sence?.

Ever understand the term dry sarcasm.....

Demar

I still do not agree with Nick's analysis however, I would like to know your understanding of the diffrence between inertia and momentum.

With regard to your second point weather you call it centrifugal or centripital force. Between 0 deg to 90 deg and 270 deg to 360 deg the force will still be in the same direction as the force produced by the reaction weights accelerating and decelerating

Nick
I agree with your hubble discription but not the no thrust part.

Regards

Jiff

NickLappos

11th Jul 2007, 20:18

Jiff,
In Newtonian Physics, your agreement is not necessary it is, however, indication of your understanding.

Jiff

12th Jul 2007, 07:49

Ouch ouch ouch

Personally I never really knew Newton however I feel I have a reasonable understanding of something that flat fith lots of flashing lights.

How about a force, phase diagram.............oh go on then you first.

Jiff

NickLappos

12th Jul 2007, 08:56

Jiff,
Remember the analogy where a boy stands on a boat and throws a rock aftward, and the boy/boat system moves forward?

What happens if the boy does everything but forgets to let go of the rock? The boy/boat system moves forward, then moves aft as the rock moves in its arc. The boat oscillates back and forth, but the CG of the total system goes nowhere.

Your elaborate system is exactly that, and even if you plot forces and phases until you turn a bit purple, the net result for a set of masses that is still attached is that all the forces must sum back to zero net force.

In your system, the boy never tosses the rock, Jiff.

Jiff

12th Jul 2007, 10:15

Nick,
Assuming that the boy dose not let go of the rock but moves the rock along the centre line of the boat in an arc, then theoretically the boat will accelerate forward while the rock is accelerating in its arc and then decelerate to a stop as the rock decelerates, but overall moving forward and stopping and not oscilating
I do not feel the system is elloborate as it lends it's self to constant speed power mechanical sourse, driving through ofset gears. An electric motor or servo per reaction weight performes well and also has excelent controllability.

Best Regards
Jiff

NickLappos

12th Jul 2007, 10:18

Jiff,
You are getting close:

The boat moves forward as the rock moves aft, then the boat moves aft as the rock moves forward. The net gain is zero. What about zero intrigues you?

Jiff

12th Jul 2007, 10:34

Nick,

After reviewing the force diagram I feel my first claim of 80% to 90% efficency is inacurate, 50% to 60% is realistic.

However you did not indicate that the rock moves back and forth, you only stated that the boy did everything but let the rock go.

The only thing about Zero that intriges me is the lack of gauges.

Regards

Jiff

IFMU

12th Jul 2007, 10:46

The boy throws, boat accellerates. Boy reaches extent of his throw, boat decellerates. Stays there. Until, in fact, he pulls his hand back to not throw again. Boy pulls arm back, boat moves back. All this assumes boy is facing aft on said boat.

If the boy has an arm 3 miles long, he can move the boat something like 3miles * mass of rock / mass of boat. As long as he only does the fake throw once.

-- IFMU

ShyTorque

12th Jul 2007, 10:57

But the boy is acting as a power supply due to stored enrgy from his breakfast. Once he becomes hungry and tired, he will stop until fed lunch and the boat will come to a halt.

(Hamsters are the same, probably worse, due to their propensity to hibernate).

Jiff

12th Jul 2007, 17:03

Deemar,

If looking at a single reaction weight your argument as stated in point #2 holds water and the vibration would be extreme for a single reaction weight.
Also with a single reaction weight force directional controlability is only acheivable from a mean perspective and therefore probbably unsuitable for a thrust producing device.
With two reaction weights force direction is exact however your second point still is reasonably significant, with the addition of two extra reaction weights your second argument loses more significance.
If you study page three you will notice that the machine is in balance therefore the centrifugal argument is almost zero if in a natural force direction produced.

Jiff

Graviman

12th Jul 2007, 17:16

Jiff, without trying to damp your enthusiasm, this device will not work - period. Essentially you are not producing any change in momentum over a complete cycle, so no thrust can be produced. This is as fundamental to physics as conservation of energy (in fact for the same reasons in Spacetime). Conservation of linear momentum comes about from the universe trying to keep the total mass centroid in the same spacial position. Gravity keeps accelerating this position downwards. Helicopters essentially throw air at the ground because gravity wants something to accelerate downwards, the helicopter powertrain makes sure this is the air and not the machine!

I don't have time for a lengthy debate about this, but in my experience Nick knows his stuff and is worth listening too...

waspy77

12th Jul 2007, 17:44

The diagrams show, as you state, a machine that will generate a linear force from rotational movement. The notation seems a little out of my experience as you seem to indicate that the centre of the 2nd moment of area is at the shaft, when it is surely nearer the weight.

Can you please elaborate the method by which you will create the torque to accelerate and decelerate the rotational motion. How will you stop this being transferred (every action has an equal and opposite reaction) to the body of the inertia machine.

NickLappos

12th Jul 2007, 19:33

waspy,
the diagrams show that the device will make linear motion, not force, from the rotary motion applied. There is no net force produced by the machine, as every mass it moves is countered by a return movement.

Another thought experiment:

Take a large rocket, and fly it into a great big bag, then tie the bag shut. Will the rocket still climb?

No, the gases blown out the back of the rocket will strike the back of the bag, and produce enough momentum capture opposite the rocket thrust to make the whole rocket-bag system just stop in the air (as the expanding gases inflate the bag so very tight that it would like to explode.)

In effect, Jiff's mechanism grabs all the stuff that was accelerated backward, and by stopping it, cancels the net force.

waspy77

12th Jul 2007, 21:34

Nick,
http://upload.wikimedia.org/math/3/2/d/32d4b39203e22b47e94d27a74be6268a.pngNewton's 2nd Law of Motion

As the velocity is increased, then a force is generated tangentially in the direction of the velocity.
As the velocity is decreased, then a force is generated tangentially in the direction opposite to the velocity.
If the acceleration is proportional to the angular position, then you end up with a variation of Harmonic Motion.
Draw tangential force lines around a circle maximum length at the bottom reducing poportionally as you approach the top, reversing directions and increasing back to the bottom. You will now see that summed around the circle there is a net force. With two running counter to each other, you will see that all lateral forces are cancelled out.
If you can disprove this mathematically then please do.

Now, the problem is that the diagram does not have the whole system bounded. Where does the torque come from that accelerates and decelerates the masses? and how is it decoupled from the body of the inertia machine (so preventing Newton's third law) which would cancel out any effect.

I have asked Jiff how he intends to do this. What he has drawn so far is basic mechanics. I'm interested to see the torque generation system that is none Newtonian, if he has an answer to that then that would be something worth seeing.

NickLappos

13th Jul 2007, 01:21

waspy,

Your second paragraph is where I started, and is the point we are both making (we agree!)

However, you are reading Newton in reverse, because the force makes the velocity change, the velocity change (actually momentum change) does not make the force.

That being said, Jiff has some studying to do, and perhaps needs to find a plug to energize his fancy vibrator.

waspy77

13th Jul 2007, 07:33

Nick, we agree practically, but not entirely
Firstly
Newton's Second law states.
The rate of change of momentum of a body is proportional to the resultant force acting on the body and is in the same direction.
In reality we feel intuitively that the force must come first, however they are instantaneous and inseparable.

In a balloon that becomes untied, pressure is used to accelerate the air in a specific direction so producing a reaction force. What came first momentum change or force?
In a jet engine fuel is burnt and gas increases in velocity so producing a reaction force. What came first the change in momentum or the force?
On a helicopter rotor disc, as the blades flap the 2nd moment of area reduces (CG moves inwards) reducing the angular momentum. A tangential lead/lag force is produced. What came first the change in angular momentum, or the force?
The answer to the above is that they both occur instantaneously.

Thought No.1
If a body is moving with constant velocity, and the mass of the body is changed instantaneously (without momentum change) then what happens to the velocity? Newton says that it will tend to increase in order to preserve momentum. If it doesn't, then a force must be present to explain the change in momentum.

