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Author Topic: Free energy from gravitation using Newtonian Physic  (Read 98106 times)

pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #15 on: April 25, 2007, 04:00:00 AM »

I am duplicating my current model so that I have an extra for demonstration purposes. I will write a step by step procedure as if we were starting up manufacturing. It will take a while, but here is what you need for materials.

About a 12 inch length of 3 inch I.D. PVC pipe, schedule 40, that is with a ? inch side wall.

A PVC pipe coupler for the 3 in. pipe, they have a 3.5 in. I.D. and a 4 in. pipe will fit on the outside of the coupler.

About a 4 in. length of 4 in. I.D. PVC pipe schedule 40. (not in the picture)

Two steel spheres WLS4480-20E Sargent ?Welch; These spheres are drilled, they are much cheaper if you can fix non-drilled spheres to a string,

I make wire loops to fit inside the drilled hole in the spheres, the hole is larger on one side than the other, this will stop the wire crimp half way through.
I use fishing wire and crimps, I will check this: my source is old.

Then I use fishing clips and 30 lb fluorocarbon fishing line, to connect it through the holes to the center.

I will submit a detailed description in the not to distant future. Thanks for your help. These machines have energy increases of over 300%. 

supersam

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Re: Free energy from gravitation using Newtonian Physic
« Reply #16 on: April 26, 2007, 01:06:14 AM »
pequaide,

i've got all that on hand, except for the steel sphaere's.  the ones i saw in the photos looked to be about 3/4" steel balls.  is this about right?  if so the bulldozer guy on my jobsite had to replace something the other day and i know where maybe a hundred of these are laying on the ground getting ready to be buried or salvaged.  i also know a machinist that can drill me any kind of hole imaginable for free.  so i guess i am sitting on go right now. can't wait for construction details to replicate.

lol
sam

pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #17 on: April 26, 2007, 04:24:41 AM »
The spheres are 1 inch in diameter; they have a mass of 66g, they have a 1/16 in. hole through the center and it is enlarged to 1/8 in. half way through.

If I did the math correctly a ? inch sphere would only have a mass of 28g. They probably would not stop the cylinder.

You could buy 1 inch spheres from MSC industrial supplies and have your friend drill the holes. They are cheaper if you can buy them solid and get them drilled (free).  I think I can have the details tomorrow.

supersam

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Re: Free energy from gravitation using Newtonian Physic
« Reply #18 on: April 26, 2007, 04:23:16 PM »
pequaid,

the 66gram figure seems to be the most important part to me. ??? 8)  is that 66grams after or before the drilling takes place.  i can do it either way.  1" spheres, no problem.

lol
sam ??? 8)

pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #19 on: April 27, 2007, 03:49:47 AM »
If you are about 66 grams (each) you will be fine, you will have to add or subtract the added mass of the pipe, no problem. Most important is the placment of the holes.

Materials list; for models with two masses added (3 in. pipe and 4 in. pipe)

1.   A length of 3 in. I.D. PVC pipe (about 10 in. per machine)
2.   A 3 in. PVC pipe coupler (one per device)
3.   A length of 4 in. inside diameter PVC pipe (about 4 in. per device)
4.   30 lb fluorocarbon fishing line (only a few inches per device but it is constantly breaking)
5.   20 lb stranded fishing wire with appropriate diameter crimps (.047 in. dia. leader sleeves)
6.   fishing connecters (two per device) interlock snaps size 3
7.   two spheres 1 in. (two per device)WLS4480-20E Sargent Welch

Tools list:
1.   #60 wire gauge drill bit .0400 in.
2.   1/16 in. drill bit
3.   1/8 in. drill bit
4.   3/16 in. drill bit
5.   ? in. drill bit
6.   saber saw blade
7.   saber saw
8.   drill press or drill motor
9.   band saw or hack saw
10.   hack saw blade
11.   calipers (you might use a compass if you have no calipers)
12.   metal scribe or awl
13.   7/8 in hole saw
14.   sandpaper; wet dry work good
15.   fibered tape to reinforce the sand paper
16.   rat tail file
17.   1 ? in. dremal cutting disk and dremal saw

