Free energy from gravitation using Newtonian Physic

[The Eskimo Quinn]

For the last post, i think do to the number of threads you should have probably used a general announcement so everyone could see, I have had them sunce being on this site also, however, conspiracy theories nuts aside, if the government was ever going to activly target a website, which websites do you think that would be??? I doubt the webmaster can do too much or is responsible

[Kator01]

Hello Eskimo Quinn,

problem is already known to stefan. But it seems to consist, as another member reported trojan-attack as he was just opening a german thread.

We may have a pattern. Maybe some threads are selectively choosen to get attacked- True conspiracy - all nuts

Since some of  these viruses are based on Java-Skript simply turn off Java and Java-Sript if you read and post here

Regards

Kator

[pequaide]

The (Kogel, Leipzig) pendulum at work is about 8.5 inches (21.6 cm) in length. It has a period of a little less than a second (55 cycles per minute). It is on a bearing and the bob is square (not aerodynamic). The bob has a mass of about .8 kilograms. We don?t even keep a dust cover on the open bearing. It will swing back and forth for at least 31 minutes. That is 55 * 4 * 31 = 6840 quarter cycles. So in the first down swing the pendulum bob will have approximately 99.985 % (1/6840) of the total energy that would be available to it.

Quinn: You quote huge energy losses for pendulums. I have no idea what you are talking about. I think your racking (post145) is an excuse.   

Let?s assume that the (Kogel, Leipzig) pendulum drops .216 meters; that would give the bob a maximum low point velocity of 2.058 m/sec, but only if it was in a vacuum and if the bearing point was friction free. Air resistant is probably greater than bearing friction, but let?s give half the resistance to each. So in a vacuum the pendulum bob would be moving 2.058 m/sec * 99.9925% = 2.0578 m/sec. This means that the resistance in the first down swing of the pendulum bob is not measurable with the equipment we are using. And the resistance is insignificant. So I think this alleged 75% loss to racking (Quinn) is an excuse some are using to pretend that energy can?t be made from gravity. 00.0075% doesn?t seem like an impediment to me.

Kator01 All that is needed for a cylinder and spheres device is a horizontally mounted wheel with a center bearing (a pulley), and low friction carts. The pulley has to be wrapped with a string that then has a mass (cart) attached to the end. When the mass on the moving pulley is released the mass will unwrap the string and apply a force in the opposite direction of the motion of the wheel. If the pulley has a mass or is being pulled by a second mass, the mass of the wheel (pulley) or the second mass will stops (unless it is greatly larger than the unwrapping mass).

Let me give you this scenario. A nine kilogram block is moving in a straight line; it is followed by a tenth kilogram a short distance away. All ten kilograms are moving 1 m/sec on dry ice. The nine kilograms has a string attached that is unwrapping from a pulley or wheel, but the pulley is at first providing no resistance. The trailing kilogram is caught by the pulley and is placed on a string that is wrapped around the wheel as in paragraph one. A string from the 9 kg block applies its motion to the pulley but only the one trailing kilogram is caught in the rotation of the wheel of the pulley.  As the one kilogram swings out and unwraps the string on the pulley the nine kilograms stops.

You started with 10 kilograms moving one meter per second in a straight line, you end with one kilogram moving in a straight line, and if you choose you can direct it to travel in the same direction.

If your machine that you propose to measure the velocity of the spheres, or swinging out mass, does not comply with The Law of Conservation of Momentum then isn?t the machine worthy of the Nobel Prize in Physics?

If you start with 10 units of momentum (in the above scenario) and end with say 3.2 units then hasn?t Newton?s Three Laws of Motion been violated. These Laws require that the momentum of a closed system remain constant. If the one kilogram object that is now moving only 3.16 m/sec slams into the 9 kilograms at rest what will the new momentum of the 10 combined kilograms be, and where did the 6.84 units of momentum go?

I am not trying to be an antagonist of your work; I earnestly await your results. I am merely pointing out that if your results show that The Law of Conservation of Momentum is false then the cylinder and spheres phenomenon is a significant violation in the world of physics. If on the other hand the one kilogram object in the above scenario moves away at 10 m/sec, and complies with The Law of Conservation of Momentum, the concept would be a free energy source.

[Homer S.]

Hello pequaide,

Kator01 told me in a German thread about your setup and your ideas of interchanging momentum between different sizes fo masses.
Contrary to Kator01 I'm still not convinced to achieve any additional energy with your stated setup ideas therefore it would be nice if you may enlighten me in some open issues.

Kator01 All that is needed for a cylinder and spheres device is a horizontally mounted wheel with a center bearing (a pulley), and low friction carts. The pulley has to be wrapped with a string that then has a mass (cart) attached to the end. When the mass on the moving pulley is released the mass will unwrap the string and apply a force in the opposite direction of the motion of the wheel. If the pulley has a mass or is being pulled by a second mass, the mass of the wheel (pulley) or the second mass will stops (unless it is greatly larger than the unwrapping mass).

1. How will the cart(s) move outwards? In my opinion if there isn't any friction between cart and surface, the released cart (mass) will move tangential away with circumferential velocity and unwrap the string. Do you agree?

2. If you agree with my assumption under (1) how can the cart (mass) apply a force in the opposite direction of the motion of the wheel? Please can you explain this assumption more deeper?

3. Why do you assume that the mass of the wheel (pulley) will stop spinning after releasing the mass (cart) placed at the circumference of the wheel?  I can not see any reason for such a behavior.

I think that are key questions about a possible functionality of your stated test setup and respectively advanced ideas. Therefore I hope you can provide satisfactory answers.

Regards,
Homer

[Kator01]

Hello pequaide,

I have attached two picture which show the principe I have in mind. In pic Drillbohrer-crankhaft.jpg you
see the main principle how rotating steel-spheres attached via strings to a spindle-nut will wind up if the spindle-axis is in a fixed position. After momentum is transfered to the spheres and the cylinder has come to rest the balls wind up the spindle-nut to a position by which you directly can measure Epot = m x g x h.

If there is a gain of 200 to 300 % in energy than friction-losses -  even if they are 50 % - is not the problem here.

The second picture shows a cylinder ( red ) with a toothed rim. In this case the rim should only have two ( 2 ) tooths at which the rope spins around at momentum-transfer. Not shown in the pic are two electromagnets attached to the spindle.Only the spindle will be accellerated by a motor.These two electromagnets hold the spheres in the cylinder-holes from inside the cylinder and by doing this will take the cylinder with them - meaning the cylinder will also be accellerated. At the desired speed you cut of the supply of both the motor and the electromagnet and do a full-break at the spindle-exis. The spindle has to come to rest immediately so that the spindle-nut can wind-up vertically reaching in the end a position with Epot = m x g x h.

Due to new business-activities I will not be able to build this but with this setup you can easily proof your concept thus avoiding complicated video-strobe-measurements.

Tell me what you think about this.

Kator