F.M.Chalkalis gravity system a donation!

[rickoff]

Hi folks, Hi rickoff, I went on that very same amusement ride in Wisconsin and it was somewhat difficult to get over the top then became easier and it was a square cage at each end of a large rotor bar.

I have never seen one with a cage at both ends (though it could certainly be built that way), and unless the occupants of both cages were working the device in perfect unison they would be working against each other and never get over the top. Even then it probably wouldn't be possible to achieve rotation if you have one end balancing the other out. It is the out of balance state that allows for development of leverage and inertia in both the Swinging Gym and the Chalkalis device.

The "fling sock" device that you mention is interesting, and displays similar properties. It takes almost nothing to spin it around, but it is amazing how much energy it develops, which can be seen when you let it fly.

Rick

[Matos de Matos]

In my Town we did not have such a nice standing swing that does 360 º.
It must be fun.
I found a paper about pumping a swing while standing.
http://audiophile.tam.cornell.edu/randpdf/swing.pdf
I suppose, if your rotation is forward after you stop at top, you have to be as high as possible and soon you start come down, incline forward and bend your knees as show in the paper.
David

[Peter Lindemann]

Incomplete Mathematical Model

@ Mr Lindemann
Have you reviewed his math been able to make any sense of it?

Thanks
Matt

Matt,

I'm not sure I have seen all of his math, but from what I have seen, it does not appear to me that he has a clear grasp of how to mathematically model his machine. Without this, he's guessing, for the most part.

Peter

[SkyWatcher]

Hi folks, well it's still on topic. I was mistaken, it's called a 'Fling Sock'.

Throws are effortless because centrifugal force magnifies your throwing power

YouTube - Fun-Attic FlingSock - Fun For Everyone
peace love light
Tyson

[Matthew Jones]

Thanks Rick and Peter.

Matt

[rickoff]

In my Town we did not have such a nice standing swing that does 360 º.
It must be fun.
I found a paper about pumping a swing while standing.
http://audiophile.tam.cornell.edu/randpdf/swing.pdf
I suppose, if your rotation is forward after you stop at top, you have to be as high as possible and soon you start come down, incline forward and bend your knees as show in the paper.
David

The diagram is somewhat similar in that any time the cage rotation is moving downward you are bending your knees, and extending your body when moving upwards, and this is what gives you the feel in your hands and arms of pulling the cage down or lifting it up. Your body weight should always be distributed as close to the arc perimeter as is possible, and so you must shift from one side of the cage to the other at both the top and the bottom of the rotation.

In the video linked here (same as in post #75) the two men are standing on opposite sides of the cage, while the woman and child are standing on just one side. Later, the men work together on one side. In all cases, there is no weight shifting from side to side at the required timings, and this is why they fail to reach, or go over, the top.
YouTube - Dennis Barber Swinging Gyms Hanbury Steam Rally Stoke Prior Fair 19th September 2009

[Cloxxki]

Hi folks, well it's still on topic. I was mistaken, it's called a 'Fling Sock'.

YouTube - Fun-Attic FlingSock - Fun For Everyone
peace love light
Tyson

Not sure I see a gain there from CF. I do believe it helps focus strength. The arm in the spin-up seeks the direction of greatest resistance. Real muscle strength is required, and pretty quick frequency, just little stroke. An efficient way to get speed into the ball with slow long range muscles. Strong short range muscle does the trick.

[Ted Ewert]

Possible math solution

I’ve been thinking about the math concerning the Chalkalis device. I’m no mathematician but this seems fairly simple to me.
I looked up the formula for acceleration of a falling object for a given distance: V = square root(2GY). Where G = 9.8 m/sec^2 and Y is the distance in meters.
I found this number to be a constant for a given distance regardless of the initial velocity. This formula says that an object will be accelerated 14 m/sec for every 10 meters it travels. That means if it’s falling at 1000 m/sec, in 10 more meters it will be traveling at 1014 m/sec.
I assume this also holds true for a pendulum, since it is a mass falling through a set distance. If you take the pendulum to the top of it’s arc, and let go, it will reach a certain velocity at the bottom of the arc. This is the velocity gravity adds each time it swings freely between the top and the bottom of it’s arc.
For example: If the velocity of the pendulum is measured at 5 m/sec at the bottom of the arc, and it started out at zero on the top, then this pendulum will always be accelerated by 5 m/sec every time it falls through that distance.
The kinetic energy for a 5 kg weight traveling at 5 m/s is approximately 63 joules. If I add another 5 m/s of velocity to the pendulum, it will cost me 63 joules of energy. Now the pendulum is traveling at 10 m/sec when it hits the bottom of the arc. The kinetic energy of a 5 kg mass traveling at 10 m/sec is 250 joules. If I subtract the 63 joules I put in I get 187 joules. If I leave 70 joules so the pendulum can get back to the top of the arc again, it means that I can use 117 joules of energy in a load.
This means for an investment of 63 joules, I not only get that investment back but I also get 117 joules free and clear.
Another example: If I add 1 m/sec to the pendulum it will cost me 2.5 joules. 5 + 1 = 6 m/sec, which equals 90 joules. Subtract the 70 to get back to the top and we have 20 left over.
This squares with the claims of Marjanovic and Milkovic for his oscillator, and for Chalkalis with his rotating pendulum.
Another point to consider is where the energy is added. Like Milkovic suggests, and Chalkalis has done, the top of the descending arc is the place to accelerate the pendulum. Gravity apparently multiplies whatever energy is injected at the start of the downward swing.
Please check my math. Like I said, I’m no math wizard but this makes logical sense to me.