Procedure:
1.   Find the smooth end of the 3 in. coupler. (one end has Genova made in USA etc. with raised printing on it)
2.   Scratch a line (with the calipers) 21.35 mm down from the smooth top end, along the side of the cylinder, all the way around the cylinder; all holes will be drilled on this line. The particular dimension (21.35 mm) is not important but that dimension needs to remain constant (always exactly the same, most calipers have a screw that will hold the dimension chosen)
3.   Make a very light indention on the line at any point, just large enough so that the caliper tip can locate it.  This is indention 1
4.   Set the calipers to 2.000 in. and mark off 2 inch intervals on the scribed line around the cylinder from indention 1, the third mark takes you to the other side (180?).  Mark off three 2 in. marks around the other side of the cylinder going the other direction from indention 1, this will leave two third marks very close together (at 180?), half way between these two marks place an indention, this is where you (later) drill the second string hole.  This is indention 2
5.   Place light indentions along the scribe line 2.538 in. counter clockwise (looking from the top) from the indentions 1 and 2. We will call these indentions 1A and 2A.
6.   Place indentions along the scribe line 3/16 in. and 1.5 in. clockwise from indentions 1 and 2. These indention will be referred to as 1B (3/16 clockwise of 1) and 1C (1.5 in. clockwise of 1), and 2B and 2C
7.   Drill1/16 in. hole toward the center line of the cylinder at indentions  1A, 1B, 1C; 2A, 2B, 2C, be careful to center the holes; 1A, and 2A are most critical
8.   Drill.0400 inch holes at 1 and 2; be careful to center the holes.
9.   Enlarge holes at 1A and 2A to 1/8 in., then 3/16 in., then ? in.  This step drilling is necessary to keep the hole from floating off center.
10.   Cut a 7/8 in. hole using the hole saw at the ? in. hole at 1A and 2A
11.   Enlarge holes at 1B, 1C and 2B, 2C to 1/8 in., then 3/16 in.
12.   Cut a slit between the holes at 1B and 1C, and 2B and 2C with a 1.25in. dremal  disk
13.   Enlarge that slit with the saber saw so that it is 3/16 wide from hole to hole.
14.   carefully extend 1B into hole1 being careful not to damage the working side of 1
15.   carefully extend 2B into 2 being careful not to damage the working side of hole 2
16.   Seat a ? in. slice of 3 in. pipe on the inside center stop ring of the coupler, this will be behind the 7/8 inch holes. This will have to be filed to seat the spheres.
17.   Place a thin sheet of plastic 7/8 in. wide 1/8 in. thick, with a centered hole, on top of the slice in step 16, the top must be level with the scribed line, and 90? to the bolt in step 18.
18.   Place a 1/16 in. (enlarge to 3.11 mm) hole between and above the 7/8 in. hole and the trailing end of the slit (1C and 2C) on the smooth top side of the cylinder for a 1/8 (by ? in.) Dia. stove bolt placement, this is used to clamp the end of the string, See photo)
19.   Make two 5/8 inch loops using the wire and leader sleeves and place them through the large side of the spheres, they will extend through about 1/8 inch.
20.   Sand and file until the spheres seat in holes 1A and 2A in the same manner.
21.   Sand and file until the slit does not catch the string.
22.   Connect the snap to the loop, connect the string to the snap, seat the sphere and feed the string through the slit, and then through the center hole of the plastic sheet seated on the plastic slice, wrap it once or twice and feed it up to the small bolt  in step 18 and clamp the string under the bolt


Pictured is a finished and an unfinished top cylinder, with a 3 in. pipe and a 4 in. pipe. I though you wanted to get started so I gave you this material; tools; and steps. I will probably change them over and over, and if you need help let me know

DrWhat

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Re: Free energy from gravitation using Newtonian Physic
« Reply #20 on: April 27, 2007, 01:37:22 PM »
I know it's asking a lot and you have written a lot of detail already, but would you be able to provide even the shortest video clip to show us what is happening.