Thought No.2
How does a helium balloon generate lift. Where does the change of momentum come from? No mass is "let go of" in fact within the "bag" of the helium ballon no energy changes form. However gravity is no respector of boundaries, it re-arranges the air around the outside of the bag, this in turn generates the necessary newtonian conditions. The sums don't work until you include the atmosphere and the mass of the planet.

I hypothetically put a rocket in a bag last night, I managed to get to the Sun. Then again I was in the bag with it, and the bag was 150,000,000km across ;)
You need to bound the system correctly. If Jiff's machine has a torque generation system that works by producing the reactive torque at a totally unconnected place in the universe, then his machine would allow us to move within that universe without Newton getting upset. This is the equivalent of the boy with the three mile arm.
Keep an open mind, remember,
Once the sun revolved around the Earth which was flat.
Manned heavier than air flight was impossible due to steam power not having sufficient power to weight.
Jiff's machine is possible, it just isn't practical at the moment.
He needs to find a bigger bag.

500e

13th Jul 2007, 09:50

I think we have cracked the problem.
The thread appears to have a momentum of its own NOW how do we harness it.:sad:

NickLappos

13th Jul 2007, 11:46

waspy,
It is not which came first, it is which causes the other. the force causes the motion. Period.

waspy77

13th Jul 2007, 13:00

Nick, you are wrong. period.

As you don't trust me,
http://www.grc.nasa.gov/WWW/K-12/airplane/newton2.html

"The important fact is that a force will cause a change in velocity; and likewise, a change in velocity will generate a force."

Force and momentum change are mutually dependant.
In a jet engine, the Newtonian representation is that the increase in velocity of the gas (and hence momentum) when directed in a certain direction generates the force.

Now back to the inertia machine the change in angular momentum is directed to produce a linear force. It doesn't work in an enclosed Newtonian frame of reference as at the other end of the driveshaft his third law is creating the opposite reaction.

NickLappos

13th Jul 2007, 13:31

Waspy,

The fact is that the inertia of a mass requires a force to change its velocity, so a force (like that exerted by the turbine blade on the air mass) is imparted on a mass and the reaction force is measured.

If a mass is left at constant velocity, it will not change velocity until an unbalancing force is applied (Dr. Newton, first law) thus the force is applied to change velocity. Force comes first, at least according to Newton.

waspy77

13th Jul 2007, 13:49

Nick,
So you are saying that in a jet engine, it is the turbine blades that accelerate the air?

added as more time to answer the specific point

Newton's first law is a special case of the second law, and was intended to counter the commonly held belief that all things return to a state of rest over time. It states that a "body" i.e. a point item of constant mass remains in constant motion unless acted upon by an external force.

I hate to use cliches, but take the example of a skater rotating arms outstretched. Upon bringing their arms in, their angular velocity increases without any external force. This appears to contradict Newton's first law, however upon closer inspection we see that the skater has reduced their moment of inertia, and hence to conserve angular momentum the angular velocity must increase. Within the rotating frame of reference, a force is experienced due to the apparent acceleration.

I still maintain the force and momentum change are instantaneous and inseparable. Further to this, whether or not there is a force or a momentum change depends upon which frame of reference you are observing from.

Back to the boy with the 3 mile long arm. Momentum change of an individual item is possible within the boundary of the system as long as momentum is conserved throughout the whole of the system.

NickLappos

13th Jul 2007, 16:40

Yes, precisely, the hot, flaming air in a gas turbine is decelerated (negative acceleration) by the blades, and in the process the air is both slowed and cooled. The blades then react (newton's opposing force) and produce the opposite force which is then reacted (eventually) against the rotor. The rotor smacks the outside air and accelerates it, causing the lift force.

For a pure turbojet, the burning fuel causes a temperature rise in the gas, which pushes against the combustion chamber (which is why rockets and jets have strong steel housings - to withstand and react the force against them), the chamber reacts this pressure force, and this reaction presses the hot gas out the open back end. The pressure is a force that accelerates the hot air.

waspy77

13th Jul 2007, 16:50

The gas is accelerated due to the introduction of energy into the system. Momentum is conserved because initially it accelerates in all directions. The shape of the chamber then changes the direction of the flow of the air, causing a momentum change. This is reacted by a force which pushes the engine, in accordance with the second law. In all practical terms, the change in momentum of the air has generated a force. The gas impinges on the turbine blades, reduces in velocity and hence momentum, and so once again a change in momentum has generated a force.

I'm sorry, but I still maintain that force and momentum change are inseparable, instantaneous, each indicates the presence of the other.

Graviman

13th Jul 2007, 20:12

As the velocity is increased, then a force is generated tangentially in the direction of the velocity.
As the velocity is decreased, then a force is generated tangentially in the direction opposite to the velocity.
If the acceleration is proportional to the angular position, then you end up with a variation of Harmonic Motion.
Draw tangential force lines around a circle maximum length at the bottom reducing poportionally as you approach the top, reversing directions and increasing back to the bottom. You will now see that summed around the circle there is a net force. With two running counter to each other, you will see that all lateral forces are cancelled out.
If you can disprove this mathematically then please do.

Waspy, i'm sorry to have to say that this really is just garbage. If you also considered the centrifugal force radial to the circle you would see that everything goes to producing no net momentum change over a cycle. Better still use lateral symmetry to project the system to 1D and you can see directly that reversing the direction at each end cancels any acceleration.

I don't have time to discuss the symantics of force impulse vs momentum (we engineers usually use whatever gives the quickest answer), but can just see this thread getting hung up on details.

IFMU

13th Jul 2007, 23:36

Can I get fries with my free lunch?

-- IFMU

NickLappos

14th Jul 2007, 02:05

waspy, I agree with Grav, and also respect your deep understanding of the concepts that we are discussing, so enough!

Suffice it to say the device Jiff will build will make some heat, some vibration, but no propulsive force!

file:///C:/DOCUME%7E1/HP_ADM%7E1/LOCALS%7E1/Temp/moz-screenshot.jpgfile:///C:/DOCUME%7E1/HP_ADM%7E1/LOCALS%7E1/Temp/moz-screenshot-1.jpg

Jed A1

14th Jul 2007, 03:34

If, on the downward stroke of Jiff's rotor system the rotor is consistently accelerated and on the upward stroke no change in angular momentum is achieved, then the machine will produce an upward thrust.

However, this will mean that the rotor will have to continuously accelerate to an ever increasing velocity.

Furthermore, by Newtons third law, any upward thrust would be equal and opposite (at best (less efficiency losses)) to any downward force generated by the propulsion mechanism.

Therefore there will be an upward thrust but never enough to elevate a machine containing the rotors generating the thrust.

Unless Newton is wrong or a bigger bag is used :hmm:

Jiff

14th Jul 2007, 09:04

Mr Lappos,

I have not had the opportunity to study the superb analisis which has been carried out on this theory.
To me the theory seems extremley simple and if we are not careful some one may well convince us that the:{ moon is really made of green cheese with another level of insight unless that insight is understood.
I am in the process of producing a working model and I will produce a force diagram shortly.

Jiff

Graviman

14th Jul 2007, 09:37

Jed, that's a nice theory but is still flawed. The direction still has to be reversed at each end so the ever increasing forces will still cancel integrated over each period.

However eventually you would have pumped enough energy in for surrounding spacetime to collapse drawing the entire planet to a point. In this respect your distance would be reduced, although the long abandoned destination crushed beyond recognition. Unfortunately it would now take the remaining time of the universe for you to get there! :ok:

I would rather hitch a lift in a blackhawk vs a blackhole!

Jiff, i respect your tanacity. Please build the machine, since i suspect no amount of analysis will convince you otherwise. But, when it doesn't do what you feel it should, don't be disappointed. You will still have gained insight into the immense power of conservation of momentum (and energy, which just describes linear momentum though time dimension).

waspy77

14th Jul 2007, 15:50

Graviman,
A piston starts at rest and ends at rest, net momentum change is zero,
we can't ascertain anything about the work done during the cycle from that.
What assumptions did you make when translating to 1D? Jiff has given us no indication of how the torque varies with angular position. The torque gives acceleration (and tangential force), which gives us velocity (and centrifugal force), which gives us position (and then torque).
I have used a simulation (stepping at 0.001s) where the torque is constant at 1 positive 0-π and negative π-2π. Using mass, radius (and therefore I) as 1, simplifies the numbers. Start the thing off with a slight clockwise velocity and round and round it goes. The result is a net force not surprisingly to me) in the direction π.