Cheers,

Ted

[Inquorate]

I’ve been thinking about the math concerning the Chalkalis device. I’m no mathematician but this seems fairly simple to me.
I looked up the formula for acceleration of a falling object for a given distance: V = square root(2GY). Where G = 9.8 m/sec^2 and Y is the distance in meters.
I found this number to be a constant for a given distance regardless of the initial velocity. This formula says that an object will be accelerated 14 m/sec for every 10 meters it travels. That means if it’s falling at 1000 m/sec, in 10 more meters it will be traveling at 1014 m/sec.
I assume this also holds true for a pendulum, since it is a mass falling through a set distance. If you take the pendulum to the top of it’s arc, and let go, it will reach a certain velocity at the bottom of the arc. This is the velocity gravity adds each time it swings freely between the top and the bottom of it’s arc.
For example: If the velocity of the pendulum is measured at 5 m/sec at the bottom of the arc, and it started out at zero on the top, then this pendulum will always be accelerated by 5 m/sec every time it falls through that distance.
The kinetic energy for a 5 kg weight traveling at 5 m/s is approximately 63 joules. If I add another 5 m/s of velocity to the pendulum, it will cost me 63 joules of energy. Now the pendulum is traveling at 10 m/sec when it hits the bottom of the arc. The kinetic energy of a 5 kg mass traveling at 10 m/sec is 250 joules. If I subtract the 63 joules I put in I get 187 joules. If I leave 70 joules so the pendulum can get back to the top of the arc again, it means that I can use 117 joules of energy in a load.
This means for an investment of 63 joules, I not only get that investment back but I also get 117 joules free and clear.
Another example: If I add 1 m/sec to the pendulum it will cost me 2.5 joules. 5 + 1 = 6 m/sec, which equals 90 joules. Subtract the 70 to get back to the top and we have 20 left over.
This squares with the claims of Marjanovic and Milkovic for his oscillator, and for Chalkalis with his rotating pendulum.
Another point to consider is where the energy is added. Like Milkovic suggests, and Chalkalis has done, the top of the descending arc is the place to accelerate the pendulum. Gravity apparently multiplies whatever energy is injected at the start of the downward swing.
Please check my math. Like I said, I’m no math wizard but this makes logical sense to me.

Cheers,

Ted

Can't see any problems, but then i'm no mathematician

[rickoff]

I’ve been thinking about the math concerning the Chalkalis device. I’m no mathematician but this seems fairly simple to me.
I looked up the formula for acceleration of a falling object for a given distance: V = square root(2GY). Where G = 9.8 m/sec^2 and Y is the distance in meters.
I found this number to be a constant for a given distance regardless of the initial velocity. This formula says that an object will be accelerated 14 m/sec for every 10 meters it travels. That means if it’s falling at 1000 m/sec, in 10 more meters it will be traveling at 1014 m/sec.
I assume this also holds true for a pendulum, since it is a mass falling through a set distance.

Hi Ted,

A pendulum acts quite differently than a free falling object. The swing of a clock pendulum, for example, from one side to the other and back again to the starting point (tic-toc), is termed the pendulum's period. In a standard pendulum clock, the length of the pendulum would be adjusted to have a period of 1 second. The starting height, given as the arc degrees above straight down vertical, is termed the amplitude. Increasing the weight has no effect upon the pendulum's swing period, and increasing the amplitude has almost no effect on the period either. Increasing the amplitude merely increases the average speed (in arc inches per second) of the oscillating mass. The length of the pendulum (from the pivot point to the center of the suspended mass) is the critical factor that determines the duration of one period, or cycle. The square of a pendulum's period varies directly with its length. A 9ft pendulum has a period 3 times longer than a 1ft pendulum. Now let's say that we lift the mass from bottom dead center (180 degrees of circular arc) to top dead center (0 degrees), give it a slight nudge clockwise and let it go. The distance that the 1ft pendulum mass travels to reach the 180 degree mark will be 37.71 inches. The 9ft pendulum would travel 9 times that arc distance, or 339.43 inches. So even though it takes the 9ft pendulum 3 times as long as the 1ft pendulum to swing 180 degrees, it covers 3 times more arc inches per second than the 1 ft pendulum.