Thanks

DrWhat

pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #21 on: May 19, 2007, 03:09:33 PM »
Lab. notes and comments
 
I was trying to build what I thought was a duplication of my last 4 inch O.D. coupler, which had a 90? stop with a 364.9g 3 in. I.D. lower pipe, added. To my amazement it took a 390g pipe to make the new cylinder stop at 90?. This is about a 4% mass increase (233.6g cylinder + 364.9g added mass; and 234.2g + 390g). Then I checked the tether length and I had inadvertently increased it by about 6%. So: the longer the tether the more mass the spheres are capable of stopping, even when the spheres mass remains constant. I had known his but had not put numbers on it before. The longer tether length apparently allows a longer period of time for the force to act. 

The top coupler (with seats for the embedded spheres at 180?) has a mass of about 234g; when it has no pipe added to its bottom opening, and the spheres masses are not added. The spheres have a mass of 132.8g. With this mass ratio of 132.8g to 234g the spheres stop the cylinder within about 4 cm from the seat. This means that the tether tightens immediately; as soon as the spheres leave the seat, and all of the force before the stop is tangent to the cylinder, because I don?t think the string (tether) has come away from the cylinder this early in the swing. The cylinder is moving strongly backwards while the string is in the slit (behind the point where the string enters the cylinder).

A 3 kilogram cylinder with two 1/2 kilogram spheres moving one meter per second has 4 units of momentum; before the spheres are released.

After the cylinder is stopped the spheres have 4 units on momentum and a velocity of 4 m/sec. The spheres can rise .8155 meters.  d = ? v?/a

If one kg (of spheres) is placed in a ten kg Attwood?s machine (with 9 kilograms balanced and the one kg of imbalance) and the spheres? mass is allowed to fall .8155 m, it will give you 12.65 units of momentum.  The spheres are now back to their original level and ready to be reloaded into the cylinder and spheres machine, but now the system has 316% the original momentum.

So you see; it doesn?t matter if the kinetic energy formula is defective or not, this system can make energy with or without the kinetic energy formula.

By using the distance formula: we know that the potential rise of 4 kg moving 1 m/sec is only .051, and the rise of 1 kg moving 4 m/sec is .8155, this is 4 times the rise as is indicated by the kinetic energy formula which predicts 4 times the energy.

Whatever the usefulness of the kinetic energy formula; the energy is still there, it is only a matter of utilizing it.

Videos: people can't get them off the DVDs I make. Can anyone help?

hartiberlin

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Re: Free energy from gravitation using Newtonian Physic
« Reply #22 on: May 19, 2007, 07:23:15 PM »
Hi,
for  converting your DVD files to MPEGs,
use VOB2MPEG, see:

http://www.videohelp.com/tools/VOB2MPG

After this you can use

virtualdubmod

http://sourceforge.net/project/showfiles.php?group_id=65889

to convert MPEGs to AVIs, e.g.with DIVX.com codec.


Or you can use also DVD2AVI for all in one in ONE step.

http://www.protectedsoft.com/download_dvd2avi.php

Regards, Stefan.


pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #23 on: November 08, 2007, 01:49:43 AM »
Let me introduce this concept for your constructive comments.

Divide a ring (thin walled hollow cylinder) into 360 equal masses distributed evenly around the circumference of the ring. The circumference of the ring is a smooth rigid circle; let?s say it is made of light modern fibers.

Let?s roll the ring down a smooth rigid incline so that it is moving across a smooth horizontal surface at one meter per second. That means the center of mass is moving one meter per second. Lets mark all the masses from 1 to 360; which is of course the degrees of the circle, but the marks are now rolling with the circumference of the circle.