That is the extent to which Jiff has defined the system.

As a thought exercise,
Imagine the masses in Jiff's machine are rockets aligned fore and aft tangentially. On the accelerating side the rocket would have to burn pointing aftward to accelerate, and on the decelerating would have to burn pointing forward to decelerate. Outside of the frame of reference, the rocket has rotated and so the rocket thrust has a cyclical lateral component that varies in sign, and a longitudinal component that is also cyclical but always positive. Lateral effects are taken out by having contra-rotating pairs. In this instance I am sure all would agree that Jiff has created an extremely complex and inefficient way of producing linear force.
What we were presented at the beginning is a very simple concept.
As I stated several posts ago, the difficulty lies in the creation of the torque (or the changing of angular momentum), and managing its reaction on the body of the thing that is creating it. If the model is built using conventional electric motors, and started up in free space, I imagine it will begin to rotate on an axis 90° to the axis of rotation. Not unlike gyroscopic precession. As the rotation of the motor body creates an equal but opposite effect at the other end of the driveshaft. I've not done the maths, but this may go someway to explain the effect of the unit on the string, as the torque changes the equilibrium between the CG and the attachment point.

In ballistic calculations, I have never included the momentum change of the Earth. Within my frame of reference the behaviour of the bullet appears none Newtonian.

I have said it before, the clever bit is not what Jiff has drawn but what he hasn't, i.e. the means of generating the torque that is none Newtonian within our frame of reference.

NickLappos

15th Jul 2007, 03:58

waspy,
I have to admit, your description leaves me breathless ("...simulation (stepping at 0.001s) where the torque is constant at 1 positive 0-π and negative π-2π.")

What the heck are you saying? That you rotationally accelerate the mass in one direction, and decelerate it in another, yet produce a force?

waspy77

15th Jul 2007, 10:01

Nick, I liked your rocket in a bag as it somes up the situation nicely.

I modelled purely the behaviour that Jiff described. The net result is that a force acts (or more importantly must be reacted). If you visualise how such a motion can be produced then the situation becomes clearer.

Imagine a two stroke engine (running with just enough fuel to overcome friction). On the power stroke a force is generated and accelerates the crankshaft (0° - 180°), it then goes into compression decelerating the crankshaft (180° - 0°). If you look at the forces on the piston (assuming 0° is at the top of the stroke) the power stroke exerts a downward force created by the expanding gases, and the compression stroke exerts a downward force on the piston by the gases resisting compression.
Now look at only the crankshaft's motion (no piston or connecting rods) without assuming how it is created. You see a rotating mass that accelerates around half its cycle and decelerates around half its cycle, this matches the description that Jiff has presented to us.
In the piston engine the torque is created via the piston, and reacted via the piston. In Jiff's machine the torque is created at the driveshaft, momentum is changing and therefore a force must be acting. Unfortunately there Jiff's description ends. Jiff has put a bag around only half of his machine.

I am not aware of any means of creating the torque, as described by Jiff, without the reactive torque (and therefore a similar force) acting on the body of the machine. However my ignorance doesn't mean it isn't possible.

I think we agree as to the impracticality of Jiff's machine, but I still maintain that the original description does what it says.

NickLappos

15th Jul 2007, 14:16

waspy,
I see what you are saying - that the "analysis" is actually a partial one, of half the cycle. That explains it all.

Thanks!

waspy77

15th Jul 2007, 15:35

That's it, strictly speaking it's half a machine.
It is not unlike drawing a rotor head without anything at the other end of the driveshaft. Eventually you face the fact that somehow you have to counter the torque.

Graviman

15th Jul 2007, 22:01

Waspy, i imagine Jiff's machine driven by a turboshaft driving a mechanism to vary the rotational speed - it's primary intent is afterall a VTOL craft. Your analysis (Excel, Delphi or Fortran i presume) cannot have included the centrifugal force (Mass*Radius*(Rad/Sec)^2) acting along the radius. If it had you would see that increasing the rotational speed would negate the force in the Pi azimuthal direction over a complete cycle period. The machine will just sit there shaking, but absolutely nothing more.

For this sort of problem keep the analysis simple. In this case it can be reduced to the fact that there is no means to provide reaction. End of story.

Jiff

16th Jul 2007, 08:59

I feel this technology would work well for STOL and VTOL applications however there are many other applications.

With regard to the analysis lets remember that there are analytical packages that will tell you that it is OK to balance a 1,000,000 ton weight on a mm diameter wire that is a mile long. Before you continue your passionate affair with your package of choice remember to keep one eye above the monitor.

Have we considered an angular X,Y force plot?

Regards

Jiff

waspy77

16th Jul 2007, 09:46

My simulation is Excel for case 1 of constant torque,
and a toolset I created in F90 many years ago for case 2 of proportional torque. This was due to a need to carry out a convergent iteration for position.

I modelled a point mass in free space with the forces necessary to create the motion described in a rotating frame of reference,
I chose to use centripetal force ~v2,
and normal to that, tangential which depends on the case.
For case 1 it was constant but swapping sign at the top of the circle. (assumption)
For case 2 ~θ similar to SHM. (assumption)
I am aware of no other forces acting on the mass to create the described motion.

The tangential force, created by the torque, causes an increase in angular velocity.
The centripetal force, necessary to maintain circular motion, grows from minimum at the bottom to maximum at the top as the mass reaches its maximum angular velocity, and then reduces back to mimimum at the bottom.

Draw out all the "centrifugal forces" acting on the mass as it travels around its circular path if you wish. Or consider that if the torque magnitude is symmetrical laterally (assumption), then all the centrifugal forces in the bottom half of the circle are always smaller then their mirror in the top half, as the velocity is always greater in the top half of the cricle than the bottom. Due to the reversal of the torque, the tangential force always has a vector horizontal or above.

Graviman, with your engineering experience, can you please point to the incorrect assumption(s) or the ommission(s) in the above analysis that lead to your statement that my analysis is "garbage". I am always willing to learn from my mistakes.

In mechanical implementation. The torque acts at the driveshaft, is transferred via the rotor arm, creating acceleration. The centripetal force is provided by the rotor arm, constraining the motion to a circular path. This force must be reacted at the driveshaft end. The changing centrifugal force (in reality the reaction of the centripetal) doesn't negate anything, but in fact is the mechanism for the linear force creation.

I suggest a turboshaft is a bit of overkill, and not the best way to create varying rotational motion. I imagine that Jiff is thinking some form of electrical motor to create the torque, with polarity changing to produce the reversal. This is all assumed implementation of course.

Jiff, creating the torque is just one (and probably the least) of your problems.

1. How do you react the opposing torque on the engine body, such that it doesn't undergo an equal and opposite motion? If not reacted this would produce an equal and opposite motion. Net change at best is rotation.
2. Instinctively I suspect that the centre of mass of the rotor is towards the lump at the end. The motion would therefore be both mass and drive mechanism rotating about this centre and causing an almost self cancelling effect. How do you intend to "anchor" the driveshaft?
3. When the body starts to translate, the motion is no longer circular, and the torque would have to be varied in order to maintain the same equations of motion, as the start and end point are no longer the same place. How do you intend to calculate and compensate for this effect?

Graviman

16th Jul 2007, 11:01

Waspy, we are just going to have to agree to disagree. I am convinced your analysis is wrong, but genuinely have no time to discuss the details (big project & second degree in physics). I can only refer you to some info about circular motion, for you to check your vector sums:

http://en.wikipedia.org/wiki/Circular_motion

To me it is very simple, nothing is being reacted against. If mech worked you require no energy to keep the weights rotating, but somehow produce kinetic energy in your craft. This is just plain wrong, so i stand by my assertion that the analysis has produced a garbage result.

waspy77

16th Jul 2007, 12:45

I understand circular motion.
I understand Keppler's laws of planetary motion that indicate that as the Cp force changes (due to varying radius of rotation) the tangential vleocity changes, such that the area swept out remains constant, i.e. a change in velocity indicates a change in force.

