Free energy from gravitation using Newtonian Physic

[Kator01]

Hello P-Motion,

yes, I understand the problem you describe. The main problems are techical ones. Once the single mass ( in our discussion m3 ) leaves the balanced system at 6 pm at v = 1.808 m/sec no momentum can be transfered by the balanced-weight momentum because all masses have the same speed.

But before I get deeper in technical discussions I would like to find in this discussion a simple way to proof the principle. I am still sceptic about the formulas although the conservation of momentum is clear.Some of the  formula-stuff is directing our attention to wrong interconnections But, as you described it : How do we transfer momentum back to m3 ?

One idea came up but I am not sure it works : lets assume m3-momentum ( including m1m2 -momentum ) will hit a spring at 6 pm in full elastic collision, compress it and then stick to it by some sort of mechanism ( detached from m1,m2 ) while the spring is locked in compressed state by another mechanism. We then have all the time to set up a test-rig where we can release the spring and measure m3 climbing the rig or directly measuring the velocity of m3 and have not to deal with more kinematic problems

The question is : does this setup deliver the answer to the momentum-transfer ? Or do we need free moving masses to do so ?

@pequaide : what you you think ?

Kator

[pequaide]

A spring is unlike gravity. Gravity has a uniform quantity of force along the entire freefall (and rise) distance; this gives you uniform acceleration (and deceleration).

In a spring: if you double the distance of compression you also double the force applied to the object performing the compression.

If you then allow the spring to unload on a less massive object, the acceleration would be very large at first. This would move the lighter object out of the compression range in a very short period of time.

Since momentum is a function of time; F = ma, a = ∆v/∆t, the quantity of momentum used to load the spring would be greater than the quantity of momentum delivered when the spring unloads.

It would be the opposite of the overbalanced wheel, because the overbalanced wheel takes longer to load than to reload (the time it takes to cast the overbalance mass back up in reverse freefall).

I think the key is not the force itself but the time over which the force acts.

In a 10kg (total mass with 1 kg overbalance and 9 kg balanced) wheel with 1kg of overbalance it takes 1.43 seconds for the one kilogram to drop one meter. In reverse freefall the 1 kg can rise 1 meter (back to the place from which it was dropped) in only .45 seconds.

This 1.4278/.4515 difference in time also gives you a proportional momentum difference of 14.007/4.429. And the energy increase is the square of this difference.  14.007/4.429 = 3.16;    3.16? = 10.00

Here is how it works: Transfer the 14.007 units of momentum that is developed by the overbalanced wheel back to the one kilogram of overbalance and it will rise 10.0 meters.  And it had only been dropped one meter.  Formula used d = ? v?/a;   v = √ (2*a*d);   d = ? at? or t = √ (2 * d / a) (these three are the same formula of course because t = v/a), F = ma was used to obtain acceleration, mv, and 1/2mv? or 1/2mt?a? because v? = a?t?.

I think the vertical and horizontal wheels would be a good choice of designs, I think this was post #77 this site (masses changed).

In the example you have 1 kg on each end (2 kg) of a one meter vertically mounted bar (the bar is very light weight, this could also be a wheel with evenly distributed mass) with a high quality center bearing.  Add an extra 1 kg to one of the end masses so that the overbalance 1 kg can rotates one half meter to the bottom. That will mean the extra mass will move from 90? to 180? (or from 3 o?clock to 6 o?clock). The acceleration rate should be one third that of gravitation 9.81* (1/3) = 3.27 m/sec/sec. This is from F = ma, a = F/m   9.81 newtons / 3 kg = 3.27 m/sec/sec

At the end of the .5 meter drop (of the extra 1 kg) all 3 kg will be moving 1.808 m/sec.  v = √ (2* .5m * 3.27 m/sec/sec) = 1.808 m/sec

As the extra (overbalanced) mass reaches the bottom release it into a light weight, .1 m radius, horizontally mounted wheel with a high quality bearing point. Now we have 2 kg moving 1.808 m/sec in a vertically mounted balanced wheel and 1 kg moving 1.808 m/sec in a horizontally mounted wheel.  Connect a string from the vertically mounted wheel to the horizontally mounted wheel so that the string is winding up on the vertical wheel and unwinding from the horizontal wheel. Now release the overbalanced mass from the horizontal wheel but keep it attached to another string that has been wrapped around the horizontal wheel (the cylinder and spheres experiment). While the overbalanced mass unwraps from the horizontal wheel it will absorb the momentum of the two 1 kg masses on the vertical wheel. This is the same phenomenon as the cylinder and spheres experiment. 

Newtonian Physics predicts that it will now be moving 1.808 * 3 = 5.42 m/sec and it will rise 1.5 m. d = ? v?/a. It was dropped only .5 m.

Add the extra mass back to the top at 90? after you have transfer 2/3 of it energy to another system. You can still start over because you have three times the original energy to work with. 

[pequaide]

Here is my definition of an overbalanced wheel.

Take a bicycle and turn it upside down, onto its seat and handle bars. The front tire of the bike is now a vertically mounted balanced wheel. Duct tape the padlock to the tire and now you have an overbalanced wheel.

If the vertically mounted balanced wheel has a mass of 9 kg and the out of balance or overbalanced mass is one kilogram then you have a 10 kilogram overbalanced wheel with an overbalance of one kilogram. The radius of rotation of all ten kilograms would be (roughly) the same. This would give you (F = ma) 9.81 newtons of force applied to 10 kilograms for a .981 m/sec/sec acceleration.  There is no need to consider lever arm length because they are all the same.

Thank you for allowing me to clarify this.

[Kator01]

Hello P-Motion,

searching for a technical solution I found the following website - unfortunately in german language.

http://doerler.gmxhome.de/

I find it helpful to visualize necessary kinematic basics in order to locate the technical problem with this approach.

Please go on the left navi-area to Impulserhaltung ( conservation of momentum ) and then to

Newtons's Wiege ( se-saw ). Here you can simulate momentum-exchange with 1,.. to 4 balls. Just change the number in the small window.
Insert number 3 and watch what happens. In this setup it is not possible to stopp all 3 masses and accelerate the left single mass to the same momentum of the tree masses swinging in from the right side.. So this is the problem to solve.

But this setup solves another problem you brought to our attention in one of your posts :

If the one weight goes around in a circle and then returns to the pendulum, it will be on the side opposite of the other 2 weights.
How would the pendulum be set up to catch it ? And how would it know when to release it ?

Choose number 1(ball) this will give you an idea how you can transfer the momentum of mass m3 leaving the wheel at 6 pm at velocity 1.8 m/s. By this you can put the horizontal wheel to a given distance away from the 6 pm - position.

The second topic is momentum-change

Left navi-area go to : Kraftstoss and then to Schau's dir an

Here you see the simulation of a car decellerated with different forces.

1) big force - very short time ( to get the car to stop )

to

3) very small force - big time  ( to get the car to stop )

I hope this helps to understand some of the basics here.

Please understand I still have not found a technical solution.



Regards

Kator

[pequaide]

Pictures: note the slit and the extention of the string. The spheres have most or all the motion, they are much lighter than the cylinder.