Rick

[Cloxxki]

I see a possible problem Ted.
If you are falling 1000m/s, subject to gravity 9.8m/ss, 10m down, you will have been subjected to gravity for 1/100th of a second. In that time frame, you will accelerate 1/100th of 9.8m/s. So you speed is rather ~1000.1m/s 10m down.

A falling object, each second, increases its AVERAGE speed from release by 5m/s. Each consecutive second, 10m MORE is added.

With G=10m/s:
Time Avg Dist Dist Chang
1s - 5m/s - 5m - 5m added
2s - 10/s - 20m - 15m added
3s - 15m/s - 45m - 25m added
4s - 20m/s - 80m - 35m added

Sorry if I misinterpreted your calculations or explanation. I will go over the rest of it later when I have sufficient time.

[elias]

I’ve been thinking about the math concerning the Chalkalis device. I’m no mathematician but this seems fairly simple to me.
I looked up the formula for acceleration of a falling object for a given distance: V = square root(2GY). Where G = 9.8 m/sec^2 and Y is the distance in meters.
I found this number to be a constant for a given distance regardless of the initial velocity. This formula says that an object will be accelerated 14 m/sec for every 10 meters it travels. That means if it’s falling at 1000 m/sec, in 10 more meters it will be traveling at 1014 m/sec.
I assume this also holds true for a pendulum, since it is a mass falling through a set distance. If you take the pendulum to the top of it’s arc, and let go, it will reach a certain velocity at the bottom of the arc. This is the velocity gravity adds each time it swings freely between the top and the bottom of it’s arc.
For example: If the velocity of the pendulum is measured at 5 m/sec at the bottom of the arc, and it started out at zero on the top, then this pendulum will always be accelerated by 5 m/sec every time it falls through that distance.
The kinetic energy for a 5 kg weight traveling at 5 m/s is approximately 63 joules. If I add another 5 m/s of velocity to the pendulum, it will cost me 63 joules of energy. Now the pendulum is traveling at 10 m/sec when it hits the bottom of the arc. The kinetic energy of a 5 kg mass traveling at 10 m/sec is 250 joules. If I subtract the 63 joules I put in I get 187 joules. If I leave 70 joules so the pendulum can get back to the top of the arc again, it means that I can use 117 joules of energy in a load.
This means for an investment of 63 joules, I not only get that investment back but I also get 117 joules free and clear.
Another example: If I add 1 m/sec to the pendulum it will cost me 2.5 joules. 5 + 1 = 6 m/sec, which equals 90 joules. Subtract the 70 to get back to the top and we have 20 left over.
This squares with the claims of Marjanovic and Milkovic for his oscillator, and for Chalkalis with his rotating pendulum.
Another point to consider is where the energy is added. Like Milkovic suggests, and Chalkalis has done, the top of the descending arc is the place to accelerate the pendulum. Gravity apparently multiplies whatever energy is injected at the start of the downward swing.
Please check my math. Like I said, I’m no math wizard but this makes logical sense to me.

Cheers,

Ted

Hi Ted,
I enjoy reading your posts. Your explanation makes sense.
Let me put it this way:
GRAVITY IS THE FREE INERTIAL PROPULSION OF THE UNIVERSE.
It just exerts PURE force.
The secret that this device is using is the fact that a pendulum loses and gains VELOCITY as it rotates, so what does this mean?
It means that we need to apply F_a to the device when the speed is the least, and let gravity add its energy on the down fall. So consider this:
the pendulum is currently rotating and it is oscillating between V1 and V2 (V1 < V2)

GOAL:
We want to increase the velocity of the pendulum V units so that the speed oscillates between (V1+V) and (V2+V).

We need to apply a Force to do this, but depending on the speed of the pendulum, the energy required is different.


1- The energy required to do this is minimum when the speed of the pendulum is at its minimum (V1):

E_min = 0.5m((V1+V)^2 - V1^2) = 0.5m(2V1*V + V^2) = m*V1*V + 0.5m*V^2.

2- When we apply our energy to increase the speed when the speed of the pendulum is at its maximum:

E_max = 0.5m((V2+V)^2 - V2^2) = 0.5m(2V2*V + V^2) = m*V2*V + 0.5m*V^2.