Visualisation of this is counterintuitive, but it helps of you imagine how to create the motion (try attached to a sealed piston of gas) If you ignore friction etc. once started the system continues without additional energy, and the linear force is seen on the piston.

If the system starts to translate through space, then the work done on one side is greater than that on the other. An effect not identical, but similar to the advancing vs retreating blade on a helicopter. Hence where the energy transfer occurs.

Good luck with the 2nd degree.

ShyTorque

16th Jul 2007, 21:32

I like the concept. I think the latest development means that if I remove the balance shaft from the engine of my motorcycle it will fly, Yippee!

I always wondered why they fitted them, I previously thought it was just to reduce horizontal vibrations caused by the vertical balancing of the crankshaft assembly.

Now, where's my socket set? ;)

NickLappos

16th Jul 2007, 23:48

waspy,
Your explanation is now finally clear, and also completely wrong.

Like the downwind turn, the concept of the machine inventing a force as it is being "swept through space" is a mis-reading of what an "inertial reference frame" is.

Please go back and look again at how Galileo and Newton defined the reference frames for their laws. A translating frame does NOT change the relationship. Otherwise our 33,000 mph trip around the sun would make a big difference in your tire wear as you turned certain corners!

Your analysis, and Jiff's device, are a lot of hooey. I think I will end my attempts to teach Physics 101 to the both of you.

Shytorque, remove that balance arm and hit them with it, please!

Graviman

17th Jul 2007, 08:37

Hehehe, i was going to comment about keeping an eye out for loose screws under the machine. I like this response better!

http://en.wikipedia.org/wiki/Inertial_frame

Waspy, a lot of folks here think you have some fundamental misunderstanding. In my experience a good grasp of the fundamentals gives good intuition about these sorts of concepts. Mine tells me that it just will not work...

waspy77

17th Jul 2007, 09:12

Back in your boxes gentlemen. I have said many times that we only have half the mechanism drawn, not only that but it is drawn for the static case (i.e. the situation changes once moving).

Shy,
In your engine all the forces can be held in balance by the fact that opposite to the piston force is a force acting on the cylinder head. By the way, why do you need any form of balancing if a rotating mass can't create lateral force at any particular instant in time?

Nick,
I responded to Graviman's intuition that says that Kinetic Energy appears to be created from nowhere. Quite rightly Graviman has looked at the machine and sees that no energy can get transferred as everything is symmetrical. Yet apparently the kinetic energy of the body increases (to a stationary observer). What I have attempted to illustrate by analogy is that the asymmetry is created by the movement (change in velocity) of the body itself. I suggest that you yourself go back and consider your application of frames of reference, they must be of constant velocity in order to be inertial. If the body is in constant velocity then there is no change in KE.

Graviman,
Waspy, a lot of folks here think you have some fundamental misunderstanding.

But nobody can actually tell me what it is. They can make belittlling comments, they can talk about intuition, but they can't show me where I have applied incorrect physics, or made an assumption or ommission. That's all I'm asking for, is a technical answer to a technical problem.

I will state my fundamental intuition simply.
If a rotating mass cannot create a lateral force, why do we bother worrying about vibration?
If a rotating mass can create a lateral force, then if we can vary the rotation, then we can vary the force. If we can vary the force cyclically, we create a net force.

I also maintain that the impracticality is not in what Jiff has drawn, but in the creation of the torque, and the managing of the reaction of the forces through the point that creates the torque.

ShyTorque

17th Jul 2007, 09:54

Shy,
In your engine all the forces can be held in balance by the fact that opposite to the piston force is a force acting on the cylinder head. By the way, why do you need any form of balancing if a rotating mass can't create lateral force at any particular instant in time?

Oh dear - I think you would benefit from some engine theory, unless of course this is a very thinly diguised attempt to "wind me up" :) .

On a plain crankshaft the mass of the pistons and con-rods would cause severe out of balance rotational forces / vibration. The crankshaft is therefore shaped/forged with balance weights to counterbalance their mass - if this wasn't done the engine would shake itself to bits.

Unfortunately, the crankshaft balance weights also go through the lateral "fore and aft" positions during their 360 degree rotation and in those positions they themselves are not opposed by the up/down movement of the pistons and con-rods. This results in fore and aft vibration. For this reason, engine designers traditionally never fully balanced a crankshaft; they "left off" some balance mass as a compromise between allowing vertical and longitudinal vibrations.

This was very obvious on single cylinder or twin cylinder motorcycles up until the 1970s, where the engine was often bolted rigidly in the frame and the rider felt those out of balance forces through his most sensitive regions.

The BSA Goldstar (500 single) was a "classic" example. At idle, the bike can be seen to rock fore and aft on it's wheels as the crankshaft mass balance goes through the fore and aft positions.

These days engines are much more sophisticated and multi cylinders prevail but as far as I am aware, the basic rules of physics have never changed.

In modern engines, a balance shaft is sometimes fitted to oppose some of the lateral vibrational forces. My Honda four cylinder motorcycle has one, unless their designers and salesman have been conning me :E

waspy77

17th Jul 2007, 10:07

Thanks ShyTorque.

When you say shake itself to bits. Are we talking about linear forces being generated???

Graviman

17th Jul 2007, 14:48

Waspy, this is what bothers me: You demonstrate a fundamental misunderstanding about simple engine balancing, then proceed to try to tell us that not only should we disregard linear momentum but that physics has been wrong for years about conservation of energy. Much to my annoyance i am no longer involved in engine design/simulation/development, but if project delivery was at stake i would beat a path to well understood principles of mechanics - every time.

http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/index.htm

This Walter Lewin lecture series on mechanics is well worth watching - if only for the great entertainment value. Please review these, since it will help you gain the grounding you need to discuss such concepts.

ShyTorque

17th Jul 2007, 14:51

Waspy,
I mean vibrational forces, the important point being that they occur in one direction, then the other at the same amplitude.

Again if not, as I joked earlier, my bike would fly (or sink into the earth if I was unlucky).

waspy77

18th Jul 2007, 07:12

Please go back and look again at how Galileo and Newton defined the reference frames for their laws. A translating frame does NOT change the relationship. Otherwise our 33,000 mph trip around the sun would make a big difference in your tire wear as you turned certain corners!
A rotational frame of reference is not inertial. The rotation of the Earth does cause effects that can be observed. We need pseudo forces, centrifugal and coriolis, to make the physics work whilst on Earth. I did not say a translational reference frame created force. I tried to explain, badly obviously, that if unrestrained (i.e. allowed to translate) then we are in an accelerating frame of reference. To visualise the asymmetry I suggested the advancing/retreating concept.

You demonstrate a fundamental misunderstanding about simple engine balancing
If you say so Graviman, what I was trying to point out, is that an engine is a closed system (in essence). We can see where the torque is generated and the forces necessary, ultimately all are accounted for and the engine does not spontaneously translate. A lot of effort is made compensating for the various momentum changes and therefore eliminating the unwanted forces.

then proceed to try to tell us that not only should we disregard linear momentum but that physics has been wrong for years about conservation of energy
Again, quite the opposite. I have attempted to explain how the laws of physics govern the various apparent interactions in Jiff's machine.

I have tried to identify the bit of Jiff's machine that appears impossible. It proved difficult firstly because it is not drawn, it is part of the assumptions.

The static case works perfectly, but then again there can be assumed to be a reaction force at the driveshaft, keeping everyone happy. Once it is free to translate then the generation of torque, also implies a generation of reaction force. If not the circular motion becomes highly erratic.

I've said before, we are seeing half of a complete engine. If you analyse a rotor head and blades, the fact that you don't inculde the generation of the torque (or torque reaction) does not make it an invalid analysis, it gives highly useful results. It just leaves something to be solved elsewhere.

Patent offices are littered with designs for machines with similar principles to Jiff's. They all fail to completely bound the system. For Jiff's system the trick is being able to generate instantaneous torque, and maintain it at a known position, for others it is the moving of the mass (apparently without having to use a force to move it.) I personally favour the use of anti-matter to increase and decrease the mass throughout a cycle! :}

Graviman

18th Jul 2007, 08:54

Waspy, i reread post #87. Not knowing your background, your knowledge can only be judged by these posts - recanting newtonian physics is not a good start!

