Free energy from gravitation using Newtonian Physic

[pequaide]

Three 1 kg masses moving  1.808 m/sec have 4.9 joules of energy. 1/2mv?

A one  1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum.

1 kg of overbalance placed on a two kilogram balanced wheel will develop a velocity of 1.808 m/sec for all three kilograms after being dropped .5 meters. The kilogram moving 5.424 m/sec will rise 1.5 m.

[AB Hammer]

Greetings

You will find, when it happens, that math will not have solved it but a dupication of a natural efect in nature. Then and only then will the math be figured out, and then math will be able to improve it.

This is my prediction

[Kator01]

Hallo pequaide,

I had an error in calculation ( 2 ). I wrongly  typed in the letter g instead of Epot = 4.9. But calculation was done with the value 4.9 correctly

your  statement :

"A one  1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum"

is wrong because 1 kg mass falling from height 0.5 has a calculated velocity = 3.13 m/sec and
not 3 times 1.8 m/sec.

and therefore  energy is :

Epot = m x g x h = 1kg x 9.81 x 0.5 m = [color=blue]4.9 [/color][kg m exp2/ sec exp2]   ( 1)

and not 14,71.

I would suggest you go back and study the relevant kinematic formulas in your science-books.
This is a basic condition for any further serious discussion. Otherwise we waste our time

I will stop here with any further explanations. Please come back with a clear cut explanation why you calculate velocity of mass m3 = 3 times 1.8 meter / second  falling from height 0,5 meter.
Explain in detail how you want to achieve this 5.424 m/s of mass m3.

I have given the scientific formulas to do the correct calculation of velocity of a mass m3 falling down 0.5 m.

Velocity is calculated by the formula :

velocity at bdp = square-root of Epot x 2 / m3 = of 3.13 meter/sec. (2)
                         = square-root of 4.9 x 2 / 1 kg = 3.13 m/s


Regards

Kator

[Kator01]

Hallo pequaide,

I had an error in calculation ( 2 ). I wrongly  typed in the letter g instead of Epot = 4.9. But calculation was done with the value 4.9 correctly

your  statement :

"A one  1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum"

is wrong because 1 kg mass falling from height 0.5 has a calculated velocity = 3.13 m/sec and
not 3 times 1.8 m/sec.

and therefore  energy is :

Epot = m x g x h = 1kg x 9.81 x 0.5 m = [color=blue]4.9 [/color][kg m exp2/ sec exp2]   ( 1)

and not 14,71.

I would suggest you go back and study the relevant kinematic formulas in your science-books.
This is a basic condition for any further serious discussion. Otherwise we waste our time

I will stop here with any further explanations. Please come back with a clear cut explanation why you calculate velocity of mass m3 = 3 times 1.8 meter / second  falling from height 0,5 meter.
Explain in detail how you want to achieve this 5.424 m/s of mass m3.

I have given the scientific formulas to do the correct calculation of velocity of a mass m3 falling down 0.5 m.

Velocity is calculated by the formula :

velocity at bdp = square-root of Epot x 2 / m3 = of 3.13 meter/sec. (2)
                         = square-root of 4.9 x 2 / 1 kg = 3.13 m/s


Regards

Kator

[pequaide]

1 kg of overbalance placed on a two kilogram balanced wheel will develop a velocity of 1.808 m/sec for all three kilograms after only the one kilogram of imbalance has been dropped .5 meters. You have three kilograms moving 1.808 m/sec. That is 5.424 units of momentum. Thats 3 * 1.808 m/sec. This is after only one kilogram has dropped .5 meters in a three to 1 over balanced wheel .

I have a machine that transfers the motion of several kg (three kg) into one kilogram (1 kg). If the motion of three kilograms moving 1.808 m/sec (3 * 1.808 = 5.424) is transfered to one kilogram the one kilogram must be moving 5.424 m/sec if it conserves momentum.  1 * 5.424 = 5.424 m/sec.

A one  1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy. 1/2mv?

If you don't like the overbalance wheel concept, let each 1 kg mass drop on its own. Each 1 kg will need to drop .1666 meters to be moving 1.808 m/sec. That is a momentum of (3 * 1.808) 5.424. Give that momentum to 1 kg and it will rise 1.5 meters. That is three times as high as .1666 m* 3.