When the mass marked 360 is on the bottom it is moving zero meters per second (in relation to the surface).  At the moment when 360 is on the bottom 180 is on the top moving 2 meters per second.  Suppose at this moment we release both the 180 mass and the 360 mass. The mass marked 360 will be at rest on the surface. The mass marked 180 can be caught on the end of a pendulum string. The mass marked 180 is now a pendulum bob and it will rise (d = ? at? or d = ? v?/a) .2039 meters. Now the center of mass of the masses marked 360 and 180 is at .1019 meters above the center of mass of the rim (180 can raise 360 .1019 meters off the surface and 180 is still .1019 meters higher than the top of the rim).

Now let?s do the same with the masses at 359 and 179, and 358 and 178, and repeat the process around the circle. This leaves the entire mass .1019 meters above the center of mass of the rolling ring; that now exists of only light modern fibers.

The question arises: How far did the ring roll down the incline to attain a velocity of one meter per second?

An object in freefall needs to only drop .051 meters to attain one meter per second velocity.

A puck on a frictionless plane need only drop .051 meters to attain one meter per second velocity.

A pendulum bob need only drop .051 meters to attain one meter per second velocity.

I would guess it is the same for a cart.

Is the ring different; or is this a free energy source?

Mr.Entropy

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Re: Free energy from gravitation using Newtonian Physic
« Reply #24 on: November 08, 2007, 03:31:40 AM »
Is the ring different; or is this a free energy source?

The ring is different, because it rolls down the incline and ends up spinning at the bottom.  Some of the work done by gravity as it descends must be spent to spin it, in addition to the portion spent to send it forward.

You have shown above that you pretty much know how to calculate how much further the ring must fall in order to end up rolling at 1m/s.

Cheers,

Mr. Entropy

pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #25 on: November 08, 2007, 12:59:19 PM »
All the motion is in the top mass; whether it is spinning or moving forward.

pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #26 on: November 11, 2007, 01:54:42 AM »
No portion of the rim is being lifted, because for every point of mass at say 23? there is an equal mass at 203?. If 23 is going up 203 is going down, and the same is true for all points around the balanced ring or rim.  Take the example of a modern wind turbine; the blades are not being lifted by the wind they are only turning. Nothing has to be lifted to achieve spin.

The center of mass is dropping because the slope allows gravity to act upon it.  And I believe you are correct; the acceleration is in relationship to the sine of the angle. Sine is of course in proportional to the length (rolling surface) of the ramp.

Classic physics would expect the rim to roll up the opposing ramp to the very same height if it were not for the existence of rolling friction and air resistance, the same rise in height that we see in pendulums.

And again: it does not matter whether the rim is spinning or translating, all the motion of two opposing masses is held in the top mass when the other is on the bottom.  If both masses have accelerated according to the relationship of F = ma, then all the momentum is held by the one mass and energy has been made.

pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #27 on: November 12, 2007, 12:56:52 PM »
Force is applied to the pendulum frame (at the point of rotation) just like force is applied to the ramp. Tilt an air table a few degrees and slide pucks on it, force is still applied to the table but it has no affect upon the experiment. These are called balanced forces and F = ma works even though these forces exist.

Paul-R

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Re: Free energy from gravitation using Newtonian Physic
« Reply #28 on: November 12, 2007, 03:56:09 PM »
  When a kid leans back on a swing, they go higher and higher. They're increasing their acceleration.
And if a weight attached to the pendulum is doing the same thing, why would it be different ?
I reckon the kid is using back muscles to impart energy to the seat of the swing.

pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #29 on: November 13, 2007, 12:52:18 AM »
You have a fundamental Law available to you that predicts free energy, stay within that Law and you have a very good chance of being successful; the Law is the Law of Conservation of Momentum.

For example: say you have two pendulums side by side that drop .051 meters, both pendulum bobs are moving 1 m/sec at the bottom. Suppose you devise a way of giving all the motion to one of the two bobs, the bob in motion will not rise .102 meters, it will rise .2038 m. You have doubled the energy. d = 1/2v?/a

I once saw a torsion pendulum, in a German clock, that dropped vertically of course but rotated horizontally.  If you then applied the cylinder and spheres phenomenon (previous posts) you would have a simple way to make energy.