The problem in your analysis lies in the fact that you have not included the reaction to the mass tangential acceleration force in the central thrust bearing. So the forces acting at the central thrust bearing should be:
1. Motor torque to drive rotational accel/deccel.
2. Centrifugal reaction force.
3. Linear reaction to tangential accel.

Force 3 in combination with the tangential accel force is what creates the need for there to be torque 1. Rerun the analysis with this in place and you will find all the force impulse balances out over a cycle. The single rotor machine just sits there and shakes...

Physics makes you pay for everything!

IFMU

18th Jul 2007, 09:46

One test is worth 1000 expert opinions.

-- IFMU

Jiff

18th Jul 2007, 10:37

IFMU,

Could not agree more, just one teeny weenie test that involves 2, 4, or eight electric motors (Waspy77 that's were the torque comes from).

If any one asks nicely I will even publish a simple electrical diagram that controls the motors.

So then lots and lots of verbal analytical gymnastics and I am the only one to publish a diagram. As soon as a force diagram is on the thread I will post mine.

Jiff

NickLappos

18th Jul 2007, 13:52

Jiff,

Your statement, "I am the only one to publish a diagram" is as foolish as your diagram. You have the bravado about your skills that is exactly matched by your lack of understanding of freshman college physics.

Build a model, and prove Newton a fool. Please.

Graviman

18th Jul 2007, 14:59

Jiff,

Remember us skeptics will only be convinced by a sustained thrust. No videos of the machine twitching, with some overemotional response. It is fair to assume it will not lift it's own weight.

You will need a demonstrably frictionless and linear system, with a spring balance to precisely determine any thrust produced. A means of gust isolation is required - and this includes cooling fans, draughts, thermals etc. The exact energy input/output will need to be measured too, including voltage and current plotted against time.

However, be prepared to be unamazed. If this machine does not do exactly what you have hyped it up as doing, you MUST sit through the MIT freshman physics course video link posted earlier. This forum will require a signed confession saying that you recant your heresy, and wish to be cleansed of your non-newtonian beliefs. Your name will forever be darkened by the shadow of fringe practices...

waspy77

18th Jul 2007, 15:39

Graviman, I think we have always agreed that the whole machine is impossible within Newtonian physics. I have at no point recanted any of the accepted laws of physics. I've been trying to find which of the assumed mechanical interactions contravene the established laws.
Why bother? Unless I can point to where my knowledge contradicts
something, I must always accept the unknown may turn round and bite me. Nobody has quickly been able to put it simply.
CF extists in a practical world, it can be varied by applying torque, and yet to create an action we need to also create a reaction.
The forces you describe at 2 and 3 are the very forces that Jiff's description claims can cause acceleration.

Firstly, as the dynamic conditions aren't fixed, I very quickly got to the point that for circular motion the driveshaft needs to be fixed, yet this contradicts that fact that if unconstrained, the machine will now be in an accelerating frame of reference. In essence we need to be able to create a remote force that can "tether" the driveshaft when we need to, and then release it for movement. If we can do that then why not just use it
directly?
Secondly, impulse mechanics quickly bounds the solution in terms of accelerations and velocities. It may even totally preclude it itself in practical terms.
Finally I combined the first and second items and have finally found what I needed.

Jiff, your force diagrams, can you publish them in terms of Force vs Time, and not Force vs Angular position.

Thanks for a very stimulating debate, now I sleep.

NickLappos

18th Jul 2007, 16:08

Let me posit why the device, which produces absolutely no thrust, can look to "weigh" less when placed an a scale. The effect is that the vibrating machine momentarily weighs "less" when the weights are being thrown downward, and momentarily weighs "more" when they are being thrown upward. Folks get fooled by the supposed different timings of the two cycles, and the idea that varying the speed/torque upward vs downward. They are the fodder that the inventor eats for breakfast. Those of us who see the whole as a box that never has more upforce than downforce - because it never does anything that is not balanced by another - know that it just vibrates a lot. Stand on a skateboard and wave a bowling ball around. If you get very good, you can roll forward and back a lot, but you will not go in any steady direction, because you are not changing momentum.

To make a compelling case, the unscrupulous inventor selects the right scale, one that has a natural vertical frequency selected to fool the fools. If the scale responds out of phase to the frequency, you can get it to "read" less than the still weight of the device. This does not mean it will fly, only that the scale and thingy system are in cahoots dynamically to point the weight needle a bit off.

Reporters and fools will then buy tickets on a flying saucer that vibrates itself upward.

Newton laughs.

Graviman

18th Jul 2007, 18:32

Waspy, you are saying that by putting the rotating weight on a non-newtonian platform it will defy gravity?
http://www.eurocosm.com/Application/images/Cuckoo-clocks/CCK-01lg.jpg

Jiff

18th Jul 2007, 18:37

So then gentlemen, no force diagram yet...............

waspy77

18th Jul 2007, 19:41

Graviman, stop trying to argue with what you think I'm saying, and actually read my last post.

I'm not over analysing, I'm exploring my own understanding, you obviously don't want to help, but please don't misrepresent me.
Waspy, you are saying that by putting the rotating weight on a flying saucer it will defy gravity?

Anybody could have said, "Jiff, no reaction mass, no change in momentum". But like all such intellectual exercises I think it is important to understand where the misunderstanding comes from.

A force acts when on one side of the machine, greater in magnitude than it does when on the other side of the machine. This is a fact.

NickLappos

18th Jul 2007, 20:11

Waspy said, "A force acts when on one side of the machine, greater in magnitude than it does when on the other side of the machine. This is a fact."

Yes, purely from a peak magnitude, it is a fact. While the peak magnitude of the force on one side is perhaps greater, the curve shape is sharper, and the AVERAGE force is exactly the same. The somewhat lesser peak force peak on the other side has a longer duration (a softer peak) so that the average force on both sides is exactly the same. The net momentum affect (force times time) is precisely the same.

Both you and Jiff are fooled by some idea that the mass can be accelerated on one side differently than the other, all while the mass is physically tied in position. Your inability to picture this trapped mass as chained to an equality of acceleration is what has prolonged this thread. Do you think that this perpetual motion trash has never been thought of?

waspy77

18th Jul 2007, 21:11

Thank you Nick,
Not fooled, you do me an injustice. I would never have bought one.:\

This is definitely not a perpetual motion machine, to class it as such is misleading.

My continued participation in this thread was due to some people trying to explain that there were reaction forces that turn out not to be there. I wanted to try and understand what they were saying, and failed because it isn't true.

My intuition was that the dynamics should lead to elliptical motion. Any help in proving this would have been welcome. It was impulse analysis that gave me the answer I wanted. Had you have posted your latest post first :(

Jiff, I was attempting to draw up graphically, what Nick has just explained. Work out the impulse around the cycle.

Graviman

19th Jul 2007, 09:08

Waspy, my annoyance stemmed from the fact that you seemed to be introducing all sorts of complexity into this. This was why i suggested projecting the 2 rotor system to a 1d line (ie consider only vector compontent along intended line of "flight"). Linear momentum is the newtonian way of stating that the system centroid does not move. Since Ft=MV then the forces will just act to maintain the centroid position.

Your post #64 included the equationhttp://upload.wikimedia.org/math/3/2/d/32d4b39203e22b47e94d27a74be6268a.png, and you also stated in post #80
I have used a simulation (stepping at 0.001s) where the torque is constant at 1 positive 0-π and negative π-2π.
so it was fair to assume you were already integrating over time...

However, all finally resolved. The machine just sits there and shakes. :ok:

500e

19th Jul 2007, 09:39

Like our dog on firework night then.:{

NickLappos

19th Jul 2007, 10:36

waspy said, "This is definitely not a perpetual motion machine, to class it as such is misleading."

Not at all. The physics of this situation are not subject to cherry-picking which ones, and which half of the cycle to apply them. If the mythical device can produce force without changing any momentum, it can change momentum by focusing that force. Those who accept the former are agreeing that the latter is also true - and they believe they have solved the quest for perpetual motion.