Now clearly the result is a pendulum rotating with a velocity oscillating between V1+V and V2 + V and is the same but the applied energy is different.

Example:
Our 1kg pendulum velocity is oscillating between 1m/s and 2m/s
We want to make it oscillate between 2m/s and 3m/s (increasing the speed by 1m/s)
E_min = 1 * 1 * 1 + 0.5 * 1 * 1 = 1.5J
E_max = 1 * 2 * 1 + 0.5 * 1 * 4 = 4J
The difference is HUGE! (more than 2.5 times less energy is required if we kick where the pendulum has the least energy)

So as the pendulum moves into the 1 o' clock position we give it a kick, thus increasing the speed and gravity adds to it until it reaches the bottom.

This means that we must draw energy from the rotating pendulum when the pendulum is at its maximum speed, otherwise we will lose everything we applied to the system. So we need to apply our force at the top and extract energy at the bottom! Thus harvesting the FREE GRAVITY.

In order to be able to get more energy from gravity in each rotation, we need to increase the pendulum length and mass.

I really hope that it works as depicted!

Thanks for your attention.

Elias

[Cloxxki]

@Elias,
Very nice post indeed, thank you.
I think I am partly understanding your explanation, and to that point I cannot but agree. However, I fear part of your example may be incorrect through assuming constant 1m/s input per down revolution.
If you want a pendulum which first oscillated between 1 and 2m/s, to do so between 2 and 3m/s, you'll need to either increase the radius, or reduce G.

What we should not forget, is that IF we add force to the pendulum at minimum speed to start the descend quicker, we will also receive A SMALLER VELOCITY INPUT from gravity itself, for the following half rotation. This is, because the weights are subjected to gravity for a shorter duration, same distance.
The higher velocity makes the pendulum travel from the end of the drive pulse to the bottom of the cycle, in less time. Gravity works on masses by time unit. Distance in this setup, is limited. Higher speed, shorter time.
I may hope there is a gain to be had in the way distance, velocity and time relate. It's certainly something that's been bugging my my whole life.

Let's set up a mathematical setup, first with a single weight pendulum.

I'll propose:
radius = 1m
mass = 1kg
rpm : at 12:00, calculate both 0rpm and 1000rpm
we'll inject once, 1J at 12:00
and calculate how much we'll have at 6:00, with or without the input.
Also, we could see what happens in we extract 0.5J or 1.0J at 6:00, to the next 12:00 pass KE.

Someone please comment on this simple math experiment so we can agree on what to look for.

Thanks,

J

[elias]

@Elias,
Very nice post indeed, thank you.
I think I am partly understanding your explanation, and to that point I cannot but agree. However, I fear part of your example may be incorrect through assuming constant 1m/s input per down revolution.
If you want a pendulum which first oscillated between 1 and 2m/s, to do so between 2 and 3m/s, you'll need to either increase the radius, or reduce G.

What we should not forget, is that IF we add force to the pendulum at minimum speed to start the descend quicker, we will also receive A SMALLER VELOCITY INPUT from gravity itself, for the following half rotation. This is, because the weights are subjected to gravity for a shorter duration, same distance.
The higher velocity makes the pendulum travel from the end of the drive pulse to the bottom of the cycle, in less time. Gravity works on masses by time unit. Distance in this setup, is limited. Higher speed, shorter time.
I may hope there is a gain to be had in the way distance, velocity and time relate. It's certainly something that's been bugging my my whole life.

Let's set up a mathematical setup, first with a single weight pendulum.

I'll propose:
radius = 1m
mass = 1kg
rpm : at 12:00, calculate both 0rpm and 1000rpm
we'll inject once, 1J at 12:00
and calculate how much we'll have at 6:00, with or without the input.
Also, we could see what happens in we extract 0.5J or 1.0J at 6:00, to the next 12:00 pass KE.

Someone please comment on this simple math experiment so we can agree on what to look for.

Thanks,

J

Hi
You are completely right cloxxki. That was a misleading example.
I need to find some time to go through much thorough calculations.

Elias

[BenTec]

I fail to get the point of this device.



First off, you loose energy while getting to the desired RPM, you try to recover invested energy on a recovery wheel that will lower the rpm again... and then it looses energy when the weight has to go up vs gravity. And that device is full of friction it has 3 freaking bearings, pendulum vs air! And you cant use magnetic bearings on the pendulum...

Why not directly have a huge Flywheel in vacuum with magnetic bearings and delivering pulsed energy to it directly makes much more sense to me.

Of course its interesting, but at the moment If you put 2 of those together on a car you got a jumping car...

Ben