Sorry if this angers you (the red-faced little emoticon says so). This is a good discussion, and you have a darn good understanding of the situation - in fact, I can't figure out where you stand on this, because you flip from an erudite discussion of F=MA into a half backed analysis of half the cycle.

Jiff

19th Jul 2007, 10:46

Still waiting.................

NickLappos

19th Jul 2007, 12:00

Jiff said, " Still waiting................."

It will take a long time for the device to fly, Jiff. Be patient!

Graviman

19th Jul 2007, 12:26

Jiff, the solution is:

When you integrate forces around the azimuth, as a function of angle, there is an apparent net force in the direction of maximum rotational speed.
However, when you integrate this force around the azimuth, as a funtion of time, the force sums to 0.0. Remember angle is a function of time.

Your machine is NoGo.

The confession need only denounce your subversive thoughts.
http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/index.htm

ShyTorque

19th Jul 2007, 13:18

Jiff, I think of your proposed machine as a bit like a rowing boat. Unfortunately, the difference is that the oar blades would be as if glued to the same piece of water. It will oscillate rather than propel.

You need to attach it to a ratchet system so once it moved in one direction it couldn't slip back to it's previous position. Wouldn't work for an aircraft though, it would all turn to rat chit.

The only way to prove / disprove your theory is to build your machine. I won't be buying shares though; I wouldn't re-mortgage your house to finance it either, if I were you.

Please film it and put it on Youtube.

Ralph Fiennes

19th Jul 2007, 13:24

Lisa Robertson is the only true Inertia Machine!

Dave_Jackson

19th Jul 2007, 17:54

For those who wish to consider this scientific discussion; at a higher level.

http://www.youtube.com/watch?v=AqnEGu8VF8Y&mode

Jiff

21st Jul 2007, 19:14

Oh the complexity of a force diagram..................

IFMU

22nd Jul 2007, 01:06

You can't expect us to do your homework for you. Scrap the napkins and get out the omega-r squareds and the f=ma's.

Or if mathematics is just a crutch for the clueless, then build it, fly it around, and we won't need a force diagram.

-- IFMU

vorticey

22nd Jul 2007, 04:11

i put 1 wheel weight of 20 grams on each of my holden comodore's wheels. one at 1200 one at 3, 6 & 9. i then went out to the local proving ground to conduct the test flight. i started of slow with a bit of vibration but pushed through feeling the exitment, i held the throttle flat to the floor boards, accelerating through terminal velocity she was all a shake, then without a word of a lie, sure enough it started to come off the ground. it was quit scarey knowing id just made history and posibly a time machine. my next test wil be all weights at 1200. stay posted.............

Jiff

23rd Jul 2007, 11:41

Vorticey,

Congratulations !!!!!!!!, I am yet to build a model, however I am thrilled to bits that yours worked.

I have spent some time refining the design and feel there is a solution to the vibration problem.

This incorporates a "Halo" gear with smaller reaction gears, which are hollowed out on one side and accelerate and decelerate as a result of the internal diameter of the “Halo” gear varying. The overall effect can be augmented by adding additional weights either to the wheel itself or attached to the wheel either directly or on an arm or via an additional geared unit.

The internal gears are held in place on a circular stator and while the halo gear is rotated the reaction gear is moved over a range equalling approximately one circumference length of a reaction wheel to enable a force direction change.

waspy77

23rd Jul 2007, 13:42

Nick, my stance on this is simple.
I think this machine is impractical, but not impossible. The red face icon indicating my frustration at people saying that I was ignoring the laws of physics, when I was trying to apply them, and understand the consequences.

I don't think this is akin to perpetual motion, as no laws of physics are broken, it needs some other form of force to achieve the motion. I suspect friction would be a good one if wanting to set it off across a surface. ShyTorque suggested a ratchet which is much the same thing. The "down" cycle needs to be reacted, which gives the "equal and opposite". The machine attempts to constrain the mass to a circular motion, which causes the mass to "drag" the machine with it. The sums show that overall momentum will be conserved. Imagine throwing a very heavy ball into the air and not letting go. Whilst you are in contact with the ground, you can react the momentum change of the ball through the Earth. Once moving though, the ball's inertia will try and drag you up with it. We are allowing the machine to vibrate in one direction only, but if it lifts then it will immediately lose the ability to react against anything, and become ballistic.

Think a bigger bag now.
Use the Earth to react the force during the "down" cycle.
Using the inertia, generate a force sufficient to accelerate the machine to escape the Earth.
Use another massive body's gravity to react the next "down" cycle.
No laws of physics broken, as the reaction forces have been generated where necessary. Numbers a bit large to contemplate though :eek:

Graviman, my posts are somewhat confusing for which I apologise, as I was attempting to find a contravention of laws of physics (people kept telling me there was one) that wasn't there. My flash of light, was when Nick put simply what I had already come to the conclusion geometrically, but couldn't quite believe.

There may come a time when this machine is actually useful. But I have to say, if we can generate and control these kinds of forces, then maybe there is a more direct way of applying them :ok:

Shawn Coyle

23rd Jul 2007, 13:50

go to:
http://jnaudin.free.fr/html/TIEwater.htm

It does work.

Shawn Coyle

23rd Jul 2007, 14:03

And here is the one with the canoe...

http://video.google.com/videoplay?docid=7426258106071720512&pl=true

waspy77

23rd Jul 2007, 14:29

I think that most have agreed now, that if another force acts too, then the machine can work, in this case the force is hydrodynamic. The water pushes back. The question is, is it practical? Could more efficient motion be produced if the motor was attached to a propeller?

NickLappos

23rd Jul 2007, 15:13

The Battle with the Scientifically Challenged is now in full swing! This is like the zombie flicks where the dead rise again and again! The silly proposition is now supported by movies!

Here is a good discussion of how the foolish are sucked in:
An oscillation thruster (also known as a stiction (http://en.wikipedia.org/wiki/Stiction) drive, internal drive or slip-stick drive) uses the motion of internal masses to create a net thrust. These thrusters include either vibrational or rotating masses, in which one portion of the cyclical motion is high-speed, and the other low-speed, or alternately high and low impulse. The result is that for some of the motion there is a high force being generated, enough to overcome friction. However on the "return stroke" the force is not high enough, and any motion occurring in the first portion is not reset. In this way the devices "steal" working mass (http://en.wikipedia.org/wiki/Working_mass) from their supporting surface, a fact that may not be apparent to casual observation.

see http://en.wikipedia.org/wiki/Reactionless_drive for the full article.

Lord, save us from this crap!

fluffy5

23rd Jul 2007, 16:57

I know the very best inertia machine, that know man made mechanic's could ever beat. My wife's mouth it just never stops it goes on and on and on and on and on and on and on and on :}

Jiff

23rd Jul 2007, 17:16

Waspy 77

"If another force acts too", seems to work OK on the flat hard surface and as a pendulum.........

I feel that the Thornson Inertial Engine produces thrust by dealing with the forces that my designs dose but with a different technique. It would be interesting to see an efficiency comparison and I intend to study that design to see if anything can be learnt from it and if so improve the design that I am working on.

In my opinion the Thornson Inertial Engine is a brilliant design and before anyone criticises it they should first understand the design, review the evidence and drop their love affair with computerised simulation that will allow a 1000000 ton weight to balance on a one millimetre diameter wire, until they remember what happens in the real world.

Nick Lappos
It’s a good job that the lord did not save us from Galileo

ShyTorque

23rd Jul 2007, 17:19

What is the point of making a machine that produces vibration in order to move, or fly (not that the latter would work)?

If there is also a requirement for friction to transmit the drive, (or a ratchet as I already mentioned), we simply could use very smooth wheels, instead.

In any event, why make a machine deliberately vibrate more than it does already for other reasons? We already have helicopters for that purpose. ;)

NickLappos

23rd Jul 2007, 18:01

Jiff,
You must be kidding, evoking Galileo's name in defense of your pablum. If Galileo were here today, he'd bend you over his knee and hit you with an oak board. He spent a lifetime trying to get dunderheads to stop believing what you are embracing!

It amazes me how folks think that the moon landings were faked and a vibratory oscillator somehow makes new physics. Jiff, exactly WHAT is your technical training and background?

HERE is another fake that you might like to espouse, it is the same mumbo-jumbo repackaged as new technology - water that burns:

http://www.youtube.com/watch?v=h6vSxR6UKFM

Mac the Knife

23rd Jul 2007, 19:19

Extraordinary claims (such as this one, which suggests that some unbalanced clockwork is somehow able to violate Newton's Third Law) demand extraordinary proofs.

I see none, either ordinary or extraordinary (I'm not impressed by bits of Tupperware drifting in someone's bathtub, or by math where half the equation is missing).

Still, it's an interesting thread, if only because it explores the borders of human credulity!

;)

Dave_Jackson

23rd Jul 2007, 19:48

Sailing boats are propelled by the movement of air. The movement of the earth's north magnetic pole is accelerating rapidly.

How about a magnetic vehicle that 'sails' the north pole? The only cost is that of slowing down the movement of the magnetic north pole. :O

To get serious for a moment and talk about helicopters;

Jiff, the owner of this thread, invoked the Lord when saying "It’s a good job that the lord did not save us from Galileo". Therefore, I take the liberty of transgressing slight from the thread to ask a very simple question.

The following question relates the current presidential helicopter.

Is the 'Jesus Nut' located in the rotor-head or in the passenger seat? :confused:

Bravo73

23rd Jul 2007, 20:25

The following question relates the current presidential helicopter.

Is the 'Jesus Nut' located in the rotor-head or in the passenger seat? :confused:

Well, thankfully, the VH-71 doesn't have a teetering head so that just leaves... ;)

NickLappos

23rd Jul 2007, 21:27

Using the ‘Beauties of Physics’ to Conquer Science Illiteracy
By CLAUDIA DREIFUS (http://topics.nytimes.com/top/reference/timestopics/people/d/claudia_dreifus/index.html?inline=nyt-per)
Published: July 17, 2007
CAMBRIDGE, Mass. — In the halls of academia, it is the rare senior professor who volunteers to teach basic science courses to undergraduates.
But Eric Mazur, the Gordon McKay Professor of Applied Physics at Harvard (http://topics.nytimes.com/top/reference/timestopics/organizations/h/harvard_university/index.html?inline=nyt-org), is driven by a passion. He wants to end science illiteracy among the nation’s college students; specifically, he strives to open them to the “great beauties of physics.”
Mazur’s own Harvard course, Physics 1b, is the kind of science class that even a literature student might love — playful, engaging, something like a trip to a science museum. Indeed, Dr. Mazur, 52, is as experimental in his classroom as he is in his research laboratory.
“It’s important to mentally engage students in what you’re teaching,” he explains. “We’re way too focused on facts and rote memorization and not on learning the process of doing science.”

balance of the article at:

http://www.nytimes.com/2007/07/17/science/17conv.html?ref=science

PS Dave, the sail works because there is a keel and some water to react against. The magnetic or solar wind sail must grab onto gravity to work. Tricky but doable.

Jiff

24th Jul 2007, 07:42

Dave Jackson,

No Dave Jackson I did not invoke the lord I was responding to Nick Lappos
"Lord, save us from this crap!”

Nick Lappos,

There are mpegs proving that a similar concept works and someone has built it and proved that it works. Hmmmmmm........
Yet nobody but nobody will produce a force diagram.

Jiff

waspy77

24th Jul 2007, 08:00

Jiff, a flat hard surface still has friction. The difference between static, and dynamic friction makes a very effective ratchet. A pendulum works because, gravity is the dominant force, and is very very close to constant throughout the stroke. If you put an "inertia machine" on a pendulum then it will produce a variable force, tune it within the cycle of the pendulum, then some very strange results will be produced. But let the pendulum run over an experimentally long time, and the net period of the pendulum would be as expected.

NickLappos

24th Jul 2007, 10:33

Jiff,
A force diagram is very very simple. Since you do not have the ability to draw one, I seriously doubt that you will understand this, but here goes:

A box floating in space: draw a line to the left that is the force produced by the magic box when it is in the peak acceleration mode (N.B. while the peak magnitude of this line is bigger, its average is the same as the other force). A line to the right for the box when it is in the deceleration mode (this line has less peak magnitude than the other one). On average, the two lines are equal length. Unless there is another force in the mix, the box doesn't move.

Now see the box on a surface, subject to gravity. In addition to the two lines above add: Weight, a force downward. The drag force, which is to the right and left, opposing motion. The drag (friction) is equal to the coefficient of friction times the weight. The drag has two magnitudes - static drag, which is larger and dynamic drag which is slightly smaller. (Note in the diagrams, weight force is not shown for clarity)

If the charlatan "scientist" chooses the situation carefully, he will select a box that has the peak force slightly higher than the dynamic drag. That way, at the peak force, the static drag is overcome and the box moves slightly left. Because the designed has chosen the right peak force to be below the threshold of static drag, the box does not move when the right force peaks. The surface has generated the opposing force, which restrains the box. The charlatan scientist never mentions that friction/drag.

http://webpages.charter.net/nlappos/inertial.jpg

scooter boy

24th Jul 2007, 11:19

Using the ‘Beauties of Physics’

As a young man I studied physics for a while but never came across any beauties :8 that I could use.
Perhaps the good professor can see things in it that I couldn't - far, far too many beards, tweed jackets, BO and bunsen burners in my recollection and absolutely no beauties in any way shape or form whatsoever.
Neither the subject nor its participants were ever sexy or beautiful.
Perhaps I was just at the wrong college, or maybe it was just the way it was being taught?

Psychology/sociology or nursing were always a better bet for beauties.:E

SB

Deemar

24th Jul 2007, 12:20

Ok, so Jiff has being crying out for it, here it is - a force diagram and some maths behind the average force over an entire cycle. The assumptions are:

1) The mass rotates through 360 degrees (2 Pi radians), and when it reaches it's original position it has the same angular velocity as when it started.

2) the forces are calculated from the point of view of the rotating mass. The force acting on the baseplate is equal and oppositte (good old Newton there).

http://picasaweb.google.com/deemar/WebImages/photo#5090733182183167090
http://i184.photobucket.com/albums/x2/deemar1/InertiaMachine.png
The maths behind the y-axis force is similar and I'll leave as an excercise for the reader (I've spent way too much time typing equations into Word tonight).

Of course this really is a hell of a lot of maths just to point out that the mass has the same momentum at the start and end of the cycle, hence the the average force over the period in question has to be 0.

Daniel

Jiff

26th Jul 2007, 07:06

Deemar,

I do not agree with you and your pic did not come out.

Nick Lappos,

My force diagram will be posted here within two days.

Jiff

BGRing

10th Aug 2007, 05:26

seeing this is an Interesting discussion. and everyone is refreshed on Centripetal and Centrifugal and torque forces . not to mention a Couple of Coupled moments.

Could some one help me on this one.

I am using my toy (DID I say toy ) Helicopter. RC.

here are the Specs.

Rotor Radius 27.5cm (oops. thats Rotor Radius, not diamiter my Boo:) )
Rotor RPM 2600
Weight at tip 5g

Given these. I am trying to get a meaningful answer.
I am Using CF = M x Vsquared over Radius.
I get a Tip speed of near 270kph
and when I get an answer using
5 x 269.547squared over 275
I get 1321.01
what dose that mean ?? IE what is the pulling power of the 5g weight on the Axis of rotation ?
is that 1.321kg or 131g or is my use of metric wrong or >>>??? ...
Please help me get a Meaningful answer

Matthew Parsons

10th Aug 2007, 05:53

BGring, lots of ways to do this. My favourite is to stick to kg, m, and s.

Convert your 270kph into 75 m/s, your 5g into 0.005kg and your 27.5cm into 0.275m.

You find the force on the 5g weight is 102.3 N (=102.3 kg m/s^2). What does that mean? Well 1 N (Newton) is about the weight of 1 apple in Earth's gravity (easy way to think of it: Newton..apple), so 102.3N is the same force as 100 apples weigh.

If you want g then you don't want a force, you want the acceleration. It's probably already been discussed in this thread, but to keep the tip weight moving in a circle it's velocity must change (in direction, not necessarily in speed). So the acceleration responsible for changing the tip weight's velocity is just the v squared over r part of the equation.

This works out to 20,454 m/s^2 or about 2085 g's !

Easy to see why balancing is so important.

BGRing

10th Aug 2007, 06:01

Post #1 Picture 2 top right diagram portion.
with the rotation velocity being Higher between points D to C than they are between Points A to B. it stands to reason that the CF is greater between the points D to C than A to B. thus . you are not getting any lift.

I remember when I was a Little Kid. Must have been 5. I ran around the house thinking I could lift myself off the ground. IE I Put one hand between my legs at front and one at back grabbed and pulled up. (:O) luckily pain set in before I either ripped my arms out of their socket or did some Damage to the Little Crown jewels. I sat down that day and Scratched my head. Must have been all those Road runner cartoons I was watching back then. (1975).

Surmise.
It is the CF at the bottom when you are allowing it to Free (Though it is not because you are allowing it to free) that counteracts any upward momentum of the Frame.

I recall a little article about using Gyros to make Anty gravity devices. Never heard any more of it. (No it is not top secret. it is just not working)

Remember in the Road runner series or any other cartoon where they had a fan on a Sail boat. or the one where they would walk a plank and nail another one at the end and then walk that one and then nail another at the end of that and walk that one and then nail another one at the end of that etc etc... just doesn't work.

My last moment of delusion was when I thought that if we had a Mag lev rail launcher on the highest mountain in Ecuador. to launch a Lifting vehicle that could then launch a rocket into orbit, but then it was done with just a lifting vehicle. and Rocket 1.

OK. here is food for your thought jiff.
what if the whole wheel was rotating on another axis. that could re direct the CF of the down side between accel and Decel. I will Leave that one to you.

<DeleteOH. Please answer my other post some one. PLEASE
:confused:Delete>

<EDIT thanks matthew. I was typing this one and didnt see your reply. thanks , scratch that last EDIT>

Dantruck

10th Aug 2007, 08:33

Oh, why all the fuss, when my “anti-gravity/purr-petual motion” device is so much more successful, doesn’t need a power source, and could easily drive the tail rotor of any conventional helicopter.
This as yet unpatented device requires one cat and a generously buttered slice of fresh, warm toast. Simply staple the toast to the cat’s back, taking care to ensure the buttered side faces uppermost, opposing gravity, when the cat stands in the normal axis, ie, with its feet glued to the workbench where, afteral, they had to be, to achieve that earlier stapling assembly technique.
Having un-glued the animal’s feet, all that remains is to charge the device by inverting it through 180 degress. Simply drop the device from a height of at least three feet for the viewer to observe how the widely held physical laws of the cat “always” landing on its feet, and the one about toast “always” landing butter side down, now come into conflict. The initial equal and opposite forces are unbalanced by the cat’s natural instinct to twist and effect a four-point landing. This supremacy over the bread’s inability to think provides the initial rotation to start the device.
Now observe as momentum maintains the direction of rotation, while the natural switching on and off of the two separate physical laws powers the device indefinitely about a horizontal axis, or at least until all the howling results in a raid by those animal protection fanatics.
In the future I hope it may also be possible to harness the perpetual levitation evident in the device as a control mechanism to prevent ground contact during hover training.
Could someone draw me a force diagram for this, please?

waspy77

10th Aug 2007, 09:06

BGRing, bit picky I know, but I think that you may have used diameter and not radius to calculate tip speed.
Matthew Parsons' calculations and explanation are spot on, but a factor of 2 out because of this. If you need to use the result then take this into account

BGRing

10th Aug 2007, 10:04

One blade is 25.3cm Blade bolt to Blade tip (Not going Diagonal, IE same distance from LE at tip). and the Bolt to Hub axis is about 2.3cm So my calculations are wrong because I used wrong length to tip.

Just so I can Check what I think I understand so far. and to Highlight a Question or two.

this is How I did It. (Now using a Rotor RPM of 2400)
2 x Pie x 27.6 = Circumference of 173.415cm
173.415 x 2400 = Distance in a Min of 416198.194cm

Previously I would times this by 60 to get per hour. then Divide by 100 to get m/h and divide by 1000 to get kph (249.718kph)

But to stick with Matthews Example, I will divide 416198.194cm/m by 60 to get cm/s and then Divide by 100 to get the working number of 69.366 m/s
Same weight at the tip of 5Grams. Or .005kg and a Radius of 0.276m

So.
Newton's is 87.167 N (= 87.167 kg m/s^2)

If you want g then you don't want a force, you want the acceleration. It's probably already been discussed in this thread, but to keep the tip weight moving in a circle it's velocity must change (in direction, not necessarily in speed). So the acceleration responsible for changing the tip weight's velocity is just the v squared over r part of the equation.

is that g as is Gravity? or grams.. ? I am guessing it is G-force as you have not implied any mass amount.

This works out to 20,454 m/s^2 or about 2085 g's !

With my new Example, I get the figure of 17433.485 m/s^2 .
Now, how did you get the next Number? (2085)

So if this is gravitation or G-force or ?. how do I use this number with my 5grams to get a pulling weight in Grams on the Blade bolt ?
is it just times. IE previous example , 2085 x 5 = 10.425kg?

Thanks in advance.
(to make it easy . I Bolded and Redden my questions. <Correct me where wrong please :) >)

waspy77

10th Aug 2007, 12:50

Sorry, your original post had rotor diameter listed as 27.5cm
If it is rotor radius that is 27.5 cm I retract my previous post.

Using your latest post I agree with your calculations that give the CF as approximately 87.17 N

(I prefer to work with angular rate which is ~ 251 rad/s
CF accn = angular rate ^2 * radius
CF accn ~ 17434 m/s^2)

Matthew Parsons

10th Aug 2007, 15:38

waspy, I should have mentioned that. Thanks. The tip speed, rotational speed and rotor length only worked if it was a radius, as it was used later in BGring's post.

BGring. The g used later on is as in G-force, which isn't a force, which unfortunately leads to more misunderstanding. 'g' is a unit of acceleration based on the gravitational pull of the earth. With non-spherical shape of the planet, the mass distribution, and the varying rotational speeds based mostly on latitude, the value of g is not constant. However, it changes so little that conventionally we use 1 g = 9.81 m/s^2. So dividing your answer of 17,433.485 by 9.81 you get 1777 g (not grams).

You can use g by itself, but to get the force you can either go back to the CF equation you started with or you can multiply g by the weight of the object. To get the weight of a 5g object, you multiply it by the acceleration due to gravity (9.81 m/s^2) and you get 0.005kg x 9.81 m/s^2 = 0.049N, or 1/20 of an apple ~one small bite.

Matthew.

BGRing

10th Aug 2007, 17:54

Thanks waspy, I changed that first post to Radius. you were correct and I thank you for Pointing it out :)

looking at you method of calculation. I understand what you are doing. you are getting the 251 by dividing the m/s by the radius. seems to come to the same answer (Similar).

Mathew. Thanks again.
here is what I think I have. is the 9.81 an exact # or an Approximation.
8.885kg
is the 9.81 because we are being lightened a little due to earth rotation?

I think dantruck has something there. Though, I think it may be easier to use two pieces of buttered bread or two cats (this would give a truly = and Opposite Force)

Dantruck

10th Aug 2007, 20:19

Interesting idea BGRing!
You won't be forgotten when I beat Jiff to the Nobel prize.:D
The 'Cat 2' version would double the howl problem, however. Meanwhile your 'two toast' version offers significant weight saving potential, I'm sure. Though we'd still need a trigger to initiate rotation. Do you think a thin film of "I Can't Believe It's Not Butter" on one of the slices would induce sufficient potential difference to achieve engine start?
Meanwhile...I am perturbed that my ideas are so popular they are already being ripped off, complete with force diagram, here...
http://en.wikipedia.org/wiki/Buttered_cat_paradox
Watch out Jiff. There's no containing brilliance in this world. And the truth will always out.:{
Dan:ok: